WO2016026359A1

WO2016026359A1 - Computer-based method and device for parsing natural language syntactic structures

Info

Publication number: WO2016026359A1
Application number: PCT/CN2015/083760
Authority: WO
Inventors: 秦一男
Original assignee: 秦一男
Priority date: 2014-08-22
Filing date: 2015-07-10
Publication date: 2016-02-25
Also published as: CN104156353B; CN104156353A

Abstract

A computer-based method and device for parsing natural language syntactic structures. Natural language syntactic structures are parsed by building a matrix model and a linear model and constructing a recursive function through the mathematical thought of a composite function according to mathematical principles of subjects comprising the abstract algebra, the set theory, the combinatorial mathematics, the computability theory, the computational linguistics and the like and corresponding computer technologies; meanwhile, methods such as the mathematical induction are comprehensively applied to proving important conclusions. A set of brand new mathematical models are built for sentences of the natural language, and the thought is basically different from that of a conventional traditional method; two overall plug-in methods comprising a single-side same-direction order preserving method and a single-side same-direction non-order-preserving method are creatively provided, and a parallel syntactic constituent generating and processing method of a set family is creatively applied; the rules of the mathematics subjects and the computer subjects are sufficiently used, and the method has high accuracy and large operation amount, and has certain technological difficulty.

Description

Method and device for analyzing natural language syntax structure based on computer

The present application claims priority to Chinese Patent Application No. 2014104196340, filed on Aug. 22, 2014, entitled "A Method and Apparatus for Resolving Computer-Based Natural Language Syntactic Structures", the entire contents of which are hereby incorporated by reference. Combined in this application.

Technical field

The present invention relates to the field of computer data processing, and in particular to a computer-based natural language syntax structure parsing method and apparatus.

Background technique

Natural language processing is an important direction in the field of computer science and artificial intelligence. It studies various theories and methods that enable effective communication between humans and computers using natural language.

Syntactic structure analysis is an important aspect of natural language processing. It automatically divides the sentence components of natural language sentences by computer to assist in the further processing of sentences. In the existing syntax structure analysis technology, Probabilistic Context Free Grammars (PCFG) is usually adopted, which is based on the complex nesting characteristics of natural language, and the rule matching probability of the calculation result of sentence and syntactic structure is calculated. The syntactic parsing result with the highest probability is selected as the final syntactic structure.

However, the method is highly complex, and the analytical accuracy of the compound sentence structure needs to be further improved.

Summary of the invention

In view of this, the present invention provides a computer-based natural language syntax structure analysis method and apparatus, which has unique ideas, ingenious methods, and detailed argumentation, and fully utilizes the laws of mathematics and computer science, and the method has high accuracy. The amount of calculation is very large and has high technical difficulty.

The invention provides a computer-based natural language syntax structure parsing method, comprising:

S1: reading a pre-processed statement data structure to be parsed, wherein the pre-processed statement data structure includes only a parallel-related word unit, a subordinate-related word unit, a predicate verb unit, a noun pronoun unit, and each word unit is The order in the preprocessed statement is numbered and labeled;

S2, for each predicate verb unit, generating a corresponding guide element, a subject element, a predicate element, and an object element;

The possible value of the guide element is one of a parallel related word unit or a dependent related word unit whose number is smaller than the corresponding predicate verb unit number, or a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and one adjacent thereto One of the associated word combination vectors composed of the dependent word unit whose number is smaller than the corresponding predicate verb unit number and whose number is greater than the parallel related word unit number, or an empty unit;

The possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than the juxtaposition included in the total parallel noun pronoun combination vector family of the corresponding predicate verb unit number. One of the noun pronoun combination vectors, or one of the syntactic vectors corresponding to the predicate element, or an empty unit;

The predicate element is a corresponding predicate verb unit;

The possible value of the object element is one of the noun pronoun units whose number is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number, or the number of the smallest word unit is greater than the corresponding predicate verb unit number. And one of the parallel noun pronoun combination vectors included in the entire parallel noun pronoun combination vector family of adjacent predicate verb unit numbers, or one of the syntactic vectors corresponding to the predicate element, or an empty unit ;

S3. Obtain all possible values of a syntax vector corresponding to each predicate verb unit according to possible values of the guide element, the subject element, the predicate element, and the object element, where the syntax vector includes a guide element, a subject element, Predicate element and object element;

S4. Generate at least one syntax structure possible matrix solution according to all possible values of all syntax vectors, where the syntax structure may be composed of syntactic vectors arranged according to the order of the predicate verb unit numbers;

S5. Verify whether the statement obtained by the possible matrix solution according to the syntax structure is identical to the preprocessed statement. If they are identical, each syntactic vector in the possible matrix solution of the syntax structure is one of the parsing result of the syntax structure;

Among them, S5 includes the following operations in order, excluding the syntactic structure that does not meet the conditions may be solved:

S5.1. If there is a sequence value that does not appear in the possible matrix solution of the syntax structure, the possible matrix solution may be excluded from the syntax structure;

S5.2. If the same order value appears in different syntax vectors or the same syntax vector appears, the possible syntax solution of the syntax structure is excluded;

S5.3. In each possible matrix solution, the syntactic vectors that are mutually substituted with other syntactic vectors are all equally substituted. If there is a contradiction between two syntactic vectors after the equal substitution, then Excluding the syntactic structure may be a matrix solution;

S5.4. In each possible matrix solution, all the syntactic vectors that have mutual substitution relationship with other syntactic vectors are equally substituted, if two position reversal order values appear after the equal substitution, then Excluding the syntactic structure may be a matrix solution;

S5.5. In any of the possible matrix solutions, if there is a syntax vector that has no substitution relationship with other syntax vectors, perform an insertion operation to obtain a possible syntax parsing structure corresponding to all the possible matrix solutions, and verify Whether the statement obtained according to the possible syntax parsing structure is identical to the pre-processed statement, further comprising:

S5.5.1, firstly perform equal-substitution of the syntactic vectors in the possible matrix solutions with the substitution relationship between them, thereby transforming the possible matrix solutions into a set of syntactic vectors with no substitution relationship between each other.

The syntactic vector in the possible matrix solution is called the first kind of syntactic vector, and the transformed syntactic vector will be transformed.

Called the second type of syntax vector;

S5.5.2, take a second type of syntax vector

Mark one by one according to the predetermined direction

The order value of each syntax element in the message; after appending the order value of the syntax element, take any

The i-th syntax element in the construct, only a unique gap is constructed on the first side of the syntax element; after the void, take a syntax vector

Second type of syntax vector

Syntactic vector in the form of overall insertion

Insert the constructed vacancy, and then generate a new syntax vector, record this new syntax vector as

The syntactic vectors obtained by inserting the whole into space are collectively referred to as the third type of syntax vector;

S5.5.3, the third type of syntax vector

Pair vector from the predetermined direction

The first syntactic element on the first side starts into the vector

Vector contained in

Each of the syntax elements up to the first syntactic element on the second side, all of which are labeled with a sequence value;

Vector contained in

The element on the first side, without the order value; the vector

The first syntax element on the second side is marked as

Will be vectored as described above

The syntactic vector part of the annotation, denoted as the iris syntax vector

After the order value is marked, take the jth syntax element in the aforementioned tail vector, and construct a unique gap only on the first side of the element; after the empty, take an unused second type of syntax vector

Syntactic vector in the form of overall insertion

Insert the constructed vacancy, and then generate a new syntax vector, then record the newly generated syntax vector as

or

Third type of syntax vector

Syntactic vector according to the predetermined direction

Each syntax element in the label is labeled with a sequential value; after the order value of the syntax element is annotated, take one

The tth syntax element in the construct, constructing a unique gap on the first side of the syntax element; after the void, taking an unused second type of syntax vector

The vector is inserted as a whole

Insert the previously constructed gap and generate a new vector, then the new vector is recorded as

S5.5.4, repeated execution of S5.5.3, the next time the emptying and insertion of the third type of syntactic vector obtained through the previous emptying and insertion steps are performed at the end of the last emptying and emptying steps. Null operation until all second type of syntax vectors will be

After all the insertions are completed, a third type of syntax vector of a single line is finally obtained, and the finally obtained third type of syntax vector is called a final single line vector;

S5.5.5, if there are two position reversal order values in all the final single row vectors corresponding to a possible syntax parsing structure, the possible syntactic parsing structure is excluded;

S5.5.6, repeat S5.5.2 to S5.5.5 until all possible syntactic parsing structures are traversed.

Further, S2 includes generating a vector family of parallel noun pronouns:

S2.1 selects two noun pronoun units that are not repeated:

A. If there are no other word units between the two noun pronoun units, the two noun pronoun units are used as a parallel noun pronoun combination vector, and the parallel noun pronoun combination vector is retained;

B. If there are other word units between the two noun pronoun units, check each word unit between the two noun pronoun units: if any between the two noun pronoun units Word units, all of which are noun pronoun units Or juxtaposed the word unit, the selected two noun pronoun units and the whole word unit between the two noun pronoun units are used as a parallel noun pronoun combination vector, and the parallel noun pronoun combination vector is retained; otherwise, no Generating a parallel noun pronoun combination vector;

S2.2 complex execution S2.1 until all combinations of noun pronoun units are traversed, and all obtained parallel noun pronoun combination vectors are generated;

S2.3 If there is a parallel noun pronoun combination vector in the possible syntactic parsing structure, all the parallel noun pronoun combination vectors are divided to form a plurality of parallel noun pronoun combination vector families, so that: in each parallel noun pronoun combination vector family Each collocated noun pronoun combination vector included in the parallel noun pronoun combination vector family all contains two common noun pronoun units.

S2.4 selects the largest number of word units contained in all noun pronoun combination vectors in each noun pronoun combination vector family, as the largest word unit of the noun pronoun combination vector family, for use in subsequent generation of the subject; The word unit with the lowest number included in all noun pronoun combination vectors is used as the smallest unit of the noun pronoun combination vector family, and is used for subsequent generation of the object.

Further, generating corresponding subject elements includes:

When the corresponding predicate verb unit number is the smallest predicate verb unit number, the possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than the corresponding One of the parallel noun pronoun combination vectors contained in the vector of the predicate verb unit number, or an empty unit.

When the corresponding predicate verb unit number is not the smallest predicate verb unit number, the possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than the corresponding predicate The verb unit number is one of the collocated noun pronoun combination vectors contained in the collocation noun pronoun combination vector family, or one of the syntactic vowel units corresponding to the predicate verb unit, or an empty unit.

Further, generating corresponding object elements includes:

When the corresponding predicate verb unit number is the largest predicate verb unit number, the possible value of the object element is one of the noun pronoun units whose number is greater than the corresponding predicate verb unit number, or the number of the smallest word unit is greater than the corresponding number. One of the parallel noun pronoun combination vectors contained in the vector of the predicate verb unit number, or an empty unit.

When the corresponding predicate verb unit number is not the largest predicate verb unit number, the possible value of the object element is a noun pronoun unit whose number is greater than the corresponding predicate verb unit number and is smaller than the adjacent predicate verb unit number. One of the collocated noun pronoun combination vectors included in one of the collocation noun pronoun combination vector numbers, or one of the smallest word units, is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number. Or one of the syntactic vectors corresponding to the predicate verb unit that appears later, or an empty unit.

Further, in the two steps S4 and S5, the possible matrix solution may be replaced by a possible linear expression solution with a syntax structure;

The syntactic structure may be equivalent to a linear expression solution of the syntactic structure;

The syntactic structure may be a linear expression solution comprising a syntactic vector expression arranged in order of predicate verb unit numbers; each of the syntactic vector expressions is a guide element, a subject element, a predicate element, an object of a corresponding syntax vector An expression in which elements are added one by one in order.

Further, the method further includes:

Each syntax vector and corresponding syntax structure relationship in the syntax structure analysis result are displayed in a human-computer interaction interface by a tree structure.

The invention also provides an apparatus for analyzing a natural language syntax structure based on a computer, comprising:

a reading component, configured to read a pre-processed statement data structure to be parsed, wherein the pre-processed statement data structure includes only a parallel-related word unit, a subordinate-related word unit, a predicate verb unit, a noun pronoun unit, and Each word unit is numbered in the order in the preprocessed statement, and the type is marked;

An element generating component, configured to generate a corresponding guide element, a subject element, a predicate element, and an object element for each predicate verb unit;

Wherein, the possible value of the guide element is one of a parallel related word unit or a dependent related word unit whose number is smaller than the corresponding predicate verb unit number, or a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and one of them One of the associated word combination vectors formed by the dependent-related word units whose neighbors are smaller than the corresponding predicate verb unit number and whose number is greater than the parallel-related word unit number, or an empty unit;

The predicate element is a corresponding predicate verb unit;

a vector generating component, configured to obtain all possible values of a syntax vector corresponding to each predicate verb unit according to possible values of the guide element, the subject element, the predicate element, and the object element, where the syntax vector includes a guide element , subject elements, predicate elements, and object elements;

a matrix generating component, configured to generate at least one syntax structure possible matrix solution according to all possible values of all syntax vectors, wherein the syntax structure may be composed of a syntax vector arranged according to a predicate verb unit number order;

a solution component for verifying whether the statement obtained by the possible matrix solution according to the syntax structure is identical to the preprocessed statement, and if they are identical, each syntactic vector in the possible matrix solution of the syntax structure is used as a syntactic structure analysis result One;

Wherein, the solving component excludes a possible syntactic structure solution by the following module operation:

a first exclusion module, if there is a sequence value that does not appear in the possible matrix solution of the syntax structure, the possible matrix solution is excluded from the syntax structure;

The second exclusion module excludes the possible matrix solution if the same sequence value appears in the different syntax vectors or the same syntax vector appears;

In the third exclusion module, in each possible matrix solution, the syntactic vectors having mutual substitution relations with other syntax vectors are all equally substituted, and if the cross-contradictions of the two syntax vectors appear after the equal-substitution, Excluding the syntactic structure possible matrix solution;

In the fourth exclusion module, in each possible matrix solution, the syntactic vectors having mutual substitution relations with other syntax vectors are all equally substituted, and if the order values of the two positions are reversed after the equal substitution, Excluding the syntactic structure possible matrix solution;

a fifth exclusion module, in any one of the possible matrix solutions, if there is a syntax vector that has no substitution relationship with other syntax vectors, performing an interpolation operation to obtain a possible syntax parsing structure corresponding to all the possible matrix solutions, and Verification of whether the statement obtained according to the possible syntax parsing structure is identical to the preprocessed statement, further comprising:

The first sub-module first performs an equal substitution of the syntactic vectors in the possible matrix solutions with the substitution relationship between them, thereby transforming the possible matrix solutions into a set of syntactic vectors without substitution relations between them.

Called the second type of syntax vector;

The second sub-module, taking a second type of syntax vector Mark one by one according to the predetermined direction

Second type of syntax vector

Syntactic vector in the form of overall insertion

Third submodule, the third type of syntax vector

Pair vector from the predetermined direction

The first syntactic element on the first side starts into the vector

Vector contained in

The element on the first side, not the order value; the vector

The first syntax element on the second side is marked as

Will be vectored as described above

The syntactic vector part of the annotation, denoted as the iris syntax vector

Syntactic vector in the form of overall insertion

or

Third type of syntax vector

Syntactic vector according to the predetermined direction

The vector is inserted as a whole

The fourth sub-module repeats the operation of the third sub-module, and each time the last nulling and emptying step ends, the third type of syntactic vector obtained through the last emptying and emptying steps is made for the next time. Empty and insert operations until all second type of syntax vectors will be

a fifth submodule, if there are two position reversal order values in all of the final single row vectors corresponding to a possible syntactic parsing structure, the possible syntactic parsing structure is excluded;

The sixth sub-module repeatedly calls the operations of the second sub-module to the fifth sub-module until all possible syntactic parsing structures are traversed.

Further, it also includes:

The result display component displays the syntax vector and the corresponding syntax structure relationship in the syntax structure analysis result on the human-computer interaction interface by using a tree structure.

DRAWINGS

The above and other objects, features and advantages of the present invention will become more apparent from

1 is a flow chart of a method for analyzing a computer-based natural language syntax structure according to an embodiment of the present invention;

2 is a schematic diagram of an apparatus for analyzing a computer-based natural language syntax structure according to an embodiment of the present invention.

Detailed ways

The invention is described below on the basis of preferred embodiments, but the invention is not limited to only these embodiments. In the following detailed description of the invention, some specific details are described in detail. The invention may be fully understood by those skilled in the art without a description of these details. In order to avoid obscuring the essence of the invention, well-known methods, procedures, components and circuits are not described in detail.

Partial partial order and partial addition on partial semigroups

Part A1 Natural language is a free semigroup of vocabulary and punctuation

According to the theory of abstract algebra and computational linguistics, natural language is a free unitary semigroup on vocabulary and punctuation. The following is explained by taking English as an example, but those skilled in the art will readily understand that the method of the present invention is also applicable to other natural languages.

Given a set A, the symbol string on A is adjacency of the elements in A, and can be repeated in the adjacency to form a finite-length linear array. For example, from the set {a, b, c}, the symbol string acbaab can be formed. This string of symbols contains three occurrences of a, two occurrences of b, and one occurrence of c, which is different from the symbol string acaabb. Although each symbol appears the same number of times, their order is different. It can be seen that the symbol string is ordered. In particular, a symbol string of length 0 is a string of 0 symbols, denoted as e. Thus, for a given finite set of symbols A, the symbol string of length n on A is a mapping from the natural number set N to A: f: N → A.

Starting with two symbol strings, we can construct a new symbol string by adjacency. For example, at the right end of the symbol string abac adjacent to the symbol string bbac, a new symbol string abacbbac is formed.

The operation of this contiguous symbol string is called: contiguous operation, referred to as contiguous.

Given a symbol string φ of length n and a symbol string 长度 of length m, where:

Φ={(1,x ₁ ), (2,x ₂ ), (3,x ₃ ), . . . , (n-1,x _n-1 ), (n,x _n )};

ψ = {(1, y ₁ ), (2, y ₂ ), (3, y ₃ ), ..., (m-1, y _m-1 ), (m, y _m )};

The connection between φ and ψ is: φ^ψ. It is of length n+m and consists of the set {(1,x ₁ ),(2,x ₂ ),(3,x ₃ ),...,(n-1,x _n-1 ),(n,x _n ), (n+1, y ₁ ), (n+2, y ₂ ), ..., (n+m, y _m )} are given symbol strings. Then, the contig is a binary operation defined on the symbol string, and the result of the operation is to get a new symbol string.

The connection between φ and ψ can also omit the contiguous mark ^, which is simplified as: φ ψ.

Then there are: φ^ψ=φψ.

The contiguous operations are combinable because for any symbol string φ, ψ, ω, there are:

Φ^(ψ^ω)=(φ^ψ)^ω

Each existing English word and English punctuation mark is defined as a symbol, then the set of all words and punctuation marks in S A={a ₁ , a ₂ , a ₃ ,..., a _n }(n∈N ) is a set of symbols.

Any given finite-length symbol string b ₁ b ₂ ... b _k (k∈N) consisting of English words and English punctuation marks, is called a word unit or a continuous word string. For a given word unit a=b ₁ b ₂ ... b _m (m∈N), a is said to be a word unit consisting of elements in A, if and only if, b ₁ , b ₂ , ..., b _m ∈A.

A unique unit of word of length 0 is called an empty unit and is denoted as e.

The set of all word units (continuous word strings) composed of elements in A is A ^s , and the statement S = a ₁ a ₂ a ₃ ... a _n , where a _n is the word unit constituting the sentence . The algebraic system (A ^s , ^, e) is a free monoid on the English word and punctuation set A.

The word units are arranged in order according to their order in the sentence, the subscripts are sequentially numbered, and τ(α) is the number of the word unit α in the sentence S.

The conditions for constructing a syntactic component order map ω, ω are as follows:

(1) ω: {a ₁ , a ₂ , a ₃ , ..., a _n } → N, N is a natural number set;

(2) For any one of a _i , a _i ∈S, there is: ω(a _i )=T(a _i ).

Obviously, ω is a single mapping.

Part A2 defines a partial order relationship

At the same time, for algebraic systems (A ^s , ^, e), define a binary relationship < _□ :

For any word unit α in A ^s , β∈A ^s , called α< _□ β, if and only if α, β number τ(α), τ(β) satisfies: τ(α)<τ(β) .

By definition, the binary relationship < _□ satisfies the following conditions:

(1) Give a∈A ^s with a≮ _□ a;

(2) For any a, b, c ∈ A ^s , if a < _□ b, then b ≮ _□ a;

(3) For any a, b, c ∈ A ^s , if a < _□ b and b < _□ c, then a < _□ c.

According to the definition of strict partial order relationship, the binary relationship < _□ is strictly partial order relationship.

Part A3 defines partial addition and syntactic order values

At the same time, on the algebraic system (A ^s , ^, e), define a new binary operation + <. Let +< be the partial addition operation defined on the strict partial order relationship < _□ in A ^s , abbreviated as partial addition, which satisfies the following characteristics: for any a, b∈A ^s , if a< _□ b, then a+< b=a^b=ab.

We can determine: for any a, b ∈ A ^s , if a < _□ b, then there is a partial addition + < and the contiguous operation ^ equivalent. The partial addition operation +< can be regarded as a contiguous operation restricted to the strict partial order relationship < _□ .

Any natural language sentence S can be regarded as a word string formula which is connected by each word unit according to a strict partial order relationship < _□ , namely: S=a ₁ +<a ₂ +<a ₃ +<...+< a _n . This feature is very beneficial for the development of mathematical processing.

In the original sentence S, in order from left to right, from the beginning of the sentence to the end of the sentence, the sequence number of the adjacent n consecutive words α ₁ , α ₂ , ..., α _{n in the} whole sentence is marked: 1,2,...,n.

In a determined label as described above, the sequence number of any given continuous string α is τ(α), and τ(α) is called the left-to-right order value of α. That is, the syntax element γ in one original sentence S is given, and the syntax order value of the syntax element γ in the original sentence S is denoted as τ(γ).

Part B Technical details

Part B1 Initial classification of language information

In the present invention, the word unit a _i constituting the sentence is recognized as a constant. The word unit a _i has its language attribute. The word units constituting the core sentence structure can be divided into four types: a parallel related word unit, a dependent related word unit, a predicate verb unit, and a noun pronoun unit. Each word unit includes at least one natural language vocabulary, which may be a word, a phrase of a particular structure, or a juxtaposition of multiple synonyms.

For the side-by-side related word unit, it may be a parallel conjunction of the parallel sentence and the parallel syntax component and, but, or, so, yet.

For a dependent related word unit, it can be a connected pronoun of a leading clause or a connecting phrase of a leading adverb and a guiding clause. For a typical guiding word, the following are listed: that, what, which, who, who, wherever, when, whoe, where, when ,why,how,whoever,whichever,while,whether,because,before,after,whatever,weverever,as,if,once,until,though,unless,although,no matter what,no matter who,no matter whom,no Matter which, in that, in order that, as though, as if, even though, even if, so that, etc. It mainly includes: a word unit that serves as a guiding clause by a word, a related word unit that serves as a guiding clause by a phrase, and a related word unit that connects a parallel sentence and a parallel sentence.

For a predicate verb unit, it can also be a verb or a verb phrase, for example, can do, do. The predicate is defined as the main action language in a natural sentence in English. The structure usually consists of two parts: the auxiliary verb + the real verb (except the main table structure). The format requirements for predicate states and voices are defined by the formula of computational linguistics as follows:

For the noun pronoun unit, it can be: a pure noun phrase (noun phrase not included in the prepositional phrase), a nominalized verb phrase (nominalized verb phrase definition: having the nature of a noun, can act as a subject or an object) Verb phrases of nominal syntactic components, including: indefinite phrases and gerund phrases, and pronouns that can be used alone. Examples of noun pronouns are as follows: food, wolf, the men, me, it, this, to do, etc.

The nominal verb phrase has a format requirement, and the formula for computational linguistics is defined as follows:

11	To+VBTo+VB	77	RB+To+VBRB+To+VB
11	To+VBTo+VB	77	RB+To+VBRB+To+VB	22	To+VB+VBNTo+VB+VBN	88	RB+To+VB+VBNRB+To+VB+VBN
33	To+VB+VBN+VBNTo+VB+VBN+VBN	99	RB+To+VB+VBN+VBNRB+To+VB+VBN+VBN	22	To+VB+VBNTo+VB+VBN	88	RB+To+VB+VBNRB+To+VB+VBN
33	To+VB+VBN+VBNTo+VB+VBN+VBN	99	RB+To+VB+VBN+VBNRB+To+VB+VBN+VBN	44	VBGVBG	1010	RB+VBGRB+VBG
55	VBG+VBNVBG+VBN	1111	RB+VBG+VBNRB+VBG+VBN	44	VBGVBG	1010	RB+VBGRB+VBG
55	VBG+VBNVBG+VBN	1111	RB+VBG+VBNRB+VBG+VBN	66	VBG+VBN+VBNVBG+VBN+VBN	1212	RB+VBG+VBN+VBNRB+VBG+VBN+VBN

{Explanation of important symbols}

rr	谓语动词单元Predicate verb unit
rr	谓语动词单元Predicate verb unit	kk	当前正在处理的谓语动词单元的顺序数The order number of the predicate verb units currently being processed
LeadLead	从属关联词单元Subordinate unit	kk
LeadLead	从属关联词单元Subordinate unit	NPINPI	纯粹名词单元Pure noun unit
ConjConj	并列关联词单元Parallel word unit	NPINPI	纯粹名词单元Pure noun unit
ConjConj	并列关联词单元Parallel word unit	VNPVNP	名词性质的动词单元Noun-like verb unit
NOMPNOMP	主格代词单元Subject pronoun unit	VNPVNP	名词性质的动词单元Noun-like verb unit
NOMPNOMP	主格代词单元Subject pronoun unit	OBJPOBJP	宾格代词单元Binger pronoun unit
NPNP	名词代词单元的统称General term for noun pronoun unit	OBJPOBJP	宾格代词单元Binger pronoun unit

In the above word unit list, the set of word units has the following relationship:

{NP}={NPI}∪{VNP}∪{NOMP}∪{OBJP}.

Part B2 defines important concepts

Description: The definition of a clause in a natural language statement is as follows: A clause is a simple sentence, that is, the most basic sentence of natural language. A clause is a set of subject-predicate collocation structure. The above three types of word units constitute the backbone of natural language sentence clauses, wherein the predicate verb unit acts as a predicate, and the noun pronoun unit acts as a subject or object.

In the present invention, the variables are defined as x, y, z, where x is the leader element, y is the subject element, z is the object element, and at the same time, r is the predicate element, then the subject-predicate structure in each statement can Expressed as:

f(x,y,r,z)=x+<Λ+<y+<σ+<r+<ρ+<z+<μ

Λ, σ, ρ, μ respectively represent any component or punctuation other than x, y, r, z, referred to as impurities, and the impurities can be removed by the existing sentence pretreatment technique. The function f(x, y, r, z) after the removal of impurities can be = x + < y + < r + < z. Expressed in the form of vectors (x, y, r, z).

The leader element x is a component of a simple sentence: when the simple sentence is a clause, the leader element is a connected pronoun of the leading clause or A connecting phrase that connects an adverb and a leading clause; when the simple sentence is a parallel sentence, the leading element is a parallel conjunction connecting the parallel sentence with other preceding parallel sentences. That is, in a simple sentence, the leader element x is a syntactic component composed of related word units for guiding subsequent simple sentences.

If a function f in S is currently being processed, then the current function f is denoted as f _k ; the order of the predicate verb units currently being processed is k. (k∈N, N is a natural number set, k≤n)

The B3 part generates three key sets: {x _k }, {y _k }, {z _k }

B 3.1 part generation {x _k }

[B 3.1.1] Preparatory work: Define the following subcollections:

1) Lead _k = {Lead|Lead< _□ r _k };

2) conj _k = {conj|conj< _□ r _k };

3) (conj _k οLead _k )=

{R _k |R _k =conj+<Lead,conj< _□ r _k ,Led< _□ r _k ,τ(Lead)=τ(conj)+1};

[B 3.1.2]{x _k } generation algorithm:

{x _k }=Lead _k ∪conj _k ∪(conj _k οLead _k )∪{e}.

Part 3.2 of 3.2 Particular Description of the Method of Generating Parallel Syntactic Components (Taking Parallel Subjects and Parallel Objects as Examples)

[B 3.2.1]Intuitive description

Description: In the following description, for the convenience of expression, the formula of the continuous string is Фt or

Contains syntax elements

Recorded as

step 1

Take the phrase of all nouns in the original sentence, and group the phrases of all nouns in the original sentence into a set, which is recorded as a set Ψ={α ₁ ,...,α _m-1 ,α _m } , m∈N,m is the number of elements in the collection.

Step 2

according to

The way, take all combinations of any two elements in the set Ψ = {α ₁ , ..., α _m-1 , α _m },

Set collection

Step 3

One of the given

By element

The syntactic order values in the original sentence S are arranged from small to large. You may wish to set

Orderly pair

Formulating a continuous string of words

among them

Is the slave in the original sentence S

To

A set of adjacent consecutive word strings or empty word strings. Exhaustion of such ordered pairs and continuous word string formulas.

Step 4

Check the formula Ф ^t if it is between the elements in the formula Ф ^t

with

The element γ between them has: γ or a noun-like phrase, or a parallel conjunction, or an empty string, then change the mark of Ф ^t to

Ф ^t generation

Set collection

then

Step 5

Any collection

If collection

Corresponding

Defining a collection family that contains collections

The whole collection of the composition. The collection family is recorded as the following expression:

Step 6

If collection

There is a corresponding collection family

Will any one of the collection families

The syntactic elements of each collection in the collection are all taken out, listed as collections

Step 7

Extract the collection separately

The largest and smallest element of the syntactic order value in the original sentence S.

Note: This method can also be used to generate other types of side-by-side components, such as generating side-by-side adjective phrases. As long as the entire NPI, the entire VNP, and the entire NOMP phrase in the method are replaced by the entire NPI, the entire VNP, and the entire NOMP phrase in the original sentence, the corresponding syntax component can be obtained.

[B 3.2.2] Formal definition

Definition: Unary function A(S), A(S) indicates that all NPI phrases, all VNP phrases, and all NOMP phrases in the original sentence S are taken out, and all NPI phrases, all VNP phrases, and all NOMP phrases in the original sentence are also taken. Listed as a set, the set is denoted by Ψ = {α ₁ , ..., α _m-1 , α _m }, _m ∈ N, m is the number of elements in the set Ψ. Then A(S)=Ψ={α ₁ ,...,α _m-1 ,α _m }.

Definition: unary function B (Ψ), B (Ψ) means according to

The way to take all combinations of any two elements in the set Ψ={α ₁ ,...,α _m-1 ,α _m },

Set collection

then

One that will be given

Recorded as

then

Definition: The binary function K(α, β), K(α, β) represents the result of the unary function B(Ψ), that is, the given one

By element

Orderly pair

Set collection

Then establish a continuous string formula

among them

Is the slave in the original sentence S

To

a set of adjacent consecutive or empty words, and

then

Definition: The unary function H(Ф ^t ), H(Ф ^t ) represents the generation of the binary function K(α, β)

Check: if the given element is γ∈Ф ^t , and

And

Both: γ=NPI or γ=VNP or γ=NOMP or γ=CONJ or γ=e, then change the mark of Ф ^t to

Ф ^t generation

Set collection

then

Definition: The binary function M(α, β), M(α, β) represents a set for any

If collection

Corresponding

Defining a collection family that contains collections

The whole set of the composition, the collection family is recorded as

then

Definition: The binary function N(α, β), N(α, β) represents the result of the binary function M(α, β)

That is, for any collection

If collection

There is a corresponding collection family

Then construct a new collection as follows

then

Definition: The unary function u(α), u(α) represents the result of the binary function N(α, β)

take

Assume

For the given element γ,

There are τ(γ) ≤ τ(δ). then

Definition: The unary function V(β), V(β) represents the result of the binary function N(α, β)

take

Assume

For the given element γ,

There are τ(δ) ≤ τ(γ). then

[B 3.2.3] Parallel subject generation algorithm:

[B 3.2.4] Parallel object generation algorithm:

[B 3.2.5] An example of the algorithm for generating parallel subjects and parallel objects

For example: the word sequence table is:

原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number
原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number	AfterAfter	从属关联词单元Subordinate unit	11
JackJack	名词代词单元Noun pronoun unit	22	AfterAfter	从属关联词单元Subordinate unit	11
JackJack	名词代词单元Noun pronoun unit	22	MaryMary	名词代词单元Noun pronoun unit	33
andAnd	并列关联词单元Parallel word unit	44	MaryMary	名词代词单元Noun pronoun unit	33
andAnd	并列关联词单元Parallel word unit	44	LindaLinda	名词代词单元Noun pronoun unit	55
leftLeft	谓语动词单元Predicate verb unit	66	LindaLinda	名词代词单元Noun pronoun unit	55
leftLeft	谓语动词单元Predicate verb unit	66	II	名词代词单元Noun pronoun unit	77
gaveGave	谓语动词单元Predicate verb unit	88	II	名词代词单元Noun pronoun unit	77
gaveGave	谓语动词单元Predicate verb unit	88	my sonMy son	名词代词单元Noun pronoun unit	99
a booka book	名词代词单元Noun pronoun unit	1010	my sonMy son	名词代词单元Noun pronoun unit	99

In the process of generating the set of subject elements {y ₁ }, {y ₂ }, the algorithm for running the parallel subject generation is as follows:

1A(S) takes out all NPI phrases, all VNP phrases, and all NOMP phrases in the original sentence, and lists all NPI phrases, all VNP phrases, and all NOMP phrases in the original sentence as a set, and records the set as Ψ= {Jack, Mary, Linda, I, my son, a book}={2,3,5,7,9,10}.

2B (Ψ) means follow

The way to take all combinations of any two elements in the set Ψ={2,3,5,7,9,10}, set the set

Then B(Ψ)={{2,3},{2,5},{2,7},{2,9},{2,10},{3,5},{3,7},{ 3,9},{3,10},{5,7},{5,9},{5,10},{7,9},{10,7},{10,9}}.

The result of 3K(α,β) versus unary function B(Ψ), that is, one given

By element

Orderly pair

then

The generated ordered pairs are:

{<2,3>,<2,5>,<2,7>,<2,9>,<2,10>,<3,5>,<3,7>,<3,9>,< 3, 10>, <5, 7>, <5, 9>, <5, 10>, <7, 9>, <7, 10>, <9, 10>}.

Set collection

Then establish a continuous string formula

among them

Is the slave in the original sentence S

To

a set of adjacent consecutive or empty words, and

then

Then Ф ¹ = 2+ < e + < 3, Ф ² = 2+ < 3 + < 4 + < 5, Ф ³ = 2+ < 3 + < 4 + < 5 + < 6 + < 7, Ф ⁴ = 2+ <3+<4+<5+<6+<7+<8+<9, Ф ⁵ =2+<3+<4+<5+<6+<7+<8+<9+<10, Ф ⁶ =3+<4+<5, Ф ⁷ =3+<4+<5+<6+<7, Ф ⁸ =3+<4+<5+<6+<7+<8+<9 , Ф ⁹ =3+<4+<5+<6+<7+<8+<9+<10, Ф ¹⁰ =5+<6+<7, Ф ¹¹ =5+<6+<7+<8+<9, Ф ¹² =5+<6+<7+<8+<9+<10, Ф ¹³ =7+<8+<9, Ф ¹⁴ =7+<8+<9+<10, Ф ¹⁵ =9+<e+<10.

4H(Ф ^t ) generated for the binary function K(α, β)

Check: if the given element is γ∈Ф ^t , and

And

Ф ^t generation

Set collection

Collection

then

5M(α,β) represents a set for any

If collection

Corresponding

Defining a collection family that contains collections

The whole set of the composition, the collection family is recorded as

then

Then M(α,β)={I ₁ ({2,3}), I ₂ ({3,5}), I ₃ ({9,10})}.

6N(α,β) results for the binary function M(α,β)

That is, for any collection

If collection

There is a corresponding collection family

Then construct a new collection as follows

Then P[I ₁ ({2,3})]={2,3,4,5}, P[I ₂ ({3,5})]={2,3,4,5}, P [I ₃ ({9,10})]={9,10}.

7u(α) results for the binary function N(α,β)

take

Assume

For the given element γ,

There are τ(γ) ≤ τ(δ). Then P ^max [I ₁ ({2, 3})] = 5, P ^max [I ₂ ({3, 5})] = 5, P ^max [I ₃ ({9, 10})] = 10.

In the process of generating the object element set {z ₁ }, {z ₂ }, the algorithm for running the guest column subject is as follows:

1A(S) takes out all NPI phrases, all VNP phrases, and all OBJP phrases in the original sentence, and lists all NPI phrases, all VNP phrases, and all OBJP phrases in the original sentence as a set, and records the set as Ψ= {Jack, Mary, Linda, I, my son, a book}={2,3,5,7,9,10}.

2B (Ψ) means follow

The result of 3K(α,β) versus unary function B(Ψ), that is, one given

By element

Orderly pair

then

The generated ordered pairs are:

Set collection

Then establish a continuous string formula

among them

Is the slave in the original sentence S

To

a set of adjacent consecutive or empty words, and

then

4H(Ф ^t ) generated for the binary function K(α, β)

Check: if the given element is γ∈Ф ^t , and

And

Ф ^t generation

Set collection

Collection

then

5M(α,β) represents a set for any

If collection

Corresponding

Defining a collection family that contains collections

The whole set of the composition, the collection family is recorded as

then

Then M(α,β)={I ₁ ({2,3}), I ₂ ({3,5}), I ₃ ({9,10})}.

6N(α,β) results for the binary function M(α,β)

That is, for any collection

If collection

There is a corresponding collection family

Then construct a new collection as follows

7V(β) represents the result of the binary function N(α, β)

take

Assume

For the given element γ,

There are τ(δ) ≤ τ(γ). Then P ^min [I ₁ ({2, 3})] = 2, P ^min [I ₂ ({3, 5})] = 2, P ^min [I ₃ ({9, 10})] = 9.

B 3.3 Part {y _k } generation method

[B 3.3.1] Preparation: Define the following sub-collections:

1) NPI _yk = {NPI|NPI< _□ r _k }.

2) VNP _yk = {VNP|VNP< _□ r _k }.

3) NOMP _k = {NOMP|NOMP< _□ r _k }.

4)

among them:

5) ry _k ={r _α |α<k, α∈N}. (N is a natural number set)

6) fy _k ={f _α |α<k, α∈N}. (N is a natural number set)

[B 3.3.2]{y _k } generation algorithm

2 is converted into: when r _k-1 is present: {y _k }=NPI _yk ∪VNP _yk ∪NOMP _k ∪G _k ∪fy _k ∪{e}, then the above equation is converted into:

B 3.4 part of the {z _k } generation method

[B 3.4.1] Preparatory work: Define the following subcollections:

among them:

5) rz _k ={r _α |k<α, α∈N}. (N is a natural number set)

6) fz _k ={f _α |k<α, α∈N}. (N is a natural number set)

[B 3.4.2]{z _k } generation algorithm

2 is converted into: when r _k+1 is present: {z _k }=NPI _zk ∪VNP _zk ∪OBJP _k ∪H _k ∪fz _k ∪{e}, then the above equation is converted into:

B 3.5 Partial matrix expressions and linear expressions

[B 3.5.1] matrix expression

Furthermore, the statement S can be expressed in a matrix form, namely:

When a function f _j acts as a subject element or an object element of another function f _k , for example, when f _k =x+<y+<r+<f _j or f _k =x+<f _j +<r+<y, _k is obtained by a composite operation. In the present invention, the compound operation is denoted as f(f).

Since the function f is also a word unit as a whole, the partial addition operation is applied to the function. If the functions f _i , f _j satisfy f _i < _□ f _j and the other function f _k can be expressed as the offset of f _i and f _j , ie f _k =f _i +<f _j , the f _{k is} subjected to the partial addition operation And got it.

Each English sentence S that does not omit the predicate verb can be regarded as a result of a finite number of compounding and partial addition operations by n functions f ₁ , ..., f _n (n is equal to the number of predicate verb units). According to this, any English sentence S that does not omit the predicate can be recorded as:

That is, any English sentence that does not omit the predicate is obtained by a composite or partial addition operation of a vector including a guide element, a subject element, a predicate element, or an object element. Next, I am faced with the problem of choosing a reasonable expression for the English natural sentence S. This expression must be able to justify all the compounding and biasing operations contained in S. The matrix form happens to have the condition that the compound operation of the function can be represented by the position of the element in a row vector, for example: f _k (f _j )=f _k (x _k ,f _j ,r _k ,z _k ) It shows the compound operation relationship between f _k and f _j ; at the same time, there is no destructive relationship between the elements: f _k =x _k +<f _j +<r _k +<z _k . In summary, in order to accurately and intuitively express the English natural sentence S, in order to better reveal the intrinsic mathematical structure of the natural sentence S, we adopt the matrix as the primary expression of the natural sentence S.

[B 3.5.2] Linear expression

At the same time, you can also use the linear form to express the statement S, namely:

With particular emphasis on:

1 Each linear expression of the English natural sentence S that does not omit the predicate contains a finite number of partial addition operations and compound operations. This paper uses a linear expression as a supplementary expression of the natural sentence S.

2 The equivalence relation between the matrix expression and the linear expression of the present invention.

3 A linear expression of an English natural sentence S, which is also naturally a linear equation with the function f ₁ , ..., f _n (n is equal to the number of predicate verb units) as an unknown quantity, therefore, the next use of this paper The process of obtaining the parsing result of the syntactic structure by the substitution method can also naturally be regarded as the process of solving the linear equations with the functions f ₁ , . . . , f _n (n is equal to the number of predicate verb units) as an unknown quantity.

B 3.6 Partial matrix substitution solver

step 1

If there are sequential values that do not appear in the possible matrix solution of the syntactic structure, then the syntactic structure may be excluded from the solution; for example, for the possible matrix solutions below

The word unit numbered 4 does not appear and is excluded.

Step 2

If the same order value appears in a different syntax vector or the same syntax vector appears, the possible syntax solution is excluded from the syntax structure;

For example, the following possible matrix solution

The word unit numbered 5 appears twice and is excluded.

Step 3

In each possible matrix solution, the syntactic vectors that find the clear position are all equally substituted. If the cross-contradictions of the two syntactic vectors appear after the equal-substitution, the possible matrix solution may be excluded.

For example, the following possible matrix solution

Substituting the above matrix, f ₂ and f ₃ appear to cross-contradict the function. Substituting: f ₂ = 3 + < e + < 6 + < (4 + < f ₂ + < 7 + < e). The f ₂ appears at both ends of the equation, and there is a logical contradiction. exclude.

Step 4

In each possible matrix solution, the syntactic vectors that find the clear position are all equally substituted. If the order values of the two positions are reversed after the equal substitution, the possible matrix solution may be excluded; this is The fundamental requirement of mathematical processing is also the essential requirement of partial addition operation defined on strict partial order relationship < _□ .

For example, the following possible matrix solution

Substituting it, f ₂ =4+<5+<6+<3+<e+<7+<e, the order is (4,5,6,3,e,7,e), and the position is reversed. Order value, exclude.

Step 5

In any possible matrix solution, if there is a syntax vector that does not have a substitution relationship with other syntax vectors, an insertion operation is performed to obtain a possible syntactic parsing structure corresponding to all the possible matrix solutions, and the parsing according to the possible syntax is verified. Whether the statement obtained by the structure is identical to the preprocessed statement, further comprising:

5.5.1. Firstly, the syntactic vectors with the substitution relationship between the possible matrix solutions are equally substituted, so that the possible matrix solutions are transformed into a set of syntactic vectors with no substitution relationship between each other.

Called the second type of syntax vector;

5.5.2, take a second type of syntax vector

Mark one by one according to the predetermined direction

Second type of syntax vector

Syntactic vector in the form of overall insertion

5.5.3, the third type of syntax vector

Pair vector from the predetermined direction

The first syntactic element on the first side starts into the vector

Vector contained in

The element on the first side, without the order value; the vector

The first syntax element on the second side is marked as

Will be vectored as described above

The syntactic vector part of the annotation, denoted as the iris syntax vector

Syntactic vector in the form of overall insertion

or

Third type of syntax vector

Syntactic vector according to the predetermined direction

The t-th syntax element in the construct, constructing a unique vacancy on one side of the syntactic element; after creating an empty space, taking an unused second-class syntactic vector

The vector is inserted as a whole

5.5.4. Repeat 5.5.3. When the last emptying and emptying steps are completed, the third type of syntactic vector obtained through the previous emptying and emptying steps is subjected to the next emptying and insertion. Null operation until all second type of syntax vectors will be

5.5.5. If there are two position reversed order values in all of the final single row vectors corresponding to a possible syntax parsing structure, the possible syntactic parsing structure is excluded;

5.5.6. Repeat 5.5.2 to 5.5.5 until all possible syntactic parsing structures are traversed.

Part B 3.7 Matrix Correction Procedure

If necessary, transfer to the correction program to correct the results of more than two syntactic structure analysis, including the following operations:

(1) The noun pronoun unit acts as a re-examination and trade-off between the subject and the object.

(2) Use the linguistic rules to check the syntactic structure. Example:

1 According to the rules of English syntactic structure, the guiding words of subject clauses cannot be omitted.

That that guides the subject clause cannot be omitted;

2 According to the rules of English syntactic structure, the subject should be consistent with the predicate in terms of person and quantity;

3 According to the nature of the verbs and the intransitive nature, determine whether the object is connected later.

(3) Re-examination and elimination of structural ambiguity.

(4) Flip, omit, and there be treated as special circumstances.

(5) Put the extracted ingredients back.

(6) Generate and output the final solution.

The correction can overcome the problem of irregular structure of some statements and improve the accuracy of analysis.

Preferably, the syntax structure can be formed into a syntax tree data structure according to the analysis result.

Part 3.8 Particular description of the two methods of insertion

[B 3.8.1] Common principles for two different insertion methods:

The order value sequence of the original sentence, 1, 2, ..., k, can be regarded as an equivalent substitution of the syntactic vector in which the explicit position is found in the possible matrix solution and the unidentified position in the possible matrix solution A finite number of global interpolations between syntax vectors. That is, the initial syntax vector corresponding to the original sentence

It can be seen as an equal-substitution of the syntactic vector in which the explicit position is found in the possible matrix solution, and then through the finite sub-interpolation between the syntactic vectors in the possible matrix solution where the clear position cannot be found. . A variety of different insertions are essentially the permutations and combinations in combinatorial mathematics.

[B 3.8.2] The first type of insertion method:

In any possible matrix solution, if there is a syntactic vector with no explicit substitution relationship with any other syntax vector, firstly, the syntactic vectors with substitutional relations with other syntax vectors in the possible matrix solution are all equal. The quantity substitution is performed, and at the same time, the syntactic vectors in the possible matrix solution and other syntactic vectors do not have an substitution relationship, and the above two aspects are integrated, and the possible matrix solutions are transformed into a group which does not exist with each other. Syntactic vector of substitution relationship

The original syntactic vectors f ₁ , f ₂ , . . . , f _{δ in} the possible matrix solution are collectively referred to as the first type of syntax vector; after the aforementioned equal-substitution, the groups transformed in the foregoing manner are mutually There is no syntactic vector for the substitution relationship

They are collectively referred to as the second type of syntactic vectors; it is emphasized that the second type of syntactic vectors are all syntactic vectors that do not have substitutional relationships with each other. When θ ≥ 2, the overall insertion is meaningful; the following discussion all preset θ ≥ 2.

Next, the single-side directional ordering overall insertion is performed, which can also be called unilateral forward ordering overall insertion: any second type of syntax vector

Label the syntax vectors one by one from right to left (and from left to right)

The order value of each syntax element in . After labeling the order value of the syntax element, take one

Syntactic element in the middle, may wish to set the syntax element is

The i-th element on the right side of the vector, only the left side of the syntax element (or only on the left side) constructs a unique gap; after the empty, take a division vector

Second type of syntax vector

Syntactic vector in the form of overall insertion

Insert the previously constructed space, and then generate a new syntax vector, record this new vector as

The syntactic vectors obtained by the overall insertion are collectively referred to as the third type of syntax vector.

Is the third type of syntactic vector; for the two syntactic vectors α and β, if the vector β is inserted into the space corresponding to the ith slot of the i-th syntax element of the right side of the syntactic vector α, a new one is obtained. The third type of syntactic vector, the newly obtained third type of syntactic vector is recorded as [α] ⁱ + <β; emphasize that the third type of syntactic vectors are all syntactic vectors that do not have substitution relations with each other. The first emptying and emptying steps are completed.

Transfer to the second emptying and inserting steps. The third type of syntactic vector obtained after the first emptying and emptying steps

In the right-to-left direction (also from left to right, but in the same direction as the last order, ie on the same side as the previous order), the slave vector

The first syntactic element in the right side of the number begins in the vector

Vector contained in

Each of the syntax elements up to the first syntax element on the left side is all labeled with a sequence value;

Vector contained in

The syntax element on the left, without the order value; the vector

The first syntax element on the left is recorded as

Will be vectored as described above

The part of the syntax vector of the label, denoted as: syntax vector

The syntactic vector is called: the tail vector. After labeling the order value, take any of the syntactic elements in the aforementioned tail vector, and let the element be the tail vector

The jth element on the right side of the element, only on the right side of the element (can also only be on the left side, but the same direction as the previous emptying, that is, on the same side as the last emptying), constructing a unique vacancy After the air is created, take a syntax vector other than that used in the first emptying and inserting steps.

with

Second type of syntax vector

Syntactic vector in the form of overall insertion

Insert the previously constructed gaps to generate a new syntax vector, then record the newly generated syntax vector as

For any given [alpha] vector syntax and β, the number of beta] vector left in the first syntax element referred to as λ (β), if present, the vector syntax [α] ⁱ + <β, the vector according to the embodiment of [α] ⁱ +<β is annotated, and the part of the syntax vector marked as described above is denoted as: vector [α _k \λ(β _k-1 )], which is called: the tail vector. The second emptying and emptying steps are completed.

According to the foregoing method, the tail vector is selected for the third type of syntactic vector obtained through the previous emptying and emptying steps, and the selected tail vector is labeled with the order value according to the foregoing method, but with the last labeling The selection direction is the same, that is, on the same side as the previous sequence; after the order value is assigned, take a syntax element in the tail vector, and construct a unique one-side vacancy according to the foregoing method, but with the last emptying The selection direction is the same, that is, on the same side as the previous emptying; after the emptying, take a second type of syntax vector other than the syntactic vector used in the previous emptying and insertion steps, to insert the entire empty space. The second type of syntax vector is inserted into the previously constructed gap, and a new syntax vector is generated; the foregoing operation is repeated: whenever the last emptying and emptying steps are completed, the method is followed according to the foregoing method. The third type of syntax vector obtained from the last emptying and insertion steps performs the next emptying and insertion operations until the second type of syntax vector

After all the insertions are completed, the third type of syntactic vector of a single line is finally obtained; the third type of syntactic vector obtained last is called the final single line vector.

The first choice of the second type of syntax vector from the first

A complete process to generate the final single-line vector is taken as a specific solution, so that each of the aforementioned emptying and emptying steps is also a step in the specific solution.

Exhaustive all programs by exhausting all possible situations for each step. Examine each of the final single-line vectors generated by the exhaustive: delete the final single-line vector in which the order values of the two positions are reversed.

If each of the final single-row vectors generated by the exhaustively appears in the order of the two positions, then the natural law is violated, and all the final single-row vectors are excluded, thereby eliminating the possible matrix solution of the syntax structure.

The final single-row vector that does not appear in the order of the two positions is in line with the natural law, and is a reasonable final single-line vector; retain the reasonable final single-line vector as one of the correct results, and retain the syntactic structure. One of the results, in order to generate a syntax tree.

[B 3.8.3] The second type of insertion method:

In any possible matrix solution, if there is a syntactic vector with no explicit substitution relationship with any other syntax vector, firstly, the syntactic vectors with substitutional relations with other syntax vectors in the possible matrix solution are all equal. Substituting the quantity, and at the same time keeping the syntactic vector of the possible matrix solution and other syntactic vectors without the substitution relationship remain unchanged. Combining the above two aspects, the possible matrix solution is transformed into a group that does not exist with each other. Syntactic vector of substitution relationship

The syntactic vectors f ₁ , f ₂ , . . . , f _{δ which} are the original ones of the possible matrix solutions are collectively referred to as the first type of syntactic vectors; after the aforementioned equivalent substitution, a set of mutual transformations will be performed in the foregoing manner. There is no syntactic vector between substitutions

It is collectively referred to as the second type of syntactic vector; it is emphasized that the second type of syntactic vectors are all syntactic vectors that do not have substitutional relationships with each other. When θ ≥ 2, the overall insertion is meaningful; the following discussion all preset θ ≥ 2.

Next, the single-side undirected unordered overall insertion is also referred to as a one-sided forward unpreserved overall insertion: any second type of syntax vector

Syntactic vector in right-to-left direction (also from left to right)

Each syntax element in the dimension is labeled one by one. After labeling the order value of the syntax element, take one

In the syntax element, you may wish to set the element to be

The mth element on the right side of the vector, only the left side of the syntax element (or only on the left side) constructs a unique gap; after the empty, take a division vector

Second type of syntax vector

Put the vector as a whole

The syntactic vectors obtained by the overall insertion are collectively referred to as the third type of syntactic vectors; for the two syntactic vectors α and β, if the vector β is inserted into the space, the mth of the right side of the vector α is inserted. The vacancy corresponding to the syntax element, and a new third type of syntax vector is obtained, and the newly obtained third type of syntax vector is recorded as (α) ^m + < β. The first emptying and emptying steps are completed.

According to the direction from right to left (can also be from left to right, but the same direction as the previous order, that is, on the same side as the previous order), the syntactic vector

In the syntax element, you may wish to set the element to be

The t-th syntax element on the right side of the text, only on the right side of the syntax element (can also be on the left side only, but the same direction as the previous emptying, ie on the same side as the last empty), the unique construction Vacancies; after the void, take a syntax vector other than the one used in the first emptying and inserting steps

with

Second type of syntax vector

The vector is inserted as a whole

The second emptying and emptying steps are completed.

According to the foregoing method, the third type of syntactic vector obtained through the previous emptying and emptying steps is labeled with a sequential value, but the same as the previous sampling order, that is, on the same side as the previous standard; After the sequence value, take a syntactic element in the third type of syntax vector, construct a unique one-sided vacancy according to the above method, but the same direction as the previous emptying, that is, on the same side as the last emptying After the emptying, take a second type of syntactic vector other than the syntactic vector used in the previous emptying and insertion steps, and insert the second type of syntactic vector into the previously constructed vacancy in a global emptying manner. , and then generate a new syntax vector; repeat the foregoing operation: every time the last emptying and emptying step ends, the third obtained after the last emptying and emptying steps are performed according to the foregoing method. The class syntax vector performs the next emptying and inserting operations until the second type of syntax vector

The first choice of the second type of syntax vector from the first

Exhaustive all programs by exhausting all possible situations for each step. Examine every final single-line vector generated by exhaustiveness: under the premise of distinguishing e at different positions in the possible matrix solution, keep two or more identical final single-row vectors one, and remove the redundant identical A single-line vector, and then the final single-line vector with the order value of the two positions reversed.

After deleting the redundant identical single-line vector, if each of the final single-line vectors has two order-reversed values, then the natural law is violated, and all the final single-row vectors are excluded, thereby eliminating the possible matrix solution of the syntax structure.

After deleting the redundant identical single-line vector, the final single-line vector with no order reversal of the two positions is in accordance with the natural law, and is a reasonable final single-line vector; retaining a reasonable final single-line vector as one of the correct results And retain the syntactic structure possible matrix solution as one of the correct results, in order to generate a syntax tree.

[B 3.8.4] Overview and comparison of the characteristics of the two methods:

It can be proved by mathematical methods that all the solutions and all the steps of the above two methods are finite, fixed, and identifiable, and can give the formula for calculating the number of all the specific schemes and all the steps, and can also give The formula for calculating the number of all final single-line vectors generated by exhaustive. Both of the above methods construct corresponding mapping full-aligned sets and corresponding recursive functions as their mathematical models. Each of the two methods described above has a rigorous mathematical basis and rigorous mathematical argumentation. The foregoing two methods are methods that fully conform to the laws of nature.

The foregoing two methods are all developed for the following principles, which are specific implementations of the following principles:

The order value sequence of the original sentence, 1, 2, ..., k, can be regarded as the equivalent substitution of the syntactic vector in the possible matrix solution to find the clear position and the search in the possible matrix solution. Obtained by a finite number of global insertions between syntactic vectors at explicit locations. That is, the initial syntax vector corresponding to the original sentence

Both of the foregoing methods satisfy the requirements of the above principles, and the final results of the two methods are completely consistent. Thus, the two methods described above are equivalent methods.

In selecting the syntactic elements used to make the space: the first method mentioned above has restrictions on the selection of the empty elements, and actually requires the order of inserting the syntax vectors; the second method described above is for emptying. There is no limit to the choice of elements. In fact, it is not required to maintain the order of inserting syntax vectors.

In terms of the final single-line vector: under the premise of distinguishing e at different positions in the possible matrix solution, the final single-line vector generated by the first method described above is a syntactic vector different from each other, and the second method described above The resulting final single-line vector may be similar, so remove the extra identical final single-line vector.

Below, this article will separately explain the above two methods.

[B 3.8.5] Specific description of the first type of insertion method

[B 3.8.5.1] Constructing a mapping as a schema pattern

Note: The parallel noun pronoun combination vector and the associated word combination vector are regarded as a whole and cannot be inserted into the whole of other syntax vectors.

The aforementioned single-sided order overall insertion method will be described in detail below. This method is able to accurately characterize each of the finite sub-interpolations between syntactic vectors that cannot find an explicit position in the possible matrix solution.

Convert the original possible matrix solution into θ new syntax vectors without finding a clear position

Recorded as:

The θ syntactic vectors are fully arranged. According to the related principles of combinatorial mathematics, such a permutation result is

That is, after such a full arrangement, a total of θ! θ element ordered group. The θ that will be obtained through such a full arrangement! The set of θ-element ordered groups is recorded as

(θ≥2)

Construct j θ element mappings ρj, j∈N, 1 ≤ j ≤ θ! Let each θ element map ρj be from the set {t ₁ , t ₂ ,...,t _θ } to the set

The mapping {t ₁ , t ₂ , ..., t _θ } is only used to represent the domain of the mapping ρj, and thus the θ element of the calibration set Φ is fully arranged, and has no other practical meaning. The construction map is as follows:

j∈N, 1 ≤ j ≤ θ! For any given j ₁ and j ₂ , j ₁ ∈N, j ₂ ∈N, ₁ ≤ j ₁ ≤ θ! , ₁ ≤ j ₂ ≤ θ! If j ₁ ≠j ₂ , then ρj ₁ ≠ρj ₂ . (θ≥2)

For the θ-element mapping ρj, the following conclusions are established:

Obviously, for any ρj(t _k ), it exists

1 ≤ δ ≤ θ, making

That is, any ρj(t _k ) is calibrated

a syntactic vector

(θ≥2)

According to the above structure, it can be known that: by θ! The finite set Ω={ρ1, ρ2, ρ3, ρ4, ρ5, ρ6,..., ρθ! } is a collection of finite sets

The θ elements are fully arranged. (θ≥2)

Set π={ρ1, ρ2, ρ3, ρ4, ρ5, ρ6,..., ρθ! }List description: (θ≥2)

Definition 1.1: The second type of syntax vector will be selected in one of the foregoing

Arranges any of the θ second-class syntax vectors used in the complete operational flow of generating the final single-line vector as a scheme pattern. Give a detailed explanation: if

As a scheme mode, it means that the first step is to select the vector.

with

And vector

Insert vector in a single-sided ordering overall insertion

The second step is to select

And

Insert the new syntax vector generated in the first step by inserting the unilaterally ordered order into the whole space, ..., the θth step is to select the vector

And

The new syntax vector generated by the (θ-1)th step is inserted in a single-sided order-slot overall insertion. Obviously, any θ element mapping ρj is a scheme mode, and all θ element mapping ρj sets

Is the collection of all program modes, then the total number of all program modes is θ! One. (θ≥2)

Definition 1.2: The operation of inserting an empty vector and generating a new vector according to any one of the scheme modes as a step of the scheme mode. For any one of the scheme modes ρj, the kth step performed according to ρj is recorded as

[n _k ] indicates that there are a total of n _k kinds of choices for the kth step.

Definition 1.3: Each step performed according to any one of the scheme modes ρj is selected in any specific case, and then each step is combined as a specific scheme. Write any specific plan as

Ρj denotes the scheme mode on which the specific scheme is based, and i _k denotes that the kth step selects the i _kth case on the step.

[B 3.8.5.2] Constructing an empty recursive function

Next, this paper constructs an emptying recursive algorithm based on any scheme mode ρj.

Through the recursive algorithm, it is possible to describe the specific operation process of the above-mentioned single-side ordering overall insertion. Before constructing the interpolation recursive algorithm, first give the following five definitions as preliminary knowledge:

The emptying recursive algorithm to be constructed below

It is the kth step performed according to the scheme mode ρj described above. Where k is the interpolation null recursive algorithm

The number of runs, that is, the number of times the aforementioned one-side ordering overall insertion operation is performed.

Definition 1.4: Give a syntax vector α, and the unary function W indicates that the syntax vector α is taken out and marked. W (α) = α _k represents a vector syntax removed α, and the vector [alpha] Syntax labeled α _k, α _k is called: input vector.

Definition 1.5: Give a syntax vector β, and the unary function Q indicates that the syntax vector β is taken out and marked. Q(β)=β _k denotes that the syntax vector β is taken out, and the syntax vector β is marked as β _k , and β _k is called: an empty vector. Run the recursive algorithm

In the process, the syntax vector β _{k is} inserted into the syntax vector α _k .

1.6 definitions: Z represents a binary function of the syntax of the vector sequence label value α _k, α _k vector syntactic counted from the right in a first syntax element denoted by ordinal value 1, and then from right to left are denoted sequential values 2,3 , ..., until the first syntax element from the left in the vector β _k-1 contained in the vector α _k is attached. The first syntax element from the left in the vector β _k-1 is denoted by λ(β _k-1 ), and the order value of λ(β _k-1 ) is denoted as n _k , then n _k is the aforementioned annotation The maximum order value. The process is characterized by: setting the syntax vector α _k = b ... λ(β _k-1 ) ... b ₂ b ₁ , and the element λ(β _k-1 ) represents the vector β _k-1 The first element in the number from the left. In the kth step, the binary function Z _k (α, β) is run for α _k to be: λ(β _k-1 )<n _k >...b ₂ <2>b ₁ <1> , this result is recorded as: vector [α _k \λ(β _k-1 )], which is called: the tail vector. mark

Representation: The maximum order value for the dovetail vector [α _k \λ(β _k-1 )] is n _k . Run the binary function Z

Definition 1.7: The binary function T represents the overall insertion operation of the tail vector. After the binary function Z is finished, the i th _kth syntax element from the right side is selected on the tail vector, and at the i _kth The right side of the syntax element constructs a unique gap, and then the vector β _k is inserted into the gap as a whole. The process is characterized by: setting the syntax vector α _k = b ... λ(β _k-1 ) ... b ₂ b ₁ , and the element λ(β _k-1 ) represents the vector β _k-1 The first element in the number from the left. Then, the dovetail vector is [α _k \λ(β _k-1 )]=λ(β _k-1 )< n _k >...b ₂ <2>b ₁ <1>. In the kth step, the binary function T _k (α, β) is run on the vector [α _k \λ(β _k-1 )] and the vector β _k , and β _k is inserted into the whole space [α _k \λ(β _k-1 )] The vacancy corresponding to the ith _kth syntax element of the right side is obtained:

Record the new vector obtained by the aforementioned insertion into

Vector

Is: output vector.

Definition 1.8: Give a syntax vector α, and record the number of syntax elements contained in α as σ[α]. If the syntax vector α contains n syntax elements, n∈N, then there is obviously: n=σ[α].

Note: In the definition of the emptying recursive algorithm below, you will see this equation:

Among them, α and β are the meanings of independent variables, which are abstract marks and can be widely used. Therefore, the above notation is not contradictory.

Next, according to the mapping ρj and the four functions described above, the interpolation null recursive algorithm is defined.

as follows:

Note: take a scheme mode ρj; collection

Special emphasis: when k = 1, the interpolation recursive algorithm

Initial condition

Yes:

Recursive algorithm

The characteristic is that the above-mentioned single-side ordering overall insertion operation is decomposed into four links: 1 to take a null vector; 2 to take an empty vector; 3 to mark a specific syntax element in the empty vector, and to intercept the sequence value, and intercept The tail vector; 4 randomly selects a syntactic element in the tail vector, and constructs a unique gap on the right side, and then inserts the inserted vector into the previously constructed gap in a global insertion manner.

[B 3.8.5.3] Insertion recursive algorithm

Operation example

(differentiating e at different locations)

Order set

θ=4, θ! =24, ρj=ρ2.

make

then

Let n _ε be the number of syntactic elements in the tail vector selected by the εth step, and i _ε is the number of element sequences corresponding to the vacancies constructed by the εth step.

Execution scheme mode ρ2, the interpolation recursive algorithm

Run 3 times. Choose a specific solution

<3,4,2>:

< ₃ , 4, 2> operates as follows: (n ₁ ∈N, i ₁ ∈N, ₁ ≤ i ₁ ≤ n ₁ )

It is easy to get n ₁ =4; take i ₁ =3, get: T ₁ (α,β)=(7 e _C (e _A 5 6 e _B )8 e _D )

< ₃ , ₄ , ₂ > operates as follows: (n ₂ ∈N, i ₂ ∈N, 1 ≤ i ₂ ≤ n ₂ )

Easy to get n ₂ =6; take i ₂ =4, get:

T ₂ (α,β)=(7 e _C (e _A 5 6(1 2 3 4)e _B )8 e _D )

< ₃ , ₄ , ₂ > operates as follows: (n ₃ ∈N, i ₃ ∈N, 1 ≤ i ₃ ≤ n ₃ )

Easy to get n ₃ =7; take i ₃ =2, get:

T ₃ (α,β)=(7 e _C (e _A 5 6(1 2 3 4)e _B )8(e _E 9 10 e _F )e _D )

[B 3.8.5.4] Description of the exhaustion and dissimilarity of the law

Conclusion 1.1: The full permutation set of the aforementioned θ-metabol map

Insertion null recursive algorithm

Exhausting all of the above possibilities for a finite number of overall insertions between syntactic vectors that cannot find an explicit position.

Conclusion 1.2: According to the aforementioned interpolation recursive algorithm

As can be seen from the definition, in any two specific schemes of the method, there are two different steps, that is, any two specific schemes are different. Therefore, under the premise of distinguishing e at different positions, all the final single-line vectors generated by all the specific schemes of the method are different syntactic vectors.

[B 3.8.5.5] The first set of calculation formulas and related proofs

Note: The following lemmas and theorems are given in any of the schemes in this paper, ρj and the interpolation null recursive algorithm.

Discussion under the premise of the discussion; all under the premise of distinguishing e at different positions in the possible matrix solution.

Definition 1.9: Run the insertion recursion calculation

The number of the tail vectors intercepted in the kth step is n, and it is possible to set the n tail vectors intercepted in the kth step to be α ₁ , α ₂ , . . . , α _n Then, the entire insertion of each of these vectors is performed in the kth step, and the sum of the number of insertions of these vectors is denoted as τ [∑(k)].

Lemma 1.1: (definition of σ, see definition 2.5)

Certificate: The following:

(1) When k≥2, according to the above definition 2.4, the tail vector of the kth step

Syntactic element

Represents the element contained in the vector α _k , <β _k-1 > represents the element of the vector β _k-1 in the vector α _k

Corresponding vacancies. According to the expression of the dovetail vector [α _k \λ(β _k-1 )], the number of elements in the syntax vector is obviously: σ[β _k ]+(i _k -1). The conclusion to be proved is established.

(2) When k=1, according to the aforementioned interpolation recursion algorithm

The definition is obtained directly.

Certificate

Lemma 1.2: Given a k, k∈N and k ≥ 1, the maximum order value of the syntax elements of the end vector vector intercepted on the kth step is:

Certificate: algorithm

The recursive definition of the function Z _k (α, β), the maximum order value of the syntax elements labeled for the tail vector [α _k \λ(β _k-1 )] intercepted at the kth step is:

According to Lemma 1.1, the number of syntactic elements contained in the tail vector [α _k \λ(β _k-1 )] of the kth step is:

The maximum order value of the syntax elements of the end vector vector intercepted on the kth step is:

Certificate

Lemma 1.3: Insertion recursive algorithm

Generated specific plan

The number of <i ₁ ,i ₂ ,...,i _k > is: (definition of σ, see definition 2.5) (k∈N,k≥1)

Proof: According to Lemma 1.2, given a k, k∈N and k ≥ 1, the maximum order value of the syntax elements of the tail vector vector intercepted in the kth step is:

Algorithm

Interpolation functions of T _k (α, β) recursive definition, i is in the range _1: 1≤i _₁ ≤n _1, i.e. 1≤i _₁ ≤σ [α _1].

Algorithm

Interpolation functions of T _k (α, β) recursive definition, k≥3 i _k-1 when the ranges _{_{are: 1≤i k-1 ≤n k-}} 1, i.e. 1≤i _k-1 ≤ σ[β _k-2 ]+(i _k-2 -1).

Let α ₁ =a _σ [α ₁ ]...a ₂ a ₁ ,β ₁ =λ(β ₁ )...b ₂ b ₁ , then according to the above results and the interpolation recursive algorithm

The definition, the total number of insertions in the first step is σ[α ₁ ], that is, β ₁ has σ[α ₁ ] insertions of α ₁ .

Let us set β ₂ =λ(β ₂ )...c ₂ c ₁ ,

Based on the aforementioned results and algorithms

The recursive definition of the interpolation function T _k (α, β) in the given case, for any given one i ₁ , ₁ ≤ i ₁ ≤ σ [α ₁ ], the number of insertions in the second step is σ [β ₁ ]+(i ₁ -1); Since the value of i ₁ is to traverse the entire natural number in the real interval [1, σ[α ₁ ]], then for all values of i ₁ , according to the summation ∑ The way to count, you can calculate: the total number of insertions in the second step is

That is, β ₂ has

Insertion of the insertion of the tail vector.

Let us set β ₃ =λ(β ₃ )...d ₂ d ₁ ,

or

Based on the aforementioned results and algorithms

The recursive definition of the interpolation function T _k (α, β) in the given step, for any given i ₂ , 1 ≤ i ₂ ≤ σ [β ₁ ] + (i ₁ -1), the insertion of the third step The number of cases is σ[β ₂ ]+(i ₂ -1); since the value of i ₂ is to traverse the entire natural number in the real interval [1, σ[β ₂ ]+(i ₂ -1)], The value of i ₁ is to traverse the entire natural number in the real interval [1, σ[α ₁ ]], and then the total value of i ₂ is counted and accumulated according to the summation ,, which can be calculated: the third step The total number of insertions is

That is, β ₃ has

Insertion of the insertion of the tail vector.

The following mathematical proof is used to prove that in the kth step, the whole of the tail vector is globally inserted, and the total number of insertions τ[∑(k)] obtained is: (k≥2)

Hypothesis: In the kth step, the entire tail vector is globally inserted, and the total number of insertions τ[∑(k)] obtained is:

Algorithm

The recursive definition of the interpolation function T _k (α, β) in the middle, the range of i _k when _k ≥ 2 is: 1 _≤ i _{k ≤} n _k , that is, 1 _{≤ i k} ≤ σ [β _k-1 ] + (i _k-1 -1).

Algorithm

The recursive definition of the interpolation function T _k (α, β), for any given one i _k , 1 _≤ i _k _≤ σ [β _k-1 ] + (i _k-1 -1), the (k The number of insertions in +1) steps is σ[β _k ]+(i _k -1); since the method of i _k is to traverse the real interval [1, σ[β _k-1 ]+(i For all the natural numbers in _k-1 -1)], for any given value of σ[β _k-1 ]+(i _k-1 -1), the count is accumulated and traversed according to the method of summation ∑ The total value of i _k is easy to calculate: the total number of insertions of the (k+1)th step obtained by accumulating the number is

That is, for any given value of σ[β _k-1 ]+(i _k-1 -1), β _k+1 has

Insertion of the insertion of the tail vector.

Further, the induction hypothesis has provided an expression of the formula σ[β _k-1 ]+(i _k-1 -1), that is, σ[β _k-1 ]+(i _k-1 - can be determined by the inductive hypothesis 1) All values. So, according to the way of summation, the count is accumulated, from the formula

Starting, based on the assumption of induction hypothesis

Traversing i _k-1, ......, i full section _2, i ₁ values, thereby erasing the parameter _{i k-1, ......, i} 2, i 1, directly calculate: a first The total number of insertions in (k+1) steps is:

That is, β _k+1 has a total

Insertion of the insertion of the tail vector. The conclusion to be proved is established and the mathematical induction method is completed.

Based on the above results, the total number of insertions τ [∑(k)] of the kth step is as follows: (k ≥ 1)

Insertion recursive algorithm

Definition, the total number of insertions in the kth step τ [∑(k)], which is the algorithm

The total number of insertions in the last step τ[∑(k)], and the algorithm

Generated specific plan

The number of the numbers is equal, combining the above results, the interpolation recursive algorithm

Generated specific plan

The number is: (the definition of σ, see definition 2.5) (k∈N, k ≥ 1)

Theorem 1.1: The specific scheme for generating any scheme pattern ρj in the method of this paper

The number of the number is Ω[ρj], then: Ω[ρj]=(the formula is as follows) (θ≥2)

Card: According to the insertion recursion algorithm

The definition of the combination, combined with the definition of the use of Lemma 1.3, the conclusion to be proved is clearly established. Certificate

Theorem 1.2: The number of the final single-row vector generated by any one of the scheme modes ρj in this method is Ω[ρj], then: Ω[ρj]=(Formula is as follows) (θ≥2)

Proof: According to the definition of the final single-line vector and the specific scheme, the number of final single-line vectors is equal to the number of specific schemes. It is also available according to Theorem 4.1.1. Certificate

[B 3.8.5.6] Example of operation of the first type of insertion method

Take collection

θ=4, θ! =24, ρj=ρ2.

take

then

The number of final single-row vectors generated by the scheme mode ρ2 is Ω[ρ2]= (the formula is as follows):

When i ₁ =1:

When i ₁ = 2:

When i ₁ = 3:

When i ₁ = 4:

The number of final single-line vectors generated by the scheme mode ρ2 is: Ω[ρ2]=140.

The number of final single-line vectors generated by all the schema modes of this method is:

[B 3.8.5.7] The second set of calculation formulas and related proofs

Recorded as:

Newly generated syntax vectors that cannot find an explicit position

They are collectively referred to as the second type of syntax vector. Any syntactic vector that is eliminated in the aforementioned equal-substitution process is called a predecessor syntax vector. For any newly generated second type of syntax vector

Will be in the aforementioned equal replacement process

The number of the predecessor syntax vector f replaced is denoted as u _ε . then

It is obtained by sub-equivalent substitution of u _ε .

For example, by vector f ₂ =e+<f ₃ +<7+<f ₅ and f ₃ =3+<e+<4+<e and f ₅ =8+<e+<9+<10 are generated by equal substitution a second class vector

It is obvious that u ₁ = 2, ie

It was obtained after two equal replacements.

Theorem 1.3: Give a second type of syntax vector

will

The number of syntax elements included in the record is

Syntactic vector

The number of the predecessor syntax vector f that is eliminated is denoted as u _ε , then

Meet the recurrence formula:

Certificate: The use of mathematical induction proves as follows:

(1), if u _ε =0, the syntax vector

The number of predecessor syntax vectors f that are eliminated is 0, that is,

Is a syntactic vector in the original possible matrix solution without equal substitution, syntactic vector

The number of syntactic elements contained in is 4, obviously the formula at this time

Established.

(2), suppose when u _ε = k

Established, at this time

The number of predecessor syntax vectors f that are eliminated is k,

The number of elements contained in is 3k+4; when u _ε =k+1, it can be regarded as

First eliminate the k predecessor syntax vector f, and then based on this,

After subtracting one of its syntactic elements, it introduces a predecessor syntax vector f, namely

Four elements were introduced while subtracting one of their own syntax elements. Then at this time

The number of elements contained in it is 3k+4+3, then the formula

Established. Based on the comprehensive (1) and (2), it can be seen that the conclusion to be proved is established.

[B 3.8.5.8] Summary of the formula for calculating the number of methods in this paper

Conclusion 1.3: Under the premise of distinguishing e at different positions, the number of specific schemes corresponding to any scheme pattern ρj in the method is the same as the number of final single-row vectors generated by the scheme pattern ρj, and is recorded as Ω. [ρj], then: Ω[ρj]=(Formula is as follows) (for the definition of σ, see definition 1.5) (θ≥2)

Conclusion 1.4: It is possible to define d _θ =σ[ρj(t _θ )], and the formula of conclusion 1.3 is transformed into: Ω[ρj]=(as follows) (for the definition of σ, see definition 1.5) (θ≥2)

Conclusion 1.5: According to theorem 1.3, there are: Ω[ρj]=(below) (for the definition of σ, see definition 2.5) (θ≥1)

Conclusion 1.6: Define g _θ =3u _θ +4, then the Ω[ρj] formula of conclusion 1.5 is transformed into: Ω[ρj]=(see below) (for definition of σ, see definition 2.5) (θ≥2)

Conclusion 1.7: Because a total of θ is generated during the implementation of the method in this paper! In the scheme mode, under the premise of distinguishing e at different positions, the number of all specific schemes generated by all scheme patterns in the method is the same as the number of all final single-row vectors generated, and the formula is:

(θ≥2)

Conclusion 1.8: Combining each of the above conclusions, a limited number of specific schemes and a limited number of final single-row vectors are generated during the implementation of the method herein. The number of specific schemes and the number of final single-row vectors are determined, and there are exact calculation formulas and corresponding proofs, which are in accordance with the laws of nature.

[B 3.8.5.9] A comprehensive demonstration of the first method

For example: take the following possible matrix solution, and firstly replace the syntactic vector in the possible matrix solution to find the clear position. Let: the possible matrix solution is as follows:

The original possible matrix solution is converted to:

set

The list is as follows: (collection

θ=3, θ! =6)

Construct the mapping ρj as follows: (θ=3,

)

Ρj:

j∈N, 1≤j≤6

The list of sets π={ρ1, ρ2, ρ3, ρ4, ρ5, ρ6} is as follows:

Execution scheme mode ρ1, to recursive algorithm

Run 2 times.

The operation is as follows: (n ₁ ∈N, i ₁ ∈N, ₁ ≤ i ₁ ≤ n ₁ )

Operates as _{_{follows: (n 2 ∈N, i 2}} ∈N, 1≤i 2 ≤n 2)

According to the scheme mode ρ1,

List:

According to the scheme mode ρ1,

List:

Execution scheme mode ρ2, to recursive algorithm

Run 2 times.

The operation is as follows: (n ₁ ∈N, i ₁ ∈N, ₁ ≤ i ₁ ≤ n ₁ )

Operates as _{_{follows: (n 2 ∈N, i 2}} ∈N, 1≤i 2 ≤n 2)

According to the scheme mode ρ2,

List:

According to the scheme mode ρ2,

List:

Execution scheme mode ρ2, to recursive algorithm

Run 2 times.

The operation is as follows: (n ₁ ∈N, i ₁ ∈N, ₁ ≤ i ₁ ≤ n ₁ )

Operates as _{_{follows: (n 2 ∈N, i 2}} ∈N, 1≤i 2 ≤n 2)

According to the scheme mode ρ3,

List:

According to the scheme mode ρ3,

List:

The process of executing the scheme mode ρ4--ρ6 is omitted.

Plug-in recursive algorithm

The list of important information is summarized as follows:

The above formula can also be converted into a form represented by the number of predecessor syntax vectors.

A list of the number of elements in the syntactic vector and the number of elements in the second type of syntax vector:

The important information of the interpolation recursive algorithm is expressed by the number of predecessor syntax vectors. The list is as follows:

Check each of the final single-line vectors for the order of the two positions reversed, omitted.

The example is fully demonstrated.

[B 3.8.6] Specific description of the second type of insertion method

The aforementioned single-sided unscheduled overall insertion method will be described in detail below. This method can accurately describe the search in the possible matrix solution. There is no finite number of global insertions between the syntactic vectors of the explicit position.

[B 3.8.6.1] Constructing an empty recursive function

Refer to the first method for the definition of the scheme model, specific schemes, and steps. The following is the difference between the second method and the first method. Constructing an emptying recursive algorithm

Through the recursive algorithm, the specific process of the above-mentioned one-side unscheduled overall insertion can be described. Before constructing the step recursive algorithm, first give the following five definitions as preliminary knowledge:

The emptying recursive algorithm to be constructed below

It is the kth step performed according to the scheme mode ρj. Where k is the interpolation null recursive algorithm

The number of runs, that is, the number of times the aforementioned one-sided unscheduled overall insertion operation is performed.

Definition 2.1: Give a syntax vector α, and the unary function W indicates that the syntax vector α is taken out and marked. W(α)=α _k denotes taking out the syntax vector α and marking the syntax vector α as α _k .

Definition 2.2: Give a syntax vector β, and the unary function Q denotes the extraction and marking of the syntax vector β. Q(β)=β _k denotes taking out the syntax vector β and marking the syntax vector β as β _k . Run the recursive algorithm

2.3 definitions: Z represents a univariate function of the syntax of sequence annotation vector α _k value, the syntax from the left vector α _k in a syntax element from a first value of an order denoted, from left to right and then successively label value 2,3, ... until all the syntax elements in the syntax vector α _k are marked. Record the maximum order value of the label as n _k . Run the unary function Z to get:

Definition 2.4: The binary function T indicates that after applying the unary function Z to the syntactic vector α _k , the m _kth element of the left is selected on the vector α _k and a unique gap is constructed on the right side of the m _kth element. Then, the syntax vector β _k is inserted into the slot in a globally inserted manner. Write the new vector obtained after inserting the empty space as:

Run the binary function T to get:

Definition 2.5: Give a syntax vector α, and record the number of syntax elements contained in α as σ[α]. If the syntax vector α Containing n syntax elements, n∈N, obviously has: n=σ[α].

As follows: (collection

)

[B 3.8.6.2] Formula for calculating the number of specific schemes and final single-line vectors

Lemma 2.1:

Card: Run recursive algorithm in accordance with the aforementioned definition

In the process, the syntactic elements in the null vector α _k and the null vector β _k are not increased or decreased, that is, for any one syntax element b:

1 if b∈α _k or b∈α _k , then

2 if

and

then

According to 1 and 2, there are obviously:

Certificate

Lemma 2.2: Let: (k∈N, k≥1)

The syntax vector ∑Ψ _k represents a syntactic vector obtained by ρj(t ₁ ), ρj(t ₂ ), ..., ρj(t _k ) sequentially passing through a single-sided unpreserved global interpolation. m ₁ , m ₂ , ..., m _k-1 respectively represent the number of any gap order of the corresponding vector. Then the following conclusions are established:

(Note:

)

Certificate: The use of mathematical induction proves as follows:

(1), if k=1,

The conclusion was established.

(2), assuming that when k = h is established, there is

When k=h+1,

Then there are:

According to Lemma 2.1, you can get:

According to the induction hypothesis:

Thus available:

Also available:

Certificate

Theorem 2.1: The number of specific schemes generated by any one of the scheme modes ρj in this method is recorded as Ω[ρj], then:

Card: According to the insertion recursion algorithm The definition, for any one of the scheme modes ρj, the number of insertions of the kth step is the same as the number of syntax elements of the nullation vector α _k of the kth step. According to the foregoing definition, it is obvious that α _k = ∑Ψ _k , according to Lemma 2.2, the kth step of the scheme mode ρj has

a situation. Since any scheme mode ρj has jurisdiction (θ-1) steps, according to the multiplication principle of combinatorial mathematics, it can be known that any scheme mode ρj of the method corresponds to

Specific programs.

Certificate

Theorem 2.2: The number of final single-row vectors generated by any one of the scheme modes ρj in this method is recorded as Ω[ρj], then:

Proof: According to the definition of the final single-line vector and the specific scheme, the number of final single-row vectors is equal to the number of specific schemes; and according to Theorem 2.1, it is available.

Conclusion 2.1: This method has a total of θ! According to the theorem 2.1, and according to the additive principle of combinatorial mathematics, the total number of specific schemes is:

Conclusion 2.2: This method has a total of θ! According to theorem 2.2, and according to the additive principle of combinatorial mathematics, the total number of final single-line vectors is:

[B 3.8.6.3] Example demonstration of the second method

The original possible matrix solution is converted to:

set

The list is as follows: collection

θ=3, θ! =6

Construct the mapping ρj as follows: (θ=3,

)

Ρj:

j∈N, 1≤j≤6

The list of sets π={ρ1, ρ2, ρ3, ρ4, ρ5, ρ6} is as follows:

Execution scheme mode ρ1, step recursive algorithm

Run 2 times.

Run as follows:

Run as follows:

Run the insertion function according to the scheme mode ρ1

List:

Run the insertion function according to the scheme mode ρ1

List:

The process of executing the scheme mode ρ2--ρ6 is omitted.

Plug-in recursive algorithm

The list of important information is summarized as follows:

One or more identical final single-row vectors are reserved for one, and the remaining identical single-row vectors are deleted, and finally 210 consecutive single-row vectors that are different from each other are completely consistent with the results of Method 1.

The example is complete.

Part C application example

C1 part example 1

Example 1: By preprocessing, you can remove the impurities in the statement and label and identify the word unit number and type in the statement. For example, for the English sentence S=“I can completely understand what what you just said really meant”, the sentence S=“I can understand what what you said meant” after removing the impurities, in which the word unit is recognized and the word After the unit type is labeled and numbered, you can get the data structure that matches the table below.

语句Statement	词单元类型Word unit type	编号Numbering
语句Statement	词单元类型Word unit type	编号Numbering	II	名词代词单元Noun pronoun unit	11
can understandCan understand	谓语动词单元Predicate verb unit	22	II	名词代词单元Noun pronoun unit	11
can understandCan understand	谓语动词单元Predicate verb unit	22	what AWhat A	从属关联词单元Subordinate unit	33
what BWhat B	从属关联词单元Subordinate unit	44	what AWhat A	从属关联词单元Subordinate unit	33
what BWhat B	从属关联词单元Subordinate unit	44	youYou	名词代词单元Noun pronoun unit	55
saidSaid	谓语动词单元Predicate verb unit	66	youYou	名词代词单元Noun pronoun unit	55
saidSaid	谓语动词单元Predicate verb unit	66	meantMe	谓语动词单元Predicate verb unit	77

The present invention is based on syntactic analysis of the pre-processed statements represented by the above data structures to obtain the component relationships of the various word units in the sentences.

1 is a flow chart of a method for parsing a computer-based natural language syntax structure according to an embodiment of the present invention. As shown in FIG. 1, the method includes:

Step 110: Read a pre-processed statement data structure to be parsed, where the pre-processed statement data structure includes only a related word unit, a predicate verb unit, and a noun pronoun unit of the sentence, and each word unit is in accordance with the The order in the preprocessed statement is numbered and labeled.

Step 120: Generate, for each predicate verb unit, a corresponding guide element, a subject element, a predicate element, and an object element; the possible value of the guide element is a parallel related word unit or subordinate whose number is smaller than the corresponding predicate verb unit number. One of the related word units, or a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and a related word combination composed of a dependent related word unit whose number is smaller than the corresponding predicate verb unit number and whose number is greater than the parallel related word unit number One of the vectors, or an empty unit;

The predicate element is a corresponding predicate verb unit;

The possible value of the object element is one of the noun pronoun units whose number is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number, or the number of the smallest word unit is greater than the corresponding predicate verb unit number. And one of the parallel noun pronoun combination vectors included in the entire parallel noun pronoun combination vector family of adjacent predicate verb unit numbers, or one of the syntactic vectors corresponding to the predicate element, or an empty unit .

Specifically, for the preprocessed statement, the total number of predicate verb units is n, and since the predicate verb unit can only be used as a predicate, each predicate verb unit corresponds to one predicate element, and each predicate verb unit is r _k , k = 1, ..., n.

After obtaining the predicate element, the corresponding guide element, subject element, and object element are generated based on the position number of each predicate element.

I, guide element

The set of related word units corresponding to each predicate verb unit r _k is:

{x _k }=Lead _k ∪conj _k ∪(conj _k o Lead _k )∪{e}

The leader element corresponding to the verb unit r _k is x _k , and its possible value set is {x _k }. Generating a possible set of values (preferred) in which the leader element corresponding to the predicate verb unit r _k is x _k includes:

The possible value of the guide element is one of a parallel related word unit or a dependent related word unit whose number is smaller than the corresponding predicate verb unit number, or a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and one adjacent thereto One of the associated word combination vectors consisting of the dependent term unit numbers whose number is smaller than the corresponding predicate verb unit number and whose number is greater than the parallel related word unit number, or an empty unit.

That is, x _k ∈Lead _k ∪conj _k ∪(conj _k o Lead _k )∪{e}.

In the above formula, the set Lead _k represents a set of dependent related word units whose number is smaller than the corresponding predicate verb unit number; conj _k represents a set of parallel related word units whose number is smaller than the corresponding predicate verb unit number; (conj _k o Lead _k ) represents a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and a related word combination vector set formed by the dependent related word unit whose number is smaller than the corresponding predicate verb unit number and whose number is greater than the parallel related word unit number, and e represents an empty unit .

For example, for the pre-processed statement S=“I can understand what what you said meant” shown in Table 1 above, there are:

r ₁ = "can understand", for r ₁ there is {x ₁ }={e}, that is, the value of the leader element corresponding to r ₁ is an empty cell.

r ₂ = "said", for r ₂ there is {x ₂ }={what A, what B, e}, and the value of the leader element corresponding to r ₂ is the first what or the second in the sentence. , that is, one of "what A" and "what B", or an empty unit.

r ₃ = "meant", for r ₃ there is {x ₃ }={what A, what B, e}, and the value of the leader element corresponding to r ₃ is the first what or the second in the sentence. , that is, one of "what A" and "what B", or an empty unit.

II, subject elements

The set of subject noun pronouns corresponding to each predicate verb unit r _k is {y _k }=NPI _yk ∪VNP _yk ∪NOMP _k ∪G _k ∪{e} or {y _k }=NPI _yk ∪VNP _yk ∪NOMP _k ∪G _k ∪fy _k ∪{e}.

The subject element corresponding to the verb unit r _k is y _k , and its possible value set is {y _k }.

Generating the corresponding subject element y _k preferably includes:

(1) When the corresponding predicate verb unit number is the smallest predicate verb unit number, the possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of its largest word unit One of the parallel noun pronoun combination vectors included in the vector of the total parallel noun pronoun combination vector of the corresponding predicate verb unit number, or an empty unit.

That is, when r _k-1 does not exist: {y _k }=NPI _yk ∪VNP _yk ∪NOMP _k ∪G _k ∪{e};

_{_{_{Whereby, y k ∈NPI yk ∪VNP yk ∪NOMP}}} k ∪G k ∪ {e}.

In the above formula, the set NPI _yk represents a pure noun unit set whose number is smaller than the corresponding predicate verb unit number; VNP _yk represents a verb unit set whose number is smaller than the noun nature of the corresponding predicate verb unit number; the NOMP _k number is smaller than the corresponding predicate The set of the main lattice pronoun units of the verb unit number; G _k represents a union of the total number of parallel unit noun pronoun combination vector numbers whose number of the largest word unit is smaller than the corresponding predicate verb unit number; e represents an empty unit.

(2) When the corresponding predicate verb unit number is not the smallest predicate verb unit number, the possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than The corresponding predicate verb unit number is one of the parallel noun pronoun combination vectors included in the common noun pronoun combination vector family, or one of the syntactic vectors corresponding to the predicate verb unit, or an empty unit.

That is, when r _k-1 is present: {y _k }=NPI _yk ∪VNP _yk ∪NOMP _k ∪G _k ∪fy _k ∪{e}.

_{_{_{Whereby, y k ∈NPI yk ∪VNP yk ∪NOMP}}} k ∪G k ∪fy k ∪ {e}.

In the above formula, the set NPI _yk represents a pure noun unit set whose number is smaller than the corresponding predicate verb unit number; VNP _yk represents a verb unit set whose number is smaller than the noun nature of the corresponding predicate verb unit number; the NOMP _k number is smaller than the corresponding predicate a set of primary lattice pronoun units of the verb unit number; G _k represents a union of the total number of parallel unit noun pronouns combined by the number of largest word units; fy _k represents the predicate verb unit corresponding to the preceding A set of syntax vectors; e represents an empty cell.

r ₁ = "can understand", which has the lowest numbered predicate verb unit for r ₁ , therefore, {y ₁ }=NOMP ₁ ∪{e}={I,e}.

r ₂ = "said", for r ₂ there is a predicate verb unit that is not the lowest number, the noun pronoun unit between r ₁ and r ₂ has only "you", and the function with number less than 2 is f ₁ , therefore, {y ₂ }=NOMP ₂ ∪fy ₂ ∪{e}={I,you}∪{f ₁ }∪{e}.

r ₃ = "meant", for r ₃ which is not the lowest numbered predicate verb unit, there is no noun pronoun unit between r ₂ and r ₃ , and the function with number less than 3 is f ₁ and f ₂ , therefore, there are: { y ₃ }=NOMP ₃ ∪fy ₃ ∪{e}={I,you}∪{f ₁ ,f ₂ }∪{e}.

III. Object elements

The set of object noun pronouns corresponding to each predicate verb unit r _k is {z _k }=NPI _zk ∪VNP _zk ∪OBJP _k ∪H _k ∪{e} or {z _k }=NPI _zk ∪VNP _zk ∪OBJP _k ∪H _k ∪fz _k ∪{e}.

At the same time, the leader element corresponding to the predicate verb unit r _k is z _k , and its possible value set is {z _k }.

Generating the corresponding object element {z _k } preferably includes:

(1) When the corresponding predicate verb unit number is the largest predicate verb unit number, the possible value of the object element is one of the noun pronoun units whose number is greater than the corresponding predicate verb unit number, or the number of its smallest word unit One of the parallel noun pronoun combination vectors, or an empty unit, contained in the vector of the entire parallel noun pronoun combination vector of the corresponding predicate verb unit number.

That is, when r _k+1 does not exist: {z _k }=NPI _zk ∪VNP _zk ∪OBJP _k ∪H _k ∪{e}.

In the above formula, the set NPI _zk represents a set of pure noun units whose number is greater than the corresponding predicate verb unit number; VNP _zk represents a set of verb units whose number is greater than the noun nature of the corresponding predicate verb unit number; OBJP _k indicates that the number is greater than the corresponding a set of binge pronoun units of the noun nature of the predicate verb unit number; H _k represents a union of the total number of parallel lexical pronoun combination vectors of the smallest word unit number greater than the corresponding predicate verb unit number; e represents an empty unit.

(2) When the corresponding predicate verb unit number is not the largest predicate verb unit number, the possible value of the object element is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number. One of the noun pronoun units, or the number of its smallest word unit is greater than the corresponding predicate verb unit number and less than the adjacent verb noun pronoun combination vector family of the collateral noun pronoun combination vector number of the adjacent predicate verb unit number One, or one of the syntactic vectors corresponding to the predicate verb unit that appears later, or an empty cell.

That is, when r _k+1 is present: {z _k }=NPI _zk ∪VNP _zk ∪OBJP _k ∪H _k ∪fz _k ∪{e}.

In the above formula, the set NPI _zk represents a pure noun unit set whose number is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number; VNP _zk indicates that the number is greater than the corresponding predicate verb unit number and a set of verb units smaller than the adjacent nouns of the predicate verb unit number; OBJP _k represents a set of binge pronoun units larger than the corresponding predicate verb unit number and smaller than the adjacent predicate verb unit number; H _k represents a union of the total number of parallel word nouns combined with the corresponding predicate verb unit number and less than the adjacent predicate verb unit number; the fz _k represents the predicate that appears later The set of syntax vectors corresponding to the verb unit; e represents the empty unit.

r ₁ = “can understand”, for r ₁ there is a predicate verb unit that is not the largest number, there is a noun pronoun unit “you” between r ₁ and r ₂ , no parallel noun pronoun combination vector, and a function with a number greater than 1 Is f ₂ , f ₃ , therefore, {z ₁ }=OBJP ₁ ∪fz ₁ ∪{e}={you}∪{f ₂ ,f ₃ }∪{e}.

r ₂ = "said", for r ₂ which is not the highest numbered predicate verb unit, there is no noun pronoun unit between r ₁ and r ₂ , and the function with number greater than 2 is f ₃ , and there is no parallel noun pronoun combination vector, Therefore, there are: {z ₂ }=fz ₂ ∪{e}={f ₃ }∪{e}.

r ₃ = "meant", r _3, which is for the maximum number of units verb, noun r ₃ after no pronouns unit, nor parallel pronouns term combination vector, and the number greater than 3, the function does not exist, therefore, {z ₃ }={e}.

Thus, through the processing in step 120, for the above example, a set of values for each element can be generated.

Step 130: Obtain all possible values of a syntax vector corresponding to each predicate verb unit according to possible values of the guide element, the subject element, the predicate element, and the object element, where the syntax vector includes a guide element and a subject element. , predicate elements, object elements.

As mentioned earlier, each subject-predicate collocation structure can be represented by a syntactic vector. According to the operation result of step 120, for the preprocessed statement S=“I can understand what what you said meant” shown in Table 1 above, there are:

{r ₁ }={can understand}

{x ₁ }={e}

{y ₁ }={I,e}

{z ₁ }={you,f ₂ ,f ₃ ,e}

Apply the principle of multiplication in combinatorial mathematics: f ₁ (x ₁ , y ₁ , r ₁ , z ₁ )= (see list below)

序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(e，I，r₁，you)f ₁ = (e, I, r ₁ , you)	(1-5)(1-5)	f₁＝(e，e，r₁，you)f ₁ =(e,e,r ₁ ,you)
(1-2)(1-2)	f₁＝(e，I，r₁，f₂)f ₁ = (e, I, r ₁ , f ₂ )	(1-6)(1-6)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-1)(1-1)	f₁＝(e，I，r₁，you)f ₁ = (e, I, r ₁ , you)	(1-5)(1-5)	f₁＝(e，e，r₁，you)f ₁ =(e,e,r ₁ ,you)
(1-2)(1-2)	f₁＝(e，I，r₁，f₂)f ₁ = (e, I, r ₁ , f ₂ )	(1-6)(1-6)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-3)(1-3)	f₁＝(e，I，r₁，f₃)f ₁ = (e, I, r ₁ , f ₃ )	(1-7)(1-7)	f₁＝(e，e，r₁，f₃)f ₁ = (e, e, r ₁ , f ₃ )
(1-4)(1-4)	f₁＝(e，I，r₁，e)f ₁ = (e, I, r ₁ , e)	(1-8)(1-8)	f₁＝(e，e，r₁，e)f ₁ = (e, e, r ₁ , e)	(1-3)(1-3)	f₁＝(e，I，r₁，f₃)f ₁ = (e, I, r ₁ , f ₃ )	(1-7)(1-7)	f₁＝(e，e，r₁，f₃)f ₁ = (e, e, r ₁ , f ₃ )

Replace the constant with the order value to get: f ₁ (x ₁ , y ₁ , r ₁ , z ₁ )= (see list below)

序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(e，1，2，5)f ₁ = (e, 1, 2, 5)	(1-5)(1-5)	f₁＝(e，e，2，5)f ₁ = (e, e, 2, 5)
(1-2)(1-2)	f₁＝(e，1，2，f₂)f ₁ = (e,1,2,f ₂ )	(1-6)(1-6)	f₁＝(e，e，2，f₂)f ₁ = (e, e, 2, f ₂ )	(1-1)(1-1)	f₁＝(e，1，2，5)f ₁ = (e, 1, 2, 5)	(1-5)(1-5)	f₁＝(e，e，2，5)f ₁ = (e, e, 2, 5)

(1-3)(1-3)	f₁＝(e，1，2，f₃)f ₁ =(e,1,2,f ₃ )	(1-7)(1-7)	f₁＝(e，e，2，f₃)f ₁ = (e, e, 2, f ₃ )
(1-3)(1-3)	f₁＝(e，1，2，f₃)f ₁ =(e,1,2,f ₃ )	(1-7)(1-7)	f₁＝(e，e，2，f₃)f ₁ = (e, e, 2, f ₃ )	(1-4)(1-4)	f₁＝(e，1，2，e)f ₁ =(e,1,2,e)	(1-8)(1-8)	f₁＝(e，e，2，e)f ₁ = (e, e, 2, e)

{r ₂ }={said}

{x ₂ }={what A,what B,e}

{y ₂ }={I,you,f ₁ ,e}

{z ₂ }={f ₃ ,e}

Apply the multiplication principle in combinatorial mathematics: f ₂ (x ₂ , y ₂ , r ₂ , z ₂ )= (see list below)

序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂
序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂	(2-1)(2-1)	f₂＝(what A，I，r₂，f₃)f ₂ =(what A, I, r ₂ , f ₃ )	(2-13)(2-13)	f₂＝(what B，f₁，r₂，f₃)f ₂ =(what B,f ₁ ,r ₂ ,f ₃ )
(2-2)(2-2)	f₂＝(what A，I，r₂，e)f ₂ =(what A,I,r ₂ ,e)	(2-14)(2-14)	f₂＝(what B，f₁，r₂，e)f ₂ =(what B,f ₁ ,r ₂ ,e)	(2-1)(2-1)	f₂＝(what A，I，r₂，f₃)f ₂ =(what A, I, r ₂ , f ₃ )	(2-13)(2-13)	f₂＝(what B，f₁，r₂，f₃)f ₂ =(what B,f ₁ ,r ₂ ,f ₃ )
(2-2)(2-2)	f₂＝(what A，I，r₂，e)f ₂ =(what A,I,r ₂ ,e)	(2-14)(2-14)	f₂＝(what B，f₁，r₂，e)f ₂ =(what B,f ₁ ,r ₂ ,e)	(2-3)(2-3)	f₂＝(what A，you，r₂，f₃)f ₂ =(what A,you,r ₂ ,f ₃ )	(2-15)(2-15)	f₂＝(what B，e，r₂，f₃)f ₂ =(what B,e,r ₂ ,f ₃ )
(2-4)(2-4)	f₂＝(what A，you，r₂，e)f ₂ =(what A,you,r ₂ ,e)	(2-16)(2-16)	f₂＝(what B，e，r₂，e)f ₂ =(what B,e,r ₂ ,e)	(2-3)(2-3)	f₂＝(what A，you，r₂，f₃)f ₂ =(what A,you,r ₂ ,f ₃ )	(2-15)(2-15)	f₂＝(what B，e，r₂，f₃)f ₂ =(what B,e,r ₂ ,f ₃ )
(2-4)(2-4)	f₂＝(what A，you，r₂，e)f ₂ =(what A,you,r ₂ ,e)	(2-16)(2-16)	f₂＝(what B，e，r₂，e)f ₂ =(what B,e,r ₂ ,e)	(2-5)(2-5)	f₂＝(what A，f₁，r₂，f₃)f ₂ =(what A,f ₁ ,r ₂ ,f ₃ )	(2-17)(2-17)	f₂＝(e，I，r₂，f₃)f ₂ = (e, I, r ₂ , f ₃ )
(2-6)(2-6)	f₂＝(what A，f₁，r₂，e)f ₂ =(what A,f ₁ ,r ₂ ,e)	(2-18)(2-18)	f₂＝(e，I，r₂，e)f ₂ = (e, I, r ₂ , e)	(2-5)(2-5)	f₂＝(what A，f₁，r₂，f₃)f ₂ =(what A,f ₁ ,r ₂ ,f ₃ )	(2-17)(2-17)	f₂＝(e，I，r₂，f₃)f ₂ = (e, I, r ₂ , f ₃ )
(2-6)(2-6)	f₂＝(what A，f₁，r₂，e)f ₂ =(what A,f ₁ ,r ₂ ,e)	(2-18)(2-18)	f₂＝(e，I，r₂，e)f ₂ = (e, I, r ₂ , e)	(2-7)(2-7)	f₂＝(what A，e，r₂，f₃)f ₂ =(what A,e,r ₂ ,f ₃ )	(2-19)(2-19)	f₂＝(e，you，r₂，f₃)f ₂ =(e,you,r ₂ ,f ₃ )
(2-8)(2-8)	f₂＝(what A，e，r₂，e)f ₂ =(what A,e,r ₂ ,e)	(2-20)(2-20)	f₂＝(e，you，r₂，e)f ₂ =(e,you,r ₂ ,e)	(2-7)(2-7)	f₂＝(what A，e，r₂，f₃)f ₂ =(what A,e,r ₂ ,f ₃ )	(2-19)(2-19)	f₂＝(e，you，r₂，f₃)f ₂ =(e,you,r ₂ ,f ₃ )
(2-8)(2-8)	f₂＝(what A，e，r₂，e)f ₂ =(what A,e,r ₂ ,e)	(2-20)(2-20)	f₂＝(e，you，r₂，e)f ₂ =(e,you,r ₂ ,e)	(2-9)(2-9)	f₂＝(what B，I，r₂，f₃)f ₂ =(what B,I,r ₂ ,f ₃ )	(2-21)(2-21)	f₂＝(e，f₁，r₂，f₃)f ₂ = (e, f ₁ , r ₂ , f ₃ )
(2-10)(2-10)	f₂＝(what B，I，r₂，e)f ₂ =(what B,I,r ₂ ,e)	(2-22)(2-22)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)	(2-9)(2-9)	f₂＝(what B，I，r₂，f₃)f ₂ =(what B,I,r ₂ ,f ₃ )	(2-21)(2-21)	f₂＝(e，f₁，r₂，f₃)f ₂ = (e, f ₁ , r ₂ , f ₃ )
(2-10)(2-10)	f₂＝(what B，I，r₂，e)f ₂ =(what B,I,r ₂ ,e)	(2-22)(2-22)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)	(2-11)(2-11)	f₂＝(what B，you，r₂，f₃)f ₂ =(what B,you,r ₂ ,f ₃ )	(2-23)(2-23)	f₂＝(e，e，r₂，f₃)f ₂ = (e, e, r ₂ , f ₃ )
(2-12)(2-12)	f₂＝(what B，you，r₂，e)f ₂ =(what B,you,r ₂ ,e)	(2-24)(2-24)	f₂＝(e，e，r₂，e)f ₂ = (e, e, r ₂ , e)	(2-11)(2-11)	f₂＝(what B，you，r₂，f₃)f ₂ =(what B,you,r ₂ ,f ₃ )	(2-23)(2-23)	f₂＝(e，e，r₂，f₃)f ₂ = (e, e, r ₂ , f ₃ )

Replace the constant with the order value to get: f ₂ (x ₂ , y ₂ , r ₂ , z ₂ )= (see list below)

序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂
序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂	(2-1)(2-1)	f₂＝(3，1，6，f₃)f ₂ = (3,1,6,f ₃ )	(2-13)(2-13)	f₂＝(4，f₁，6，f₃)f ₂ =(4,f ₁ ,6,f ₃ )
(2-2)(2-2)	f₂＝(3，1，6，e)f ₂ = (3,1,6,e)	(2-14)(2-14)	f₂＝(4，f₁，6，e)f ₂ =(4,f ₁ ,6,e)	(2-1)(2-1)	f₂＝(3，1，6，f₃)f ₂ = (3,1,6,f ₃ )	(2-13)(2-13)	f₂＝(4，f₁，6，f₃)f ₂ =(4,f ₁ ,6,f ₃ )
(2-2)(2-2)	f₂＝(3，1，6，e)f ₂ = (3,1,6,e)	(2-14)(2-14)	f₂＝(4，f₁，6，e)f ₂ =(4,f ₁ ,6,e)	(2-3)(2-3)	f₂＝(3，5，6，f₃)f ₂ = (3,5,6,f ₃ )	(2-15)(2-15)	f₂＝(4，e，6，f₃)f ₂ =(4,e,6,f ₃ )
(2-4)(2-4)	f₂＝(3，5，6，e)f ₂ = (3, 5, 6, e)	(2-16)(2-16)	f₂＝(4，e，6，e)f ₂ = (4, e, 6, e)	(2-3)(2-3)	f₂＝(3，5，6，f₃)f ₂ = (3,5,6,f ₃ )	(2-15)(2-15)	f₂＝(4，e，6，f₃)f ₂ =(4,e,6,f ₃ )
(2-4)(2-4)	f₂＝(3，5，6，e)f ₂ = (3, 5, 6, e)	(2-16)(2-16)	f₂＝(4，e，6，e)f ₂ = (4, e, 6, e)	(2-5)(2-5)	f₂＝(3，f₁，6，f₃)f ₂ = (3, f ₁ , 6, f ₃ )	(2-17)(2-17)	f₂＝(e，1，6，f₃)f ₂ = (e,1,6,f ₃ )
(2-6)(2-6)	f₂＝(3，f₁，6，e)f ₂ = (3, f ₁ , 6, e)	(2-18)(2-18)	f₂＝(e，1，6，e)f ₂ = (e, 1, 6, e)	(2-5)(2-5)	f₂＝(3，f₁，6，f₃)f ₂ = (3, f ₁ , 6, f ₃ )	(2-17)(2-17)	f₂＝(e，1，6，f₃)f ₂ = (e,1,6,f ₃ )
(2-6)(2-6)	f₂＝(3，f₁，6，e)f ₂ = (3, f ₁ , 6, e)	(2-18)(2-18)	f₂＝(e，1，6，e)f ₂ = (e, 1, 6, e)	(2-7)(2-7)	f₂＝(3，e，6，f₃)f ₂ = (3, e, 6, f ₃ )	(2-19)(2-19)	f₂＝(e，5，6，f₃)f ₂ = (e, 5, 6, f ₃ )
(2-8)(2-8)	f₂＝(3，e，6，e)f ₂ = (3, e, 6, e)	(2-20)(2-20)	f₂＝(e，5，6，e)f ₂ = (e, 5, 6, e)	(2-7)(2-7)	f₂＝(3，e，6，f₃)f ₂ = (3, e, 6, f ₃ )	(2-19)(2-19)	f₂＝(e，5，6，f₃)f ₂ = (e, 5, 6, f ₃ )
(2-8)(2-8)	f₂＝(3，e，6，e)f ₂ = (3, e, 6, e)	(2-20)(2-20)	f₂＝(e，5，6，e)f ₂ = (e, 5, 6, e)	(2-9)(2-9)	f₂＝(4，1，6，f₃)f ₂ = (4,1,6,f ₃ )	(2-21)(2-21)	f₂＝(e，f₁，6，f₃)f ₂ = (e, f ₁ , 6, f ₃ )
(2-10)(2-10)	f₂＝(4，1，6，e)f ₂ =(4,1,6,e)	(2-22)(2-22)	f₂＝(e，f₁，6，e)f ₂ = (e, f ₁ , 6, e)	(2-9)(2-9)	f₂＝(4，1，6，f₃)f ₂ = (4,1,6,f ₃ )	(2-21)(2-21)	f₂＝(e，f₁，6，f₃)f ₂ = (e, f ₁ , 6, f ₃ )
(2-10)(2-10)	f₂＝(4，1，6，e)f ₂ =(4,1,6,e)	(2-22)(2-22)	f₂＝(e，f₁，6，e)f ₂ = (e, f ₁ , 6, e)	(2-11)(2-11)	f₂＝(4，5，6，f₃)f ₂ =(4,5,6,f ₃ )	(2-23)(2-23)	f₂＝(e，e，6，f₃)f ₂ = (e, e, 6, f ₃ )
(2-12)(2-12)	f₂＝(4，5，6，e)f ₂ = (4, 5, 6, e)	(2-24)(2-24)	f₂＝(e，e，6，e)f ₂ = (e, e, 6, e)	(2-11)(2-11)	f₂＝(4，5，6，f₃)f ₂ =(4,5,6,f ₃ )	(2-23)(2-23)	f₂＝(e，e，6，f₃)f ₂ = (e, e, 6, f ₃ )

{r ₃ }={meant}

{x ₃ }={what A,what B,e}

{y ₃ }={I,you,f ₁ ,f ₂ ,e}

{z ₃ }={e}

Apply the multiplication principle in combinatorial mathematics: f ₃ (x ₃ , y ₃ , r ₃ , z ₃ )= (see list below)

序号Serial number	行矩阵f₃ Row matrix f ₃	(3-8)(3-8)	f₃＝(what B，f₁，r₃，e)f ₃ =(what B,f ₁ ,r ₃ ,e)
序号Serial number	行矩阵f₃ Row matrix f ₃	(3-8)(3-8)	f₃＝(what B，f₁，r₃，e)f ₃ =(what B,f ₁ ,r ₃ ,e)	(3-1)(3-1)	f₃＝(what A，I，r₃，e)f ₃ =(what A,I,r ₃ ,e)	(3-9)(3-9)	f₃＝(what B，f₂，r₃，e)f ₃ =(what B,f ₂ ,r ₃ ,e)
(3-2)(3-2)	f₃＝(what A，you，r₃，e)f ₃ =(what A,you,r ₃ ,e)	(3-10)(3-10)	f₃＝(what B，e，r₃，e)f ₃ =(what B,e,r ₃ ,e)	(3-1)(3-1)	f₃＝(what A，I，r₃，e)f ₃ =(what A,I,r ₃ ,e)	(3-9)(3-9)	f₃＝(what B，f₂，r₃，e)f ₃ =(what B,f ₂ ,r ₃ ,e)

(3-3)(3-3)	f₃＝(what A，f₁，r₃，e)f ₃ =(what A,f ₁ ,r ₃ ,e)	(3-11)(3-11)	f₃＝(e，I，r₃，e)f ₃ = (e, I, r ₃ , e)
(3-3)(3-3)	f₃＝(what A，f₁，r₃，e)f ₃ =(what A,f ₁ ,r ₃ ,e)	(3-11)(3-11)	f₃＝(e，I，r₃，e)f ₃ = (e, I, r ₃ , e)	(3-4)(3-4)	f₃＝(what A，f₂，r₃，e)f ₃ =(what A,f ₂ ,r ₃ ,e)	(3-12)(3-12)	f₃＝(e，you，r₃，e)f ₃ = (e,you,r ₃ ,e)
(3-5)(3-5)	f₃＝(what A，e，r₃，e)f ₃ =(what A,e,r ₃ ,e)	(3-13)(3-13)	f₃＝(e，f₁，r₃，e)f ₃ = (e, f ₁ , r ₃ , e)	(3-4)(3-4)	f₃＝(what A，f₂，r₃，e)f ₃ =(what A,f ₂ ,r ₃ ,e)	(3-12)(3-12)	f₃＝(e，you，r₃，e)f ₃ = (e,you,r ₃ ,e)
(3-5)(3-5)	f₃＝(what A，e，r₃，e)f ₃ =(what A,e,r ₃ ,e)	(3-13)(3-13)	f₃＝(e，f₁，r₃，e)f ₃ = (e, f ₁ , r ₃ , e)	(3-6)(3-6)	f₃＝(what B，I，r₃，e)f ₃ =(what B,I,r ₃ ,e)	(3-14)(3-14)	f₃＝(e，f₂，r₃，e)f ₃ = (e, f ₂ , r ₃ , e)
(3-7)(3-7)	f₃＝(what B，you，r₃，e)f ₃ =(what B,you,r ₃ ,e)	(3-15)(3-15)	f₃＝(e，e，r₃，e)f ₃ = (e, e, r ₃ , e)	(3-6)(3-6)	f₃＝(what B，I，r₃，e)f ₃ =(what B,I,r ₃ ,e)	(3-14)(3-14)	f₃＝(e，f₂，r₃，e)f ₃ = (e, f ₂ , r ₃ , e)

Replace the constant with the order value to get: f ₃ (x ₃ , y ₃ , r ₃ , z ₃ )= (see list below)

序号Serial number	行矩阵f₃ Row matrix f ₃	(3-8)(3-8)	f₃＝(4，f₁，7，e)f ₃ = (4, f ₁ , 7, e)
序号Serial number	行矩阵f₃ Row matrix f ₃	(3-8)(3-8)	f₃＝(4，f₁，7，e)f ₃ = (4, f ₁ , 7, e)	(3-1)(3-1)	f₃＝(3，1，7，e)f ₃ = (3,1,7,e)	(3-9)(3-9)	f₃＝(4，f₂，7，e)f ₃ = (4, f ₂ , 7, e)
(3-2)(3-2)	f₃＝(3，5，7，e)f ₃ = (3, 5, 7, e)	(3-10)(3-10)	f₃＝(4，e，7，e)f ₃ = (4, e, 7, e)	(3-1)(3-1)	f₃＝(3，1，7，e)f ₃ = (3,1,7,e)	(3-9)(3-9)	f₃＝(4，f₂，7，e)f ₃ = (4, f ₂ , 7, e)
(3-2)(3-2)	f₃＝(3，5，7，e)f ₃ = (3, 5, 7, e)	(3-10)(3-10)	f₃＝(4，e，7，e)f ₃ = (4, e, 7, e)	(3-3)(3-3)	f₃＝(3，f₁，7，e)f ₃ = (3, f ₁ , 7, e)	(3-11)(3-11)	f₃＝(e，1，7，e)f ₃ = (e, 1, 7, e)
(3-4)(3-4)	f₃＝(3，f₂，7，e)f ₃ = (3, f ₂ , 7, e)	(3-12)(3-12)	f₃＝(e，5，7，e)f ₃ = (e, 5, 7, e)	(3-3)(3-3)	f₃＝(3，f₁，7，e)f ₃ = (3, f ₁ , 7, e)	(3-11)(3-11)	f₃＝(e，1，7，e)f ₃ = (e, 1, 7, e)
(3-4)(3-4)	f₃＝(3，f₂，7，e)f ₃ = (3, f ₂ , 7, e)	(3-12)(3-12)	f₃＝(e，5，7，e)f ₃ = (e, 5, 7, e)	(3-5)(3-5)	f₃＝(3，e，7，e)f ₃ = (3, e, 7, e)	(3-13)(3-13)	f₃＝(e，f₁，7，e)f ₃ = (e, f ₁ , 7, e)
(3-6)(3-6)	f₃＝(4，1，7，e)f ₃ = (4,1,7,e)	(3-14)(3-14)	f₃＝(e，f₂，7，e)f ₃ = (e, f ₂ , 7, e)	(3-5)(3-5)	f₃＝(3，e，7，e)f ₃ = (3, e, 7, e)	(3-13)(3-13)	f₃＝(e，f₁，7，e)f ₃ = (e, f ₁ , 7, e)
(3-6)(3-6)	f₃＝(4，1，7，e)f ₃ = (4,1,7,e)	(3-14)(3-14)	f₃＝(e，f₂，7，e)f ₃ = (e, f ₂ , 7, e)	(3-7)(3-7)	f₃＝(4，5，7，e)f ₃ = (4, 5, 7, e)	(3-15)(3-15)	f₃＝(e，e，7，e)f ₃ = (e, e, 7, e)

Apply the principle of multiplication in combinatorial mathematics:

|S|=|f ₁ |×|f ₂ |×|f ₃ |=8×24×15=2880

A total of 2880 possible matrix solutions are generated.

Step 140: Generate at least one syntax structure possible matrix solution according to all possible values of all syntax vectors, where the syntax structure may be composed of syntax vectors arranged in order of predicate verb unit numbers.

For the pre-processed statement S = "I can understand what what you said meant" shown in Table 1 above, based on the possible values of f ₁ , f ₂ and f ₃ , a plurality of possible matrix solutions can be obtained.

Step 150: Verify whether the statement obtained by the possible matrix solution according to the syntax structure is exactly the same as the preprocessed statement. If they are identical, the syntactic vector may be outputted in the possible matrix solution and used as a syntactic structure analysis result. one.

Preferably, the word unit number is used instead of the word unit for equal-substitution, overall insertion, and partial addition operations, and then it is determined whether the sequence of sentences is in the same order as the pre-processed statement based on whether the obtained sequence of sentences is a sequentially increasing number sequence.

Step 150 can include the following steps:

Step 151: If there is a sequence value that does not appear in the possible matrix solution of the syntax structure, the possible matrix solution may be excluded from the syntax structure; for example, for the following possible matrix solution:

The word unit numbered 4 does not appear and is excluded.

Step 152: If the same order value appears in different syntax vectors or the same syntax vector appears, the possible syntax solution of the syntax structure is excluded; for example, for the following possible matrix solutions:

The word unit numbered 5 appears twice and is excluded.

Step 153: In each possible matrix solution, the syntactic vectors having mutual substitution relations with other syntactic vectors are all equally substituted, and if cross-contradictions of two syntactic vectors appear after equal-substitution, the exclusion is excluded. The syntax structure may be a matrix solution;

For example, the following possible matrix solutions:

Step 154: In each possible matrix solution, the syntactic vectors having mutual substitution relations with other syntax vectors are all replaced by equal amounts. If two position reversal order values appear after the equal amount substitution, the exclusion is performed. The syntax structure may be a matrix solution;

For example, the following possible matrix solutions:

Step 155: In any one of the possible matrix solutions, if there is a syntax vector that does not have a mutual substitution relationship with other syntax vectors, perform an insertion operation to obtain a possible syntax parsing structure corresponding to all the possible matrix solutions, and verify the basis Whether the statement obtained by the possible syntax parsing structure is identical to the pre-processed statement, further comprising:

In step 155.1, the syntactic vector having the substitution relationship between the possible matrix solutions is firstly substituted, thereby transforming the possible matrix solution into a set of syntactic vectors which do not have an substitution relationship with each other.

Called the second type of syntax vector;

Step 155.2, take a second type of syntax vector

Mark one by one according to the predetermined direction

Second type of syntax vector

Syntactic vector in the form of overall insertion

Step 155.3, for the third type of syntax vector

Pair vector from the predetermined direction

The first syntactic element on the first side starts into the vector

Vector contained in

The element on the first side, without the order value; the vector

The first syntax element on the second side is marked as

Will be vectored as described above

The syntactic vector part of the annotation, denoted as the iris syntax vector

Syntactic vector in the form of overall insertion

or

Third type of syntax vector

Syntactic vector according to the predetermined direction

The vector is inserted as a whole

In step 155.4, step 155.3 is repeatedly executed. When the last emptying and emptying steps are completed, the third type of syntactic vector obtained through the previous emptying and emptying steps is subjected to the next emptying and emptying operation. Until all the second type of syntax vector

Step 155.5, if there are two position reversal order values in all the final single row vectors corresponding to a possible syntax parsing structure, the possible syntactic parsing structure is excluded;

In step 155.6, steps 155.2 through 155.5 are repeated until all possible syntactic parsing structures are traversed.

For example, for the example described above, a syntactic structure might be solved as:

Convert the above matrix into a linear expression:

After the aforementioned insertion operation, there are two position reversal order values in each final single line vector, and are excluded.

For the example described above, a syntactic structure might be solved as:

You can convert a matrix to a linear expression:

Perform an equal amount of substitution operations to get the statement:

α=e+<1+<2+<(3+<(4+<5+<6+<e)+<7+<e)

Remove the empty unit e and get:

α=1+<2+<(3+<(4+<5+<6)+<7)

It is the same as the pre-processed statement, which is one of the parsing results of the syntax structure.

Substituting the word unit constant into the above matrix, the syntax structure matrix solution can be expressed as:

The linear expression of S corresponding to this matrix expression is as follows:

According to this, the syntactic structure of the sentence "I can understand what what you said means" is: I is the subject of the main sentence, can understand as the predicate of the main clause, and the clause "what what you said meant" is the object clause of the main clause. In the clause, the first what is the clause of the clause, "what you said" is the subject of the clause, the mean is the predicate of the object clause, the object clause itself has no object; for the "what you said" clause, it acts as the object clause The subject clauses nested inside, what is the guiding word, you are the main language, and the said is the predicate.

Further, the method may further include a displaying step of displaying each syntax vector in the syntax structure analysis result and the corresponding syntax structure relationship in a human-computer interaction interface by using a tree structure.

Part C2 Example 2

Example 2: As another example, the following describes the parsing process of the method of the present embodiment for a complicated structure such as "That men who were appointed didn't bother the liberals wash't remarked upon by the press."

The above statement is preprocessed to remove impurities and the numbered word sequence is:

原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number
原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number	ThatThat	从属关联词单元Subordinate unit	11
menMen	名词代词单元Noun pronoun unit	22	ThatThat	从属关联词单元Subordinate unit	11
menMen	名词代词单元Noun pronoun unit	22	whoWho	从属关联词单元Subordinate unit	33
were appointedWere appointed	谓语动词单元Predicate verb unit	44	whoWho	从属关联词单元Subordinate unit	33
were appointedWere appointed	谓语动词单元Predicate verb unit	44	didn’t botherDid’t bother	谓语动词单元Predicate verb unit	55
the liberalsThe liberals	名词代词单元Noun pronoun unit	66	didn’t botherDid’t bother	谓语动词单元Predicate verb unit	55

Wasn’t remarked

Predicate verb unit

7

There are three predicate verb units in the sentence, which are denoted as r ₁ , r ₂ and r _{3 respectively} .

For r ₁ there, {r ₁ }={were appointed}

{x ₁ }={That,who,e} (e is an empty string)

{y ₁ }={men,e}

{z ₁ }={f ₂ ,f ₃ ,e}

序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(That，men，r₁，f₂)f ₁ =(That,men,r ₁ ,f ₂ )	(1-10)(1-10)	f₁＝(That，e，r₁，f₃)f ₁ =(That,e,r ₁ ,f ₃ )
(1-2)(1-2)	f₁＝(who，men，r₁，f₂)f ₁ =(who,men,r ₁ ,f ₂ )	(1-11)(1-11)	f₁＝(who，e，r₁，f₃)f ₁ =(who,e,r ₁ ,f ₃ )	(1-1)(1-1)	f₁＝(That，men，r₁，f₂)f ₁ =(That,men,r ₁ ,f ₂ )	(1-10)(1-10)	f₁＝(That，e，r₁，f₃)f ₁ =(That,e,r ₁ ,f ₃ )
(1-2)(1-2)	f₁＝(who，men，r₁，f₂)f ₁ =(who,men,r ₁ ,f ₂ )	(1-11)(1-11)	f₁＝(who，e，r₁，f₃)f ₁ =(who,e,r ₁ ,f ₃ )	(1-3)(1-3)	f₁＝(e，men，r₁，f₂)f ₁ =(e,men,r ₁ ,f ₂ )	(1-12)(1-12)	f₁＝(e，e，r₁，f₃)f ₁ = (e, e, r ₁ , f ₃ )
(1-4)(1-4)	f₁＝(That，e，r₁，f₂)f ₁ =(That,e,r ₁ ,f ₂ )	(1-13)(1-13)	f₁＝(That，men，r₁，e)f ₁ =(That,men,r ₁ ,e)	(1-3)(1-3)	f₁＝(e，men，r₁，f₂)f ₁ =(e,men,r ₁ ,f ₂ )	(1-12)(1-12)	f₁＝(e，e，r₁，f₃)f ₁ = (e, e, r ₁ , f ₃ )
(1-4)(1-4)	f₁＝(That，e，r₁，f₂)f ₁ =(That,e,r ₁ ,f ₂ )	(1-13)(1-13)	f₁＝(That，men，r₁，e)f ₁ =(That,men,r ₁ ,e)	(1-5)(1-5)	f₁＝(who，e，r₁，f₂)f ₁ =(who,e,r ₁ ,f ₂ )	(1-14)(1-14)	f₁＝(who，men，r₁，e)f ₁ =(who,men,r ₁ ,e)
(1-6)(1-6)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-15)(1-15)	f₁＝(e，men，r₁，e)f ₁ =(e,men,r ₁ ,e)	(1-5)(1-5)	f₁＝(who，e，r₁，f₂)f ₁ =(who,e,r ₁ ,f ₂ )	(1-14)(1-14)	f₁＝(who，men，r₁，e)f ₁ =(who,men,r ₁ ,e)
(1-6)(1-6)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-15)(1-15)	f₁＝(e，men，r₁，e)f ₁ =(e,men,r ₁ ,e)	(1-7)(1-7)	f₁＝(That，men，r₁，f₃)f ₁ =(That,men,r ₁ ,f ₃ )	(1-16)(1-16)	f₁＝(That，e，r₁，e)f ₁ =(That,e,r ₁ ,e)
(1-8)(1-8)	f₁＝(who，men，r₁，f₃)f ₁ =(who,men,r ₁ ,f ₃ )	(1-17)(1-17)	f₁＝(who，e，r₁，e)f ₁ =(who,e,r ₁ ,e)	(1-7)(1-7)	f₁＝(That，men，r₁，f₃)f ₁ =(That,men,r ₁ ,f ₃ )	(1-16)(1-16)	f₁＝(That，e，r₁，e)f ₁ =(That,e,r ₁ ,e)
(1-8)(1-8)	f₁＝(who，men，r₁，f₃)f ₁ =(who,men,r ₁ ,f ₃ )	(1-17)(1-17)	f₁＝(who，e，r₁，e)f ₁ =(who,e,r ₁ ,e)	(1-9)(1-9)	f₁＝(e，men，r₁，f₃)f ₁ =(e,men,r ₁ ,f ₃ )	(1-18)(1-18)	f₁＝(e，e，r₁，e)f ₁ = (e, e, r ₁ , e)

序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(1，2，4，f₂)f ₁ = (1, ₂ , 4, f ₂ )	(1-10)(1-10)	f₁＝(1，e，4，f₃)f ₁ = (1, e, 4, f ₃ )
(1-2)(1-2)	f₁＝(3，2，4，f₂)f ₁ = (3, ₂ , 4, f ₂ )	(1-11)(1-11)	f₁＝(3，e，4，f₃)f ₁ = (3, e, 4, f ₃ )	(1-1)(1-1)	f₁＝(1，2，4，f₂)f ₁ = (1, ₂ , 4, f ₂ )	(1-10)(1-10)	f₁＝(1，e，4，f₃)f ₁ = (1, e, 4, f ₃ )
(1-2)(1-2)	f₁＝(3，2，4，f₂)f ₁ = (3, ₂ , 4, f ₂ )	(1-11)(1-11)	f₁＝(3，e，4，f₃)f ₁ = (3, e, 4, f ₃ )	(1-3)(1-3)	f₁＝(e，2，4，f₂)f ₁ = (e, 2, 4, f ₂ )	(1-12)(1-12)	f₁＝(e，e，4，f₃)f ₁ = (e, e, 4, f ₃ )
(1-4)(1-4)	f₁＝(1，e，4，f₂)f ₁ = (1, e, 4, f ₂ )	(1-13)(1-13)	f₁＝(1，2，4，e)f ₁ = (1, 2, 4, e)	(1-3)(1-3)	f₁＝(e，2，4，f₂)f ₁ = (e, 2, 4, f ₂ )	(1-12)(1-12)	f₁＝(e，e，4，f₃)f ₁ = (e, e, 4, f ₃ )
(1-4)(1-4)	f₁＝(1，e，4，f₂)f ₁ = (1, e, 4, f ₂ )	(1-13)(1-13)	f₁＝(1，2，4，e)f ₁ = (1, 2, 4, e)	(1-5)(1-5)	f₁＝(3，e，4，f₂)f ₁ = (3, e, 4, f ₂ )	(1-14)(1-14)	f₁＝(3，2，4，e)f ₁ = (3, 2, 4, e)
(1-6)(1-6)	f₁＝(e，e，4，f₂)f ₁ = (e, e, 4, f ₂ )	(1-15)(1-15)	f₁＝(e，2，4，e)f ₁ = (e, 2, 4, e)	(1-5)(1-5)	f₁＝(3，e，4，f₂)f ₁ = (3, e, 4, f ₂ )	(1-14)(1-14)	f₁＝(3，2，4，e)f ₁ = (3, 2, 4, e)
(1-6)(1-6)	f₁＝(e，e，4，f₂)f ₁ = (e, e, 4, f ₂ )	(1-15)(1-15)	f₁＝(e，2，4，e)f ₁ = (e, 2, 4, e)	(1-7)(1-7)	f₁＝(1，2，4，f₃)f ₁ = (1, 2, 4, f ₃ )	(1-16)(1-16)	f₁＝(1，e，4，e)f ₁ = (1, e, 4, e)
(1-8)(1-8)	f₁＝(3，2，4，f₃)f ₁ = (3, 2, ₄ , f ₃ )	(1-17)(1-17)	f₁＝(3，e，4，e)f ₁ = (3, e, 4, e)	(1-7)(1-7)	f₁＝(1，2，4，f₃)f ₁ = (1, 2, 4, f ₃ )	(1-16)(1-16)	f₁＝(1，e，4，e)f ₁ = (1, e, 4, e)

(1-9)

f ₁ = (e, 2, 4, f ₃ )

(1-18)

f ₁ = (e, e, 4, e)

For r ₂ , {r ₂ }={didn't bother}

{x ₂ }={That,who,e} (e is an empty string)

{y ₂ }={men,f ₁ ,e}

{z ₂ }={the liberals,f ₃ ,e}

序号Serial number	行矩阵f₂ Row matrix f ₂	(2-14)(2-14)	f₂＝(who，f₁，r₂，f₃)f ₂ =(who,f ₁ ,r ₂ ,f ₃ )
序号Serial number	行矩阵f₂ Row matrix f ₂	(2-14)(2-14)	f₂＝(who，f₁，r₂，f₃)f ₂ =(who,f ₁ ,r ₂ ,f ₃ )	(2-1)(2-1)	f₂＝(That，men，r₂，the liberals)f ₂ =(That,men,r ₂ ,the liberals)	(2-15)(2-15)	f₂＝(e，f₁，r₂，f₃)f ₂ = (e, f ₁ , r ₂ , f ₃ )
(2-2)(2-2)	f₂＝(who，men，r₂，the liberals)f ₂ =(who,men,r ₂ ,the liberals)	(2-16)(2-16)	f₂＝(That，e，r₂，f₃)f ₂ =(That,e,r ₂ ,f ₃ )	(2-1)(2-1)		(2-15)(2-15)	f₂＝(e，f₁，r₂，f₃)f ₂ = (e, f ₁ , r ₂ , f ₃ )
(2-2)(2-2)		(2-16)(2-16)	f₂＝(That，e，r₂，f₃)f ₂ =(That,e,r ₂ ,f ₃ )	(2-3)(2-3)	f₂＝(e，men，r₂，the liberals)f ₂ =(e,men,r ₂ ,the liberals)	(2-17)(2-17)	f₂＝(who，e，r₂，f₃)f ₂ =(who,e,r ₂ ,f ₃ )
(2-4)(2-4)	f₂＝(That，f₁，r₂，the liberals)f ₂ =(That,f ₁ ,r ₂ ,the liberals)	(2-18)(2-18)	f₂＝(e，e，r₂，f₃)f ₂ = (e, e, r ₂ , f ₃ )	(2-3)(2-3)	f₂＝(e，men，r₂，the liberals)f ₂ =(e,men,r ₂ ,the liberals)	(2-17)(2-17)	f₂＝(who，e，r₂，f₃)f ₂ =(who,e,r ₂ ,f ₃ )
(2-4)(2-4)		(2-18)(2-18)	f₂＝(e，e，r₂，f₃)f ₂ = (e, e, r ₂ , f ₃ )	(2-5)(2-5)	f₂＝(who，f₁，r₂，the liberals)f ₂ =(who,f ₁ ,r ₂ ,the liberals)	(2-19)(2-19)	f₂＝(That，men，r₂，e)f ₂ =(That,men,r ₂ ,e)
(2-6)(2-6)	f₂＝(e，f₁，r₂，the liberals)f ₂ =(e,f ₁ ,r ₂ ,the liberals)	(2-20)(2-20)	f₂＝(who，men，r₂，e)f ₂ =(who,men,r ₂ ,e)	(2-5)(2-5)		(2-19)(2-19)	f₂＝(That，men，r₂，e)f ₂ =(That,men,r ₂ ,e)
(2-6)(2-6)	f₂＝(e，f₁，r₂，the liberals)f ₂ =(e,f ₁ ,r ₂ ,the liberals)	(2-20)(2-20)	f₂＝(who，men，r₂，e)f ₂ =(who,men,r ₂ ,e)	(2-7)(2-7)	f₂＝(That，e，r₂，the liberals)f ₂ =(That,e,r ₂ ,the liberals)	(2-21)(2-21)	f₂＝(e，men，r₂，e)f ₂ =(e,men,r ₂ ,e)
(2-8)(2-8)	f₂＝(who，e，r₂，the liberals)f ₂ =(who,e,r ₂ ,the liberals)	(2-22)(2-22)	f₂＝(That，f₁，r₂，e)f ₂ =(That,f ₁ ,r ₂ ,e)	(2-7)(2-7)	f₂＝(That，e，r₂，the liberals)f ₂ =(That,e,r ₂ ,the liberals)	(2-21)(2-21)	f₂＝(e，men，r₂，e)f ₂ =(e,men,r ₂ ,e)
(2-8)(2-8)	f₂＝(who，e，r₂，the liberals)f ₂ =(who,e,r ₂ ,the liberals)	(2-22)(2-22)	f₂＝(That，f₁，r₂，e)f ₂ =(That,f ₁ ,r ₂ ,e)	(2-9)(2-9)	f₂＝(e，e，r₂，the liberals)f ₂ =(e,e,r ₂ ,the liberals)	(2-23)(2-23)	f₂＝(who，f₁，r₂，e)f ₂ =(who,f ₁ ,r ₂ ,e)
(2-10)(2-10)	f₂＝(That，men，r₂，f₃)f ₂ =(That,men,r ₂ ,f ₃ )	(2-24)(2-24)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)	(2-9)(2-9)	f₂＝(e，e，r₂，the liberals)f ₂ =(e,e,r ₂ ,the liberals)	(2-23)(2-23)	f₂＝(who，f₁，r₂，e)f ₂ =(who,f ₁ ,r ₂ ,e)
(2-10)(2-10)	f₂＝(That，men，r₂，f₃)f ₂ =(That,men,r ₂ ,f ₃ )	(2-24)(2-24)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)	(2-11)(2-11)	f₂＝(who，men，r₂，f₃)f ₂ =(who,men,r ₂ ,f ₃ )	(2-25)(2-25)	f₂＝(That，e，r₂，e)f ₂ =(That,e,r ₂ ,e)
(2-12)(2-12)	f₂＝(e，men，r₂，f₃)f ₂ =(e,men,r ₂ ,f ₃ )	(2-26)(2-26)	f₂＝(who，e，r₂，e)f ₂ =(who,e,r ₂ ,e)	(2-11)(2-11)	f₂＝(who，men，r₂，f₃)f ₂ =(who,men,r ₂ ,f ₃ )	(2-25)(2-25)	f₂＝(That，e，r₂，e)f ₂ =(That,e,r ₂ ,e)
(2-12)(2-12)	f₂＝(e，men，r₂，f₃)f ₂ =(e,men,r ₂ ,f ₃ )	(2-26)(2-26)	f₂＝(who，e，r₂，e)f ₂ =(who,e,r ₂ ,e)	(2-13)(2-13)	f₂＝(That，f₁，r₂，f₃)f ₂ =(That,f ₁ ,r ₂ ,f ₃ )	(2-27)(2-27)	f₂＝(e，e，r₂，e)f ₂ = (e, e, r ₂ , e)

序号Serial number	行矩阵f₂ Row matrix f ₂	(2-14)(2-14)	f₂＝(3，f₁，5，f₃)f ₂ = (3, f ₁ , 5, f ₃ )
序号Serial number	行矩阵f₂ Row matrix f ₂	(2-14)(2-14)	f₂＝(3，f₁，5，f₃)f ₂ = (3, f ₁ , 5, f ₃ )	(2-1)(2-1)	f₂＝(1，2，5，6)f ₂ = (1, 2, 5, 6)	(2-15)(2-15)	f₂＝(e，f₁，5，f₃)f ₂ = (e, f ₁ , 5, f ₃ )
(2-2)(2-2)	f₂＝(3，2，5，6)f ₂ = (3, _{2, 5} , 6)	(2-16)(2-16)	f₂＝(1，e，5，f₃)f ₂ = (1, e, 5, f ₃ )	(2-1)(2-1)	f₂＝(1，2，5，6)f ₂ = (1, 2, 5, 6)	(2-15)(2-15)	f₂＝(e，f₁，5，f₃)f ₂ = (e, f ₁ , 5, f ₃ )
(2-2)(2-2)	f₂＝(3，2，5，6)f ₂ = (3, _{2, 5} , 6)	(2-16)(2-16)	f₂＝(1，e，5，f₃)f ₂ = (1, e, 5, f ₃ )	(2-3)(2-3)	f₂＝(e，2，5，6)f ₂ = (e, 2, 5, 6)	(2-17)(2-17)	f₂＝(3，e，5，f₃)f ₂ = (3, e, 5, f ₃ )
(2-4)(2-4)	f₂＝(1，f₁，5，6)f ₂ = (1, f ₁ , 5, 6)	(2-18)(2-18)	f₂＝(e，e，5，f₃)f ₂ = (e, e, 5, f ₃ )	(2-3)(2-3)	f₂＝(e，2，5，6)f ₂ = (e, 2, 5, 6)	(2-17)(2-17)	f₂＝(3，e，5，f₃)f ₂ = (3, e, 5, f ₃ )
(2-4)(2-4)	f₂＝(1，f₁，5，6)f ₂ = (1, f ₁ , 5, 6)	(2-18)(2-18)	f₂＝(e，e，5，f₃)f ₂ = (e, e, 5, f ₃ )	(2-5)(2-5)	f₂＝(3，f₁，5，6)f ₂ = (3, f ₁ , 5, 6)	(2-19)(2-19)	f₂＝(1，2，5，e)f ₂ = (1, 2, 5, e)

(2-6)(2-6)	f₂＝(e，f₁，5，6)f ₂ = (e, f ₁ , 5, 6)	(2-20)(2-20)	f₂＝(3，2，5，e)f ₂ = (3, ₂ , 5, e)
(2-6)(2-6)	f₂＝(e，f₁，5，6)f ₂ = (e, f ₁ , 5, 6)	(2-20)(2-20)	f₂＝(3，2，5，e)f ₂ = (3, ₂ , 5, e)	(2-7)(2-7)	f₂＝(1，e，5，6)f ₂ = (1, e, 5, 6)	(2-21)(2-21)	f₂＝(e，2，5，e)f ₂ = (e, 2, 5, e)
(2-8)(2-8)	f₂＝(3，e，5，6)f ₂ = (3, e, 5, 6)	(2-22)(2-22)	f₂＝(1，f₁，5，e)f ₂ = (1, f ₁ , 5, e)	(2-7)(2-7)	f₂＝(1，e，5，6)f ₂ = (1, e, 5, 6)	(2-21)(2-21)	f₂＝(e，2，5，e)f ₂ = (e, 2, 5, e)
(2-8)(2-8)	f₂＝(3，e，5，6)f ₂ = (3, e, 5, 6)	(2-22)(2-22)	f₂＝(1，f₁，5，e)f ₂ = (1, f ₁ , 5, e)	(2-9)(2-9)	f₂＝(e，e，5，6)f ₂ = (e, e, 5, 6)	(2-23)(2-23)	f₂＝(3，f₁，5，e)f ₂ = (3, f ₁ , 5, e)
(2-10)(2-10)	f₂＝(1，2，5，f₃)f ₂ = (1, 2, 5, f ₃ )	(2-24)(2-24)	f₂＝(e，f₁，5，e)f ₂ = (e, f ₁ , 5, e)	(2-9)(2-9)	f₂＝(e，e，5，6)f ₂ = (e, e, 5, 6)	(2-23)(2-23)	f₂＝(3，f₁，5，e)f ₂ = (3, f ₁ , 5, e)
(2-10)(2-10)	f₂＝(1，2，5，f₃)f ₂ = (1, 2, 5, f ₃ )	(2-24)(2-24)	f₂＝(e，f₁，5，e)f ₂ = (e, f ₁ , 5, e)	(2-11)(2-11)	f₂＝(3，2，5，f₃)f ₂ = (3, ₂ , 5, f ₃ )	(2-25)(2-25)	f₂＝(1，e，5，e)f ₂ = (1, e, 5, e)
(2-12)(2-12)	f₂＝(e，2，5，f₃)f ₂ = (e, 2, 5, f ₃ )	(2-26)(2-26)	f₂＝(3，e，5，e)f ₂ = (3, e, 5, e)	(2-11)(2-11)	f₂＝(3，2，5，f₃)f ₂ = (3, ₂ , 5, f ₃ )	(2-25)(2-25)	f₂＝(1，e，5，e)f ₂ = (1, e, 5, e)
(2-12)(2-12)	f₂＝(e，2，5，f₃)f ₂ = (e, 2, 5, f ₃ )	(2-26)(2-26)	f₂＝(3，e，5，e)f ₂ = (3, e, 5, e)	(2-13)(2-13)	f₂＝(1，f₁，5，f₃)f ₂ = (1, f ₁ , 5, f ₃ )	(2-27)(2-27)	f₂＝(e，e，5，e)f ₂ = (e, e, 5, e)

For r ₃ there are: {r ₃ }={wasn't remarked}

{x ₃ }={That,who,e}

{y ₃ }={men,the liberals,f ₁ ,f ₂ ,e}

{z ₃ }={e}

序号Serial number	行矩阵f₃ Row matrix f ₃	(3-8)(3-8)	f₃＝(who，f₁，r₃，e)f ₃ =(who,f ₁ ,r ₃ ,e)
序号Serial number	行矩阵f₃ Row matrix f ₃	(3-8)(3-8)	f₃＝(who，f₁，r₃，e)f ₃ =(who,f ₁ ,r ₃ ,e)	(3-1)(3-1)	f₃＝(That，men，r₃，e)f ₃ =(That,men,r ₃ ,e)	(3-9)(3-9)	f₃＝(e，f₁，r₃，e)f ₃ = (e, f ₁ , r ₃ , e)
(3-2)(3-2)	f₃＝(who，men，r₃，e)f ₃ =(who,men,r ₃ ,e)	(3-10)(3-10)	f₃＝(That，f₂，r₃，e)f ₃ =(That,f ₂ ,r ₃ ,e)	(3-1)(3-1)	f₃＝(That，men，r₃，e)f ₃ =(That,men,r ₃ ,e)	(3-9)(3-9)	f₃＝(e，f₁，r₃，e)f ₃ = (e, f ₁ , r ₃ , e)
(3-2)(3-2)	f₃＝(who，men，r₃，e)f ₃ =(who,men,r ₃ ,e)	(3-10)(3-10)	f₃＝(That，f₂，r₃，e)f ₃ =(That,f ₂ ,r ₃ ,e)	(3-3)(3-3)	f₃＝(e，men，r₃，e)f ₃ =(e,men,r ₃ ,e)	(3-11)(3-11)	f₃＝(who，f₂，r₃，e)f ₃ =(who,f ₂ ,r ₃ ,e)
(3-4)(3-4)	f₃＝(That，the liberals，r₃，e)f ₃ =(That, the liberals,r ₃ ,e)	(3-12)(3-12)	f₃＝(e，f₂，r₃，e)f ₃ = (e, f ₂ , r ₃ , e)	(3-3)(3-3)	f₃＝(e，men，r₃，e)f ₃ =(e,men,r ₃ ,e)	(3-11)(3-11)	f₃＝(who，f₂，r₃，e)f ₃ =(who,f ₂ ,r ₃ ,e)
(3-4)(3-4)	f₃＝(That，the liberals，r₃，e)f ₃ =(That, the liberals,r ₃ ,e)	(3-12)(3-12)	f₃＝(e，f₂，r₃，e)f ₃ = (e, f ₂ , r ₃ , e)	(3-5)(3-5)	f₃＝(who，the liberals，r₃，e)f ₃ =(who,the liberals,r ₃ ,e)	(3-13)(3-13)	f₃＝(That，e，r₃，e)f ₃ =(That,e,r ₃ ,e)
(3-6)(3-6)	f₃＝(e，the liberals，r₃，e)f ₃ = (e, the liberals, r ₃ , e)	(3-14)(3-14)	f₃＝(who，e，r₃，e)f ₃ =(who,e,r ₃ ,e)	(3-5)(3-5)	f₃＝(who，the liberals，r₃，e)f ₃ =(who,the liberals,r ₃ ,e)	(3-13)(3-13)	f₃＝(That，e，r₃，e)f ₃ =(That,e,r ₃ ,e)
(3-6)(3-6)	f₃＝(e，the liberals，r₃，e)f ₃ = (e, the liberals, r ₃ , e)	(3-14)(3-14)	f₃＝(who，e，r₃，e)f ₃ =(who,e,r ₃ ,e)	(3-7)(3-7)	f₃＝(That，f₁，r₃，e)f ₃ =(That,f ₁ ,r ₃ ,e)	(3-15)(3-15)	f₃＝(e，e，r₃，e)f ₃ = (e, e, r ₃ , e)

Replace the constant with a sequence value, f ₃ (x ₃ , y ₃ , r ₃ , z ₃ )= (see list below)

Serial number

Row matrix f ₃

(3-8)

f ₃ = (3, f ₁ , 7, e)

(3-1)(3-1)	f₃＝(1，2，7，e)f ₃ = (1, 2, 7, e)	(3-9)(3-9)	f₃＝(e，f₁，7，e)f ₃ = (e, f ₁ , 7, e)
(3-1)(3-1)	f₃＝(1，2，7，e)f ₃ = (1, 2, 7, e)	(3-9)(3-9)	f₃＝(e，f₁，7，e)f ₃ = (e, f ₁ , 7, e)	(3-2)(3-2)	f₃＝(3，2，7，e)f ₃ = ( ₃ , 2, 7, e)	(3-10)(3-10)	f₃＝(1，f₂，7，e)f ₃ = (1, f ₂ , 7, e)
(3-3)(3-3)	f₃＝(e，2，7，e)f ₃ = (e, 2, 7, e)	(3-11)(3-11)	f₃＝(3，f₂，7，e)f ₃ = (3, f ₂ , 7, e)	(3-2)(3-2)	f₃＝(3，2，7，e)f ₃ = ( ₃ , 2, 7, e)	(3-10)(3-10)	f₃＝(1，f₂，7，e)f ₃ = (1, f ₂ , 7, e)
(3-3)(3-3)	f₃＝(e，2，7，e)f ₃ = (e, 2, 7, e)	(3-11)(3-11)	f₃＝(3，f₂，7，e)f ₃ = (3, f ₂ , 7, e)	(3-4)(3-4)	f₃＝(1，6，7，e)f ₃ = (1,6,7,e)	(3-12)(3-12)	f₃＝(e，f₂，7，e)f ₃ = (e, f ₂ , 7, e)
(3-5)(3-5)	f₃＝(3，6，7，e)f ₃ = (3,6,7,e)	(3-13)(3-13)	f₃＝(1，e，7，e)f ₃ = (1, e, 7, e)	(3-4)(3-4)	f₃＝(1，6，7，e)f ₃ = (1,6,7,e)	(3-12)(3-12)	f₃＝(e，f₂，7，e)f ₃ = (e, f ₂ , 7, e)
(3-5)(3-5)	f₃＝(3，6，7，e)f ₃ = (3,6,7,e)	(3-13)(3-13)	f₃＝(1，e，7，e)f ₃ = (1, e, 7, e)	(3-6)(3-6)	f₃＝(e，6，7，e)f ₃ = (e, 6, 7, e)	(3-14)(3-14)	f₃＝(3，e，7，e)f ₃ = (3, e, 7, e)
(3-7)(3-7)	f₃＝(1，f₁，7，e)f ₃ = (1, f ₁ , 7, e)	(3-15)(3-15)	f₃＝(e，e，7，e)f ₃ = (e, e, 7, e)	(3-6)(3-6)	f₃＝(e，6，7，e)f ₃ = (e, 6, 7, e)	(3-14)(3-14)	f₃＝(3，e，7，e)f ₃ = (3, e, 7, e)

Apply the principle of multiplication in combinatorial mathematics:

|S|=|f ₁ |×|f ₂ |×|f ₃ |=18×27×15=7290

A total of 7290 possible matrix solutions are generated.

For all syntactic structures, the matrix solution may be solved, and the running matrix is substituted into the solver and the structural correction program to obtain the possible matrix solution as the final result of the syntactic structure analysis:

This example sentence is a typical example sentence successfully processed by the overall interpolation method in this paper. After the overall insertion processing described above, one of the overall null insertion results of the above possible matrix solution is a final single-row vector as follows: e+<(1+<2+<(3+<e+<4+<e)+ <5+<6)+<7+<e. This final single-line vector does not have a reverse order number and is a reasonable final single-line vector. This final single-line vector is identical to the original sentence. This possible matrix solution is also the correct syntactic structure analysis result of this example sentence.

Reverting the number of the above possible matrix solution to a word unit yields the following form:

Convert this matrix to a linear expression:

Remove e:

Thus, the correct interpretation of the above sentence example is obtained, that is, f ₃ is the main sentence, that is, the core sentence; f ₂ is the subject of f ₃ , that is, the subject clause; f ₁ is the attributive clause, and the men is modified.

This example can better show the superiority of the method. In response to the above statement, two of the world's most advanced natural language syntax structure parsing devices recognized by the current computer industry

- Berkeley Parser and Stanford Parser, at the time of submission, still give the wrong result. The results given by these two devices are identical. The results are as follows:

1That men didn’t bother;

2who were appointed;

3the liberals wasn’t remarked upon by the press.

1 is the main sentence, which is the core sentence; 3 is the object of 1 , that is, the object clause; 2 is the attributive clause, modifying the men; That is the qualifier, modifying the men.

In English, if the subject clause is at the beginning of the sentence and it is guided by that, then that can't be omitted, even if it is spoken. In the method of the present invention, since the sentence is processed into a syntax vector, the subject clause is included. The part that did’t bother the liberals, in the process of parsing, reserved sufficient space to fully protect its possibility of being generated as a complete clause.

The major technical loopholes in the analysis of the subject clauses that led to the error often failed to make up for the two world-leading natural language syntax structure parsing devices.

Part C3 Example 3

Example 3: As another example, the parsing process of the method of the present embodiment for a complicated structure such as "Jack who has a beautiful car is a businessman." is explained below. The above statement is preprocessed to remove impurities and the numbered word sequence is:

原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number
原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number	JackJack	名词代词单元Noun pronoun unit	11
whoWho	从属关联词单元Subordinate unit	22	JackJack	名词代词单元Noun pronoun unit	11
whoWho	从属关联词单元Subordinate unit	22	hasHas	谓语动词单元Predicate verb unit	33
a cara car	名词代词单元Noun pronoun unit	44	hasHas	谓语动词单元Predicate verb unit	33
a cara car	名词代词单元Noun pronoun unit	44	isIs	谓语动词单元Predicate verb unit	55
a businessmana businessman	名词代词单元Noun pronoun unit	66	isIs	谓语动词单元Predicate verb unit	55

There are two predicate verb units in the sentence, which are denoted as r ₁ and r _{2 respectively} .

For r ₁ there, {r ₁ }={has}

{x ₁ }={who,e} (e is an empty string)

{y ₁ }={Jack,e}

{z ₁ }={a car,f ₂ ,e}

序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(who，Jack，r₁，a car)f ₁ =(who,Jack,r ₁ ,a car)	(1-7)(1-7)	f₁＝(who，e，r₁，f₂)f ₁ =(who,e,r ₁ ,f ₂ )

(1-2)(1-2)	f₁＝(e，Jack，r₁，a car)f ₁ = (e, Jack, r ₁ , a car)	(1-8)(1-8)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )
(1-2)(1-2)	f₁＝(e，Jack，r₁，a car)f ₁ = (e, Jack, r ₁ , a car)	(1-8)(1-8)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-3)(1-3)	f₁＝(who，e，r₁，a car)f ₁ =(who,e,r ₁ ,a car)	(1-9)(1-9)	f₁＝(who，Jack，r₁，e)f ₁ =(who,Jack,r ₁ ,e)
(1-4)(1-4)	f₁＝(e，e，r₁，a car)f ₁ =(e,e,r ₁ ,a car)	(1-10)(1-10)	f₁＝(e，Jack，r₁，e)f ₁ = (e, Jack, r ₁ , e)	(1-3)(1-3)	f₁＝(who，e，r₁，a car)f ₁ =(who,e,r ₁ ,a car)	(1-9)(1-9)	f₁＝(who，Jack，r₁，e)f ₁ =(who,Jack,r ₁ ,e)
(1-4)(1-4)	f₁＝(e，e，r₁，a car)f ₁ =(e,e,r ₁ ,a car)	(1-10)(1-10)	f₁＝(e，Jack，r₁，e)f ₁ = (e, Jack, r ₁ , e)	(1-5)(1-5)	f₁＝(who，Jack，r₁，f₂)f ₁ =(who,Jack,r ₁ ,f ₂ )	(1-11)(1-11)	f₁＝(who，e，r₁，e)f ₁ =(who,e,r ₁ ,e)
(1-6)(1-6)	f₁＝(e，Jack，r₁，f₂)f ₁ = (e, Jack, r ₁ , f ₂ )	(1-12)(1-12)	f₁＝(e，e，r₁，e)f ₁ = (e, e, r ₁ , e)	(1-5)(1-5)	f₁＝(who，Jack，r₁，f₂)f ₁ =(who,Jack,r ₁ ,f ₂ )	(1-11)(1-11)	f₁＝(who，e，r₁，e)f ₁ =(who,e,r ₁ ,e)

Replace the constant with a sequence value, f ₁ (x ₁ , y ₁ , r ₁ , z ₁ )= (see list below)

序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(2，1，3，4)f ₁ =(2,1,3,4)	(1-7)(1-7)	f₁＝(2，e，3，f₂)f ₁ =(2,e,3,f ₂ )
(1-2)(1-2)	f₁＝(e，1，3，4)f ₁ = (e, 1, 3, 4)	(1-8)(1-8)	f₁＝(e，e，3，f₂)f ₁ = (e, e, 3, f ₂ )	(1-1)(1-1)	f₁＝(2，1，3，4)f ₁ =(2,1,3,4)	(1-7)(1-7)	f₁＝(2，e，3，f₂)f ₁ =(2,e,3,f ₂ )
(1-2)(1-2)	f₁＝(e，1，3，4)f ₁ = (e, 1, 3, 4)	(1-8)(1-8)	f₁＝(e，e，3，f₂)f ₁ = (e, e, 3, f ₂ )	(1-3)(1-3)	f₁＝(2，e，3，4)f ₁ = (2, e, 3, 4)	(1-9)(1-9)	f₁＝(2，1，3，e)f ₁ = (2,1,3,e)
(1-4)(1-4)	f₁＝(e，e，3，4)f ₁ = (e, e, 3, 4)	(1-10)(1-10)	f₁＝(e，1，3，e)f ₁ = (e, 1, 3, e)	(1-3)(1-3)	f₁＝(2，e，3，4)f ₁ = (2, e, 3, 4)	(1-9)(1-9)	f₁＝(2，1，3，e)f ₁ = (2,1,3,e)
(1-4)(1-4)	f₁＝(e，e，3，4)f ₁ = (e, e, 3, 4)	(1-10)(1-10)	f₁＝(e，1，3，e)f ₁ = (e, 1, 3, e)	(1-5)(1-5)	f₁＝(2，1，3，f₂)f ₁ = (2,1,3,f ₂ )	(1-11)(1-11)	f₁＝(2，e，3，e)f ₁ = (2, e, 3, e)
(1-6)(1-6)	f₁＝(e，1，3，f₂)f ₁ =(e,1,3,f ₂ )	(1-12)(1-12)	f₁＝(e，e，3，e)f ₁ = (e, e, 3, e)	(1-5)(1-5)	f₁＝(2，1，3，f₂)f ₁ = (2,1,3,f ₂ )	(1-11)(1-11)	f₁＝(2，e，3，e)f ₁ = (2, e, 3, e)

For r ₂ , {r ₂ }={is}

{x ₂ }={who,e} (e is an empty string)

{y ₂ }={Jack,a car,f ₁ ,e}

{z ₂ }={a businessman,e}

序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂
序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂	(2-1)(2-1)	f₂＝(who，Jack，r₂，a businessman)f ₂ =(who,Jack,r ₂ ,a businessman)	(2-9)(2-9)	f₂＝(who，Jack，r₂，e)f ₂ =(who,Jack,r ₂ ,e)
(2-2)(2-2)	f₂＝(e，Jack，r₂，a businessman)f ₂ = (e, Jack, r ₂ , a businessman)	(2-10)(2-10)	f₂＝(e，Jack，r₂，e)f ₂ = (e, Jack, r ₂ , e)	(2-1)(2-1)		(2-9)(2-9)	f₂＝(who，Jack，r₂，e)f ₂ =(who,Jack,r ₂ ,e)

(2-3)(2-3)	f₂＝(who，a car，r₂，a businessman)f ₂ =(who,a car,r ₂ ,a businessman)	(2-11)(2-11)	f₂＝(who，a car，r₂，e)f ₂ =(who,a car,r ₂ ,e)
(2-3)(2-3)		(2-11)(2-11)	f₂＝(who，a car，r₂，e)f ₂ =(who,a car,r ₂ ,e)	(2-4)(2-4)	f₂＝(e，a car，r₂，a businessman)f ₂ =(e,a car,r ₂ ,a businessman)	(2-12)(2-12)	f₂＝(e，a car，r₂，e)f ₂ =(e,a car,r ₂ ,e)
(2-5)(2-5)	f₂＝(who，f₁，r₂，a businessman)f ₂ =(who,f ₁ ,r ₂ ,a businessman)	(2-13)(2-13)	f₂＝(who，f₁，r₂，e)f ₂ =(who,f ₁ ,r ₂ ,e)	(2-4)(2-4)		(2-12)(2-12)	f₂＝(e，a car，r₂，e)f ₂ =(e,a car,r ₂ ,e)
(2-5)(2-5)		(2-13)(2-13)	f₂＝(who，f₁，r₂，e)f ₂ =(who,f ₁ ,r ₂ ,e)	(2-6)(2-6)	f₂＝(e，f₁，r₂，a businessman)f ₂ =(e,f ₁ ,r ₂ ,a businessman)	(2-14)(2-14)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)
(2-7)(2-7)	f₂＝(who，e，r₂，a businessman)f ₂ =(who,e,r ₂ ,a businessman)	(2-15)(2-15)	f₂＝(who，e，r₂，e)f ₂ =(who,e,r ₂ ,e)	(2-6)(2-6)	f₂＝(e，f₁，r₂，a businessman)f ₂ =(e,f ₁ ,r ₂ ,a businessman)	(2-14)(2-14)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)
(2-7)(2-7)	f₂＝(who，e，r₂，a businessman)f ₂ =(who,e,r ₂ ,a businessman)	(2-15)(2-15)	f₂＝(who，e，r₂，e)f ₂ =(who,e,r ₂ ,e)	(2-8)(2-8)	f₂＝(e，e，r₂，a businessman)f ₂ =(e,e,r ₂ ,a businessman)	(2-16)(2-16)	f₂＝(e，e，r₂，e)f ₂ = (e, e, r ₂ , e)

Replace the constant with a sequence value, f ₂ (x ₂ , y ₂ , r ₂ , z ₂ )= (see list below)

序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂
序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂	(2-1)(2-1)	f₂＝(2，1，5，6)f ₂ = (2,1,5,6)	(2-9)(2-9)	f₂＝(2，1，5，e)f ₂ = (2,1,5,e)
(2-2)(2-2)	f₂＝(e，1，5，6)f ₂ = (e, 1, 5, 6)	(2-10)(2-10)	f₂＝(e，1，5，e)f ₂ = (e, 1, 5, e)	(2-1)(2-1)	f₂＝(2，1，5，6)f ₂ = (2,1,5,6)	(2-9)(2-9)	f₂＝(2，1，5，e)f ₂ = (2,1,5,e)
(2-2)(2-2)	f₂＝(e，1，5，6)f ₂ = (e, 1, 5, 6)	(2-10)(2-10)	f₂＝(e，1，5，e)f ₂ = (e, 1, 5, e)	(2-3)(2-3)	f₂＝(2，4，5，6)f ₂ = (2, 4, 5, 6)	(2-11)(2-11)	f₂＝(2，4，5，e)f ₂ = (2, 4, 5, e)
(2-4)(2-4)	f₂＝(e，4，5，6)f ₂ = (e, 4, 5, 6)	(2-12)(2-12)	f₂＝(e，4，5，e)f ₂ = (e, 4, 5, e)	(2-3)(2-3)	f₂＝(2，4，5，6)f ₂ = (2, 4, 5, 6)	(2-11)(2-11)	f₂＝(2，4，5，e)f ₂ = (2, 4, 5, e)
(2-4)(2-4)	f₂＝(e，4，5，6)f ₂ = (e, 4, 5, 6)	(2-12)(2-12)	f₂＝(e，4，5，e)f ₂ = (e, 4, 5, e)	(2-5)(2-5)	f₂＝(2，f₁，5，6)f ₂ = (2, f ₁ , 5, 6)	(2-13)(2-13)	f₁＝(2，f₁，5，e)f ₁ =(2,f ₁ ,5,e)
(2-6)(2-6)	f₂＝(e，f₁，5，6)f ₂ = (e, f ₁ , 5, 6)	(2-14)(2-14)	f₂＝(e，f₁，5，e)f ₂ = (e, f ₁ , 5, e)	(2-5)(2-5)	f₂＝(2，f₁，5，6)f ₂ = (2, f ₁ , 5, 6)	(2-13)(2-13)	f₁＝(2，f₁，5，e)f ₁ =(2,f ₁ ,5,e)
(2-6)(2-6)	f₂＝(e，f₁，5，6)f ₂ = (e, f ₁ , 5, 6)	(2-14)(2-14)	f₂＝(e，f₁，5，e)f ₂ = (e, f ₁ , 5, e)	(2-7)(2-7)	f₂＝(2，e，5，6)f ₂ = (2, e, 5, 6)	(2-15)(2-15)	f₂＝(2，e，5，e)f ₂ = (2, e, 5, e)
(2-8)(2-8)	f₂＝(e，e，5，6)f ₂ = (e, e, 5, 6)	(2-16)(2-16)	f₂＝(e，e，5，e)f ₂ = (e, e, 5, e)	(2-7)(2-7)	f₂＝(2，e，5，6)f ₂ = (2, e, 5, 6)	(2-15)(2-15)	f₂＝(2，e，5，e)f ₂ = (2, e, 5, e)

Apply the principle of multiplication in combinatorial mathematics:

|S|=|f ₁ |×|f ₂ |=12×16=192

A total of 192 possible matrix solutions are generated.

A possible matrix solution can be obtained as the final result of the parsing of the syntax structure:

This example sentence is a typical example sentence successfully processed by the overall interpolation method in this paper. After the overall interpolation process described above, the above possible matrix solution yields the only final single-row vector without the inverse number: e+<1+<(2+<e+<3+<4)+<5+<6 . This final single-line vector is a reasonable final single-line vector. The syntactic sequence value of this final single-line vector is exactly the same as the original sentence. This possible matrix solution is also the correct syntactic structure analysis result of this example sentence.

Convert this matrix to a linear expression:

Remove e:

Part C4 Example 4

Example 4: As another example, the parsing process of the method of the present embodiment for a sentence of a parallel structure such as "After Jack, Mary and Linda left, I gave my son a new book." will be described below.

原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number
原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number	AfterAfter	从属关联词单元Subordinate unit	11

JackJack	名词代词单元Noun pronoun unit	22
JackJack	名词代词单元Noun pronoun unit	22	MaryMary	名词代词单元Noun pronoun unit	33
andAnd	并列关联词单元Parallel word unit	44	MaryMary	名词代词单元Noun pronoun unit	33
andAnd	并列关联词单元Parallel word unit	44	LindaLinda	名词代词单元Noun pronoun unit	55
leftLeft	谓语动词单元Predicate verb unit	66	LindaLinda	名词代词单元Noun pronoun unit	55
leftLeft	谓语动词单元Predicate verb unit	66	II	名词代词单元Noun pronoun unit	77
gaveGave	谓语动词单元Predicate verb unit	88	II	名词代词单元Noun pronoun unit	77
gaveGave	谓语动词单元Predicate verb unit	88	my sonMy son	名词代词单元Noun pronoun unit	99
a booka book	名词代词单元Noun pronoun unit	1010	my sonMy son	名词代词单元Noun pronoun unit	99

The following steps include generating a parallel lexical pronoun combination vector family:

S2.1 selects two noun pronoun units that are not repeated:

B. If there are other word units between the two noun pronoun units, check each word unit between the two noun pronoun units: if any between the two noun pronoun units Word units, all of which are noun pronoun units or side-by-side related word units, then use the selected two noun pronoun units and the whole word unit between the two noun pronoun units as a parallel noun pronoun combination vector, and retain the juxtaposition Noun pronoun combination vector; otherwise, no parallel noun pronoun combination vector is generated;

1A(S) takes out all NPI phrases, all VNP phrases, and all NOMP phrases in the original sentence, and lists all NPI phrases, all VNP phrases, and all NOMP phrases in the original sentence as a set, and records the set as Ψ={Jack, Mary, Linda, I, my son, a book}={2,3,5,7,9,10}.

2B (Ψ) means follow

The result of 3K(α,β) versus unary function B(Ψ), that is, one given

By element

Orderly pair

then

The generated ordered pairs are:

Set collection

Then establish a continuous string formula

among them

Is the slave in the original sentence S

To

a set of adjacent consecutive or empty words, and

then

Then Φ ¹ = 2+ < e + < 3, Φ ² = 2+ < 3 + < 4 + < 5, Φ ³ = 2+ < 3 + < 4 + < 5 + < 6 + < 7, Φ ⁴ = 2+ <3+<4+<5+<6+<7+<8+<9, Φ ⁵ =2+<3+<4+<5+<6+<7+<8+<9+<10, Φ ⁶ =3+<4+<5, Φ ⁷ =3+<4+<5+<6+<7, Φ ⁸ =3+<4+<5+<6+<7+<8+<9 , Φ ⁹ =3+<4+<5+<6+<7+<8+<9+<10, Φ ¹⁰ =5+<6+<7, Φ ¹¹ =5+<6+<7+<8+<9,Φ ¹² =5+<6+<7+<8+<9+<10, Φ ¹³ =7+<8+<9, Φ ¹⁴ =7+<8+<9+<10, Φ ¹⁵ =9+<e+<10.

4H(Φ ^t ) generated for the binary function K(α, β)

Check: if the given element is γ∈Φ ^t , and

And

Both: γ=NPI or γ=VNP or γ=NOMP or γ=CONJ or γ=e, then change the mark of Φ ^t to

Φ ^t generation

Set collection

Collection

then

5M(α,β) represents a set for any

If collection

Corresponding

Defining a collection family that contains collections

The whole set of the composition, the collection family is recorded as

then

Then M(α,β)={I ₁ ({2,3}), I ₂ ({3,5}), I ₃ ({9,10})}.

6N(α,β) results for the binary function M(α,β)

That is, for any collection

If collection

There is a corresponding collection family

Then construct a new collection as follows

7U(α) results for the binary function N(α,β)

take

Assume

For the given element γ,

For r ₁ there is {r ₁ }={left}, numbered 6. Then the corresponding subject is selected as follows: when there is no r _k-1 : {y _k }=NPI _yk ∪VNP _yk ∪NOMP _k ∪G _k ∪{e};

among them:

In the above formula, G _k denotes a union of the total number of collocated noun pronouns whose maximum value is smaller than the corresponding predicate verb unit number.

Then there are:

Then the set of subject elements corresponding to r ₁ is:

In the process of generating the set of object elements {z ₁ }, {z ₂ }, the algorithm for running the parallel object is as follows:

2B (Ψ) means follow

The result of 3K(α,β) versus unary function B(Ψ), that is, one given

By element

Orderly pair

then

The generated ordered pairs are:

Set collection

Then establish a continuous string formula

among them

Is the slave in the original sentence S

To

a set of adjacent consecutive or empty words, and

then

4H(Φ ^t ) generated for the binary function K(α, β)

Check: if the given element is γ∈Φ ^t , and

And

Φ ^t generation

Set collection

Collection

then

5M(α,β) represents a set for any

If collection

Corresponding

Defining a collection family that contains collections

The whole set of the composition, the collection family is recorded as

then

Then M(α,β)={I ₁ ({2,3}), I ₂ ({3,5}), I ₃ ({9,10})}.

6N(α,β) results for the binary function M(α,β)

That is, for any collection

If collection

There is a corresponding set brigade

Then construct a new collection as follows

7V(β) represents the result of the binary function N(α, β)

take

Assume

For the given element γ,

For r ₂ there is {r ₂ }={gave}, numbered 8. Then, the corresponding object is selected by: when there is no r _k+1 : {z _k }=NPI _zk ∪VNP _zk ∪OBJP _k ∪H _k ∪{e};

among them:

Where: when r _k+1 does not exist:

In the above formula, H _k represents a union of the total number of collocated noun pronouns of the corresponding predicate verb unit.

Then there are:

Then the set of object elements corresponding to r ₂ is:

Note: In the process of processing, the parallel noun pronoun combination vector is treated as a whole; the parallel noun pronoun combination can not be inserted into other syntax vectors; when checking the order value, directly combine the parallel noun pronouns into the included syntactic order Values can be substituted.

For r ₁ there is {r ₁ }={left}

{x ₁ }={After,and,e} (e is an empty string)

{z ₁ }={f ₂ ,e}

For r ₂ there is {r ₂ }={gave}

{x ₂ }={After,and,e} (e is an empty string)

Replace the constant with the sequence value, slightly.

Apply the principle of multiplication in combinatorial mathematics:

|S|=|f ₁ |×|f ₂ |=42×108=4536

A total of 4536 possible matrix solutions are generated.

The above possible matrix solution is further reduced to obtain the following form:

Note: This result is obtained by the overall insertion method.

Convert this matrix to a linear expression:

Remove e:

Part C5 Example 5

Example 5: As another example, the parsing process of the method of the present embodiment for a sentence of a parallel structure such as "Linda was singing, and Mary was dancing." will be described below.

原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number
原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number	LindaLinda	名词代词单元Noun pronoun unit	11
was singingWas singing	谓语动词单元Predicate verb unit	22	LindaLinda	名词代词单元Noun pronoun unit	11
was singingWas singing	谓语动词单元Predicate verb unit	22	andAnd	并列关联词单元Parallel word unit	33
MaryMary	名词代词单元Noun pronoun unit	44	andAnd	并列关联词单元Parallel word unit	33
MaryMary	名词代词单元Noun pronoun unit	44	was dancingWas dancing	谓语动词单元Predicate verb unit	55

For r ₁ there is {r ₁ }={was singing}

{x ₁ }={e} (e is an empty string)

{y ₁ }={Linda,e}

{z ₁ }={Mary,f ₂ ,e}

序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(e，Linda，r₁，Mary)f ₁ = (e, Linda, r ₁ , Mary)
(1-2)(1-2)	f₁＝(e，e，r₁，Mary)f ₁ = (e, e, r ₁ , Mary)	(1-1)(1-1)	f₁＝(e，Linda，r₁，Mary)f ₁ = (e, Linda, r ₁ , Mary)
(1-2)(1-2)	f₁＝(e，e，r₁，Mary)f ₁ = (e, e, r ₁ , Mary)	(1-3)(1-3)	f₁＝(e，Linda，r₁，f₂)f ₁ = (e, Linda, r ₁ , f ₂ )
(1-4)(1-4)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-3)(1-3)	f₁＝(e，Linda，r₁，f₂)f ₁ = (e, Linda, r ₁ , f ₂ )
(1-4)(1-4)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-5)(1-5)	f₁＝(e，Linda，r₁，e)f ₁ = (e, Linda, r ₁ , e)
(1-6)(1-6)	f₁＝(e，e，r₁，e)f ₁ = (e, e, r ₁ , e)	(1-5)(1-5)	f₁＝(e，Linda，r₁，e)f ₁ = (e, Linda, r ₁ , e)

For r ₂ there is {r ₂ }={was dancing}

{x ₂ }={and,e} (e is an empty string)

{y ₂ }={Linda,Mary,f ₁ ,e}

{z ₂ }={e}

序号Serial number	行矩阵f₂ Row matrix f ₂
序号Serial number	行矩阵f₂ Row matrix f ₂	(2-1)(2-1)	f₂＝(and，Linda，r₂，e)f ₂ =(and,Linda,r ₂ ,e)
(2-2)(2-2)	f₂＝(and，Mary，r₂，e)f ₂ =(and,Mary,r ₂ ,e)	(2-1)(2-1)	f₂＝(and，Linda，r₂，e)f ₂ =(and,Linda,r ₂ ,e)
(2-2)(2-2)	f₂＝(and，Mary，r₂，e)f ₂ =(and,Mary,r ₂ ,e)	(2-3)(2-3)	f₂＝(and，f₁，r₂，e)f ₂ =(and,f ₁ ,r ₂ ,e)
(2-4)(2-4)	f₂＝(and，e，r₂，e)f ₂ =(and,e,r ₂ ,e)	(2-3)(2-3)	f₂＝(and，f₁，r₂，e)f ₂ =(and,f ₁ ,r ₂ ,e)
(2-4)(2-4)	f₂＝(and，e，r₂，e)f ₂ =(and,e,r ₂ ,e)	(2-5)(2-5)	f₂＝(e，Linda，r₂，e)f ₂ = (e, Linda, r ₂ , e)
(2-6)(2-6)	f₂＝(e，Mary，r₂，e)f ₂ =(e,Mary,r ₂ ,e)	(2-5)(2-5)	f₂＝(e，Linda，r₂，e)f ₂ = (e, Linda, r ₂ , e)
(2-6)(2-6)	f₂＝(e，Mary，r₂，e)f ₂ =(e,Mary,r ₂ ,e)	(2-7)(2-7)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)
(2-8)(2-8)	f₂＝(e，e，r₂，e)f ₂ = (e, e, r ₂ , e)	(2-7)(2-7)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)

Replace the constant with the sequence value, slightly.

Apply the principle of multiplication in combinatorial mathematics:

|S|=|f ₁ |×|f ₂ |=6×8=48

A total of 48 possible matrix solutions are generated.

Convert this matrix to a linear expression:

Remove e:

Part C6 Example 6

Example 6: As another example, the parsing process of the method of the present embodiment for a sentence of a parallel structure such as "I know that you have a car and that he has a bike." will be described below.

原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number
原句短语Original sentence	短语类型Phrase type	顺序编号Sequence number	II	名词代词单元Noun pronoun unit	11
knowKnow	谓语动词单元Predicate verb unit	22	II	名词代词单元Noun pronoun unit	11
knowKnow	谓语动词单元Predicate verb unit	22	that AThat A	从属关联词单元Subordinate unit	33
youYou	名词代词单元Noun pronoun unit	44	that AThat A	从属关联词单元Subordinate unit	33
youYou	名词代词单元Noun pronoun unit	44	haveHave	谓语动词单元Predicate verb unit	55
a cara car	名词代词单元Noun pronoun unit	66	haveHave	谓语动词单元Predicate verb unit	55
a cara car	名词代词单元Noun pronoun unit	66	andAnd	并列关联词单元Parallel word unit	77
that BThat B	从属关联词单元Subordinate unit	88	andAnd	并列关联词单元Parallel word unit	77
that BThat B	从属关联词单元Subordinate unit	88	heHe	名词代词单元Noun pronoun unit	99
hasHas	谓语动词单元Predicate verb unit	1010	heHe	名词代词单元Noun pronoun unit	99
hasHas	谓语动词单元Predicate verb unit	1010	a bikea bike	名词代词单元Noun pronoun unit	1111

The set of guide elements corresponding to each predicate verb unit r _k is:

{x _k }=Lead _k ∪conj _k ∪(conj _k οLead _k )∪{e}

The leader element corresponding to the verb unit r _k is x _k , and its possible value set is {x _k }. The set of possible elements of the predicate verb unit r _k is x _k :

{x ₂ }={Lead ₂ }∪{e}={that A,e};

{x ₃ }=Lead ₃ ∪conj ₃ ∪(conj ₃ οLead ₃ )∪{e}={that A,and,that B,Ψ,e}.

The above two formulas are derived from the algorithm for generating the leader element: {x _k }=Lead _k ∪conj _k ∪(conj _k οLead _k )∪{e}. Where (con _jk οLead _k )={R _k |R _k =conj+<Lead,conj< _□ r _k ,Lead< _□ r _k ,τ(Lead)=τ(conj)+1}; (conj _k οLead _k And a set of related word combination vectors consisting of a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and a dependent related word unit whose number is smaller than the corresponding predicate verb unit number and whose number is greater than the parallel related word unit number.

For r ₁ there is {r ₁ }={know}

{x ₁ }={e} (e is an empty string)

{y ₁ }={I,e}

{z ₁ }={you,f ₂ ,f ₃ ,e}

序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁
序号Serial number	行矩阵f₁ Row matrix f ₁	序号Serial number	行矩阵f₁ Row matrix f ₁	(1-1)(1-1)	f₁＝(e，I，r₁，you)f ₁ = (e, I, r ₁ , you)	(1-5)(1-5)	f₁＝(e，I，r₁，f₃)f ₁ = (e, I, r ₁ , f ₃ )
(1-2)(1-2)	f₁＝(e，e，r₁，you)f ₁ =(e,e,r ₁ ,you)	(1-6)(1-6)	f₁＝(e，e，r₁，f₃)f ₁ = (e, e, r ₁ , f ₃ )	(1-1)(1-1)	f₁＝(e，I，r₁，you)f ₁ = (e, I, r ₁ , you)	(1-5)(1-5)	f₁＝(e，I，r₁，f₃)f ₁ = (e, I, r ₁ , f ₃ )
(1-2)(1-2)	f₁＝(e，e，r₁，you)f ₁ =(e,e,r ₁ ,you)	(1-6)(1-6)	f₁＝(e，e，r₁，f₃)f ₁ = (e, e, r ₁ , f ₃ )	(1-3)(1-3)	f₁＝(e，I，r₁，f₂)f ₁ = (e, I, r ₁ , f ₂ )	(1-7)(1-7)	f₁＝(e，I，r₁，e)f ₁ = (e, I, r ₁ , e)
(1-4)(1-4)	f₁＝(e，e，r₁，f₂)f ₁ = (e, e, r ₁ , f ₂ )	(1-8)(1-8)	f₁＝(e，e，r₁，e)f ₁ = (e, e, r ₁ , e)	(1-3)(1-3)	f₁＝(e，I，r₁，f₂)f ₁ = (e, I, r ₁ , f ₂ )	(1-7)(1-7)	f₁＝(e，I，r₁，e)f ₁ = (e, I, r ₁ , e)

For r ₂ there is {r ₂ }={have}

{x ₂ }={that A,e} (e is an empty string)

{y ₂ }={you,I,f ₁ ,e}

{z ₂ }={a car,f ₃ ,e}

序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂
序号Serial number	行矩阵f₂ Row matrix f ₂	序号Serial number	行矩阵f₂ Row matrix f ₂	(2-1)(2-1)	f₂＝(that A，you，r₂，a car)f ₂ =(that A,you,r ₂ ,a car)	(2-13)(2-13)	f₂＝(that A，f₁，r₂，f₃)f ₂ =(that A,f ₁ ,r ₂ ,f ₃ )
(2-2)(2-2)	f₂＝(e，you，r₂，a car)f ₂ =(e,you,r ₂ ,a car)	(2-14)(2-14)	f₂＝(e，f₁，r₂，f₃)f ₂ = (e, f ₁ , r ₂ , f ₃ )	(2-1)(2-1)	f₂＝(that A，you，r₂，a car)f ₂ =(that A,you,r ₂ ,a car)	(2-13)(2-13)	f₂＝(that A，f₁，r₂，f₃)f ₂ =(that A,f ₁ ,r ₂ ,f ₃ )
(2-2)(2-2)	f₂＝(e，you，r₂，a car)f ₂ =(e,you,r ₂ ,a car)	(2-14)(2-14)	f₂＝(e，f₁，r₂，f₃)f ₂ = (e, f ₁ , r ₂ , f ₃ )	(2-3)(2-3)	f₂＝(that A，I，r₂，a car)f ₂ =(that A,I,r ₂ ,a car)	(2-15)(2-15)	f₂＝(that A，e，r₂，f₃)f ₂ =(that A,e,r ₂ ,f ₃ )
(2-4)(2-4)	f₂＝(e，I，r₂，a car)f ₂ =(e,I,r ₂ ,a car)	(2-16)(2-16)	f₂＝(e，e，r₂，f₃)f ₂ = (e, e, r ₂ , f ₃ )	(2-3)(2-3)	f₂＝(that A，I，r₂，a car)f ₂ =(that A,I,r ₂ ,a car)	(2-15)(2-15)	f₂＝(that A，e，r₂，f₃)f ₂ =(that A,e,r ₂ ,f ₃ )

(2-5)(2-5)	f₂＝(that A，f₁，r₂，a car)f ₂ =(that A,f ₁ ,r ₂ ,a car)	(2-17)(2-17)	f₂＝(that A，you，r₂，e)f ₂ =(that A,you,r ₂ ,e)
(2-5)(2-5)	f₂＝(that A，f₁，r₂，a car)f ₂ =(that A,f ₁ ,r ₂ ,a car)	(2-17)(2-17)	f₂＝(that A，you，r₂，e)f ₂ =(that A,you,r ₂ ,e)	(2-6)(2-6)	f₂＝(e，f₁，r₂，a car)f ₂ =(e,f ₁ ,r ₂ ,a car)	(2-18)(2-18)	f₂＝(e，you，r₂，e)f ₂ =(e,you,r ₂ ,e)
(2-7)(2-7)	f₂＝(that A，e，r₂，a car)f ₂ =(that A,e,r ₂ ,a car)	(2-19)(2-19)	f₂＝(that A，I，r₂，e)f ₂ =(that A,I,r ₂ ,e)	(2-6)(2-6)	f₂＝(e，f₁，r₂，a car)f ₂ =(e,f ₁ ,r ₂ ,a car)	(2-18)(2-18)	f₂＝(e，you，r₂，e)f ₂ =(e,you,r ₂ ,e)
(2-7)(2-7)	f₂＝(that A，e，r₂，a car)f ₂ =(that A,e,r ₂ ,a car)	(2-19)(2-19)	f₂＝(that A，I，r₂，e)f ₂ =(that A,I,r ₂ ,e)	(2-8)(2-8)	f₂＝(e，e，r₂，a car)f ₂ =(e,e,r ₂ ,a car)	(2-20)(2-20)	f₂＝(e，I，r₂，e)f ₂ = (e, I, r ₂ , e)
(2-9)(2-9)	f₂＝(that A，you，r₂，f₃)f ₂ =(that A,you,r ₂ ,f ₃ )	(2-21)(2-21)	f₂＝(that A，f₁，r₂，e)f ₂ =(that A,f ₁ ,r ₂ ,e)	(2-8)(2-8)	f₂＝(e，e，r₂，a car)f ₂ =(e,e,r ₂ ,a car)	(2-20)(2-20)	f₂＝(e，I，r₂，e)f ₂ = (e, I, r ₂ , e)
(2-9)(2-9)	f₂＝(that A，you，r₂，f₃)f ₂ =(that A,you,r ₂ ,f ₃ )	(2-21)(2-21)	f₂＝(that A，f₁，r₂，e)f ₂ =(that A,f ₁ ,r ₂ ,e)	(2-10)(2-10)	f₂＝(e，you，r₂，f₃)f ₂ =(e,you,r ₂ ,f ₃ )	(2-22)(2-22)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)
(2-11)(2-11)	f₂＝(that A，I，r₂，f₃)f ₂ =(that A,I,r ₂ ,f ₃ )	(2-23)(2-23)	f₂＝(that A，e，r₂，e)f ₂ =(that A,e,r ₂ ,e)	(2-10)(2-10)	f₂＝(e，you，r₂，f₃)f ₂ =(e,you,r ₂ ,f ₃ )	(2-22)(2-22)	f₂＝(e，f₁，r₂，e)f ₂ =(e,f ₁ ,r ₂ ,e)
(2-11)(2-11)	f₂＝(that A，I，r₂，f₃)f ₂ =(that A,I,r ₂ ,f ₃ )	(2-23)(2-23)	f₂＝(that A，e，r₂，e)f ₂ =(that A,e,r ₂ ,e)	(2-12)(2-12)	f₂＝(e，I，r₂，f₃)f ₂ = (e, I, r ₂ , f ₃ )	(2-24)(2-24)	f₂＝(e，e，r₂，e)f ₂ = (e, e, r ₂ , e)

For r ₃ there is {r ₃ }={has}

{x ₃ }={that A,that B,and,Ψ,e} (e is an empty string)

{y ₃ }={you,I,a car,he,f ₁ ,f ₂ ,e}

{z ₃ }={a bike,e}

序号Serial number	行矩阵f₃ Row matrix f ₃	序号Serial number	行矩阵f₃ Row matrix f ₃
序号Serial number	行矩阵f₃ Row matrix f ₃	序号Serial number	行矩阵f₃ Row matrix f ₃	(3-1)(3-1)	f₃＝(that A，you，r₃，a bike)f ₃ =(that A,you,r ₃ ,a bike)	(3-36)(3-36)	f₃＝(that A，you，r₃，e)f ₃ =(that A,you,r ₃ ,e)
(3-2)(3-2)	f₃＝(that B，you，r₃，a bike)f ₃ =(that B,you,r ₃ ,a bike)	(3-37)(3-37)	f₃＝(that B，you，r₃，e)f ₃ =(that B,you,r ₃ ,e)	(3-1)(3-1)	f₃＝(that A，you，r₃，a bike)f ₃ =(that A,you,r ₃ ,a bike)	(3-36)(3-36)	f₃＝(that A，you，r₃，e)f ₃ =(that A,you,r ₃ ,e)
(3-2)(3-2)	f₃＝(that B，you，r₃，a bike)f ₃ =(that B,you,r ₃ ,a bike)	(3-37)(3-37)	f₃＝(that B，you，r₃，e)f ₃ =(that B,you,r ₃ ,e)	(3-3)(3-3)	f₃＝(and，you，r₃，a bike)f ₃ =(and,you,r ₃ ,a bike)	(3-38)(3-38)	f₃＝(and，you，r₃，e)f ₃ =(and,you,r ₃ ,e)
(3-4)(3-4)	f₃＝(Ψ，you，r₃，a bike)f ₃ =(Ψ,you,r ₃ ,a bike)	(3-39)(3-39)	f₃＝(Ψ，you，r₃，e)f ₃ =(Ψ,you,r ₃ ,e)	(3-3)(3-3)	f₃＝(and，you，r₃，a bike)f ₃ =(and,you,r ₃ ,a bike)	(3-38)(3-38)	f₃＝(and，you，r₃，e)f ₃ =(and,you,r ₃ ,e)
(3-4)(3-4)	f₃＝(Ψ，you，r₃，a bike)f ₃ =(Ψ,you,r ₃ ,a bike)	(3-39)(3-39)	f₃＝(Ψ，you，r₃，e)f ₃ =(Ψ,you,r ₃ ,e)	(3-5)(3-5)	f₃＝(e，you，r₃，a bike)f ₃ = (e,you,r ₃ ,a bike)	(3-40)(3-40)	f₃＝(e，you，r₃，e)f ₃ = (e,you,r ₃ ,e)
(3-6)(3-6)	f₃＝(that A，I，r₃，a bike)f ₃ =(that A,I,r ₃ ,a bike)	(3-41)(3-41)	f₃＝(that A，I，r₃，e)f ₃ =(that A,I,r ₃ ,e)	(3-5)(3-5)	f₃＝(e，you，r₃，a bike)f ₃ = (e,you,r ₃ ,a bike)	(3-40)(3-40)	f₃＝(e，you，r₃，e)f ₃ = (e,you,r ₃ ,e)
(3-6)(3-6)	f₃＝(that A，I，r₃，a bike)f ₃ =(that A,I,r ₃ ,a bike)	(3-41)(3-41)	f₃＝(that A，I，r₃，e)f ₃ =(that A,I,r ₃ ,e)	(3-7)(3-7)	f₃＝(that B，I，r₃，a bike)f ₃ =(that B,I,r ₃ ,a bike)	(3-42)(3-42)	f₃＝(that B，I，r₃，e)f ₃ =(that B,I,r ₃ ,e)

(3-8)(3-8)	f₃＝(and，I，r₃，a bike)f ₃ =(and,I,r ₃ ,a bike)	(3-43)(3-43)	f₃＝(and，I，r₃，e)f ₃ =(and,I,r ₃ ,e)
(3-8)(3-8)	f₃＝(and，I，r₃，a bike)f ₃ =(and,I,r ₃ ,a bike)	(3-43)(3-43)	f₃＝(and，I，r₃，e)f ₃ =(and,I,r ₃ ,e)	(3-9)(3-9)	f₃＝(Ψ，I，r₃，a bike)f ₃ = (Ψ, I, r ₃ , a bike)	(3-44)(3-44)	f₃＝(Ψ，I，r₃，e)f ₃ = (Ψ, I, r ₃ , e)
(3-10)(3-10)	f₃＝(e，I，r₃，a bike)f ₃ = (e, I, r ₃ , a bike)	(3-45)(3-45)	f₃＝(e，I，r₃，e)f ₃ = (e, I, r ₃ , e)	(3-9)(3-9)	f₃＝(Ψ，I，r₃，a bike)f ₃ = (Ψ, I, r ₃ , a bike)	(3-44)(3-44)	f₃＝(Ψ，I，r₃，e)f ₃ = (Ψ, I, r ₃ , e)
(3-10)(3-10)	f₃＝(e，I，r₃，a bike)f ₃ = (e, I, r ₃ , a bike)	(3-45)(3-45)	f₃＝(e，I，r₃，e)f ₃ = (e, I, r ₃ , e)	(3-11)(3-11)	f₃＝(that A，a car，r₃，a bike)f ₃ =(that A,a car,r ₃ ,a bike)	(3-46)(3-46)	f₃＝(that A，a car，r₃，e)f ₃ =(that A,a car,r ₃ ,e)
(3-12)(3-12)	f₃＝(that B，a car，r₃，a bike)f ₃ =(that B,a car,r ₃ ,a bike)	(3-47)(3-47)	f₃＝(that B，a car，r₃，e)f ₃ =(that B,a car,r ₃ ,e)	(3-11)(3-11)	f₃＝(that A，a car，r₃，a bike)f ₃ =(that A,a car,r ₃ ,a bike)	(3-46)(3-46)	f₃＝(that A，a car，r₃，e)f ₃ =(that A,a car,r ₃ ,e)
(3-12)(3-12)	f₃＝(that B，a car，r₃，a bike)f ₃ =(that B,a car,r ₃ ,a bike)	(3-47)(3-47)	f₃＝(that B，a car，r₃，e)f ₃ =(that B,a car,r ₃ ,e)	(3-13)(3-13)	f₃＝(and，a car，r₃，a bike)f ₃ =(and,a car,r ₃ ,a bike)	(3-48)(3-48)	f₃＝(and，a car，r₃，e)f ₃ =(and,a car,r ₃ ,e)
(3-14)(3-14)	f₃＝(Ψ，a car，r₃，a bike)f ₃ = (Ψ, a car, r ₃ , a bike)	(3-49)(3-49)	f₃＝(Ψ，a car，r₃，e)f ₃ =(Ψ, a car,r ₃ ,e)	(3-13)(3-13)	f₃＝(and，a car，r₃，a bike)f ₃ =(and,a car,r ₃ ,a bike)	(3-48)(3-48)	f₃＝(and，a car，r₃，e)f ₃ =(and,a car,r ₃ ,e)
(3-14)(3-14)	f₃＝(Ψ，a car，r₃，a bike)f ₃ = (Ψ, a car, r ₃ , a bike)	(3-49)(3-49)	f₃＝(Ψ，a car，r₃，e)f ₃ =(Ψ, a car,r ₃ ,e)	(3-15)(3-15)	f₃＝(e，a car，r₃，a bike)f ₃ =(e,a car,r ₃ ,a bike)	(3-50)(3-50)	f₃＝(e，a car，r₃，e)f ₃ = (e, a car, r ₃ , e)
(3-16)(3-16)	f₃＝(that A，he，r₃，a bike)f ₃ =(that A,he,r ₃ ,a bike)	(3-51)(3-51)	f₃＝(that A，he，r₃，e)f ₃ =(that A,he,r ₃ ,e)	(3-15)(3-15)	f₃＝(e，a car，r₃，a bike)f ₃ =(e,a car,r ₃ ,a bike)	(3-50)(3-50)	f₃＝(e，a car，r₃，e)f ₃ = (e, a car, r ₃ , e)
(3-16)(3-16)	f₃＝(that A，he，r₃，a bike)f ₃ =(that A,he,r ₃ ,a bike)	(3-51)(3-51)	f₃＝(that A，he，r₃，e)f ₃ =(that A,he,r ₃ ,e)	(3-17)(3-17)	f₃＝(that B，he，r₃，a bike)f ₃ =(that B,he,r ₃ ,a bike)	(3-52)(3-52)	f₃＝(that B，he，r₃，e)f ₃ =(that B,he,r ₃ ,e)
(3-18)(3-18)	f₃＝(and，he，r₃，a bike)f ₃ =(and,he,r ₃ ,a bike)	(3-53)(3-53)	f₃＝(and，he，r₃，e)f ₃ =(and,he,r ₃ ,e)	(3-17)(3-17)	f₃＝(that B，he，r₃，a bike)f ₃ =(that B,he,r ₃ ,a bike)	(3-52)(3-52)	f₃＝(that B，he，r₃，e)f ₃ =(that B,he,r ₃ ,e)
(3-18)(3-18)	f₃＝(and，he，r₃，a bike)f ₃ =(and,he,r ₃ ,a bike)	(3-53)(3-53)	f₃＝(and，he，r₃，e)f ₃ =(and,he,r ₃ ,e)	(3-19)(3-19)	f₃＝(Ψ，he，r₃，a bike)f ₃ =(Ψ,he,r ₃ ,a bike)	(3-54)(3-54)	f₃＝(Ψ，he，r₃，e)f ₃ =(Ψ,he,r ₃ ,e)
(3-20)(3-20)	f₃＝(e，he，r₃，a bike)f ₃ =(e,he,r ₃ ,a bike)	(3-55)(3-55)	f₃＝(e，he，r₃，e)f ₃ =(e,he,r ₃ ,e)	(3-19)(3-19)	f₃＝(Ψ，he，r₃，a bike)f ₃ =(Ψ,he,r ₃ ,a bike)	(3-54)(3-54)	f₃＝(Ψ，he，r₃，e)f ₃ =(Ψ,he,r ₃ ,e)
(3-20)(3-20)	f₃＝(e，he，r₃，a bike)f ₃ =(e,he,r ₃ ,a bike)	(3-55)(3-55)	f₃＝(e，he，r₃，e)f ₃ =(e,he,r ₃ ,e)	(3-21)(3-21)	f₃＝(that A，f₁，r₃，a bike)f ₃ =(that A,f ₁ ,r ₃ ,a bike)	(3-56)(3-56)	f₃＝(that A，f₁，r₃，e)f ₃ =(that A,f ₁ ,r ₃ ,e)
(3-22)(3-22)	f₃＝(that B，f₁，r₃，a bike)f ₃ =(that B,f ₁ ,r ₃ ,a bike)	(3-57)(3-57)	f₃＝(that B，f₁，r₃，e)f ₃ =(that B,f ₁ ,r ₃ ,e)	(3-21)(3-21)	f₃＝(that A，f₁，r₃，a bike)f ₃ =(that A,f ₁ ,r ₃ ,a bike)	(3-56)(3-56)	f₃＝(that A，f₁，r₃，e)f ₃ =(that A,f ₁ ,r ₃ ,e)
(3-22)(3-22)	f₃＝(that B，f₁，r₃，a bike)f ₃ =(that B,f ₁ ,r ₃ ,a bike)	(3-57)(3-57)	f₃＝(that B，f₁，r₃，e)f ₃ =(that B,f ₁ ,r ₃ ,e)	(3-23)(3-23)	f₃＝(and，f₁，r₃，a bike)f ₃ =(and,f ₁ ,r ₃ ,a bike)	(3-58)(3-58)	f₃＝(and，f₁，r₃，e)f ₃ =(and,f ₁ ,r ₃ ,e)
(3-24)(3-24)	f₃＝(Ψ，f₁，r₃，a bike)f ₃ = (Ψ, f ₁ , r ₃ , a bike)	(3-59)(3-59)	f₃＝(Ψ，f₁，r₃，e)f ₃ = (Ψ, f ₁ , r ₃ , e)	(3-23)(3-23)	f₃＝(and，f₁，r₃，a bike)f ₃ =(and,f ₁ ,r ₃ ,a bike)	(3-58)(3-58)	f₃＝(and，f₁，r₃，e)f ₃ =(and,f ₁ ,r ₃ ,e)
(3-24)(3-24)	f₃＝(Ψ，f₁，r₃，a bike)f ₃ = (Ψ, f ₁ , r ₃ , a bike)	(3-59)(3-59)	f₃＝(Ψ，f₁，r₃，e)f ₃ = (Ψ, f ₁ , r ₃ , e)	(3-25)(3-25)	f₃＝(e，f₁，r₃，a bike)f ₃ = (e, f ₁ , r ₃ , a bike)	(3-60)(3-60)	f₃＝(e，f₁，r₃，e)f ₃ = (e, f ₁ , r ₃ , e)
(3-26)(3-26)	f₃＝(that A，f₂，r₃，a bike)f ₃ =(that A,f ₂ ,r ₃ ,a bike)	(3-61)(3-61)	f₃＝(that A，f₂，r₃，e)f ₃ =(that A,f ₂ ,r ₃ ,e)	(3-25)(3-25)	f₃＝(e，f₁，r₃，a bike)f ₃ = (e, f ₁ , r ₃ , a bike)	(3-60)(3-60)	f₃＝(e，f₁，r₃，e)f ₃ = (e, f ₁ , r ₃ , e)
(3-26)(3-26)	f₃＝(that A，f₂，r₃，a bike)f ₃ =(that A,f ₂ ,r ₃ ,a bike)	(3-61)(3-61)	f₃＝(that A，f₂，r₃，e)f ₃ =(that A,f ₂ ,r ₃ ,e)	(3-27)(3-27)	f₃＝(that B，f₂，r₃，a bike)f ₃ =(that B,f ₂ ,r ₃ ,a bike)	(3-62)(3-62)	f₃＝(that B，f₂，r₃，e)f ₃ =(that B,f ₂ ,r ₃ ,e)

(3-28)(3-28)	f₃＝(and，f₂，r₃，a bike)f ₃ =(and,f ₂ ,r ₃ ,a bike)	(3-63)(3-63)	f₃＝(and，f₂，r₃，e)f ₃ =(and,f ₂ ,r ₃ ,e)
(3-28)(3-28)	f₃＝(and，f₂，r₃，a bike)f ₃ =(and,f ₂ ,r ₃ ,a bike)	(3-63)(3-63)	f₃＝(and，f₂，r₃，e)f ₃ =(and,f ₂ ,r ₃ ,e)	(3-29)(3-29)	f₃＝(Ψ，f₂，r₃，a bike)f ₃ = (Ψ, f ₂ , r ₃ , a bike)	(3-64)(3-64)	f₃＝(Ψ，f₂，r₃，e)f ₃ = (Ψ, f ₂ , r ₃ , e)
(3-30)(3-30)	f₃＝(e，f₂，r₃，a bike)f ₃ = (e, f ₂ , r ₃ , a bike)	(3-65)(3-65)	f₃＝(e，f₂，r₃，e)f ₃ = (e, f ₂ , r ₃ , e)	(3-29)(3-29)	f₃＝(Ψ，f₂，r₃，a bike)f ₃ = (Ψ, f ₂ , r ₃ , a bike)	(3-64)(3-64)	f₃＝(Ψ，f₂，r₃，e)f ₃ = (Ψ, f ₂ , r ₃ , e)
(3-30)(3-30)	f₃＝(e，f₂，r₃，a bike)f ₃ = (e, f ₂ , r ₃ , a bike)	(3-65)(3-65)	f₃＝(e，f₂，r₃，e)f ₃ = (e, f ₂ , r ₃ , e)	(3-31)(3-31)	f₃＝(that A，e，r₃，a bike)f ₃ =(that A,e,r ₃ ,a bike)	(3-66)(3-66)	f₃＝(that A，e，r₃，e)f ₃ =(that A,e,r ₃ ,e)
(3-32)(3-32)	f₃＝(that B，e，r₃，a bike)f ₃ =(that B,e,r ₃ ,a bike)	(3-67)(3-67)	f₃＝(that B，e，r₃，e)f ₃ =(that B,e,r ₃ ,e)	(3-31)(3-31)	f₃＝(that A，e，r₃，a bike)f ₃ =(that A,e,r ₃ ,a bike)	(3-66)(3-66)	f₃＝(that A，e，r₃，e)f ₃ =(that A,e,r ₃ ,e)
(3-32)(3-32)	f₃＝(that B，e，r₃，a bike)f ₃ =(that B,e,r ₃ ,a bike)	(3-67)(3-67)	f₃＝(that B，e，r₃，e)f ₃ =(that B,e,r ₃ ,e)	(3-33)(3-33)	f₃＝(and，e，r₃，a bike)f ₃ =(and,e,r ₃ ,a bike)	(3-68)(3-68)	f₃＝(and，e，r₃，e)f ₃ =(and,e,r ₃ ,e)
(3-34)(3-34)	f₃＝(Ψ，e，r₃，a bike)f ₃ =(Ψ,e,r ₃ ,a bike)	(3-69)(3-69)	f₃＝(Ψ，e，r₃，e)f ₃ =(Ψ,e,r ₃ ,e)	(3-33)(3-33)	f₃＝(and，e，r₃，a bike)f ₃ =(and,e,r ₃ ,a bike)	(3-68)(3-68)	f₃＝(and，e，r₃，e)f ₃ =(and,e,r ₃ ,e)
(3-34)(3-34)	f₃＝(Ψ，e，r₃，a bike)f ₃ =(Ψ,e,r ₃ ,a bike)	(3-69)(3-69)	f₃＝(Ψ，e，r₃，e)f ₃ =(Ψ,e,r ₃ ,e)	(3-35)(3-35)	f₃＝(e，e，r₃，a bike)f ₃ = (e, e, r ₃ , a bike)	(3-70)(3-70)	f₃＝(e，e，f₃，e)f ₃ = (e, e, f ₃ , e)

Replace the constant with the sequence value, slightly.

Apply the principle of multiplication in combinatorial mathematics:

|S|=|f ₁ |×|f ₂ |×|f ₃ |=8×24×70=13440

A total of 13440 possible matrix solutions are generated.

The correct structure of the example sentence is: I know as the main sentence; that A you have a car is the first object clause of the predicate know of the main clause; and that B hehas a bike is juxtaposed with the first object clause Two object clauses; the dependent word units that A and that B respectively guide two object clauses; the two object clauses are connected by the side-by-side unit and the conjunction; in the process of processing, the associated word combination vector Ψ=and that B as a whole To handle

The associated word combination vector Ψ cannot be inserted into other syntax vectors; when checking the order value, the two syntactic sequence values included in the associated word combination vector can be directly substituted. The end result is that the second object clause is considered to be inserted into the empty space at the end of the first object clause.

2 is a schematic diagram of an apparatus for analyzing a computer-based natural language syntax structure according to the present invention, the apparatus shown:

The reading component 21 is configured to read the pre-processed statement data structure to be parsed, and the pre-processed statement data structure includes only the parallel-associated word unit, the subordinate-related word unit, the predicate verb unit, and the noun pronoun unit of the statement. And each word unit is numbered in the order in the preprocessed statement, and the type is marked;

The element generating component 22 is configured to generate, for each predicate verb unit, a corresponding guiding element element, a subject element, a predicate element, and an object element;

The predicate element is a corresponding predicate verb unit;

a vector generating component 23, configured to obtain, according to possible values of the leader element, the subject element, the predicate element, and the object element, all possible values of a syntax vector corresponding to each predicate verb unit, where the syntax vector includes a guide language Elements, subject elements, predicate elements, and object elements;

a matrix generating component 24, configured to generate at least one syntax structure possible matrix solution according to all possible values of all syntax vectors, wherein the syntax structure possible matrix solution consists of a syntax vector arranged according to a predicate verb unit number order;

The solving component 25 is configured to verify whether the statement obtained by the possible matrix solution according to the syntax structure is identical to the preprocessed statement. If they are identical, each syntactic vector in the possible matrix solution of the syntactic structure is used as a syntactic structure analysis result. one;

Wherein, the solving component 25 excludes possible syntactic structure solutions by the following module operations:

Called the second type of syntax vector;

The second sub-module, taking a second type of syntax vector

Mark one by one according to the predetermined direction

Second type of syntax vector

Syntactic vector in the form of overall insertion

Third submodule, the third type of syntax vector

Pair vector from the predetermined direction

The first syntactic element on the first side starts into the vector

Vector contained in

The element on the first side, not the order value; the vector

The first syntax element on the second side is marked as

Will be vectored as described above

The syntactic vector part of the annotation, denoted as the iris syntax vector

Syntactic vector in the form of overall insertion

or

Third type of syntax vector

Syntactic vector according to the predetermined direction

The vector is inserted as a whole

Further, the device may further include:

The present invention focuses on solving the problem of accurate parsing of compound sentence structures in natural language. The most important features of the present invention are: 1 fully utilizing the properties of the composite function; 2 using a matrix model and a linear model to describe the syntactic formula; 3 using the related principles of combinatorial mathematics to generate a matrix model and a linear model. By using the invention, the accuracy of the natural language syntax structure analysis can be improved.

From a mathematical point of view, natural language has discrete characteristics, and this is the difficulty in the parsing of syntactic structures. The invention effectively combines the syntactic vector with the matrix form, without destroying the integrity of the sentence structure, and does not hinder the analysis of the intrinsic components and the relationship between the words in each sentence. The invention adopts a matrix model and a linear model to characterize the sentence formula, which not only conforms to the discrete characteristics of natural language, but also effectively reveals the information association on the syntactic structure.

From the perspective of computer technology, the present invention adopts a matrix model and a linear model to convert a single-line natural language sentence into a hierarchical linear nested form, thereby largely avoiding the computer directly labeling the original sentence of the natural language. And partition structure The resulting confusion makes the computer's program tasks clearer and more concise. The matrix model and the linear model used in the present invention are equivalent to drawing a plurality of parallel runways for natural language sentences, so that the natural language sentences start at the same time on a plurality of parallel runways, and then the correct results are screened therefrom; Provide multiple planes for natural language statements, process natural language statements on multiple planes, and then filter the correct results.

In the process of generating a matrix, the present invention uses the correlation principle of combinatorial mathematics to generate all matrices, and then excludes them one by one, and finally obtains at least one possible correct syntactic structure parsing result. In this process, only the mathematical principle and information coding are needed. Only the values of the real numbers need to be processed. Each step is finally implemented to check whether the value of the syntax vector is in ascending order, that is, the size of the real number is not involved. Language information in English itself.

At the same time, the present invention requires a large amount of mathematical operations, and therefore must be realized by the computing power of the computer.

In summary, the present invention is based on mathematical principles of abstract algebra, set theory, combinatorial mathematics, computability theory and computational linguistics, and corresponding computer techniques, using mathematical ideas of complex functions, by establishing matrix models and linear models, constructing The recursive function is used to analyze the natural language syntactic structure. At the same time, the important conclusions are proved by the methods of mathematical induction.

The invention has unique concept, ingenious method and detailed argumentation, and fully utilizes the laws of mathematics and computer science, and the method has high accuracy and high technical difficulty.

Claims

A computer-based method for parsing natural language syntactic structures, including:

S1: reading a pre-processed statement data structure to be parsed, wherein the pre-processed statement data structure includes only a parallel-related word unit, a subordinate-related word unit, a predicate verb unit, a noun pronoun unit, and each word unit is The order in the preprocessed statement is numbered and labeled;

S2, for each predicate verb unit, generating a corresponding guide element, a subject element, a predicate element, and an object element;

The possible value of the guide element is one of a parallel related word unit or a dependent related word unit whose number is smaller than the corresponding predicate verb unit number, or a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and one adjacent thereto One of the associated word combination vectors composed of the dependent word unit whose number is smaller than the corresponding predicate verb unit number and whose number is greater than the parallel related word unit number, or an empty unit;

The possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than the juxtaposition included in the total parallel noun pronoun combination vector family of the corresponding predicate verb unit number. One of the noun pronoun combination vectors, or one of the syntactic vectors corresponding to the predicate element, or an empty unit;

The predicate element is a corresponding predicate verb unit;

The possible value of the object element is one of the noun pronoun units whose number is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number, or the number of the smallest word unit is greater than the corresponding predicate verb unit number. And one of the parallel noun pronoun combination vectors included in the entire parallel noun pronoun combination vector family of adjacent predicate verb unit numbers, or one of the syntactic vectors corresponding to the predicate element, or an empty unit ;

S3. Obtain all possible values of a syntax vector corresponding to each predicate verb unit according to possible values of the guide element, the subject element, the predicate element, and the object element, where the syntax vector includes a guide element, a subject element, Predicate element and object element;

S4. Generate at least one syntax structure possible matrix solution according to all possible values of all syntax vectors, where the syntax structure may be composed of syntactic vectors arranged according to the order of the predicate verb unit numbers;

S5. Verify whether the statement obtained by the possible matrix solution according to the syntax structure is identical to the preprocessed statement. If they are identical, each syntactic vector in the possible matrix solution of the syntax structure is one of the parsing result of the syntax structure;

Among them, S5 includes the following operations in order, excluding the syntactic structure that does not meet the conditions may be solved:

S5.1. If there is a sequence value that does not appear in the possible matrix solution of the syntax structure, the possible matrix solution may be excluded from the syntax structure;

S5.2. If the same order value appears in different syntax vectors or the same syntax vector appears, the possible syntax solution of the syntax structure is excluded;

S5.3. In each possible matrix solution, the syntactic vectors that are mutually substituted with other syntactic vectors are all equally substituted. If there is a contradiction between two syntactic vectors after the equal substitution, then Excluding the syntactic structure may be a matrix solution;

S5.4. In each possible matrix solution, all the syntactic vectors that have mutual substitution relationship with other syntactic vectors are equally substituted, if two position reversal order values appear after the equal substitution, then Excluding the syntactic structure may be a matrix solution;

S5.5. In any of the possible matrix solutions, if there is a syntax vector that has no substitution relationship with other syntax vectors, perform an insertion operation to obtain a possible syntax parsing structure corresponding to all the possible matrix solutions, and verify Whether the statement obtained according to the possible syntax parsing structure is identical to the pre-processed statement, further comprising:

S5.5.1, firstly perform equal-substitution of the syntactic vectors in the possible matrix solutions with the substitution relationship between them, thereby transforming the possible matrix solutions into a set of syntactic vectors with no substitution relationship between each other.
The syntactic vector in the possible matrix solution is called the first kind of syntactic vector, and the transformed syntactic vector will be transformed.
Called the second type of syntax vector;

S5.5.2, take a second type of syntax vector
Mark one by one according to the predetermined direction
The order value of each syntax element in the message; after appending the order value of the syntax element, take any
The i-th syntax element in the construct, only a unique gap is constructed on the first side of the syntax element; after the void, take a syntax vector
Second type of syntax vector
Syntactic vector in the form of overall insertion
Insert the constructed vacancy, and then generate a new syntax vector, record this new syntax vector as
The syntactic vectors obtained by inserting the whole into space are collectively referred to as the third type of syntax vector;

S5.5.3, the third type of syntax vector
Pair vector from the predetermined direction
The first syntactic element on the first side starts into the vector
Vector contained in
Each of the syntax elements up to the first syntactic element on the second side, all of which are labeled with a sequence value;
Vector contained in
The element on the first side, without the order value; the vector
The first syntax element on the second side is marked as
Will be vectored as described above
The syntactic vector part of the annotation, denoted as the iris syntax vector
After the order value is marked, take the jth syntax element in the aforementioned tail vector, and construct a unique gap only on the first side of the element; after the empty, take an unused second type of syntax vector
Syntactic vector in the form of overall insertion
Insert the constructed vacancy, and then generate a new syntax vector, then record the newly generated syntax vector as
or

Third type of syntax vector
Syntactic vector according to the predetermined direction
Each syntax element in the label is labeled with a sequential value; after the order value of the syntax element is annotated, take one
The tth syntax element in the construct, constructing a unique gap on the first side of the syntax element; after the void, taking an unused second type of syntax vector
The vector is inserted as a whole
Insert the previously constructed gap and generate a new vector, then the new vector is recorded as

S5.5.4, repeated execution of S5.5.3, the next time the emptying and insertion of the third type of syntactic vector obtained through the previous emptying and insertion steps are performed at the end of the last emptying and emptying steps. Null operation until all second type of syntax vectors will be
After all the insertions are completed, a third type of syntax vector of a single line is finally obtained, and the finally obtained third type of syntax vector is called a final single line vector;

S5.5.5, if there are two position reversal order values in all the final single row vectors corresponding to a possible syntax parsing structure, the possible syntactic parsing structure is excluded;

S5.5.6, repeat S5.5.2 to S5.5.5 until all possible syntactic parsing structures are traversed.
The computer-based natural language syntax structure parsing method according to claim 1, wherein S2 comprises generating a parallel noun pronoun combination vector family:

S2.1 selects two noun pronoun units that are not repeated:

A. If there are no other word units between the two noun pronoun units, the two noun pronoun units are used as a parallel noun pronoun combination vector, and the parallel noun pronoun combination vector is retained;

B. If there are other word units between the two noun pronoun units, check each word unit between the two noun pronoun units: if any between the two noun pronoun units Word units, all of which are noun pronoun units or side-by-side related word units, then use the selected two noun pronoun units and the whole word unit between the two noun pronoun units as a parallel noun pronoun combination vector, and retain the juxtaposition Noun pronoun combination vector; otherwise, no parallel noun pronoun combination vector is generated;

S2.2 complex execution S2.1 until all combinations of noun pronoun units are traversed, and all obtained parallel noun pronoun combination vectors are generated;

S2.3 If there is a parallel noun pronoun combination vector in the possible syntactic parsing structure, all the parallel noun pronoun combination vectors are divided to form a plurality of parallel noun pronoun combination vector families, so that: in each parallel noun pronoun combination vector family Each collocated noun pronoun combination vector included in the parallel noun pronoun combination vector family all contains two common noun pronoun units.

S2.4 selects the largest number of word units contained in all noun pronoun combination vectors in each noun pronoun combination vector family, as the largest word unit of the noun pronoun combination vector family, for use in subsequent generation of the subject; The word unit with the lowest number included in all noun pronoun combination vectors is used as the smallest unit of the noun pronoun combination vector family, and is used for subsequent generation of the object.
The computer-based natural language syntax structure parsing method according to claim 1, wherein the generating the corresponding subject element comprises:

When the corresponding predicate verb unit number is the smallest predicate verb unit number, the possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than the corresponding Predicate The unit of the word unit number is one of the parallel noun pronoun combination vectors contained in the vector group of noun pronouns, or an empty unit.

When the corresponding predicate verb unit number is not the smallest predicate verb unit number, the possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than the corresponding predicate The verb unit number is one of the collocated noun pronoun combination vectors contained in the collocation noun pronoun combination vector family, or one of the syntactic vowel units corresponding to the predicate verb unit, or an empty unit.
The method for parsing a computer-based natural language syntax structure according to claim 1, wherein the generating the corresponding object element comprises:

When the corresponding predicate verb unit number is the largest predicate verb unit number, the possible value of the object element is one of the noun pronoun units whose number is greater than the corresponding predicate verb unit number, or the number of the smallest word unit is greater than the corresponding number. One of the parallel noun pronoun combination vectors contained in the vector of the predicate verb unit number, or an empty unit.

When the corresponding predicate verb unit number is not the largest predicate verb unit number, the possible value of the object element is a noun pronoun unit whose number is greater than the corresponding predicate verb unit number and is smaller than the adjacent predicate verb unit number. One of the collocated noun pronoun combination vectors included in one of the collocation noun pronoun combination vector numbers, or one of the smallest word units, is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number. Or one of the syntactic vectors corresponding to the predicate verb unit that appears later, or an empty unit.
The method for parsing a computer-based natural language syntax structure according to claim 1, wherein in the two steps S4 and S5, the possible matrix solution is replaced by a possible linear expression solution with a syntax structure;

The syntactic structure may be equivalent to a linear expression solution of the syntactic structure;

The syntactic structure may be a linear expression solution comprising a syntactic vector expression arranged in order of predicate verb unit numbers; each of the syntactic vector expressions is a guide element, a subject element, a predicate element, an object of a corresponding syntax vector An expression in which elements are added one by one in order.
The method for parsing a computer-based natural language syntax structure according to claim 1, wherein the method further comprises:

Each syntax vector and corresponding syntax structure relationship in the syntax structure analysis result are displayed in a human-computer interaction interface by a tree structure.
A computer-based natural language syntax structure parsing apparatus, comprising:

a reading component, configured to read a pre-processed statement data structure to be parsed, wherein the pre-processed statement data structure includes only a parallel-related word unit, a subordinate-related word unit, a predicate verb unit, a noun pronoun unit, and Each word unit is numbered in the order in the preprocessed statement, and the type is marked;

An element generation component for generating a corresponding guide element, a subject element, and a predicate element for each predicate verb unit And object elements;

Wherein, the possible value of the guide element is one of a parallel related word unit or a dependent related word unit whose number is smaller than the corresponding predicate verb unit number, or a parallel related word unit whose number is smaller than the corresponding predicate verb unit number and one of them One of the associated word combination vectors formed by the dependent-related word units whose neighbors are smaller than the corresponding predicate verb unit number and whose number is greater than the parallel-related word unit number, or an empty unit;

The possible value of the subject element is one of the noun pronoun units whose number is smaller than the corresponding predicate verb unit number, or the number of the largest word unit is smaller than the juxtaposition included in the total parallel noun pronoun combination vector family of the corresponding predicate verb unit number. One of the noun pronoun combination vectors, or one of the syntactic vectors corresponding to the predicate element, or an empty unit;

The predicate element is a corresponding predicate verb unit;

The possible value of the object element is one of the noun pronoun units whose number is greater than the corresponding predicate verb unit number and less than the adjacent predicate verb unit number, or the number of the smallest word unit is greater than the corresponding predicate verb unit number. And one of the parallel noun pronoun combination vectors included in the entire parallel noun pronoun combination vector family of adjacent predicate verb unit numbers, or one of the syntactic vectors corresponding to the predicate element, or an empty unit ;

a vector generating component, configured to obtain all possible values of a syntax vector corresponding to each predicate verb unit according to possible values of the guide element, the subject element, the predicate element, and the object element, where the syntax vector includes a guide element , subject elements, predicate elements, and object elements;

a matrix generating component, configured to generate at least one syntax structure possible matrix solution according to all possible values of all syntax vectors, wherein the syntax structure may be composed of a syntax vector arranged according to a predicate verb unit number order;

a solution component for verifying whether the statement obtained by the possible matrix solution according to the syntax structure is identical to the preprocessed statement, and if they are identical, each syntactic vector in the possible matrix solution of the syntax structure is used as a syntactic structure analysis result One;

Wherein, the solving component excludes a possible syntactic structure solution by the following module operation:

a first exclusion module, if there is a sequence value that does not appear in the possible matrix solution of the syntax structure, the possible matrix solution is excluded from the syntax structure;

The second exclusion module excludes the possible matrix solution if the same sequence value appears in the different syntax vectors or the same syntax vector appears;

In the third exclusion module, in each possible matrix solution, the syntactic vectors having mutual substitution relations with other syntax vectors are all equally substituted, and if the cross-contradictions of the two syntax vectors appear after the equal-substitution, Excluding the syntactic structure possible matrix solution;

In the fourth exclusion module, in each possible matrix solution, the syntactic vectors having mutual substitution relations with other syntax vectors are all equally substituted, and if the order values of the two positions are reversed after the equal substitution, Exclude the syntactic structure Energy matrix solution

a fifth exclusion module, in any one of the possible matrix solutions, if there is a syntax vector that has no substitution relationship with other syntax vectors, performing an interpolation operation to obtain a possible syntax parsing structure corresponding to all the possible matrix solutions, and Verification of whether the statement obtained according to the possible syntax parsing structure is identical to the preprocessed statement, further comprising:

The first sub-module first performs an equal substitution of the syntactic vectors in the possible matrix solutions with the substitution relationship between them, thereby transforming the possible matrix solutions into a set of syntactic vectors without substitution relations between them.
The syntactic vector in the possible matrix solution is called the first kind of syntactic vector, and the transformed syntactic vector will be transformed.
Called the second type of syntax vector;

The second sub-module, taking a second type of syntax vector
Mark one by one according to the predetermined direction
The order value of each syntax element in the message; after appending the order value of the syntax element, take any
The i-th syntax element in the construct, only a unique gap is constructed on the first side of the syntax element; after the void, take a syntax vector
Second type of syntax vector
Syntactic vector in the form of overall insertion
Insert the constructed vacancy, and then generate a new syntax vector, record this new syntax vector as
The syntactic vectors obtained by inserting the whole into space are collectively referred to as the third type of syntax vector;

Third submodule, the third type of syntax vector
Pair vector from the predetermined direction
The first syntactic element on the first side starts into the vector
Vector contained in
Each of the syntax elements up to the first syntactic element on the second side, all of which are labeled with a sequence value;
Vector contained in
The element on the first side, without the order value; the vector
The first syntax element on the second side is marked as
Will be vectored as described above
The syntactic vector part of the annotation, denoted as the iris syntax vector
After the order value is marked, take the jth syntax element in the aforementioned tail vector, and construct a unique gap only on the first side of the element; after the empty, take an unused second type of syntax vector
Syntactic vector in the form of overall insertion
Insert the constructed vacancy, and then generate a new syntax vector, then record the newly generated syntax vector as
or

Third type of syntax vector
Syntactic vector according to the predetermined direction
Each syntax element in the label is labeled with a sequential value; after the order value of the syntax element is annotated, take one
The tth syntax element in the construct, constructing a unique gap on the first side of the syntax element; after the void, taking an unused second type of syntax vector
The vector is inserted as a whole
Insert the previously constructed gap and generate a new vector, then the new vector is recorded as

The fourth sub-module repeats the operation of the third sub-module, and each time the last nulling and emptying step ends, the third type of syntactic vector obtained through the last emptying and emptying steps is made for the next time. Empty and insert operations until all second type of syntax vectors will be
After all the insertions are completed, a third type of syntax vector of a single line is finally obtained, and the finally obtained third type of syntax vector is called a final single line vector;

a fifth submodule, if there are two position reversal order values in all of the final single row vectors corresponding to a possible syntactic parsing structure, the possible syntactic parsing structure is excluded;

The sixth sub-module repeatedly calls the operations of the second sub-module to the fifth sub-module until all possible syntactic parsing structures are traversed.
The apparatus for parsing a computer-based natural language syntax structure according to claim 7, further comprising:

The result display component displays the syntax vector and the corresponding syntax structure relationship in the syntax structure analysis result on the human-computer interaction interface by using a tree structure.