Finding a Heaviest Triangle is not Harder than Matrix Multiplication

Artur Czumaj & Andrzej Lingas

overview

• The problem

• Previous solutions

• New solution– Idea– Flow– Proof

• Consequences & improvements

• Summary of results

The Problem

• Finding a fixed subgraph

• Finding a triangle - shortest path

• Known to be no more than matrix multiplication

• Rectangular matrix mult is O(n^w), w< 2.376

• So: Finding a heaviest triangle

Previous solutions

• V. Vassilevska and R. Williams. Finding a maximum weight triangle in n3−δ time, with applications. Proc. 38th Annual ACM Symposium on Theory of Computing (STOC’06), pp. 225–231, 2006. [12]

• V. Vassilevska, R. Williams, R. Yuster. Finding the smallest H-subgraph in real weighted graphs and related problems. To appear in Proc. 33rd International Colloquium on Automata, Languages and Programming (ICALP’06), 2006. [13]

• Maximum witness of Boolean matrix product

appetizer

New solution (HT) – idea

• Based on triangles having three vertices

• Generate set of where triangles might be found “potentials” – not set of triangles

• Cut down set

• Recursion – set is hierarchically defined

• Start with set of n^3 potential triangles

• At each step reduce to O(ξ2) sets of (n/ξ)3 triplets

HT – flow - definition

• procedure HTξ(G, I,K, J)• Input: A graph G = (V,E) with vertex weights (G

is given as adjacency matrix)• vertices are numbered in non-decreasing weight

order from 1 to n• subintervals I, K, and J of [1, . . . , n], of the

same length κ assumed to be a power of ξ• Output: Maximum-weight triangle (i, j, k), if any,

such that i I, j J, and k K∈ ∈ ∈

Flow - outline

• If set is of size 1 and it has a triangle, return it

• Break down intervals into sections• Generate matrixes indicating paths

between intervals• Find triplets of interval sections which are

interconnected• Prune set of triplets• Recursion on all triplets of interval sections

Flow detail 1 – stopping condition and definitions

• if κ = 1 then• if (i0 + 1, j0 + 1, k0 + 1) is a (non-

degenerate) triangle in G then return (i0 + 1, j0 + 1, k0 + 1);

• stop• let i0, j0, and k0 be such that• I = [i0 + 1, i0 + κ], K = [k0 + 1, k0 + κ], and

J = [j0 + 1, j0 + κ]• ℓ = κ/ξ

Flow detail 2 - generate path indication matrices

Flow detail 3 – find triplets of sections

Flow detail 4 – prune set

Flow detail 5 - recursion

HT – Proof of correctness

• All triangles potentials are added in

• Only subsumed triangle potentials are removed

• Note vertex numbering

HT proof of time

• Define chains

• Dilworth’s lemma– Largest antichain includes one element from

each chain

HT proof of time - cont

HT proof of time cont 2

Consequences – sparse matrixes

• Sparse matrixes – m is small• Can split into two parts, one as above, one m-

bound• Choose clever splitting point

Consequences - Kh clique

• How about finding a maximum subgrapgh of more than 3 vertexes?

• Insight – can make a hierarchy with a new graph where we look fro triangle, and where each vertex is a third of the clique we are looking for

• What to do when not a multiple of 3

Consequences - more

• Instead of triangle or general sets of linked vertexes, relate to particular connection patterns – isomorphic graphs

• Can use fast rectangular matrix multiplication

Consequences – results summary