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

2. 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

3. 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

4. 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

5. 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!

6. 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

7. 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

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

9. 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

10. 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’

11. 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

12. 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:

13. 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

14. 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)

15. 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

16. 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

17. 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)

18. 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!

19. 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

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