Solving Graph Isomorphism Problem for a Special Case

Home , Complete (complexity), Degree (graph theory), Graph isomorphism

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Vertex from graph G1 Vertex from graph G2 1 G 2 F 2 3 E 4 D 5 C 6 B 3 4 7 A TABLE II: dsv(i), for all i ∈ V of the graph G1 Vertex i degree(i) dsv(i) 6 5 1 1 {(2,3)} 2 3 {(1,1),(3,2),(4,4)} 3 2 {(2,3),(4,4)} 7 4 4 {(5,1),(3,2),(6,2),(2,3)} 5 1 {(4,4)} Fig. 1: Graph G1 6 2 {(7,1),(4,4)} A B 7 1 {(6,2)}

TABLE III: list1 of the graph G1 Vertex list list list list C D elements elements elements elements 1 2 - - - 2 1 3 4 - 3 2 4 - - 4 2 3 5 6 5 4 - - - E F 6 4 7 - - 7 6 - - -

1) Preprocessing 2) Checking the input G 3) Generate UIDs 4) Map Isomorphism Fig. 2: Graph G2 A. Preprocessing We will process on both the graphs and store the results for later use. Here deg seqi is list of degree sequences of the V : Set of vertices , |V| = n neighbours of the vertex i. This step will be performed on E : Set of edges (Ordered pair of vertices), every vertex i, i ∈ V for both the graphs. For understanding, E ⊂ (V ×V), and for a given vi ∈ V and deg seq4 is shown in the Table IV. Si = {x | (x,vi) ∈ E} , ∀v j,vk ∈ S, v j 6= vk =⇒ dsv(v j) 6= Algorithm 1 dsv(vk) . Preprocessiong Where dsv(p) is the neighbourhood degree list of a 1: deg seqi ← 0 given vertex p with the vertices. For example, dsv(4) = 2: S = {x | (x,i) ∈ E} {(5,1),(3,2),(6,2),(2,3)}. It is a list of tuples, where the first 3: for each v ∈ S do element is the vertex and second is its degree. The dsv is sorted 4: deg seqi.append(dsv(v)) by the degree (the second element). 5: end for We will solve this problem using the proposed method. The input is two graphs G1 and G2 in their adjacency lists B. Checking the input form, list1,list2 respectively. The algorithm will first check that whether these graphs follow the Definition of Permissible In this module the algorithm will check whether the given Graphs. or not. Table III shows the adjacency list, list1 of the graphs fits in the above definition or not. After getting the graph G1. This algorithm consists of 3 sub modules: deg seqi, for all the i ∈ V, the algorithm sorts the elements TABLE IV: degree sequence list, deg seq4 of the graph G1 vertices. Function generate UID is called for each vertex of vertex v dsv(v) of the neighbours v of the vertex 4 graphs G1 and G2. This function takes count and the vertex i 2 {(1,1),(3,2),(4,4)} as its input. Bool array element will change its value to one, 3 {(2,3),(4,4)} if that vertex is visited. The function will be called for every 6 {(7,1),(4,4)} vertex i with initial condition set to generate UID (n, i). Now, 5 {(4,4)} we will perform this operation on the vertex 4 of Graph G1. Algorithm 2 Whether this method is applicable or not Now, we know from Table I that vertex D from graph G2 is isomorphic to the vertex 4 from graph G1. 1: Sort the lists in deg seqi by the length of the lists and then lexicographically. D. Map Isomorphism 2: Sort dsv(i) according to the order of lists in the deg seqi. After generating UIDs from both the graph G1 and G2 the true 3: result ← algorithm will try to map the UIDs of graph G1 to UIDs of 4: for each i V do p ∈ Graph G2 and will report the isomorphism I. Here, UID j is 5: for each x,y ∈ deg seqi do UID of graph p and of jth vertex. UIDp is the list of all the 6: if x=y then UIDs of all the vertex of the graph p. The map Isomorphism 7: result ← false function will be called for UID of every vertex of the graph 8: end if G1. The iso is and array which maps vertex of one graph to 9: end for the vertex of the other graph. If for all the UIDs the value of 10: end for flag is 1 and all the iso arrays give the same result then the two graphs are isomorphic and the isomorphism function is given by iso. The algorithm uses the constraint given in the of this list. In the step 1. of the Algorithm 2, lists in the Definition of Permissible Graphs., which is why the algorithm deg seqi are sorted by lexicographically (according to the revolves around the degree sequence of the vertices. second element in the tuple) and then by the length of the list. Now, we will perform this step on Vertices of the graph 1 Algorithm 4 map Isomorphism(UIDi ) G . 1 2 2 If value of the result is true then we can apply our algorithm. 1: for each UID j ∈ UID do And it can go thorough out next stage. 2: iso ←−1 3: ind ← 0 C. Generate UIDs 4: flag ← 0 1 2 Here we will introduce an encoding technique, in which we 5: if len(UIDi )= len(UID j ) then 1 will assign different ids to all the vertices. And after that we 6: while ind < len(UIDi ) do will compare this ids in both the graphs among all the vertices. 7: ind ← ind + 1 1 These unique ids (UID) are independent of the order in which 8: if iso[UIDi [ind][0]] 6= −1 and 1 2 we calculated these UIDs. THe bool array is of size |V| and iso[UIDi [ind][0]] 6= UID j [ind][0] then 9: flag ← 0 Algorithm 3 generate UID (count, i) 10: break 11: end if 1: tmp1=dsv(i)[0] 12: if UID1[ind][1] 6= UID2[ind][1] then 2: while counter > 0 do i j 13: flag ← 0 3: tmp2=[] 14: break 4: for each x ∈ tmp1 do 15: else 5: if bool[x] = 0 then 16: iso[UID1[ind][0]] ← UID2[ind][0] 6: UID.append(dsv(x)) i j 17: flag ← 1 7: tmp2.append(dsv(x)[0]) 18: end if 8: count ← count − 1 19: end while 9: else 20: if flag=1 then 10: UID.append((x,−1)) 21: break 11: end if 22: end if 12: end for 23: end if 13: UID.append((−2,−2)) 24: end for 14: end while Any vertex of a permissible graphs can not have more than it stores whether all the vertex are included or not. The array one UID. is initialized to 0 for all the elements. The count variable is initialized with total number of vertices, |V|. It will decrement Proof. The UID consist of UIDd (degrees) and UIDn (nodes). its value by one as the algorithm explores the dsv of all the • UIDn[1]= v and UIDd[1]= deg(v) . TABLE V: sorted degree sequence list, deg seqi for i ∈ Vof the graph G1

vertex i deg seqi changed remarks order of dsv(i) 1 [ {(1,1),(3,2),(4,4)}] {(2,3)} - 2 [ {(2,3)}, {(2,3),(4,4)}, {(5,1),(3,2),(6,2),(2,3)} ] {(1,1),(3,2),(4,4)} - 3 [ {(1,1),(3,2),(4,4)}, {(5,1),(3,2),(6,2),(2,3)}] {(2,3),(4,4)} - 4 [ {(4,4)}, {(2,3),(4,4)}, {(7,1),(4,4)}, {(1,1),(3,2),(4,4)}] {(5,1),(6,2),(3,2),(2,3)} changed 5 [ {(5,1),(3,2),(6,2),(2,3)}] {(4,4)} - 6 [ {(6,2)}, {(5,1),(3,2),(6,2),(2,3)}] {(7,1),(4,4)} - 7 [ {(7,1),(4,4)}] {(6,2)} -

TABLE VI: UID of vertex 4 of Graph G1 vertex 4 -2 5 6 3 2 -2 4 7 4 2 4 1 3 4 -2 4 6 4 2 4 2 3 4 -2 degree 4 -2 1 2 2 3 -2 4 1 4 3 4 1 2 4 -2 -1 2 -1 -1 -1 3 -1 -1 -2

TABLE VII: UID of vertex D of Graph G2 vertex D -2 C B E F -2 D A D F D G E D -2 D 6 D F D F E D -2 degree D -2 1 2 2 3 -2 4 1 4 3 4 1 2 4 -2 -1 2 -1 -1 -1 3 -1 -1 -2

TABLE VIII: Practical results • UIDn[2]= −2 and UIDd[2]= −2 • The next deg(v) elements will be of dsn(v). Number Total Connected Trees Planar Permissible Fraction • Between any two i and j such that, UIDn[i] = −2, of vertices graphs graphs graphs graphs UIDd[i] = −2 and UIDn[ j] = −2, UIDd[ j] = −2 and 1 1 1 1 1 1 1 (trivial) 2 2 1 1 2 1 0.5 there is no k, where i < k < j, such that UIDn[k]= −2, 3 4 2 1 4 0 0 UIDd[k]= −2: 4 11 6 2 11 1 0.0909 5 34 21 3 33 0 0 6 156 112 6 142 6 0.038 The graphs described in Deﬁnition of Permissible Graphs 7 1044 853 11 822 62 0.05938 contains non-planar graphs. 8 12346 11117 23 6966 1024 0.0829 9 274668 261080 47 79853 29285 0.1066 Proof. We will prove it by example. Consider Fig. 3, number

1 2 3 3) Generate UIDs : The maximum length of any UID is of the order of the number of O(n2). This function will be called n times, making this step run in O(n3) time. 4) Try to fit Isomorphism: For finding the isomorphism, the Algorithm 4 will compare UID of one graph with all the 4 5 6 UIDs of the other graph. This is total n×n2 comparisons. This is O(n3). The algorithm will be called n times. So, the overall complexity becomes n × n3 making the complexity O(n4). Overall, the algorithm runs in O(n4) time. 7 8 The code was written in python and c++. In the Table VIII the number of permissible graphs for number of vertices ranging Fig. 3: Graph G 3 from 1 to 9 is given. We generated these graphs using SAGE [15] and nauty [16]. For n = 9, 10.66% of the graphs lie under of vertices is 8 and number of edges is 19. The graph has a the Definition of Permissible Graphs. There are only 0.0171% triangle in it. According to Euler’s formula [14] this graph is of graphs are tree for n = 9. The Theorem II-D states that non-planar graph (3v − 6 <= e). Once can easily verify that there are non-planar graphs, which can also be solved by the this graph can also be solved using the proposed method. proposed method.

III. TIME COMPLEXITY AND PRACTICAL RESULTS IV. CONCLUSION 1) The Prerprocessing step : Degree of each vertex can be The proposed algorithm solves GI problem for a special found in O(n2) and after that their degree sequence can case. Before running any slower algorithm it is better to be calculated in O(n3). Overall, this step takes O(n3) check the graphs using this algorithm. The time complexity is time. O(n4), but these bounds are loose bounds. Because generating 2) Checking the input: Time omplexity O(n4). UID and matching the UIDs can be done faster, if we have used fast methods to sort them using the property that every value is bounded by v, making the bounds tighter. More importantly, this is polynomial time solution for this type of graphs.This algorithm is embarrassingly parallel making it useful in practical purposes.

ACKNOWLEDGMENT The authors would like to thank Dhruv Patel for helping in writing the software in python.

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