Combinatorial invariants for graph isomorphism problem

Com binatorial in v arian ts for graph isomorphism problem. Jarek Duda Jagiel lonian University, R eymonta 4, 30-05 9 Kr ak´ ow, Pol and, email: dudaj@interia. pl Abstract Presen ted approac h in p olynomial time calc ulates large n umb er of in v ari- an ts for eac h v ertex, which w on’t c hange w ith graph isomorphism and should fully determine the graph. F or example n um b er s of closed paths of length k for giv en starting v ertex, w hat can b e though as th e diagonal terms of k -th p o w er of the adjacency matrix. F or k = 2 we would get degree of ve rities inv a riant, higher describ es lo cal top ology deep er. No w if tw o graphs are isomorp hic, they ha v e the same set of suc h v ectors of in v arian ts - w e can sort theses v ec- tors lexicographicall y and compare them. If they agree, p ermutations f rom sorting allo w to reconstruct the isomorph ism. I’m presen ting argumen ts that these inv arian ts should fu lly d etermine the graph, b ut unfortunately I can’t pro ve it in this moment . This approac h can giv e hop e, that maybe P=NP - instead of c hecking all instances, we should mak e arithmetics on these large n umb ers. 1 In tro d uction W e ha v e some undirected gra phs, giv en b y the adjacency matrix G 1 , G 2 ∈ { 0 , 1 } n × n W e w ould like to c hec k if there is a p erm utation matrix: P ∈ { 0 , 1 } n × n : P T P = 1 n , G 1 = P G 2 P T So w e hav e to c hec k if G 1 , G 2 are similar and if the similarit y matrix is p ermutation. The similarit y matrix could b e fo und using num erical metho ds, whic h are asymp- totic - the pro blem is to estimate when to b e sure if w e w on’t get p ermutation. 1 2 ALGORITHM 2 T o chec k if the matrixes are similar, w e can compare their c haracteristic p olynomials, what can b e done in p olynomial time. If not - the graphs are not isomorphic, but if y es - we still don’t kno w if the similarit y matrix is p erm utation, but it seems unlik ely that similarit y matrix b et w een tw o { 0 , 1 } matrixes isn’t p ermutation. 2 Algorith m T o get safer algorithm we will fo cus on, we can compare diagonals of p ow ers of the adjacency mat r ixes. There is kno wn and easy to c hec k com binatorial prop erty , that: ( G k ) ij = n um b er of paths from i to j of length k where in path edges and v ertices can b e rep eated. Without loss of generalities, w e can assume that graph is connected, so from F ro b enius-Pe rron theorem it has unique dominan t eigen v alue ( ≤ n ) and corr e- sp onding eigen v ector is nonnegativ e - the diagonal terms of p o w ers of the adja cency matrixes will increase exp onen tially (length of n umbers gro ws linearly) and in the limit has distribution as the eigen v ector. If the graphs are isomorphic, diagonals of a b ov e p ow ers ha s to b e the same up to p erm utation. This time the isomorphism is suggested by large nu mbers on the diagonal - w e can just sort them. Sometimes differen t v ertices can give the same in v arian ts - it suggests some symmetry in the g raph. In this case we ha v e to b e careful if w e w ould lik e to reconstruct the isomorphism - w e should build it neigh b or b y neigh b or. F or second p ow er the in v aria nts ensure that degrees of v ertices ag rees. Higher p ow ers c hec ks lo cal top ology of v ertex deep er and deeper. The length of n um b ers in p o we rs of matrixes gro ws linearly with the p ow er, so calculating p ow ers can b e done in p olynomial time. The a lg orithm: For graph G : For i, k = 1 , .., n calculate d k i = ( G k ) ii sort vectors { ( d k i ) k } i lexicograph ically giv es us n 2 in v a rian ts in p olynomial time. 3 RECONSTR UCTION 3 If gr aphs are isomorphic, ab ov e in v arian ts has to agree. But if they a gree, are gra phs isomorphic? Do they determine gra ph uniquely? I’ll show that in ’generic’ case it’s true - w e can ev en reconstruct the matrix. Unfortunately I cannot prov e in this momen t that there are no graphs that starting from ab ov e inv arian ts, then ev en tually using some standard tec hniques, w e couldn’t determine in p olynomial time if they are isomorphic, but it lo oks highly unpro v able. In practice w e can mak e arithmetics mo dulo some large n um b er and j ust c hec k a few steps G → G 2 and tr y to reconstruct isomorphism p o w er by p o w er. If something’s wrong, w e should see it early , if not we should quic kly get isomor- phism to c hec k. 3 Reconstru ction W e w ould lik e to hav e some nice com binatorial pro cedure to uniquely reconstruct the gra ph. Unfortunately I couldn’t find it. I will sho w algebraic construction, whic h is rather unpractical, but should give unique gr a ph in practically all - ’generic’ cases. Generally - there is some symmetric matrix A ∈ R n × n , but w e only know d k i := ( A k ) ii for i, k = 1 , .., n. The matrix is real, symmetric so w e can diag onalize it - there exists V , D : A = V D V T where V T V = V V T = 1 n , D ij = λ i δ ij where matrix V is made of eigen ve ctors: Av i = λ i v i . Observ e that X i d k i = T r A k = X i λ k i so we can reconstruct the characteristic p olynomials determining the sp ectrum. W e can presen t canonical base in the base of eigen v ectors: e i = P j W ij v j . W riting this r elat io n in columns w e get: 1 n = W V , so W = V T , e i = X j V j i v j Finally we hav e (for k = 0 w e ha v e A 0 = 1 n ): d k i = e T i A k e i = X j V j i v T j ! X l λ k l V li v l ! = X j λ k j ( V j i ) 2 3 RECONSTR UCTION 4 W e already know the eigenv alues - for each i w e get in terp olatio n problem. Assume t hat there are no t w o equal eigen v alues - it’s one o f generic prop ert y w e w ould need. In this case, b ecause the V andermonde matrix is rev ersible, we can find all ( V ij ) 2 . W e also see that chec king more than n p ow ers do esn’t bring any new infor mat ion. If some eigen v alues rep eats, we w ould find smaller nu mber of co efficien t and hav e freedom to distribute o ur squares of terms b et w een them, what w ould complicate the next step. W e see that w e ha v e another problem - determine signs for nonzero terms. Remem b er that V is orthogonal: ∀ ij X k V ik V j k = δ ij and tha t in f a ct w e are in terested only in A : X k V ik V j k λ k = A ij W e see tha t m ultiplying whole column b y − 1 do esn’t c hange A - w e can fix signs in the first row of V as w e wan t. No w using ab ov e tw o equations, and assuming that A ij ∈ { 0 , 1 } and t here are no zero rows /columns in A (graph is connected), we ha v e to determine the rest of signs. It can b e thought as choosing sings for 2-dimensional ve ctors, so tha t they sum up to one of t w o p oints in R 2 . W e kno w that there is one suc h assignme nt. Co ordinates of these ve ctors are some real num bers - in generic case there shouldn’t b e second one. So in ’g eneric’ case - t ha t there are no tw o the same eigen v alues and that the signs can b e assigned in unique w ay , the in v arian ts determine the g r aph up to isomorphism. I’v e also found some relations b etw een d k i and c haracteristic p olynomials o f the matrix with remov ed column a nd ro w of the same n um b er - some kind of generalized Newton’s identities . They could b e helpful for reconstructing the ma t rix. The deriv ation is practically exactly D an Kalman’s deriv ation of Newton’s identities [3], so I’m presen ting it shortly and referring to the pap er for details. Denote X := x 1 n , p ( x ) = x n + a n − 1 x n − 1 + ... + a 0 = det ( X − A ) - the c haracteristic p olynomial of A , p i ( x ) - c haracteristic p olynomials of A with remov ed i -th ro w and column. F ro m Ceyley-Hamiltonian theorem, we kno w that p ( A ) = 0. Dividing it by X − A , w e will get: ( X − A ) − 1 p ( X ) = X n − 1 + ( A + a n − 1 1 n ) X n − 2 + ... + ( A n − 1 + a n − 1 A n − 2 + ... + a 1 1 n ) 1 n 4 MORE INV ARIANTS 5 In this moment in [3] is take n t race of b oth side, the left o ccurs to b e p ′ ( x ) - w e get Newton’s iden tities. W e can also b e more subtle - tak e for example diagonal elemen ts ( i -th) - us ing form ula for inv erse ma t r ix, w e get p i on the left side, so: X i p i ( x ) = p ′ ( x ) Using the righ t side, w e get: p i ( x ) = x n − 1 + ( A ii + a n − 1 ) x n − 2 + ( ( A 2 ) ii + a n − 1 A ii ) x n − 3 + ... ... + (( A n − 1 ) ii + a n − 1 ( A n − 2 ) ii + ... + a 1 ) Finally ha ving ( d k i ) k ,i =1 ..n , summing ov er i and using Newton’s iden tities w e can find the c haracteristic p o lynomial and using ab ov e relations find ( p i ) i =1 ..n and sp ectrum of A with remov ed i -th row and column. On the other hand hav ing ( p i ) i =1 ..n w e can sum them to g et a 1 , ..., a n − 1 and finally d k i using ab ov e iden tities. W e don’t get the determinan t ( a 0 ) this w ay . 4 More inv arian ts In this momen t w e ha v e n indep enden t in v ariants for each v ertex, wh ich usually should determine the graph and needed n − 1 multiplications of large matrixes with large num b ers. W e see that if w e w ould need less p ow ers, the algorithm w ould b e muc h fa ster. W e should achie ve it using mor e in v arian ts. Unfortunately I still didn’t prov e fully determining of graph, but using more in v a rian ts mak es it ev en more probable. In previous sections, for v ertex i from N k i = (( A k ) ij ) j w e to ok only the dia g onal term, b ecause the rest p erm ute in not kno wn w a y . W e see that w e could tak e the rest of terms of N k i , but w e should forget ab out their order, but w e should r emem b er that for differen t k , j denotes the same v ertex. F or example fo r eac h v ertex i w e should sort lexicographically: p i := {  A ij , ( A 2 ) ij , .., ( A n ) ij  : j = 1 , .., n, j 6 = i } F or ev ery p o w er of A w e get this w a y n ( n − 1) inv a rian ts. Using the method from the end of previous section, we see that these in v a rian ts are equiv alen t kno wing for eac h i list of c haracteristic p olynomials of A with remo v ed i -th row and one column. It suggest that aga in there is no p oint in using more than n p ow ers. REFERENCES 6 The other w ay of constructing in v a rian ts is using no t only n um b er of pathes from giv en v ertex, but also inv a rian ts o f it’s neigh b ors. There is h uge num ber of p o ssi- bilities now - f o r example tak e sum of some in v ariants of ev ery neighbor of the v ertex. There hav e to b e plen ty of r elations b etw een these in v a r ian ts. W e should now c ho ose some in v arian ts whic h fully determine the graph and uses as small p ow ers as p ossible. T o summarize - I didn’t excluded cases that tw o graphs has the same inv a rian ts and they are not isomorphic, but it lo oks extremely improbable and proba bly it should b e corrected in p olynomial time by trying to reconstruct t he isomorphism using orders from sorting. References [1] M.R. Gar ey , D.S. Johnson Computers and Intr actabil ity: A Guide to the The ory of NP-Completeness , W. H. F reeman, 1979. [2] J. Kabler, U. Sc haning and J. T oran, The Gr aph Isomorphism Pr oblem: Its Structur al Complexity , Birkhauser, 1993. [3] D. Kalman , A matrix pr o of of Newton ’s I dentities , Mathematics Magazine vol. 73/4, 2000