On Voting Caterpillars - PowerPoint Presentation

Slide1

On Voting Caterpillars

Felix Fischer, Ariel D.

Procaccia

and Alex

SamorodnitskySlide2

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.

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Voting trees

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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?Slide5

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

)  .Slide6

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’).

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¾ deterministic upper bound

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 = 5Slide8

5/6 randomized upper bound

On board.Slide9

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. Slide10

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.

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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. Slide12

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.Slide13

k-RC and Markov chains

Define

M

=

M

(T).  = A.Initial distribution is uniform. P(i,j) =

i=j: (si+1)/mjTi: 1/miTj: 0

(k) = q(k+1)

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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, p

i(k) prob. of i in k-RSC,

qi(k) prob. of i

in k-RC. Then |pi(k) - q

i(k)|  m / e

k-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 s

i  (m-1)/2.Lemma 4.6. Let T. M(T) is rapidly mixing.

Proof of Lemma 4.2 follows (on board). Slide15

Stability of caterpillars

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.Slide16

Second order Copeland

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. Slide17

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). Slide18

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.