On Subgraph Isomorphism

On Subgraph Isomorphism Sergey Gubin ∗ Ab str act —Article explicitl y expresses Subgr aph Isomorphism b y a polynomia l size asymmetric l inear system. Keywor ds: Sub gr aph Isomorphism, Line ar mo deling , Algor ithm, Computational Complexi ty, NP-c omplete In tro duction In 1988, Y annak akis prov ed [1] that the T rav eling Sales- man Problem’s (TSP) p olyto pe ca nnot b e ex pressed by a po lynomial size symmetric linear prog ram, whe r e sym- metry means that the p olytop e is a n inv ar iant under no de relab eling. Because TSP is a NP-complete proble m [2 ], the theorem holds for all NP-complete problems. The question ab out the s ize of a s ymmetric linear mo dels was left op en in [1] and it has r emained op en s ince. This article answers that question. W e present an explicit p olynomial size asymmetric linear mo del for Subgraph Iso morphism (SubGI). Since SubGI is a NP- complete problem [3], this r esult is complimentary to the Y annak a kis theor em. The po lynomial size asymmetric linear system is built based on an arbitrar y but fixed la b eling of gra phs in- volv e d - hence the system’s asymmetry . The p o lynomial size for the system is achiev ed by immer sing the problem in a spa ce of higher dimension, wher e v ar iables pr esent relab eling p o ssibilities for vertex couples. W e illus tr ate our metho d with several examples. Partic- ularly , we explicitly present po lynomial size as ymmetric linear prog rams for TSP and for the Satisfiability Prob- lem for conjunctive normal forms (SA T). 1 Subgraph Isomorphism Let G b e a given graph - we will call it an input. Let S b e another given gr aph - we will ca ll it a pattern. The problem is whether G cont ains a subgraph whic h is iso- morphic to S . F or any given couple of g raphs ( G, S ), this decision pro blem is a SubGI ins tance. Its size can b e es- timated by the num ber of vertices in graph G . An y g raph may be seen as a rela tion. So, SubGI may be seen as a finite version of the fo llowing gener a l pr ob- lem: wheth er a given rela tion p osseses a given pro p erty . That explains the theore tical and pr actical importa nce of SubGI. SubGI is a NP-co mplete pro blem [3]. The Ullmann algo- ∗ sgubin@genesyslab.com rithm [4] is the bes t k nown metho d to solve the problem. Y et it and other known gene r al metho ds a r e inefficient. Up to date, the efficien t metho ds w er e k nown only for particular types of gra ph couples ( G, S ) [5, 6, 7, and oth- ers]. This article describ es a reduction of SubGI to a sy stem of linear equations and inequalities. The reduction’s com- putational co mplexity and the res ulting sys tem’s size are po lynomial over the size of SubGI. F o r a given couple o f graphs ( G, S ), the resulting system has solutions iff input G contains a s ubg raph isomor phic to pattern S . As w ell as for graphs, our r eduction works for (multi) digraphs with (multi) lo ops. W e will present the reduc- tion fo r the multi digraph v ersion of SubGI whic h is, in many cases, more practical. So, input G and pattern S are (m ulti) digraphs with (multi) lo o ps ev erywhere b elow in this article. Because our system co ntains a polynomial num b er of linear equatio ns and inequa lities with a p olyno mial num- ber of unknowns, it can be solved in p olyno mial time by , for exa mple, the Khachiyan ellips oid algo rithm [8, 9]. 2 Base polytop e Let n b e a natural num ber . Let the following v ariables be unknowns: x ij µν = x j iν µ : i, j, µ, ν = 1 , 2 , . . . , n i 6 = j µ 6 = ν y iiµµ = y j j ν ν : i, j, µ, ν = 1 , 2 , . . . , n In the case of n = 1, v a riables x ij µν are miss ing indeed. Let’s co nsider the fo llowing linea r system:                                    x ij µν = x j iν µ , x ij µν ≥ 0 - where i, j, µ, ν = 1 , 2 , . . . , n, i 6 = j, µ 6 = ν P n µ =1 , µ 6 = ν x ij µν = y j j ν ν , - where i, j, ν = 1 , 2 , . . . , n, i 6 = j P n i =1 , i 6 = j x ij µν = y j j ν ν , - where j, µ, ν = 1 , 2 , . . . , n, µ 6 = ν P n ν =1 y j j ν ν = 1 , y j j ν ν ≥ 0 - where j = 1 , 2 , . . . , n (1) The system can b e descr ibe d with the following box ma- trix of size n × n : B =       Y 1 , 1 X 1 , 2 . . . X 1 ,n X 2 , 1 Y 2 , 2 . . . X 2 ,n . . . . . . . . . . . . X n, 1 X n, 2 . . . Y n,n       n × n The i -th diagonal b ox in b ox matrix B is the fo llowing diagonal matrix : Y ii = diag( y i,i, 1 , 1 , y i,i, 2 , 2 , . . . , y i,i,ν,ν , . . . , y i,i,n,n ) The ( i, j ) -th off-dia gonal b ox in box matrix B is the fol- lowing matrix: X ij =       0 x i,j, 1 , 2 . . . x i,j, 1 ,n x i,j, 2 , 1 0 . . . x i,j, 2 ,n . . . . . . . . . . . . x i,j,n, 1 x i,j,n, 2 . . . 0       n × n System 1 reflects the following relations betw een elemen ts of b ox matrix B : (1) B is a symmetric matrix: X ij = X T j i ; (2) The tota l of each column in ma tr ix X ij do es not de- pend on i but o nly on b ox column j and on the column in this box column. The total is the appropr iate element in diag onal matrix Y j j ; (3) The total over i of elements x ij µν do es not dep end on µ but only on b ox column j and on the c olumn ν in this b ox column. This total is the a ppropriate element in matrix Y j j - element y j j ν ν ; (4) The ( j, ν )-th columns in off-diagonal b oxes X ij of box matr ix B cons titute a doubly stochastic matrix m ul- tiplied by element y j j ν ν . (5) The total o f all elements in ma tr ix Y j j is equal 1; (6) Due to the matrix’s symmetry , all of the ab ov e is true in the horizontal direction, to o; System 1 always has so lutio ns. The following solution is minimal in the sense of Euclidean norm - we call it a c enter : x ij µν ≡ 1 n ( n − 1) , y j j ν ν ≡ 1 n Obviously , the set of all solutions of system 1 is a con vex set. Also, b eca use system 1 is a linear s y stem, the set is a p olytop e. W e call this po ly top e a b ase p olytop e . The following solution of system 1 is a vertex of the base po lytop e: ther e is one a nd o nly one non-zer o element in each b ox Y ii and X ij . Obviously , all non-zero elements in the boxes are equal 1 and they are a rrang ed in a grid of elements in matr ix B , one elemen t p er b ox. W e call any suc h solution of system 1 a solution grid . The follo wing le mma shows that all v ertices o f the base po lytop e are solution g r ids. Lemma 1 An y solution of sy s tem 1 is a conv ex combi- nation of solution gr ids. Pro of System 1 co nsists of linear eq ua tions and the fol- lowing ineq ualities: x ij µν ≥ 0 , y j j ν ν ≥ 0 - where a ll indexes a re in their appropria te ranges. The linea r e quations have solutions - the cen ter, fo r example. The solutions constitute a linear subspace in the linear space of all n 2 × n 2 matrices with r eal elements. Thus, vertices of bas e p olytop e are thos e po ints in t he linear subspac e where the num ber of v ariables which equal 0 is maxima l p ossible. It s o happ ens that these p o ints are the solutio n g rids, QED. 3 Compatibilit y matrix Let digraphs G and S b e the giv en SubGI instance ( G, S ). Let V G and V S be vertex sets of the inp ut and pattern appro priately . Obviously , | V G | ≥ | V S | - the instance would hav e resolution “NO” o therwise. Now, let’s add | V G | − | V S | isolated vertices to pattern S . Let’s preserve notion S for the resulting patter n. Let n be the n umber of vertices in the input and the pattern after the addition o f isolated vertices: n = | V G | = | V S | Obviously , the SubGI instance ( G, S ) emerging after the addition o f isolated vertices has th e same r esolution a s the or iginal instance indeed. Let’s ar bitrarily lab el/enumerate vertices in input G and pattern S . Let A G and A S be the adjacency matri- ces of the input and pa tter n a ppr opriate to the lab eling. Obviously , SubGI instance ( G, S ) ha s resolution “YES” iff there exists s uch a relab eling of pattern S that all ele - men ts of matrix A S emerging after that rela b eling will b e less than or equal to the appro priate elements of matrix A G . In other words, SubGI insta nce ( G, S ) ha s reso lu- tion “YES” iff the following in tegral quadratic s y stem has solutions 1 : A G ≥ X A S X T (2) - where X is the unknown p er mutation matrix of size n × n . Permutation matrix X pre s ents the unknown v er- tex r elab eling o f pa ttern S after which the ex istence of an input’s subgraph iso morphic to S has to b eco me self- evident. Obviously , suc h a relab eling o f S exists iff G has at least o ne s ubgraph isomor phic to S . T o solve system 2, let’s build the following matrix which we call a c omp atibili ty matrix . 1 F or t wo matrices A = ( a ij ) and B = ( b ij ) of th e same size, relation A ≥ B means that ∀ i, j ( a ij ≥ b ij ) . Let the input and p attern’s adjacency matrices be as follows: A G = ( g µν ) n × n , A S = ( s ij ) n × n F or each couple of pattern’s v ertices, let’s build a c omp at- ibility b ox . The compatibility b ox for vertices with labels i and j is the following matrix C ij = ( e ij µν ) n × n : e ij µν =  1 , s ij ≤ g µν ∧ s j i ≤ g ν µ 0 , s ij > g µν ∨ s j i > g ν µ (3) Compatibility b ox C ij shows all pos sible re-enumerations for the pattern’s vertices i and j with disreg ard to the rest of the pattern’s vertices. Ob viously , compatibility b oxes C ii are dia gonal matrices. And all diag onal ele ment s in compatibility boxes C ij , i 6 = j , ar e equal 0. The c o mpatibility matrix for SubGI ins tance ( G, S ) is the following b ox matrix: C = ( C ij ) n × n The co mpatibility matrix agg r egates all co mpatibility boxes in accor dance with their indexes. Obviously , integral quadratic system 2 has a so lution iff in the compa tibilit y matrix there is a gr id o f elements, one element per co mpatibilit y b ox, in which all elemen ts are equa l 1: γ = { e ij µν = 1 | µ = µ ( i ) , ν = ν ( j ) } - wher e γ is the grid. Any such grid of elements in com- patibilit y ma trix C we call a solution grid 2 , to o . Lemma 2 SubGI instance ( G, S ) has reso lution “ YES” iff compatibility matrix C co ntains a solution grid. Pro of An y solution gr id de fines a vertex relab eling of S which satisfies system 2, QED. 4 Linear model for SubGI The similarities b etw een compatibility matrix C and the ba s e p olytop e B are obvious. Due to lemmas 1 and 2, we can decide ab o ut the existence/a bsence of solution grids in matrix C searching ma trix B fo r so lutio n grids sub ject to the following constrains: x i 0 j 0 µ 0 ν 0 = 0 , y j 0 j 0 ν 0 ν 0 = 0 (4) - where indexes are the indexes of all those elements of compatibility matrix C w hich ar e equal 0, e i 0 j 0 µ 0 ν 0 = e j 0 j 0 ν 0 ν 0 = 0 Then, lemmas 1 and 2 imply the follo wing polynomial size asy mmetr ic linear mo del for SubGI. 2 In the next sect ion, we will sho w that the solution gr ids from this section and the solution grids f rom the previous section are th e same. Theorem SubGI instance ( G, S ) has r e solution “YES” iff the aggreg ated system 1 and 4 has a solution. Pro of An y solution of the aggr egated system is a co nv ex hu ll of solution grids. There is a solution grid iff the resolution for instance ( G, S ) is “YE S” , QED. System 1, 4 co nsists of O ( n 4 ) linear equa tions and in- equalities with O ( n 4 ) unknowns. The existence/absence of the sys tem’s solutions ca n be detected using the el- lipsoid algorithm [8, 9]. Because all coefficients of the system ar e 0 or 1, the ellipso id algo rithm will solve this system in strongly p olyno mia l time. Let’s notice that constrains 4 explicitly in v olve the input and pattern’s v ertex labeling trough their adjacency ma- trices - see definition 3 of the compatibility b oxes. Thus, system 1, 4 is an asymmetric linear system. It ca n b e seen that the system’s solutions constitute a con v ex sub- set of the Birkhoff p olytop e [10] in R n 4 . V ertex relab eling of dig raphs G and S will ro tate that subset all over the po lytop e. 5 Examples Let’s use our metho d a nd r esolve the following SubGI instances. V ertex vs v ertex: Le t input and pattern hav e just one vertex eac h: A G = ( g 1 , 1 ) 1 × 1 , A S = ( s 1 , 1 ) 1 × 1 System 1 for n = 1 lo oks as follows: y 1 , 1 , 1 , 1 = 1 Constrains 4 for the instance lo ok as follows: y 1 , 1 , 1 , 1 =  1 , s 1 , 1 ≤ g 1 , 1 0 , s 1 , 1 > g 1 , 1 Thu s, the res olution for this SubGI instance is “YES” iff there is no exces s of lo o ps in the pattern: s 1 , 1 ≤ g 1 , 1 Arc vs arc: Let input and pattern b e just ar cs: G : 1 → 2 , S : 1 ← 2 F or n = 2, sy stem 1 lo o ks as follows: x 1 , 2 , 1 , 2 = y 2 , 2 , 2 , 2 x 1 , 2 , 2 , 1 = y 2 , 2 , 1 , 1 x 2 , 1 , 1 , 2 = y 1 , 1 , 2 , 2 x 2 , 1 , 2 , 1 = y 1 , 1 , 1 , 1 y 1 , 1 , 1 , 1 + y 1 , 1 , 2 , 2 = 1 y 2 , 2 , 1 , 1 + y 2 , 2 , 2 , 2 = 1 (5) Constrains 4 for the giv en input a nd pattern may be presented as follows: y 1 , 1 , 1 , 1 = 0 , y 2 , 2 , 2 , 2 = 0 The a g grega ted system has a s olution: x 1 , 2 , 2 , 1 = y 2 , 2 , 1 , 1 = x 2 , 1 , 1 , 2 = y 1 , 1 , 2 , 2 = 1 x 1 , 2 , 1 , 2 = y 2 , 2 , 2 , 2 = x 2 , 1 , 2 , 1 = y 1 , 1 , 1 , 1 = 0 Thu s, the resolution for the given SubGI instance is “YES”. The appropriate relab eling of the pattern is transp osition (1 , 2). Arc vs lo op: Let input and pattern be an arc and a lo op appr o priately: G : 1 → 2 , S : 1 → 1 Adding to S one isolated vertex with index 2 will pro duce the case o f n = 2. Sy stem 1 for the case is system 5, and co nstrains 4 for the instance ma y b e presented as follows: y 1 , 1 , 1 , 1 = 0 , y 1 , 1 , 2 , 2 = 0 The a g grega ted system has no s olutions: 1 = y 1 , 1 , 1 , 1 + y 1 , 1 , 2 , 2 = 0 Thu s, the resolution for the given SubGI instance is “NO”. Arc/lo op vs lo op/ arc: Let input and pa ttern b e the following digraphs: G : 1 → 2 → 2 , S : 1 → 1 → 2 System 1 for the ca s e is system 5 , and constr ains 4 for the instance may b e prese nted as follows: y 1 , 1 , 1 , 1 = 0 , x 1 , 2 , 2 , 1 = 0 The a g grega ted system has no s olutions: 1 = y 1 , 1 , 2 , 2 = x 1 , 2 , 2 , 1 = 0 Thu s, the resolution for the given SubGI instance is “NO”. Edge vs arc: Let input a nd pattern be the following di- graphs: G : 1 → 2 → 1 , S : 1 → 2 System 1 for the ca se is system 5, and there a re no constrains 4 for the instance. Thus, the aggrega ted system consists of system 5 alone. The center of its solutions is the following p oint: ∀ i, j, µ, ν ( x ij µν = y iiµµ = 1 / 2) Thu s, the resolution for the given SubGI instance is “YES”. Cycle vs edge: Let input and pattern be the following digraphs: G : 1 → 2 → 3 → 1 , S : 1 → 2 → 1 Compatibility matrix C for this SubGI insta nce lo oks as follows: 1 0 0 0 0 0 0 1 1 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 0 1 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 Compatibility boxes entirely filled with 0 will pro- duce constrains 4 incompatible with system 1 , i.e. the agg regated system 1 and 4 will hav e no solutions. Thu s, the resolution for the given SubGI instance is “NO”. Cycle vs path: Let input and pa ttern b e the following digraphs: G : 1 → 2 → 3 → 1 , S : 1 → 2 → 3 Compatibility matrix C for this SubGI insta nce lo oks as follows: 1 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 0 1 1 0 0 0 1 1 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 1 0 0 0 1 Constrains 4 pro duced by this compatibilit y matrix are compatible with system 1 , i.e. the aggrega ted system has solutions. Two of the three s olution g r ids of the system a re shown in the ab ov e matr ix in italic and in b old. Thus, the reso lution for the given SubGI instance is “YES”. Cycle vs cycle: Let input and pattern be the following digraphs: G : 1 → 2 → 3 → 4 → 1 , S : 1 → 2 → 3 → 1 Compatibility matrix C for this SubGI insta nce lo oks a s follows: 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 1 0 1 1 0 1 0 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 1 1 0 0 1 0 1 0 0 0 0 1 0 0 1 0 1 1 0 0 0 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 0 0 1 Constrains 4 pro duced b y this compatibility ma- trix are incompatible with system 1, i.e. the a g- grega ted system has no solutions. T o see that, le t’s apply system 1 to the compatibility matrix a s con- strains o n its ele ment s. T o satisfy these cons trains at least partially , the forth b ox column a nd the forth box ro w of the compatibilit y matrix have to be trimmed/depleted as follows: 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 After this depletion, the fact that the fourth box column contradicts with the third group of e q uations in system 1 beco mes obvious. Thus, the resolution for the given SubGI instance is “NO”. 6 Linear program f or TSP Let input G b e an a r c-weigh ted digr aph, i.e. let each arc in G hav e a weight. TSP is a pro blem of finding a Hamiltonian cycle in G with the minimal total weigh t 3 . That is a NP-co mplete problem [2 ]. 3 Because G is a digraph, we actually consider here the Asym- metric T rav eling Salesman Pr oblem (A TSP). The pattern S for TSP is any circular p er mut ation matrix, for example: S =      0 0 0 . . . 1 1 0 0 . . . 0 0 1 0 . . . 0 . . . . . . . . . . . . . . .      n × n Let’s construct system 4 fo r SubGI instance ( G, S ). Then, a ggreg ated linear sy s tem 1 and 4 will ex- press the Ha miltonian Cycle Problem which is a NP- complete problem [2], a s well. Let w ( µ, ν ) b e a weigh t function - the weigh t of the arc from vertex µ in to v ertex ν in input G . As usual, let w ( µ, ν ) = + ∞ for non-a djacent vertices. Then, the following asy mmetr ic polynomia l size linear pro- gram will express TSP: X i,j,µ,ν w ( µ, ν ) x ij µν → min - sub ject to co nstrains 1 and 4. F rom the practical pers p ec tive, let’s notice that w e do not require function w ( µ, ν ) to be p ositive. 7 Linear model for SA T In 1971, Cook [3] found that with a po lynomial nu m be r of op erations any no n-deterministic T ur ing machine (NDTM) can be expressed by the appropri- ate conjunctive normal form (CNF): the ques tion of whether ther e is an acceptable input is a question of whether the a ppropria te CNF is satisfiable. That made SA T the first NP- complete pro blem, b ecause it is a NP-pro ble m and the very w ords “NP-pro ble m” mean a problem which can be solved by NDTM in po lynomial time. In 1 973, Levin [1 1] indep endently rep eated the result in terms of sea rch. In 1972 , Ka rp [2] selected SA T as a ro ot of NP-co mpleteness the- ory: a pr oblem is NP-complete if SA T can b e reduced to that problem in po lynomial time, and v isa versa. Let f be a given CNF: f = c 1 ∧ c 2 ∧ . . . ∧ c m - where clause c i is a disjunction of k i literals - some Bo olean v ariables or their negations. F ormula f de- fines an instance o f SA T: whether there is such a true-assig nment to the inv olved Boolean v ariables which w o uld make f = tru e . Ultimately , we could apply the distributive laws and r ewrite fo rmula f in a disjunctive form (DF). That w ould reduce SA T to an existence pro blem for implicants in the emerging DF. This last problem can b e easily expr essed as a SubGI instance. Let’s enumerate liter als in ea ch of the clauses in formula f . F o r each couple o f cla uses ( c i , c j ), let’s build a co mpatibilit y b ox: the ( α, β )-elemen t in the matrix is 0 or 1 depending o n whether the α -th literal in c la use c i and the β -th literal in clause c j are com- plimen tary . Let’s aggr egate all these compatibilit y boxes in a b ox matrix. O bviously , there is an impli- cant in the DF o f f iff there is a grid o f elements in the b ox matrix, one element per compatibility b ox, whose all elemen ts are equal 1. Each s uch g rid of elements consists of the couples of liter a ls which par - ticipate in an implicant. The box matrix built in such a w ay may b e seen as input G . Then, pattern S ma y b e a b ox ma tr ix of the s a me structur e as G but whose boxes are en tirely filled with 0 except their upp er-left-cor ner ele ments, which are equal 1: S = ( S ij ) m × m , S ij =    1 0 . . . 0 0 . . . . . . . . . . . .    k i × k j There is one obvious restriction on the rela b eling of S : the elemen ts of b oxes S ij are not allo wed to lea ve their boxes. This restr iction can b e accommo dated in system 4 with a p o lynomial num b er of additional linear co nstrains. Conclusion W e describ ed a p oly nomial time reduction o f SubGI to a po lynomial s ize asymmetric linear sy s tem. The system consists of sys tems 1 and 4. Subsystem 1 depe nds on the size of SubGI instance, only . Sub- system 4 describ es the structure of the given input and pa ttern. The s ystem’s asymmetry is due to the explicit in volvemen t of the input a nd pattern’s adja- cency matrices in the constructio n of system 4 - see definition 3. So, the res ult may b e seen as c o mpli- men tary to the Y a nnak akis theorem [1]. Linear system 1, 4 defines a s ub-p olytop e in the Birkhoff po lytop e. V ertices o f this sub-p olytop e are those permutation matrices which satisfy quadra tic int egral sys tem 2. Relab eling o f the input and pat- tern rotates this sub-p olytop e all ov er the Birk ho ff po lytop e. Ultimately , system 1, 4 may b e seen as a para llel testing of all guesses, where guesses are n × n p ermu- tation matrices - the unknowns in s y stem 2. Basi- cally , this para llelization was achieved with enco ding SubGI in the contradictions betw een relab eling p os- sibilities fo r different vertices. Obviously , the de s crib ed “contin uous” solution of SubGI is not unique. Also, w e could dev elop a p oly- nomial time discrete algorithm which would search the compatibility matrix for the solution grids as, for exa mple, it was do ne in [12] for 3SA T whic h is a NP-complete pro blem [3], to o 4 . 4 F or 3SA T, see a demo at ht tp://www.timescube.com References [1] Mihalis Y annak a kis, Expr essing c ombinatoria l opti- mization pr oblems by line ar pr o gr ams , In P r o c. of the t wen tieth annual ACM S ympo s. on Theory of computing, Chic a go, Illinois, pp. 2 23 - 22 8, 1988 [2] Ric har d M. Kar p, R e ducibility A mong Combina to- rial Pr oblems , In Complexity of C o mputer Compu- tations, P ro c. Sympos. IBM, Thomas J. W atso n Res. Cent er, Y o r ktown Heigh ts, N.Y. 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Schrijv er, Ge ometric A l- gorithms and Combinatorial Optimizatio n , Spring e r, Berlin, 1988 [10] G. Birkhoff, T res observaciones sobr e el algeb r a lin- e al , Univ. Nac. T ucumn Rev, Ser. A, no. 5, (1946) p. 14 7-151 [11] Levin, Leonid, (197 3). Universal se ar ch pr oblems (Russian: Unive rsal’nye p er eb ornye zadach i) , Prob- lems o f Information T ra nsmission (Russian:, Pro b- lemy Peredac hi Info r matsii) 9 (3): 265266, 1973 (Russian); transla ted into English b y T rakhten- brot, B. A. (1984). A survey of R ussian appr o aches to p er eb or (brute- for c e se ar ches) algorithms , An- nals of the Histo r y of Computing 6 (4): 3844 00. doi:10.11 09/MAHC.19 8 4.100 36. [12] Sergey Gubin, Polynomial time algorithm for 3-SA T , arXiv:cs/0 7010 23 [cs.CC]