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• W4200547882 abstract "In Section 4.5, “Assertion 4-2 based on Theorem 1, 2, 3, 4: Two undirected and simple graphs G1 and G2. The number of nodes for the two graphs is the same as n. The number of edges for the two graphs is the same as m. There are no isolated nodes in both graphs. The vertex and edge adjacency matrices of two simple undirected graphs are generated as V1 and V2, E1 and E2. The sums of rows/columns produce four sets of an array as aV1, bV2, aE1, bE1 based on V1 and V2, E1 and E2. If and only if ∑n k=1 ak V1 = ∑n k=1 bk V2, and ∑n k=1 (ak V1)2 = ∑n k=1 (bk V2)2, then aV1 and bV2 are a permutation of another set; ∑m k=1 ak E1 = ∑m k=1 bk E2, and ∑m k=1 (ak E1)2 = ∑m k=1 (bk E2)2, then aE1 and bE2, are a permutation of another set, G1 and G2 are isomorphic.” should be “Assertion 4-2 based on Theorems 1–4: Two undirected and simple graphs G1 and G2. The number of nodes for the two graphs is the same as n. The number of edges for the two graphs is the same as m. There are no isolated nodes in both graphs. The vertex and edge adjacency matrices of two simple undirected graphs are generated as V1 and V2, E1 and E2. The sums of rows/columns produce four sets of an array as aV1, bV2, aE1, bE2 based on V1 and V2, E1 and E2. If and only if , , …, and and is a permutation of another set, , , …, and and , is a permutation of another set, and the eigenvalue and the maximum linearly independent system of left and right singular vector of both vertex and edge adjacency matrices are equinumerous,2 and are isomorphic. In Section 4.7 “Theorem 4…, we still need to” should be replaced by “Mathematical proof for Theorem 4. The “only if” of the theorem (necessary condition) is simple because the permutation array group is bijective. They always have the equivalent arrays as = , and , , , , , , if two arrays are bijective. The “if” (sufficient condition) requires the following three main lemmas from fundamental theorem of arithmetic2 for n = 2. That is, given two natural number sets of arrays and , , , If and only if , and , A is a permutation of B and vice versa. (n = 2 case) If and , then we have If there is any equals to 1, n = 2 case holds. The proof is as follows. Suppose , we have and . Thus , , , Therefore, either or . When , we have and , . When , we have and , . Therefore, n = 2 case holds. If is a positive integer and larger than 1, according to the fundamental theorem of arithmetic, either is a prime number itself or can be represented as the product of prime numbers; moreover, this representation is unique, up to (except for) the order of the factors. Then Repeat the previous proof. divides at least one of . Then if we rearrange , we have . Because is prime, factors are 1 or . Then we have . Then we remove it from both sides of the equation. Continue this process until all of and are removed. If all of the pi is removed, the left side of the equality is 1, so there is no left . Similarly, if all of the qi is removed, the right side of the equality is 1. The number of is equal to . Then we have proved, is either solution A or solution B, which is a solution of (1). since it makes all two equations valid. The word “system” indicates that the equations are to be considered collectively, rather than individually. Because , , then we have (. Because , then . If (2) is not unique, there must exist another four , holds (1), but belong to = . Suppose k is a integer , because , then we could construct , , , , where , , , . holds. Because and . To make , we must have , and then . Because , then we have , and then , . could be equal to , this results in a contraction since . Therefore, the initial assumption as (2) is not unique must be false. Thus, n = 2 case is proved. Therefore, Theorem 4, where n = 2 has been proved. Our mathematical proof with is proved by the following induction-based method. Questions for n = 3, n = 4, …, and for extended permutation theorem. When = 3, if and only if , , and , the set of is a permutation of the set of . When = 4, if and only if and , , and , the set of is a permutation of the set of . is established, if and only if and , …, the set of is a permutation of the set of . Mathematical proof: Let be the statement of permutation theorem. We give proof by induction on N. Base case. The statement holds for n = 1 and n = 2. is easily seen to be true, and is true by the above-mentioned proof when n = 2. Inductive step. The following steps will show that for any that if holds, then also holds. This can be done as follows. Assume the induction hypothesis that is true (for some unspecified value of ), that is, if and only if and , …, the set of is a permutation of the set of . In Section 5.1, “(4) check if … in Figure 23.” should be replaced by “(3) Check if . If so, go to the next step. If not, graphs and are not isomorphic. (4) Check if . If so, go to the next step. If not, graphs and are not isomorphic. (5) Generate the array of row sum of the edge adjacency matrix and produce two sets of an array as , . Compute and check if , until m step. If so, go to the next step. If not, graphs and are not isomorphic. (6) Compute and check if . If not, graph and are not isomorphic. (7) Compute and check if . If not, graph and are not isomorphic. (8) Continue to compute until n step and check if . If not, graph and are not isomorphic. (9) Continue to compute until m step and check if . If not, produce the results that graphs and are not isomorphic. Until m steps, go to the next step. (10) Implement the singular vector composition (SVD) on both vertex and edge adjacency matrices for two graphs. Check if the corresponding eigenvalue is equinumerous.1, 2 If not, produce the results that graphs and are not isomorphic. (11) Check if the maximum linearly independent system of the left and right singular vector of both vertex and edge adjacency matrices is equinumerous.2 If not, produce the results that graphs and are not isomorphic. If so, produce the results that graphs and are isomorphic.” Regarding the solution proposed as a graph isomorphism algorithm in this article, two arrays A and B are equinumerous if and only if the cardinality of the set of same elements is equivalent, and the number of the set of same elements is equivalent. In other words, there exists a one-to-one correspondence (a bijection) between them, that is, if there exists a function from A to B such that for every element y of B there is exactly one element x of A with f(x) = y. For example, two arrays 1, 2, 2, 5, 3, 3, 3 and 2, 3, 3, 4, 5, 5, 5, the first array has two 2, three 3 equal, and the second array has two 3 and three 5 equal. Therefore, the two arrays are equinumerous. For two vector spaces, if the corresponding row/column is equinumerous to another, the two vector spaces are equinumerous. Weak equinumerosity. If one array contains all of the equinumerosity of another array, they are weak equinumerous. For example, in two arrays 1, 1, 1, 5, 2, 3, 4 and 2, 2, 3, 4, 5, 1, 6, the first array contains all of the equinumerosity of the second array. They are weak equinumerous. Mathematical proof for equinumerosity theorem.1 Without loss of generality, the columns of and the rows of are defined and used in this article, which are called left and right singular vectors in this article. Definition 4A.P-multiple eigenvalues. An order matrix has eigenvalues. If there are eigenvalues that are the same, then these eigenvalues are called multiple eigenvalues. Definition 5B.A maximally independent vector set. It is defined as: Let be a vector group, if it satisfies: (1) is linearly independent; (2) if any other vector in the space can be expressed as a linear combination of elements of a maximal set—the basis . Then A set of vectors is maximally linearly independent if including any other vector in the vector space would make it linearly dependent. Definition 6C.A maximally independent vector system. Under the linear transformation, the maximally linear independent subset has been transferred to have only one 1, and the others are 0. The current format of the original vector set is called the maximally independent vector system. Property 4A. ([1])Let be a real symmetric matrix. There exist an orthogonal symmetric matrix and a diagonal matrix such that , where the diagonal element of is the eigenvalue of and the column vector of is the eigenvector of . Proof.By mathematical induction, it is obviously true for first-dimension square matrices. Suppose the above proposition be true for a square matrix with the dimension of . Then for the dimension square matrix , it obviously has at least one eigenvalue , and the eigenvector corresponding to λ is , and is extended to a set of the orthogonal basis of , and arranged into a matrix . Then there is (here ). Whereas is a real symmetric matrix of dimension , the orthogonal matrix is assumed such that , then . Let be an orthogonal symmetric matrix, . Since the eigenvalues of and are the same, the eigenvalues of are also the eigenvalues of . Finally, it is only necessary to prove that the column vector of is the eigenvector of . Set and substitute , then , compared with each other, is the eigenvector of . Proof.Both and are similar to the diagonal matrix (the element at the diagonal is the eigenvalue sequence). Suppose there are different eigenvalues. Then, is invertible. Then could be the linearly independent eigenvector of substitute in (1) by , then: Put (3) into (2), and . If the left singular vector set also applies, then . Under the same principle, for edge adjacency matrix , therefore two graphs are isomorphic. (Theorem 4B) Theorem 5C. ([1])There exist the row interchange matrix , and and satisfy if and only if they are equinumerous. Similarly, there exist the column interchange matrix , and and satisfies if and only if they are equinumerous. Proof.The necessary proof is obvious. For sufficient proof. From Definition 8, both the vector set and are corresponding to the equinumerous sequence , then , so , (), both , and are row interchange operation matrix, then , then , it is still the elementary row interchange matrix. Then the sufficient proof is completed." @default.
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• W4200547882 date "2021-09-27" @default.
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• W4200547882 title "A polynomial‐time algorithm for simple undirected graph isomorphism" @default.
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