Felix Fischer, Ariel D. Procaccia and Alex Samorodnitsky
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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.
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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?
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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 )  .
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.
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(T)C  (T’)C.
2. (T)A\C(T)=(T’).
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m = 3k, k odd.
T is 3 cycle of regular
components.
In T, i si = k + (k-1)/2.
One component in T’ is
transitive.
In T’ i s.t. si = k + (k-1),
winner doesn’t change.
The ratio tends to ¾.
T’
k=5
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On board.
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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.
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.
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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.
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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.
Define M = M(T).
  = A.
 Initial distribution is
uniform.
 P(i,j) =
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 i=j: (si+1)/m
 jTi: 1/m
 iTj: 0
 (k)
= q(k+1)
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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 i si  (m-1)/2.
Lemma 4.6. Let T. M(T) is rapidly mixing.
Proof of Lemma 4.2 follows (on board).
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Example with ratio ½ + o(1). Consists of an
almost regular tournament.
Theorem 4.7. Let , ’ = (4)1/4. If
i i si = (m-1)/2 + m,
then
#{i: |si-m/2| > 3’m/2}  ’m.
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Second order score of i is j:iTj sj.
Theorem 4.8. k-RSC with k polynomial gives
an approximation ratio of ½+(1/m).
Lemma 4.9. i (i j:iTj sj)  m2/4 – m/2 (on
board).
 Max second order score is (m-1)(m-2)/2.
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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).
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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.