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This research paper discusses algorithms for finding cycles in graphs using rectangular matrix multiplication and dynamic programming. It explores the complexity of different algorithms and presents improved solutions for finding triangles, longer cycles, and paths of a given length.
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Finding cycles using rectangular matrix multiplication and dynamic programming Raphael Yuster Haifa University Uri Zwick Tel Aviv University …
Rectangular Matrix multiplication c c b = a a b a c a = b c b M(a,b,c) = M(b,c,a) = M(c,a,b) = ...
Rectangular Matrix multiplication c b c = a a There are also improved specially designed rectangular matrix multiplication algorithms
Chain Matrix Product • Let Ai be an ni ni+1 matrix. • Compute A1A2…Ak. • Matrix multiplication is associative, so all association orders give the same result, but may have vastly different cost. • Easy to solve using dynamic programming.
Chain Matrix Product (cont.) Let P(i,j) be the cost of computing Ai…Aj. Complexity of naïve solution is O(k3). For M(a,b,c)=abc, there is a much more complicated O(k log k) algorithm.
Finding triangles A triangle A graph Does the graph contain a triangle? Yes!
Algorithms for finding trianglesLet A be the adjacency matrix of a graph G. Then, G contains a triangle iff A2AT≠0.
Finding triangles in O(m2 /(+1)) time [AYZ ’97] = m(-1) /(+1) Let be a parameter. .High degree vertices: vertices of degree .Low degree vertices: vertices of degree < . There are at most 2m/high degree vertices =
Finding longer cycles A graph G contains a Ck iff Ak-1AT≠0 ? We want simple cycles!
Colorcoding[AYZ ’95] Assign each vertex v a random number c(v) from {0,1,...,k-1}. Remove all edges (u,v) for which c(v)≠c(u)+1 (mod k). All cycles of length k in the graph are now simple. If a graph contains a Ck then with a probability of at least k-k it still contains a Ckafter this process. An improved version works with probability 2-O(k). Can be derandomized at a logarithmic cost.
Finding C2k in O(m2-1/k) time [AYZ ’97] = m1/k Let be a parameter. .High degree vertices: vertices of degree .Low degree vertices: vertices of degree < . There are at most 2m/high degree vertices = Finding paths of length k Finding a cycle passing through a high degree vertex
Algorithms for finding a Ck [AYZ ’97]:“We have not been able to use fast matrix multiplication to obtain faster algorithms for finding Ck, for k4, in sparse graphs.”
Finding even cycles even faster [YZ ’97] In undirected graphs, a C2k, for any fixed k, can be found in O(n2) time! We are still working on:Finding odd cycles עוד יותר מהר
Finding C4’s in O(m2-2/ n1/) time [Eisendbrand, Grandoni ’03] Let be a parameter. .High degree vertices: vertices of degree .Low degree vertices: vertices of degree < . There are at most 2m/high degree vertices
Finding C4’s in O(m(4-1)/(2+1)) time [YZ ’03] Let be a parameter. High degree vertices: ≤ deg(v)Medium degree vertices: 1/2 ≤ deg(v) < Low degree vertices: deg(v) < 1/2 There are at most 2m/high degree vertices.There are at most 2m/1/2medium degree vertices.
Where do we go from here? • Why just three degree classes? • How many cases will we have for C5, C6, …?
The general case • Partition the vertices into log n degree classes: Vi={ vV | 2i ≤ deg(v) ≤ 2i+1 } • Number of cycle classes is only (log n)k. • What is the most efficient way of handling each class?
Handling a given cycle class Choose i and j. Find all paths of length j-i that pass through vertices of degrees di ,…, dj. Find all paths of length k-(j-i) that pass through vertices of degrees dj ,…, di. Check whether the graph contains a cycle from the class.
Finding paths Degree: Sizeof class: We get a chain matrix product problem! We can use the sparsity of the matrices.
Finding paths (cont.) It is convenient to express everything as a power of m. Degree: Sizeof class:
Finding cycles Theorem: There is an time algorithm for finding Ck’s in directed graphs.
Finding C5’s Theorem: Requires a non-constant number of degree classes. The worst-case running time is obtained for regular graphs.
Finding C6’s Conjecture: If the conjecture is true, then c6<1.65.
Finding odd cycles Conjecture: for odd k3. The conjecture holds for k=3,5.
Open problems • Is it possible to extend the technique presented to obtain improved algorithms? • Find a feasible way for rigorously computing c6, c7, etc. • Other applications?