Halls B G Theorem E(H) L(H) R(H) Hall.1 Hall.2 Albert R - PowerPoint PPT Presentation

Hall graph H Mathematics for Computer Science MIT 6.042J/18.062J Hall’s B G Theorem E(H) L(H) R(H) Hall.1 Hall.2 Albert R Meyer. April 3, 2013 Albert R Meyer. April 3, 2013 Hall graph H Hall graph H A match is a A match is a total injective function total injective function m:G � B m:G � B g — m(g) ∈ E that follows edges Albert R Meyer. April 3, 2013 Hall.3 Albert R Meyer. April 3, 2013 Hall.4 1

Hall’s Theorem Hall graph H Hall’s condition A match is a If |S| ≤ |E(S)| for all total injective function sets of girls, S, m:G � B � then there is a match. graph(m) ⊆ E Hall.5 Hall.6 Albert R Meyer. April 3, 2013 Albert R Meyer. April 3, 2013 How to verify no bottlenecks? How to verify no bottlenecks? If every girl likes ≥ d boys, fairly efficient matching and every boy likes ≤ d girls, procedure is known then no bottlenecks. (explained in algorithms subjects) …but there is a special a degree-constrained situation that ensures a Hall graph match… Albert R Meyer. April 3, 2013 Hall.7 Albert R Meyer. April 3, 2013 Hall.8 2

How to verify no bottlenecks? How to verify no bottlenecks? If every girl likes ≥ d boys, If every girl likes ≥ d boys, and every boy likes ≤ d girls, and every boy likes ≤ d girls, then no bottlenecks. then no bottlenecks. proof: say set S of girls has e proof: incident edges: |S| ≤ e ≤ d |E(S)| d so |S| |E(S)| ≤ no bottleneck Albert R Meyer. April 3, 2013 Hall.9 Albert R Meyer. April 3, 2013 Hall.10 Proof of Hall’s Theorem Proof of Hall’s Theorem Suppose no bottlenecks. Suppose no bottlenecks. Lemma: If S a set of girls with Lemma: No bottlenecks |S| = |E(S)|, within any set S of girls. then no bottlenecks between obviously s and E(S) either Albert R Meyer. April 3, 2013 Hall.11 Albert R Meyer. April 3, 2013 Hall.12 3

s E(S) Proof of Hall’s Theorem ? bottleneck between & No bottlenecks implies T bottleneck s E(S) there is a perfect match. proof: E(S) S by strong induction on # girls then T ∪ S X is a bottleneck Hall.13 Hall.14 Albert R Meyer. April 3, 2013 Albert R Meyer. April 3, 2013 Proof of Hall’s Theorem Proof of Hall’s Theorem by induction, match Case 1: there is a nonempty (S, E(S)) proper subset S of girls with (S, E(S)) and |S|=|E(S)|. separately. by Lemmas , no bottlenecks in Hall graph (S, E(S)), and none in (S, E(S)) Albert R Meyer. April 3, 2013 Hall.15 Albert R Meyer. April 3, 2013 Hall.16 4

Hall’s Theorem Proof of Hall’s Theorem by induction, match Case 2: |S| < |E(S)| for all (S, E(S)) (S, E(S)) and nonempty proper subsets S. separately. Matchings Pick a girl, g. don’t overlap, so union is a complete matching. Hall.17 Hall.18 Albert R Meyer. April 3, 2013 Albert R Meyer. April 3, 2013 Hall’s Theorem Hall’s Theorem Case 2: |S| < |E(S)| for all Case 2: |S| < |E(S)| for all nonempty proper subsets S. nonempty proper subsets S. Pick a girl, g. She must be Match g with b. compatible with some boy, b (in fact, at least 2 boys). Albert R Meyer. April 3, 2013 Hall.19 Albert R Meyer. April 3, 2013 Hall.20 5

Hall’s Theorem Hall’s Theorem Case 2: |S| < |E(S)| for all Case 2: |S| < |E(S)| for all nonempty proper subsets S. nonempty proper subsets S. By induction, can match Match g with b. Removing b remaining girls & boys. This still leaves |S| ≤ |E(S)|, so no match along with g—b is bottlenecks. complete match. QED Hall.21 Hall.22 Albert R Meyer. April 3, 2013 Albert R Meyer. April 3, 2013 6

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