Presented by Yuval Shimron Course

 Find solutions to sub-cases of the Subgraph Isomorphism Problem in polynomial time.  Find More efficient solutions to some sub- cases that already had polynomial time solutions.  Find simple paths and cycles of specific length k.  This was the initial goal of the authors… 2

 (1) For a fixed k, if G=(V,E) contains a cycle of length k it can be found in O(V ω ) expected time or O(V ω logV) worst-case time (ω<2.376 is the exponent of matrix multiplication).  (2) For a fixed k, if a planar graph G=(V,E) contains a cycle of length k it can be found in O(V) expected time or O(VlogV) worst-case time (Applies also to any non-trivial minor- closed family of graphs). 3

 (3) If G=(V,E) contains a subgraph isomorphic to a bounded tree-width graph H=(V H,E H ) where |V H | = O(logV), then such a subgraph can be found in polynomial time.  Was not previously known even if H were just a simple path of length O(logV).  Shows that the LOG PATH problem is in NC (and not just in P). 4

 Randomized method  Vertices are randomly colored using k = |V H | colors.  If |V H | = O(logV), then with a small (but only polynomial small) probability all the vertices of the (isomorphic to H) subgraph are colored in distinct colors.  Makes the task of finding this ‘color-coded’ subgraph much easier. ▪ Be patient… 5

 De-randomized algorithm?  Needs a family of colorings of G, such that every subset of k vertices of G is assigned with distinct colors by at least one of these coloring. ▪ In other words, a family of perfect hash functions from {1, 2, …, |V|} to {1, 2, …, k}.  Only “small” loss of efficiency. 6

 If acyclic – simple  O(E) time for a simple algorithm.  So eliminate cycles:  Choose a random permutation.  Build by using : ▪ Direct the edges: 7

 Every directed path of length k in G’ is a simple path of length k in G.  Every simple path of length k in G has a 2/(k+1)! chance of becoming a directed path in G’.  So if no path of length k was found in G’ repeat the process.  The expected number of times this process is repeated is at most (k+1)!/2. 8

 So we get O(E(k+1)!) time complexity.  This is also the result for the directed case. ▪ Delete edges that don’t agree with.  Use the following fact + DFS to reduce it to O(V(k+1)!) for the undirected case:  Every graph with V vertices and at least k|V| edges contains a path of length k.  So first run a DFS on the original graph.  Apply the above algorithm only if no vertex of depth k was found (answered in O(k|V|) time). 9

 Choose random acyclic orientation G’.  Raise the adjacency matrix of G’ to the power of k-1 using O(logk) matrix multiplications.  This gives all the pairs of vertices connected by a path of length k-1.  Check if any of these pairs are connected.  If so.  If not, repeat the process. ▪ Expected number of at most k!/2 time.  Complexity: O(k!(logk)V ω )=O(V ω ) for a fixed k. 10

 To find a path of length k-1 in a graph G we can choose a random coloring of the vertices of G in k colors.  Every simple path of length k-1 in G has a chance of k!/k k > e -k to become colorful.  Each vertex is colored with a different color.  We can find it using lemma

 Use Color-Coding to find a colorful path of length k-1 in 2 O(k) E worst case time (if exists).  Actually it finds a path of length k that starts at a specific vertex s. ▪ but we can always add some vertex s to G (with a new color).  The algorithm uses a given (random) coloring c : V {1, 2, … k}  The algorithm uses a dynamic programming approach. 12

 Suppose we’ve found for each vertex v the sets of colors on colorful paths of length i that connects s and v.  A collection of at most color sets.  For that we only need to record the color sets appearing on i-length paths.  And not the path themselves…  We inspect every color set C of that collection. 13

 We also inspect every edge (v,u) in E.  If we add to the collection of u that corresponds to colorful paths of length i+1.  The graph G contains a colorful path of length k-1 iff the final collection, corresponds to paths of length k-1, of at least one vertex is non-empty. 14

 The number of operations is at most.  The proof holds for both directed and undirected graphs. 15

 We can find all pairs of vertices connected by path of length k-1 in or worst case time.  To get time simply run 3.1 algorithm |V| times, from each vertex of G=(V,E).  Use recursive approach to get time. 16

 Keep all partitions of {1,2,…,k} into two subsets C 1,C 2 of size k/2 each. There are such partitions.  For each partition, split G into two graphs derived from C 1, C 2 coloring.  Recursively find pairs of vertices connected by paths of k/2-1.  Store the results in Boolean matrices A 1,A 2. 17

 Define B to be a Boolean matrix of adjacency relations between V 1,V 2 vertices.  Compute A 1 BA 2.  You get all pairs connected by paths of length k-1 ▪ First k/2 vertices are colored by colors from C 1 ▪ Last k/2 vertices are colored by colors from C 2  By OR-ing all the matrices obtained from all the partitions you get your answer.  Time complexity? 18

 A simple path of length k-1 in a directed / undirected graph G=(V,E) can be found (if exists) in:  expected time for undirected graph. ▪ DFS…  expected time for directed graph.  A simple cycle of size k in a directed / undirected graph G=(V,E) can be found (if exists) in either or expected time.  Simply use lemma

 The previous randomized algorithms can be derandomized with a loss of efficiency.  Extra logV factor to the complexity.  What we need is a family of k-perfect hash functions from {1, 2, …, |V|} to {1, 2, …, k}.  If we use these hash functions we know that for every subset of k vertices there exists a coloring that gives each vertex in it, a distinct color. 20

 There exists an algorithm that constructs a k-perfect family of hash functions from {1, 2,..., n} to {1, 2,..., k}.  But its size is.  There also exists an algorithm that constructs a k-perfect family of hash functions from {1, 2,..., n} to {1, 2,..., k 2 } that its size is. 21

 So we use 2-level hashing:  Mapping from {1, 2,..., n} to {1, 2,..., k 2 } by using the second algorithm.  Mapping from {1, 2,..., k 2 } to {1, 2,..., k} by using the first algorithm..  And we get just the promised extra O(logV) time.  The value of each element can be evaluated in O(1) time. 22

 Use k-perfect hash coloring functions.  Choose a random coloring (ant not a permutation) c : V --> {1, 2, … k}  Remove edges (u,v) s.t..  Direct remaining edges (u,v) from u to v.  Again G’, the obtained graph, is acyclic.  Simple path of length k in G has a probability of 2k -k to become a directed path in G’.  Different from the Color-Coding method. 23

 An undirected graph G is d-degenerate if every subgraph of it has a vertex of degree at most d.  Smallest such d is called the degeneracy or the max-min degree of G.  Maximum over the minimum degrees of all sub-graphs of G.  If G is d-degenerate then clearly. 24

 Let G be a connected undirected graph.  An acyclic orientation of G=(V,E) such that for every v we have can be found in O(E) time. 25

 A graph H is a minor of undirected graph G if it can be obtained from G by the removal and the contraction of edges.  A family C of graphs is minor-closed if a minor of any graph in it is also a member of the family.  If such C is non-trivial then all graphs in C are of bounded degeneracy.  s.t.. 26

 Consider the family of planar graphs C planar  It is minor-closed.  Each planar graph has a vertex whose degree is at most 5. . 27

 Let C be a non-trivial minor-closed family of graphs and let be a fixed parameter. There exists a randomized algorithm that given an undirected graph in C finds a C k - cycle of size k in it if one exists, in O(V) expected time.  Proof:  Let G = (V,E) be a graph in C that contains a C k.  Choose a random coloring c : V -> {1, 2, 3, …, k}.  C k is considered well-colored if colored in a consecutive way by the colors 1, 2, …, k. 28

 The C k in G has a chance of 2/k k-1 to be well-colored.  Can we find it efficiently?  Yes, but with some probability…  Assume that the degeneracy of C is d = O(1).  We describe a randomized algorithm that given a coloring c, finds C k with probability of 1/(2d) k.  Combining both gives a probability of at least so the expected time is. 29

 We can assume all edges of G connect vertices that are colored by consecutive colors (mod k).  Edges that don’t may be safely removed.  We orientate the graph so that the out-degree of all the vertices is at most d.  This takes only O(V) time.  The algorithm tries to find the edge that connects the vertices in C k colored by k and k-1: v k,v k-1. It “flipps coins” to guess it’s orientation and index – 2d possible combinations. 30

 For each guess of such index i  If the orientation is from v k-1 to v k : ▪ All edges that leave v k-1 but whose index is not i are removed.  Otherwise does the opposite. ▪ (for edges that leave v k )  Result is the graph G’ that contains a C k with a probability of at least 1/(2d).  A forest of rooted stars. 31

 Each such star is contracted into a single vertex and assigned with the color k-1.  The obtained graph is denoted by G’’.  G’’ contains a well-colored C k-1 iff G’ contains a well-colored C k.  Since each edge of G’ and therefore G’’ connects consecutively colored vertices.  G’’ is also a graph in the minor-closed family C.  So we recursively look for C k-1. 32

 It will take us O((k-1)V) expected time.  And yields C k-1 with a probability of at least 1/(2d) k-1.  Obviously it’s easy to reconstruct C k from C k-1.  We can stop the recursion when k=3 and use an existing algorithm for finding triangles in a general graph in time.  Any triangle in a three-colored graph is well- colored.  is in our case. 33

 There exists a determinist algorithm that given a graph in C, finds C k if exist, in O(VlogV) WC time.  Proof:  Instead of using random coloring we exhaust a list of k O(k) logV colorings that has this property: ▪ Every sequence of k vertices is consecutively colored by 1,2,…,k by at list one coloring of the list.  Instead of guessing the direction and index of each edge in the C k we exhaust for each coloring all the (2d) k possible choices. ▪ If G contains a C k then at least one C k will be found this way. 34

 A graph G 1 is said to be isomorphic to a graph G 2 if there exists a bijection: f : V(G 1 ) -> V(G 2 ) such that any two vertices u and v of G 1 are adjacent in G 1 iff ƒ(u) and ƒ(v) are adjacent in G 2. 35

 Let F be a directed/undirected forest on k vertices. Let G be a directed/undirected graph.  A sub-graph isomorphic to F can be found if exists in:  expected time in the directed case.  expected time in the undirected case. 36

 Proof:  Start as usual, by choosing a random coloring: c : V -> {1, …, k} of G.  With a probability of at least e -k the copy of F in G becomes colorful. ▪ Meaning, each vertex is assigned with a different color.  Suppose that F is composed of l (directed) trees T 1, T 2, …, T l with k 1, k 2, …, k l vertices each.  Let F i be the (directed) forest composed of T 1, T 2, …, T i. 37

 For each we find the color sets that appear on colorful copies of T i in G.  Note that copies of T i, T j with disjoint color sets are necessarily disjoint.  Then, in 2 O(k) time we find the color sets that appear on colorful copies of F i for.  If the collection corresponding to F=F l is not empty then G contains a colorful copy of F.  How do we find it…? 38

 How do we find the color sets that appear on colorful copies of T i in G?  Let t be an arbitrary vertex in T i =T.  For each vertex v in G we find the color sets that appear on copies of T in which v plays the role of r.  If T is a singe vertex then it’s easily done…  Otherwise let e=(r,r’) be a (directed) edge in T. ▪ We break T into two (directed) sub-trees T’, T’’. 39

 We recursively find, for each vertex v in G, the color sets in copies of T’ and T’’ in which v plays the role of r and then of r’.  For every (directed) edge (u,v) we update u’s collection with v’s collection if they are disjoint.  The complexity of this recursive algorithm is as required.  For the undirected case we use the fact that a graph with at least k|V| edges contains as a subgraph any forest on k vertices. 40

 Remember tree-width of a graph G?  The minimum tree-width over all possible tree- decomposition of G to (X,T).  T = (I, F) is a tree.  X = { X i : i I} is a set of subsets of V such that: ▪ The union of all X i s equals to V. ▪ For every edge (u,v) of G there exists an i such that u,v are in X i. ▪ If, and j is on the path from i to k in T then: 41

 Let H be a directed or undirected graph on k vertices with tree-width t. Let G be a directed or undirected graph.  A sub-graph of G, isomorphic to H, if one exists, can be found in expected time and in worst case time.  Proof is similar to that of Theorem 6.1.  So we will skip it... 42

 In [RS86b] it is shown that if C is a minor closed family of graphs that excludes at least one planar graph G’ then there exists a (huge) constant c G’ such that every graph in C has a tree-width of at most c G’.  So we can use 6.3 wherever |V H | = O(logV) and H excludes at least one planar graph.  and decide in polynomial time whether G contains a graph isomorphic to H. 43

 As a very special case of Theorem 6.3 we get that the LOG PATH problem is in P  A path of logV vertices is a tree.  In addition, all the algorithms we described are easily parallelizable.  So we get that the LOG PATH problem and other problems are in NC. 44

 The Color-Coding method efficiently finds k- vertex simple paths, k-vertex cycles, and other small sub-graphs within a given graph using probabilistic algorithms.  The Color-Coding method is a good example of demonstrating de-randomization techniques.  Algorithms presented can be easily parallelized.  Yielding efficient NC algorithms. 45