Systems of distinct representations

If A is a subset of the vertices of a graph, then define Consider a bipartite graph G(X∪Y, E). A matching in G is a subset E1⊆E such that no vertex is incident with more than one edge in E1. X Y

Thm 5.1. (Hall’s marriage theorem) Philip Hall, 1904 - 1982 A complete matching (perfect matching) from X to Y is a matching such that every vertex in X is incident with an edge in E1. Thm 5.1. (Hall’s marriage theorem) G(X∪Y, E) has a complete matching iff for every A⊆X. Pf: “ ” : clear. “ ” : Assume that |(A)| ≥ |A| for every A⊆X. Let |X|=n, m<n, and M be a matching of size m.

x0 Want to show a larger matching exists. Call the edge of M red and all other edges blue. Let x0∈X be a vertex not incident with an edge in M. Claim there exists a simple path (of odd length) starting with x0 and a blue edge, using red and blue alternately and terminating with a blue edge and a vertex y not incident with any red edge.

x1 x2 x0 If we find such a path P, we are done because we obtain a matching with m+1 edges by deleting the red edges of P from M and adding the blue edges of P. y1 y2 y3

Since |({x0})| ≥ 1, there is a vertex y1 adjacent to x0. If y1 is not incident with any red edge, then we’re done. If y1 is incident with an red edge, let x1 be the other end of the red edge.

If x0, x1,…, xk and y1,…, yk have been defined, then since there exist a vertex yk+1, distinct from the previous yi’s, that is adjacent to at least one vertex in {x0, x1,…, xk}. If yk+1 is not incident with a red edge; stop; o/w, let xk+1 be the other end of that red edge.

When the procedure stops, we construct the path P starting with yk+1 and the blue edge joining it to, say , i1<k+1. Then add the red edge . By construction is joined by a blue edge to some , i2<i1. Then add the red edge : continue till x0 is reached. ▧

There are 52*51*50*49*48/5! possible choices of 5 cards and there are 52*51*50*49 ordered 4 distinct cards. Each ordered 4-tuple relates to 48 unordered 5-tuples and each unordered 5-tuple relates to 5*4! ordered 4-tuples. Then by Problem 5A. E.g. Consider a standard deck of 52 cards. You’re dealt five cards at random. You keep one and put the other 4 (in a specific order) into an envelope which is taken to your partner in another room, who looks at these and announces the name of the fifth card. How? (c1, c2, c3, c4, c5)↔ (c1, c2, c3, c4)

Def. Property H (Hall’s condition): Consider subsets A0, A1,…, An-1 of a finite set S. We say Ai’s have property H if (for all k) the union of any k-tuple of subsets Ai has at least k elements. If it is exactly k, then we call this k-tuple a critical block.

Def. SDR (System of distinct representation) of the sets A0,…, An-1 to be a sequence of n distinct elements a0,…, an-1, with ai∈Ai, 0 ≤ i ≤ n-1.

Def. Let m0 ≤ m1 ≤ … ≤ mn-1. Define where and

Lemma 5.2. For n ≥ 1, let fn: Zn → N be defined by mi’s are sorted non-decreasingly from ai’s. Then fn is non-decreasing with respect to each of ai.

Pf: then Let m0≤…≤mk-1≤mk=ai≤mk+1≤…≤ml≤ml+1≤…≤mn-1 If ai’ ≥ ai and m0≤...≤mk-1≤mk+1≤…≤ml≤ai’ ≤ml+1≤…≤mn-1 then ▧

Denote N(A0,…, An-1) the number of SDRs of (A0,…, An-1). Thm 5.3. Let (A0,…, An-1) be a sequence of subsets of S. Let mi = |Ai| (i=0,…,n-1) and let m0 ≤ m1 ≤…≤ mn-1. If the sequence has property H, then N(A0,…, An-1) ≥ Fn(m0,…, mn-1).

Pf: Prove by induction. It is clear for n=1. case1: No critical block. Choose any element a of A0 as its representative and then remove a from all the other sets. This yields sets, A1(a),…, An-1(a), which still hold H property.

By ind. hypothesis, we have

Case2: There is a critical block with v0 <...< vk-1 and 0 < k < n. In this case, delete all elements of from all the other sets Ai which produces where

By ind. Hypothesis, we have Note that . Therefore

Thus, The product of the above two terms is equal to Fn(m0,…, mn-1). ▧

Dénes Kőnig, (1884 –1944), a Hungarian mathematician Thm 5.4. (König’s theorem) Let A be a 0-1 matrix. The minimum number of lines (row or column) of A that contain all the 1’s of A is equal to the maximum number of 1’s in A, no two on a line.

Pf: m: the minimum number of lines of A containing all the 1’s of A. M: the maximum number of 1’s , no two on a line. Clearly, m ≥ M.

If some k-tuple of the Ai’s had < k elements, then we Let the min covering by lines consist of r rows and s columns (r+s=m). WLOG, assume they’re the first r rows and s columns. Define sets Ai, 1 ≤ i ≤ r, by Ai={j>s : aij=1}. If some k-tuple of the Ai’s had < k elements, then we could replace the corresponding k rows by k-1 columns. s r ……..... ..1..1..

Since, by assumption, this is impossible. Thus, Ai’s satisfy property H. So Ai’s have an SDR. This means there are r 1’s, no two on a line, in the first r rows and not in the first s columns. By the same argument, there are s 1’s, no two on a line, in the first s columns and not in the first r rows. This shows that M ≥ r+s =m. ▧

Garrett Birkhoff 1911 - 1996 Thm 5.5 (Birkhoff) Let A=(aij) be an n x n matrix with nonnegative integers as entries, such that every row and column of A has sum L. Then A is the sum of L permutation matrices.

Let Ai={j : aij>0}, i=1…n. Pf: Let Ai={j : aij>0}, i=1…n. For any k-tuple of the Ai’s, the sum of the corresponding rows of A is kL. The non-zero entries in the chosen k rows must be in at least k columns. Hence the Ai’s satisfy property H. An SDR of the Ai’s corresponds to a permutation matrix P=(pij) such that aij>0 if pij=1. The theorem follows by induction (on what?). ▧