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On Voting Caterpillars

Felix Fischer, Ariel D. Procaccia and Alex Samorodnitsky. On Voting Caterpillars. Tournaments. A = {1,...,m}: set of candidates. A tournament is a complete and asymmetric relation T on A. T (A) set of tournaments. Related to voting. The Copeland score of i is its outdegree .

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On Voting Caterpillars

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  1. Felix Fischer, Ariel D. Procaccia and Alex Samorodnitsky On Voting Caterpillars

  2. Tournaments • A = {1,...,m}: set of candidates. • A tournament is a complete and asymmetric relation T on A. T(A) set of tournaments. • Related to voting. • The Copeland score of i is its outdegree. • Copeland Winner: max Copeland score. 1 2 3 4 5 6

  3. Voting trees 1 2 ? 3 ? ? 3 1 3 ? 2 1

  4. Implementation by Voting trees • A candidate can appear multiple times in leaves of tree, or not appear (not surjective!). • Which functions f:T(A)A can be implemented by voting trees? Many papers (since the 1960’s) but no characterization. • [Moulin 86] Copeland cannot be implemented when m  8. • [Srivastava and Trick 96] ... but can be implemented when m  7. • Can Copeland be approximated by trees?

  5. The two models • Deterministic model: a voting tree  has an -approx ratio if T, (s(T) / maxisi )  . • Randomized model: • Randomizations over voting trees. • Randomization is admissible if its support contains only surjective trees. • Dist.  over trees has an -approx ratio if T, (E[s(T)] / maxisi )  .

  6. Components • C  A is a component of T if  i,jC, kC, iTkjTk. • Lemma [Moulin 86]: T and T’ differ only inside a component C,  a voting tree. • (T)C  (T’)C. • (T)A\C(T)=(T’). 1 2 3 4 5 6

  7. ¾ deterministic upper bound • m = 3k, k odd. • T is 3 cycle of regular components. • In T, isi = k + (k-1)/2. • One component in T’ is transitive. • In T’ is.t. si = k + (k-1), winner doesn’t change. • The ratio tends to ¾. ’ T k = 5

  8. 5/6 randomized upper bound • On board.

  9. Randomized lower bound • Theorem 4.1.  admissible randomization over voting trees of polynomial size with an approximation ratio of ½-O(1/m). • Inadmissible randomization that achieves ratio ½ is trivial. • Important to keep the trees small.

  10. Spot the fake caterpillar • 1-Caterpillar is a singleton tree. • k-Caterpillar is a binary tree where left child of root is (k-1)-caterpillar, and right child is a leaf. • Voting k-caterpillar is a k-caterpillar whose leaves are labeled by A. ? ? ? 1 ? 5 ? 4 3 4 2

  11. Randomized voting caterpillars • k-RSC: uniform distribution over surjective voting k-caterpillars. • Lemma 4.2. Let T, pi(k) the prob. of i winning in k-RSC. Then  k=k(m,) polynomial such that i pi(k)si (m-1)/2-. • Theorem follows with =1 since si m-1.

  12. k-RC • k-RC: assign alternatives independently and uniformly to leaves of k-caterpillar. • k-RC is inadmissible. • Lemma 4.3. Let T, pi(k) prob. of i in k-RSC, qi(k) prob. of i in k-RC. Then |pi(k) - qi(k)|  m / ek-m.

  13. k-RC and Markov chains • Define M = M(T). •  = A. • Initial distribution is uniform. • P(i,j) = • i=j: (si+1)/m • jTi: 1/m • iTj: 0 • (k) = q(k+1) 2 3 3 2 1 3 3 1 5 1 5 2 4 3

  14. Extra sketchy proof of Lem. 4.2 • Lemma 4.2. Let T, pi(k) the prob. of i winning in k-RSC. Then  k=k(m,) polynomial such that i pi(k)si (m-1)/2-. • Lemma 4.3. Let T, pi(k) prob. of i in k-RSC, qi(k) prob. of i in k-RC. Then |pi(k) - qi(k)|  m / ek-m. • Lemma 4.4. Let T. M(T) converges to a unique stationary distribution . • i = i(si+1)/m + (j: iTj j) / m • Lemma 4.5. Let T. i isi  (m-1)/2. • Lemma 4.6. Let T.M(T) is rapidly mixing. • Proof of Lemma 4.2 follows (on board).

  15. Stability of caterpillars • Example with ratio ½ + o(1). Consists of an almost regular tournament. • Theorem 4.7. Let , ’ = (4)1/4. If i isi = (m-1)/2 + m, then #{i: |si-m/2| > 3’m/2}  ’m.

  16. Second order Copeland • Second order score of i is j:iTjsj. • Theorem 4.8. k-RSC with k polynomial gives an approximation ratio of ½+(1/m). • Lemma 4.9. i (ij:iTjsj)  m2/4 – m/2 (on board). • Max second order score is (m-1)(m-2)/2.

  17. גם בארזים נפלה שלהבת • Permutation trees give (log(m)/m)-approx. • Huge randomized balanced trees intuitively do very well. • k-RPT: every leaf at depth k, labels assigned uniformly at random. • Theorem 5.1. K K’  K s.t. K’-RPT gives an approx ratio of at most O(1/m).

  18. Open problems • Randomized model: gap between LB of ½ (admissible, small) and UB of 5/6 (even inadmissible and large) • Deterministic: enigmatic gap between LB of (logm/m) and UB of ¾.

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