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### Matching

1. 1. HALL’S MATCHING THEOREM 1. Perfect Matching in Bipartite Graphs A bipartite graph is a graph G = (V, E) whose vertex set V may be partitioned into two disjoint set VI , VO in such a way that every edge e ∈ E has one endpoint in VI and one endpoint in VO . The sets VI and VO in this partition will be referred to as the input set and the output set, respectively. Deﬁne a perfect matching in a bipartite graph G to be an injective mapping f : VI → VO such that for every x ∈ VI there is an edge e ∈ E with endpoints x and f (x). For any subset A ⊂ VI , deﬁne ∂A to be the set of all vertices y ∈ VO that are endpoints of edges with one endpoint in A. Theorem 1. (Hall’s Matching Theorem) Let G be a bipartite graph with input set VI , output set VO , and edge set E. There exists a perfect matching f : VI → VO if and only if for every subset A ⊂ VI , (1) |∂A| ≥ |A|. Proof. By induction on the cardinality of VI . If |VI | = 1 the result is trivially true. Suppose, then, that the result is true if |VI | ≤ n, and consider a bipartite graph G whose input set VI has cardinality n + 1. There are two possibilities: either (1) for every proper subset A ⊂ VI , the cardinality of ∂A is at least one greater than the cardinality of A; or (2) there exists a proper subset A ⊂ VI such that |∂A| = |A|. Case 1: Choose any x ∈ VI and any y ∈ ∂{x} (by hypothesis, ∂{x} has at least one ∗ element). Let G∗ be the bipartite graph with input set VI∗ = VI − {x}, output set VO = VO − {y}, and whose edges are the same as those of G, but with edges incident to either x or y deleted. The bipartite graph G∗ satisﬁes the hypothesis (1), because in Case 1 every proper subset A ⊂ VI has |∂A| ≥ |A| + 1, so deleting the single vertex y from ∂A still leaves at least |A| vertices. By the induction hypothesis, there is a perfect matching in G∗ ; this perfect matching extends to a perfect matching in the original graph G by setting f (x) = y. Case 2: Let A ⊂ VI be a proper subset of VI such that |∂A| = |A|. Construct bipartite ∗ graphs G∗ and G∗∗ with input sets VI∗ = A and VI∗∗ = VI − A, output sets VO = ∂A and ∗∗ = V − ∂A, and edges inherited from the original graph G. We shall use the induction VO O hypothesis to show that there is a perfect matching in each of the bipartite graphs G∗ and G∗∗ . If this is so, then a perfect matching in G may be obtained by taking the joins of the perfect matchings in G∗ and G∗∗ . Observe that the vertex set VI∗ and VI∗∗ have cardinalities no greater than n, because A = VI∗ is a proper subset of VI . Thus, the induction hypothesis will guarantee the existence of perfect matchings in G∗ and G∗∗ provided it is shown that the hypothesis (1) is satisﬁed for each of these graphs. Consider ﬁrst the graph G∗ : For any subset B ⊆ A = VI∗ , the boundary ∂ ∗ B in the graph G∗ coincides with ∂B in G. Consequently, G∗ satisﬁes (1). Now consider G∗∗ : If there were a subset B ⊆ VI∗∗ = VI − A whose boundary ∂ ∗∗ B in the graph 1
2. 2. 2 HALL’S MATCHING THEOREM G∗∗ had fewer than |B| elements, then in the graph G the boundary ∂(B ∪ A) would have at most |B ∪ A| − 1 elements, because ∂(B ∪ A) = ∂ ∗∗ B ∪ ∂A. This is impossible, because the graph G satisﬁes (1). Hence, G∗∗ also satisﬁes (1). 2. The Birkhoff-von Neumann Theorem A doubly stochastic matrix is a square matrix with nonnegative entries whose row sums and column sums are all 1. A magic square is a square matrix with nonnegative integer entries whose row sums and column sums are all equal; the common value of the row sums and column sums is called the weight of the square. Observe that if T is a magic square of weight d ≥ 1, then one obtains a doubly stochastic matrix by dividing all entries of T by d. Conversely, if P is a doubly stochastic matrix with rational entries, then one may obtain a magic square by multiplying all entries by their least common denominator. The magic squares of weight 1 are called permutation matrices: for any m × m permutation matrix T , there exists a permutation σ of the set [m] such that (2) Ti,σ(i) = 1 Ti,j = 0 for all i ∈ [m], and if j = σ(i). Theorem 2. Every doubly stochastic matrix is a convex combination (weighted average) of permutation matrices. Every magic square of weight d is the sum of d (not necessarily distinct) permutation matrices. Proof. We shall consider only the assertion about magic squares; the assertion about doubly stochastic matrices may be proved by similar arguments.) By deﬁnition, every magic square of weight 1 is a permutation matrix. Let T be an m × m magic square of weight d > 1. Consider the bipartite graph with VI = VO = [m] such that, for any pair (i, j) ∈ VI × VO , there is an edge from i to j if and only if Ti,j > 0. Claim: The hypothesis (1) of the Matching Theorem is satisﬁed. Proof. Let B be a subset of VI with r ≤ m elements. Since T is a magic square of weight d, the sum of all entries Ti,j such that i ∈ B must be rd. The positive entries among these must all lie in the columns indexed by elements of ∂B; consequently, the sum of the entries Ti,j such that j ∈ ∂B must be at least rd. But this sum cannot exceed d|∂B|, since the column sums of T are all d. The Matching Theorem now implies that there is a perfect matching in the bipartite graph. Since VI = VO = [m], this perfect matching must be a permutation σ of the set [m]. By construction, the permutation matrix T σ deﬁned by equations (2) is dominated (entry by entry) by the magic square T , so the diﬀerence T − T σ is a magic square of weight d − 1. Thus, the assertion follows by induction on d.
3. 3. HALL’S MATCHING THEOREM 3 3. Strassen’s Monotone Coupling Theorem A poset is a partially ordered set (X , ≤). Recall that a partial order ≤ must satisfy the following properties: for all x, y, z ∈ X , (3) x ≤ x; (4) x ≤ y & y ≤ x =⇒ x = y; (5) x ≤ y & y ≤ z =⇒ x ≤ z. Posets occur frequently as state spaces in statistical mechanics and elsewhere. An important example is the conﬁguration space Σ = {0, 1}V of a spin system: here V is a set of sites, often the vertices of a lattice, and the elements of Σ are assignments of zeros and ones (“spins”) to the sites (“conﬁgurations”). The partial order ≤ is deﬁned as follows: x≤y iﬀ xs ≤ ys ∀ s ∈ V. 1 An ideal of a poset (X , ≤) is a subset J ⊂ X with the property that if x ∈ J and x ≤ y then y ∈ J . If µ and ν are two probability distributions on X , say that ν stochastically dominates µ (and write µ ≤ ν) if for every ideal J , (6) µ(J ) ≤ ν(J ). Theorem 3. (Strassen) Let (X , ≤) be a ﬁnite poset, and let µ, ν be probability distributions on X . If µ ≤ ν then on some probability space (in fact, on any probability space supporting a random variable uniformly distributed on the unit interval) are deﬁned X −valued random variables M, N with distributions µ, ν, resepectively, such that (7) M ≤ N. Proof. We shall only consider the case where the probability distributions µ, ν assign rational probabilities k/N (with a common denominator N ) to the elements of the poset X . The general case may be deduced from this by an approximation argument, which the reader will supply (Exercise!). Case A: For every x ∈ X , the probabilities µ(x) and ν(x) are either 0 or 1/N . Consider the bipartite graph with VI = {x ∈ X : µ(x) = 1/N } and VI = {x ∈ X : ν(x) = 1/N }, where x ∈ VI and y ∈ VO are connected by an edge if and only if x ≤ y. The hypothesis that µ is stochastically dominated by ν implies that the hypothesis (1) of the Matching Theorem is satisﬁed. Consequently, there is a perfect matching f : VI → VO . Let M be an X −valued random variable M with distribution µ (such a random variable will exist on any probability space supporting a uniform-[0,1] random variable). Deﬁne N = f (M ). Then the pair (M, N ) satisﬁes M ≤ N , and the marginal distributions of M and N are µ and ν, as the reader will easily check. Case B: For every x ∈ X , the probabilities µ(x) and ν(x) are integer multiples of 1/N . For each x ∈ X , if µ(x) = k/N then construct k “copies” x1 , x2 , . . . , xk of x, and for each such copy set πI (xi ) = x. Deﬁne VI to be the set of all such copies, where x ranges over X . Similarly, for each y ∈ X , if ν(y) = m/N then construct m copies y1 , y2 , . . . , ym of y, 1sometimes called an upper corner
4. 4. 4 HALL’S MATCHING THEOREM and for each such copy set πO (yi ) = y. Deﬁne VO to be the set of all such copies, where y ranges over X . For each pair xi ∈ VI and yj ∈ VO , put an edge from xi to yj if and only if πI (xi ) ≤ πO (yj ). Once again, it is easily veriﬁed that hypothesis (1) of the Matching Theorem is satisﬁed, since µ ≤ ν. Consequently, there is a perfect matching f : VI → VO . Let U be a random variable that is uniformly distributed on VI – such a random variable exists on any probability space supporting a Uniform-[0,1] random variable. Set M = πI (U ) and N = πO (f (U )); then the marginal distributions of M and N are µ and ν, respectively, and M ≤ N , by construction. 4. Enumeration of 3 × 3 Magic Squares The enumeration of magic squares is a classical problem in combinatorics, dating to McMahon in the early 20th century (or earlier). McMahon obtained an explicit formula for the number h(n) of 3 × 3 magic squares of weight n: (8) h(n) = 3 n+3 n+2 + 4 2 Much later Richard Stanley proved that for every m ≥ 2, the number of m × m magic squares of weight n is a polynomial function of n, and that the degree of the polynomial is (m − 1)2 . This implies that the function is determined by (m − 1)2 values, by the Lagrange interpolation formula. In this section we shall outline a proof of McMahon’s formula; in the homework exercises an appproach to Stanley’s theorem will be outlined. The Birkhoﬀ-von Neumann Theorem is the key to the enumeration of magic squares. This theorem asserts that every magic square R of weight d is the sum of d permutation matrices. The 3 × 3 permutation matrices are       1 0 0 0 1 0 0 0 1 I = 0 1 0 , S = 0 0 1 , S 2 = 1 0 0 , 0 0 1 1 0 0 0 1 0       0 1 0 0 0 1 1 0 0 Tab = 1 0 0 , Tac = 0 1 0 , Tbc = 0 0 1 . 0 0 1 1 0 0 0 1 0 These 6 matrices satisfy the relation (9) I + S + S 2 = Tab + Tac + Tbc . Proposition 4. Every 3 × 3 magic square R has a unique representation as (10) R = m1 I + m2 S + m3 S 2 + m4 Tab + m5 Tac + m6 Tbc where mi are nonnegative integers whose sum is the weight of the square and are such that at least one of m1 , m2 , m3 is 0. Proof. That there is such a representation for every 3 × 3 magic square follows from the Birkhoﬀ-von Neumann theorem and the relation (9). Suppose that some magic square R had two such representations m = (m1 , . . . , m6 ) and n = (n1 , . . . , n6 ). By adding multiples
5. 5. HALL’S MATCHING THEOREM 5 of either I + S + S 2 or Tab + Tac + Tbc to the relations m and n, respectively, one may obtain a relation p1 I + p2 S + p3 S 2 + p4 Tab + p5 Tac + p6 Tbc =q1 I + q2 S + q3 S 2 + q4 Tab + q5 Tac + q6 Tbc where pi − qi = 0 for at least one of i = 1, 2, or 3 but not all of the diﬀerences pj − qj are 0. Suppose, for instance, that this is the case with pi = qi with i = 2. Then r1 I + r3 S 2 = −r4 Tab − r5 Tac − r6 Tbc where ri = pi − qi . But the matrix on the left  ∗ 0 ∗ ∗ 0 ∗ side has the form  ∗ 0 . ∗ This implies that r4 = r5 = r6 = 0, and hence that r1 = r3 = 0. This contradicts the hypothesis that there are two distinct representations. Using the uniqueness of the rpresentation (10), we may obtain an explicit formula for the generating function H(z) := h(n)z n of the sequence h(n). Observe that the sum n for each square of weight n. By Proposition 4, each such square deﬁning H(z) counts z has a unique representation (10), and the weight n is the sum of the six integers mi in the representation. Consequently, ∞ (11) h(n)z n = H(z) := z m1 +m2 +m3 +m4 +m5 +m6 m n=0 where the sum m is over all six-tuples m of nonnegative integers such that at least one of the entries m1 , m2 , m3 is 0. Deﬁne A1 , A2 , A3 , A12 , A13 , A23 , A123 to be the sets of all six-tuples m of nonnegative integers such that (a) mi = 0 for Ai ; (b)mi = mj = 0 for Aij ; and (c)m1 = m2 = m3 = 0 for A123 . Then by Inclusion-Exclusion, H(z) = (12) + + A1 − A2 − A12 + A3 − A13 A23 . A123 Now each of the seven sums in the last expression is a product of geometric series: for instance, 3 ∞ (13) z A123 m1 +m2 +m3 +m4 +m5 +m6 z = m4 ,m5 ,m6 ≥0 m4 +m5 +m6 = z m=0 m = (1 − z)−3 .
6. 6. 6 HALL’S MATCHING THEOREM Thus, (14) H(z) = 3(1 − z)−5 − 3(1 − z)−4 + (1 − z)−3 . McMahon’s result now follows by expanding each of these terms in a Taylor series and gathering terms.