Lecture 15: Matching Problems. (Reading: AM&O Chapter 12)

Transcription

1 Lecture 15: Matching Problems (Reading: AM& Chapter 12)

2 The General Matching Problem Given: Undirected graph G = (N, A), with costs c ij on each arc (i, j) of G. Find: A matching (set of disjoint arcs) in G having minimum weight.

3 Special Cases bipartite matching: G is a bipartite graph, that is, N = S T, with S and T disjoint and all arcs having one end in S and one end in T. cardinality matching: all costs are 1, that is, we want the maximum number of arcs in the matching perfect matching: we insist that every node be incident to a matching arc, that is, the matching should have cardinality exactly n/2 Note: We can transform any min cost perfect matching problem to a min cost partial matching problem by subtracting a large number M (e.g. sum of the c ij s) from each of the arc weights. (Why?)

4 Applications Assignment Problem: A firm has hired n graduates to fill m n vacant positions (positions may remain unfilled, but all graduates must be assigned). Based on qualifications, the firm has assigned a profitability index p ij representing the expected yearly profit for the company of assigning graduate i to job j. How can the firm most profitably assign the n graduates to these m positions? Nonbipartite Assignments: A college wishes to pair incoming freshmen for rooming assignments. ach pair of freshment is given a compatibility measure C ij, based on home town, background, and common interests (and gender). The college wants to pair these students in such a way as to maximize total compatibility.

5 ptimal Depletion of Inventory: A set of p items is stored in a stockpile. These items are perishable, in that the value of the item decreases as the item gets older. A degradation function v(a) measures how valuable an item of age a is. ach item i is currently of age a i. ne item will be needed at each of the times t 1,..., t p, and it is desired to determine which item should be chosen at each of these time periods in order to maximize total value of the items at time of use.

6 Scheduling on Parallel Machines: A set of w jobs must be assigned for processing on one of r machines. ach job i has a processing time p ij on machine j. Any job can be assigned to any machine, and the machines work simultaneously, although each can only work on one job at a time until that job is finished (no preempting). The flow time of a job is the total time the job is in the system before it processed, that is, the total time for jobs scheduled before it to finish, plus the time to process the job itself. It is desired to schedule the w jobs on the r machines in such a way as to minimize the total of all flow times for the jobs.

7 number from last M 11 M 12 1p 11 p 2 11 machine 1 J1 wp 11 1p 1r 2p 1r wp 1r M 1w jobs 1p w1 2p w1 wp w1 1p wr M r1 Jw 2 p wr M r2 wp wr machine r M rw Matching Problem for Parallel Scheduling

8 Cardinality Matchings Suppose that we are solving the cardinality matching on a general graph, and have a current matching M. An augmenting path with respect to M is a path P a : v 0, (v 0, v 1 ), v 1, (v 1, v 2 ),..., v 2k, (v 2k, v 2k+1 ), v 2k+1 where v 0 and v 2k+1 are exposed, that is, incident to no arc of M (v 0, v 1 ), (v 2, v 3 ),..., (v 2k, v 2k+1 ) are not in M (v 1, v 2 ), (v 3, v 4 ),..., (v 2k 1, v 2k ) are in M v v v v v v k-1 2k 2k+1 exposed An Augmenting Path v exposed

9 Augmenting Path Theorem Theorem (Berge): A matching M is optimal in G if and only if there is no augmenting path in G with respect to M. Proof: Let M be a matching in G. Suppose first that there exists an augmenting path P a in G with respect to M. Then by reversing the status of each arc in this path as to being in and out of M, we obtain a new matching M in G having one more arc than M, so that M is not optimal. Conversely, suppose there exists matching M of larger cardinality than M. Consider the symmetric difference M M = (M M ) \ (M M ) of M and M. This will comprise a node disjoint union of cycles and paths, each consisting of alternating elements of M and M (why?). Since M > M, then there must be at least one component of this union that has more arcs of M than M in it. This in turn must be an augmenting path.

10 Bipartite Cardinality Matching Algorithm The Bipartite Matching Algorithm starts with the empty matching, and then attempts to find an augmenting path with respect to the current matching, until there is no further augmenting path to be found. From the Augmenting Path Theorem the current matching is then maximum cardinality.

11 Bipartite Matching Algorithm Initialize: M = while an augmenting path is found do mark all nodes unlabeled, unscanned while there is at least one unscanned exposed node do choose unscanned exposed node i 0. Give i 0 the label even, set pred(i) = 0, i V while there are labeled unscanned nodes do choose labeled unscanned node i if i has label odd then if i is exposed then An augmenting path has been found. Identify augmenting path P a, using pred, and augment M by setting M = M P a. Mark all nodes scanned. else Let (i, j) be the (unique) matching arc adjacent to i. If j is unlabeled then give j label even and set pred(j) = i end if else (i has label even ) for each nonmatching arc (i, j) adjacent to i with j unlabeled do give j label odd, set pred(j) = i end if mark i scanned end while end while end while

12 xample Matching Problem with Nonoptimal Matching 8

13 xample (cont d.) Augmenting Path Found

14 xample (cont d.) Min Cardinality Cover Found

15 Correctness of Algorithm This follows from the fact that this is mimicking the Max Flow Algorithm. In fact, we have the following proof of the König-gervary Graph Theorem. Lemma: Suppose no augmenting path can be found with respect to the matching M at the end of a labeling stage in the Bipartite Matching Algorithm performed on connected bipartite graph G. Assign to any remaining unlabeled nodes an even label if it is in S and an odd label if it is in T. Then the set of odd labeled nodes comprise a node cover of G of the same cardinality as M. Proof: Note that all of the unlabeled nodes remaining after the algorithm ends have to be adjacent to a matching arc. But this means that every odd labeled node must be adjacent to a matching arc, since otherwise an augmenting path would be found. No two even labeled nodes are adjacent, otherwise G would not be bipartite. Therefore the set of odd labeled nodes is the same cardinality as M and cover all arcs of G.

16 Complexity ach labeling stage is simply a graph search, and takes (m) time. Since each nonterminal labeling stage finds an augmenting path and the maximum number of augmentations is n/2, then there are at most n/2 graph search stages. Thus the total complexity is (nm). Best Improvement: Use Dinic s Algorithm (modified to fit the matching problem) to augment along all shortest augmenting paths at a time. Complexity: (m n).

17 Consequences of the König-gervary Graph Theorem König-gervary Graph Theorem: In a bipartite graph, the maximum cardinality of a matching is equal to the minimum cardinality of a node cover. Perfect Matching Version: Let G be be a bipartite graph with S = T = k. Then there is a perfect matching in G if and only if there is no node cover of G of size less than k. Neighborhood of a set V of nodes: The set Γ(V ) of all nodes that are adjacent to at least one node in V. Hall s Theorem: Let G be be a bipartite graph with S = T = k. Then G has a perfect matching if and only if there is no set S 0 S such that Γ(V ) < V. Proof: xercise.

18 Cardinality Matchings in General Graphs Problem with Bipartite Matching Algorithm: In general graphs the algorithm may not find an augmenting path when in fact there is one. This is always the result of finding a... blossom: the cycle in the search tree encountered when either an odd labeled node is encountered from an odd labeled node, or an even labeled node is encountered from an even labeled node. A Blossom

19 Solution: whenever an even-even or odd-odd arc is encountered in the bipartite search procedure, then a blossom has been found. Shrink the blossom, that is, contract all of its arcs to a single node. Now proceed as if the node were a part of the search tree (with an even label). A Blossom

20 General Matching Algorithm Initialize: M = while an augmenting path is found do mark all nodes unlabeled, unscanned while there is at least one unscanned exposed node do choose unscanned exposed node i 0. Give i 0 the label even and set pred(i) = 0, i V while there are labeled unscanned nodes do choose labeled unscanned node i if i has label odd then if i is exposed then An augmenting path has been found. Identify augmenting path P a, using pred, expanding blossoms where necessary, and augment M by setting M = M P a. Mark all nodes scanned. else Let (i, j) be the (unique) matching arc adjacent to i. if j is unlabeled then give j label even and set pred(j) = i else if j has label odd then A blossom b has been detected. Contract the blossom, giving it label even. end if end if else (i has label even ) for each nonmatching arc (i, j) adjacent to i do if j is unlabeled then give j label odd and set pred(j) = i else if j has label even then A blossom b has been detected. Contract the blossom, giving it label even. end if end if mark i scanned end while end while end while

21 xample A nonoptimal matching

22 xample (cont d.) First labeling pass 12

23 xample (cont d.) First blossom found 12

24 xample (cont d.) Second blossom found

25 xample (cont d.) Third blossom found

26 xample (cont d.) xposed node found

27 xample (cont d.) Augmenting path

28 xample (cont d.) ptimal matching (exercise)

29 Implementation Details contracting a blossom: If a blossom b has been detected, proceed as follows: search back through the tree to identify the blossom. form a doubly-linked list of blossom nodes that traverse the blossom clockwise and counterclockwise. update the forward stars of b and all adjacent nodes to reflect the contraction of the blossom. set pred(b)=pred(root of blossom). expanding a blossom: When a blossom label is encountered, reconstruct the blossom, and traverse the blossom in the direction which starts with the matching arc. Recursively expand any nodes that continue to be blossoms.

30 Complexity number of augmentations: at most n/2 search procedure (ignoring contractions): (m) number of contractions: At most n/2 (blossoms have at least 3 nodes) complexity of each contraction: the bottleneck is updating the adjacency lists, which takes (n) per node, for a total of (n 2 ) Total complexity: (n 4 )

31 Improving complexity to (n 3 ): The adjacency list updating can be improved to (n 2 ) over all contractions by using the following technique for updating the adjacency list of a blossom: 1. Declare all nodes unmarked 2. For each (pseudo)node in the blossom, mark all of its adjacent nodes 3. After all blossom nodes are processed, scan all nodes and place them in the forward star of the new pseudonode. Now ach (pseudo)node is processed at most once in the entire algorithm, since once it is processed it is replaced by the pseudonode it is a part of. There are at most 3n/2 nodes and pseudonodes appearing in the entire algorithm. The cost of processing a blossom is order n (the number of nodes in the blossom) + n (for the final pass). Thus the total time to update the adjacency lists is (n 2 ), for a total of (n 3 ) for the whole algorithm.

32 Weighted Matchings bipartite case: this can be solved as a min cost flow problem by attaching source and sink nodes as for the cardinality matching problem with cost 0, and arc costs on the original arcs as given for the matching problem. complexity: Using the Successive Shortest Path Method (and noting that there are at most n/2 flow augmentations) we get complexity (ns(n, m, C)), where S(n, m, C) is the complexity of the best shortest path algorithm using nonnegative costs. nonbipartite case: Can be solved using a primaldual algorithm on a more involved matching LP. (We will not do this here.) Best (dataindependent) time is (n(m + n log n)).