New PDF release: American Mathematical Monthly, volume 117, number 1, january

By Daniel J. Velleman

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Extra resources for American Mathematical Monthly, volume 117, number 1, january 2010

Example text

A society S is containment-free if no voter’s approval set contains another voter’s approval set. 3. For any society S, there is a containment-free society T such that S T. Proof. If S is containment-free, then we are done, so suppose that S is not containmentfree. Then A y ⊆ A x for some x = y, and (after rotating if necessary) the string rep46 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 117 resentation of S contains x ρ y σ y τ x , where ρ, σ , and τ are (possibly empty) strings of intervening endpoints.

Brams and P. C. Fishburn, Going from theory to practice: The mixed success of approval voting, Social Choice and Welfare 25 (2005) 457–474. 1007/s00355-005-0013-y 4. M. Chudnovsky, N. Robertson, P. Seymour, and R. Thomas, The strong perfect graph theorem, Ann. of Math. 164 (2006) 51–229. 51 5. , Progress on perfect graphs, Math. Program. Ser. B 97 (2003) 405–422. 6. V. Chv´atal. On certain polytopes associated with graphs, J. Combin. Theory Ser. B 13 (1975) 138–154. 1016/0095-8956(75)90041-6 7.

An be its voter approval sets. Let us call a set I ⊆ {1, 2, . . , n} good if |I | = d + 1 and k n m i∈I Ai = ∅. By Theorem 9 it suffices to show that there are at least d+1 d+1 d+1 good sets. We will do this by counting in two different ways the number N of all pairs (I, J ), where I ⊆ J ⊆ {1, 2, . . , n}, I is good, and |J | = m. Let g be the number of good sets. Since every good set is of size d + 1 and extends to an m-element n−d−1 n−d−1 subset of {1, 2, . . , n} in m−d−1 ways, we have N = g m−d−1 .

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American Mathematical Monthly, volume 117, number 1, january 2010 by Daniel J. Velleman

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