Rank Maximal Matching By Robert W. Irving, Telikepalli Kavitha, Kurt Mehlhorn, Dimitrios Michail.

Introduce the problem Let be a bipartite graph and εbe the edge set, each edge (a,p) has a rank, find the best matching for G A rank maximum matching is : In a matching M, you can’t find another matching mi that | mi | > |Mi| with respect to edges of rank i. Just like allocation jobs to students or probationary jobs to trainee teachers

Definitions P(M) : The signature of a Matching M, which is represented in a r-tuple form (x1, x2, … , xr), where xi denotes the size of a match for an edge set of rank i. We also define a total order that if (x1,x2,…,xr) (y1,y2,…,yr) for some k <= r, xk < yk and for all j < k, xk = yk Mi : A subset of a rank maximum matching M having rank maximum matching with rank <= i, a subset of Mi-1

Algorithms Previous achives: Today’s goal: O(min(n+C, C )m) xc= x1+……+xk Today’s goal: O(min(n+C, C )m) Another solution works in O(Cmn) C: The maximum rank of all nodes m: The number of edges n: The number of vertexes

Solutions: A combinational algorithm Shortcut: Delete those who will not be used anymore. Use E, O, and U disjoint sets to partition the graph. Alternating path!

Definitions: E, O, and U(1/2) Three disjoint sets: E/O: Nodes that can be reached by a free node by an even/odd-numbered path in terns of alternating path 1 1 1 2 2 Free

Definitions: E, O and U(2/2) U: nodes can not be reached by any free nodes using alternating path. Lemma 2.1:The sets E, U, O are pair wise disjoint, every maximum matching in G pairs the nodes in U, matches all nodes in O, matches each node in O and another in E, and have cardinality |O| + |E|/2. There is no edge in G connecting a node in E with a node in U, or between 2 nodes in E, no maximum matching in G uses an edge connecting 2 nodes in O or one in O and another in U

A combinational algorithm(1/2) Gi: Mi: A greedy matching with signature(s1,s2,…,si) Initial state: G1’ = G1, M1=maximum matching of G1’

A combinational aogorithm(2/2) For i = 1 to r-1 do Partition Gi’ into node sets Ei, Oi, and Ui with respect to Mi Delete all edges incident to Oi or Ui which has rank > I Also delete all edges that connect 2 nodes in O, or one in O and another in U Add εi into Gi’, said Gi+1’ Augmenting Mi to get Mi+1

Prove the correctness(1/4) Lemma 2.2:If every greedy matching of Gi is a maximum matching of Gi’, then every greedy matching of Gi+1 is contained in Gi+1’ Let Ni+1 be a greedy matching Ni = Ni+1 ε≦I  Ni is a greedy matching in Gi  Ni is a maximum matching in Gi  Ni does not use any edge of Gi’, so will never use in Ni+1  Ni+1 is contained in Gi+1’

Prove the correctness(2/4) Lemma 2.3: For every i, j such that j > I, the number of edges at most i is the same in Mi and Mj Since Mj is obtained from Mi by successive augmenting, every vertex matched by Mi is also matched by Mj  all nodes in Ui and Oi are matched in Mj  #(ε≦i) > #(ε≦j) #(ε≦i) in Mj < #(ε≦i) in Mi by definition

Prove the correctness(3/4) By induction prove: Every greedy matching in Gk is a maximum matching in Gk’ Mk is a greedy matching in Gk Case k = 1  a greedy matching in G1 is a maximum match in G1’, trivially. Assume case k = n is true Consider case k = n+1:

Prove the correctness(4/4) Mi is a greedy matching in Gi with signature (s1,s2,…,si), suppose Gi+1’s signature is (r1,r2,…,ri,ri+1) By lemma 2.3  s1 + s2 + … + si = r1 + r2 + … + ri  Mi+1 = (s1,s2,…,si,ri+1) with ri+1 < si+1 By lemma 2.2  Mi is a maximum matching  Mi+1 is contained in Gi+1’  Mi+1 is a maximum matching of Gi+1  its cardinality is at least s1 + s2 + … + si+1, so si+1 = ri+1 By induction hypothesis, any greedy matching Ni+1 in Gi+1is contained in Gi+1’ and Ni has the same cardinality of Mi+1  Ni+1 is a maximum matching

Time complexity Partition the nodes and reduce the edge set takes O(m) Augmenting the path takes O(min(n^1/2, |Mi+1|-|Mi| + 1) * m) In total r phases takes O(min(r*n^1/2, n+ r) * m) Replace r by C  check in each phase if no more size expantion happened, O(m) time O(min(C *n^1/2, n+C) * m)

Reduced to weighted Matching A primer-dual linear programming problem. A reduced weighting function for each edge (e) = v(a) + v(b) – w(e) For all a A, b P Main idea: Give a wide ranged weight to each node. E.g. Give nr-I for any rank = i edges

An Overview of Algorithm For i = 1, M1 can be calculated in O(n1/2m) All nodes in A and all free nodes in P is assigned 0 Matched nodes in P is assigned potential 1. Scaling: All edge scale n times Each node potential multiply by n Each node in A add 1

Lemmas for dual solution(1/2) Lemma 3.1 Mi has the same number of edges of all ranks j, j < i as Mi-1 Lemma 3.2 Let πi be the sum of potential, then n*πi-1 = n*w(Mi-1)≦w(Mi)≦ πi ≦n* πi-1+ n Lemma 3.3 Assume all arithmetic operations can be done in constant time, then a phase takes O(m*n)

Lemmas for dual solution(2/2) Lemma 3.4 If the reduces cost of an edge c(e) is more than n at the beginning of a phase, this edge can not become tight during the execution of this phase Lemma 3.5 Any edge which is not tight at the end of a phase, will never become tight!

The scaling algorithm(1/2) Initial state: π(a) = 0 for all a belong to A, π(b) = 0 for b belong to P with no match, 1 for matched nodes. Algo: For i = 2 to r do if Mi-1 is a perfect matching, then output Mi-1, else define π(v) = n*π(v) + 1 , for all v belongs to A π(v) = n*π(v) , for all v belongs to P Add edges in εi , for each new edges e, if either and of e is in the ∞ category, delete it. For any remaining edge, compute their reduced weight by setting w(e) = 1 Use primer dual algo for maximum matching in above graph, update each nodes’ potential, if π(v) > n+1,delete it

The scaling algorithm(2/2) Delete all non tight edges, and if π(v) > 2, move v to ∞ category