1 The Arthur-Merlin Story

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1 Comp 260: Advanced Algorithms Tufts University, Spring 2018 Prof. Lenore Cowen Scribe: Elias Marcopoulos Lecture 1a: Bipartite Perfect Matching 1 The Arthur-Merlin Story In the land ruled by the legendary King Arthur, there existed one hundred knights and one hundred tasks for the knights to complete. King Arthur proclaimed that all knights should complete exactly one task. He soon asked each of the knights to write down a list of which of the tasks they would be willing to complete, and then hand him their preferences. King Arthur took all of the knights preferences and went back to his castle. He called for his magician, the legendary Merlin, who had access to all the elves, fairies, and other mystical creatures. Arthur ordered Merlin to find a pairing of knights to tasks such that there would be a perfect matching of the one hundred knights to the one hundred tasks, so that each knight is matched to a task they find acceptable. Definition A bipartite graph is a graph consisting of two disjoint sets of nodes X and Y, and a set of edges connecting pairs of nodes (a, b), with a X and b Y. Definition A perfect matching (or perfect marriage) on a bipartite graph G = (X, Y, E) is a subset S E of the edges such that if (a, b) S then (a, c) S for any b c and (c, b) S for any c a. Soon enough, by way of magic, Merlin obtained a solution and handed the task assignment list to Arthur. Arthur, suspicious of Merlin s abilities, wanted to check that the claimed solution from Merlin was valid. To do so, he used the following algorithm: ArthursAlgorithm(): For each knight: Check that they are only matched to one task. Check that the task they are matched to is on their list. 1

2 Definition An algorithm is polynomial-time if its running time for an input of length n is O(n c ) for some constant c. ArthursAlgorithm() takes polynomial time in the number of knights and tasks, since for each knight it takes at most O(n) time to check the tasks and that they have only one task assigned to them. If there are O(n) knights, then it takes O(n 2 ) total time. Now suppose that Merlin was not so lucky. When handed the knights preference lists, he was not able to find a perfect matching despite utilizing the powers of fairies, elves, creatures, and the clouds, since no perfect matching existed that would suit the knights. Returning to Arthur, Merlin explained his situation, but Arthur was unbelieving: Merlin, you must find a way to match these knights and tasks. If you cannot find a matching, then you shall lose your head. Merlin offered to bring in all the work that his creatures had done. Each creature worked on a different possible assignment of knights to tasks and each creature found that either a knight was assigned to a task they found unacceptable, that multiple knights were assigned to the same task, or that some knights were not assigned a task. This was way too much work for Arthur to check, though! Arthur told Merlin that if he can show proof that there exists no matching to Arthur that can be verified in polynomial-time, he would not lose his head. 1.1 The Decision Problem The decision question solved by Merlin in the story is: Given a bipartite graph with n knights and n tasks does there exist a perfect matching? We can use ArthursAlgorithm() to verify a proposed solution in polynomialtime. If a given solution is verified as valid, then the graph contains a perfect matching, answering the decision problem above. Thus we have shown that bipartite perfect matching is in NP: Definition NP is the set of problems for which Merlin can present a proof to Arthur that the solution is yes, such that Arthur can verify the proof in polynomial-time. Note that the above definition is equivalent to the definition of NP given 2

3 in Comp 170 involving using nondeterministic Turing machines to solve a problem in polynomial-time. Also note that there is an asymmetry between yes instances and no instances. Using ArthursAlgorithm(), a proposed solution that is invalid does not mean there does not exist a solution, meaning the answer to the decision problem is not necessarily no. If an algorithm van verify no instances in polynomial time, then the corresponding problem is said to be in conp. So saving Merlin s head involves proving that perfect matching is in conp. For some problems, Arthur does not need Merlin: he can solve the problem himself in polynomial-time. Definition P is the set of problems Arthur can solve in polynomialtime (i.e. can do without Merlin). Arthur, who is hesitant to pay a wizard s salary and subsidize the fairies, elves, creatures, and other expenses, asks what sort of problems he can pass off to Merlin to solve and which he can still verify the solutions to in polynomial-time. Arthur sometimes dreams of not needing Merlin at all in order to keep the kingdom s amassed gold, and thinks maybe the problems which he can verify in polynomial-time he can also solve himself in polynomial-time without magic. If Arthur s dream came true, then P = N P. On the other hand, if there exists some problem that is in NP but that Arthur cannot solve on his own in polynomial-time then P NP. The P = NP? question is a famous open problem that has remained unsolved for over 50 years. All we do know is that there is a specific set of problems, called the NP-Hard problems, that if we succeed in constructing a polynomial time algorithm for one of these problems implies Arthur can solve all NP problems by himself. Definition A problem X is NP-Hard if X P P = NP. Definition A problem X is NP-Complete if BOTH X NP and X is NP-Hard. While there is no proof that P NP, we suspect that this is true. Therefore, we suspect there exists no polynomial-time solver for the NP- Hard problems, and that Arthur needs Merlin for these problems. 3

4 1.2 Saving Merlin s Head Merlin s head will be saved if he can give Arthur a proof that no perfect matching exists. Given the above definitions, we can see why Merlin will have trouble proving to Arthur that there exists no perfect matching between tasks and knights. Showing there is a perfect matching is easily verifiable, but showing there isn t a perfect matching is not necessarily simple. Merlin, in the worst case, would need to compile all of the different possible task assignments and show for each one that it does not satisfy all of the knights preferences. Since there are O(n!) ways to rearrange tasks, this would take Arthur much longer than polynomial-time to verify. Therefore Merlin will need to be clever to come up with a more efficient method of proof. Definition Co-NP is the set of problems for which Merlin can present a proof to Arthur that the solution is no, such that Arthur can verify the proof in polynomial-time. We show that such a proof is possible by presenting a characterization of when perfect matchings are possible. This characterization, called Hall s Theorem, allows the construction of proofs that Arthur can verify. Definition For any set S of vertices in G, the neighbor set of G is all vertices adjacent to vertices in S, written N G (S). Definition A vertex is called saturated if it is matched and unsaturated otherwise. Theorem (Hall s Theorem). Let G be a bipartite graph with bipartition (X, Y ). Then G contains a matching that saturates all the vertices of X if and only if N G (S) S for all S X. Proof. One direction is obvious. Clearly if there is a set of k knights who only like k-1 tasks between them, there can be no task assignment. For the other direction, we must show that if G contains no matching that assigns tasks to all the knights then there must exist a set S of knights such that S N G (S). We prove this now by contradiction. 4

5 Suppose G is a bipartite graph such that N G (S) S for all S X, but G contains no matching saturating all the vertices in X. Let M be a maximum-cardinality matching in G, and let u X be unmatched in M. Definition An alternating path is a path of edges such that exactly every other edge is contained in M. Definition An augmenting path is an alternating path in which the first and last edges are unmatched. The matching size can be increased by one by swapping all the edges of an augmenting path from matched to unmatched and vice versa. Let S X be the set of all vertices reached by alternating paths from u. Let T = N G (S). Claim M matches T perfectly with S \ {u}. The alternating paths reach Y along edges not in M and reach u along the edges in M. In other words, every vertex of S \ {u} is reached along an edge in M from a vertex in T, so S \ {u} is all matched. But T is matched too, because if there was an unsaturated vertex in T, we would have an alternating path from u to the vertex, contradicting the maximality. When the number of knights is equal to the number of tasks, as in our story, then a matching that matches all the knights must automatically match all the tasks. Thus we get the following corollary to Hall s theorem: Corollary (Hall s Theorem: Perfect matching). Let G be a bipartite graph with bipartition (X, Y ) and X = Y. Then G contains a perfect matching if and only if N G (S) S for all S X. Because Hall s Theorem contains states only if, it implies that if no perfect matching exists for n knights and n tasks then Merlin can exhibit a set of k knights that between them only like j knights where j < k n. This set of k knights can be shown to only like j tasks in polynomial-time, and thus Arthur can verify Merlin s no. This means that Bipartite Perfect Matching P conp, and so Merlin s head is saved. It turns out, however, that Bipartite Perfect Matching is in P, which is not too hard to show. This means that to match knights to tasks, Arthur doesn t need Merlin s magic at all. We present the algorithm below: 5

6 Theorem There exists a polynomial time algorithm to find a perfect matching. Proof. By construction. We present an algorithm in this section which is based on the proof of Hall s Theorem. The Augmenting Path Algorithm Input: A bipartite graph G = (X, Y, E) with a partition of X (knights) and Y (tasks), and a matching M with k edges in G. Output: A matching M with at least one more edge than M or a proof that G violates Hall s condition [?]. The Algorithm: Let u U be any unmatched vertex in X. Grow a BFS tree from u U. Consider only unmatched edges from X to Y and only matched edges from Y to X. If an augmenting path is found, create a new matching M by swapping the matched/unmatched edges along this path and output M. If no augmenting path is found, then output S, N G (S). Since no augmenting path is found, T = S \ {u} and thus N G (S) < S. Time complexity: In each step we add one more edge. We need n edges for the final matching and therefore there are n steps. In each step a BFS tree is created. Each vertex and edge is visited once. Therefore the total time complexity is O(n(n + n 2 )) = O(n 3 ). In fact, since then new algorithms have been discovered that are more efficient. We may cover these later in the semester. So Merlin s head is saved! And not only is he saved, but he also can tell Arthur to figure out the task assignments himself next time, as doing so doesn t require any magic at all. In fact, there are quite a number of variations on the perfect matching problems. We considered only the case that we have tasks who are okay with some tasks and not with others. Imagine we gave weightings to these preferences. Imagine some tasks are better suited for certain knights, so that they have a preference for certain knights making the problem go both ways. These additions change the problem in a significant way, and have many 6

7 applications in real life such as giving puppies and pet owners a mutually acceptable arrangement. In the next section we will consider a different kind of matching for hospitals and residents. References [1] P. Hall, On representations of distinct subsets Journal of the London Math. Society, Vol 10(26-30),