Lecture 4 Impartial Selection; PageRank; Facility Location CSC2556 - - PowerPoint PPT Presentation

CSC2556 Lecture 4 Impartial Selection; PageRank; Facility Location

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Announcements

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• Hope to add a homework question by next lecture
• Proposal tentatively due around Feb end

➢ But it will help to decide the topic earlier, and start

working.

• I’ll put up a list of possible project ideas (in case

you cannot find something related to your research)

➢ Will also be available to have more meetings during the

next two months to help select projects

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Impartial Selection

Impartial Selection

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• “How can we select 𝑙 people out of 𝑜 people?”

➢ Applications: electing a student representation committee,

selecting 𝑙 out of 𝑜 grant applications to fund using peer review, …

• Model

➢ Input: a directed graph 𝐻 = (𝑊, 𝐹) ➢ Nodes 𝑊 = {𝑤1, … , 𝑤𝑜} are the 𝑜 people ➢ Edge 𝑓 = 𝑤𝑗, 𝑤𝑘 ∈ 𝐹: 𝑤𝑗 supports/approves of 𝑤𝑘

• We do not allow or ignore self-edges (𝑤𝑗, 𝑤𝑗)

➢ Output: a subset 𝑊′ ⊆ 𝑊 with 𝑊′ = 𝑙 ➢ 𝑙 ∈ {1, … , 𝑜 − 1} is given

Impartial Selection

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• Impartiality: A 𝑙-selection rule 𝑔 is impartial if 𝑤𝑗 ∈

𝑔(𝐻) does not depend on the outgoing edges of 𝑤𝑗

➢ 𝑤𝑗 cannot manipulate his outgoing edges to get selected ➢ Q: But the definition says 𝑤𝑗 can neither go from 𝑤𝑗 ∉ 𝑔(𝐻)

to 𝑤𝑗 ∈ 𝑔(𝐻), nor from 𝑤𝑗 ∈ 𝑔(𝐻) to 𝑤𝑗 ∉ 𝑔(𝐻). Why?

• Societal goal: maximize the sum of in-degrees of

selected agents σ𝑤∈𝑔 𝐻 𝑗𝑜 𝑤

➢ 𝑗𝑜(𝑤) = set of nodes that have an edge to 𝑤 ➢ 𝑝𝑣𝑢 𝑤 = set of nodes that 𝑤 has an edge to ➢ Note: OPT will pick the 𝑙 nodes with the highest indegrees

Optimal ≠ Impartial

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• An optimal 1-selecton rule must select 𝑤1 or 𝑤2
• The other node can remove his edge to the winner,

and make sure the optimal rule selects him instead

• This violates impartiality

𝑤1 𝑤2 𝑤3 𝑤𝑜

…

Goal: Approximately Optimal

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• 𝛽-approximation: We want a 𝑙-selection system

that always returns a set with total indegree at least 𝛽 times the total indegree of the optimal set

• Q: For 𝑙 = 1, what about the following rule?

Rule: “Select the lowest index vertex in 𝑝𝑣𝑢 𝑤1 . If 𝑝𝑣𝑢 𝑤1 = ∅, select 𝑤2.”

➢ A. Impartial + constant approximation ➢ B. Impartial + bad approximation ➢ C. Not impartial + constant approximation ➢ D. Not impartial + bad approximation

No Finite Approximation 

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• Theorem [Alon et al. 2011]

For every 𝑙 ∈ {1, … , 𝑜 − 1}, there is no impartial 𝑙- selection rule with a finite approximation ratio.

• Proof:

➢ For small 𝑙, this is trivial. E.g., consider 𝑙 = 1.

• What if 𝐻 has two nodes 𝑤1 and 𝑤2 that point to each other, and

there are no other edges?

• For finite approximation, the rule must choose either 𝑤1 or 𝑤2
• Say it chooses 𝑤1. If 𝑤2 now removes his edge to 𝑤1, the rule must

choose 𝑤2 for any finite approximation.

• Same argument as before. But applies to any “finite approximation

rule”, and not just the optimal rule.

No Finite Approximation 

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• Theorem [Alon et al. 2011]

For every 𝑙 ∈ {1, … , 𝑜 − 1}, there is no impartial 𝑙- selection rule with a finite approximation ratio.

• Proof:

➢ Proof is more intricate for larger 𝑙. Let’s do 𝑙 = 𝑜 − 1.

• 𝑙 = 𝑜 − 1: given a graph, “eliminate” a node.

➢ Suppose for contradiction that there is such a rule 𝑔. ➢ W.l.o.g., say 𝑤𝑜 is eliminated in the empty graph. ➢ Consider a family of graphs in which a subset of

{𝑤1, … , 𝑤𝑜−1} have edges to 𝑤𝑜.

No Finite Approximation 

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• Proof (𝑙 = 𝑜 − 1 continued):

➢ Consider star graphs in which a non-empty

subset of {𝑤1, … , 𝑤𝑜−1} have edge to 𝑤𝑜, and there are no other edges

• Represented by bit strings 0,1 𝑜−1\{0}

➢ 𝑤𝑜 cannot be eliminated in any star graph

• Otherwise we have infinite approximation

➢ 𝑔 maps 0,1 𝑜−1\{0} to {1, … , 𝑜 − 1}

• “Who will be eliminated?”

➢ Impartiality: 𝑔 Ԧ

𝑦 = 𝑗 ⇔ 𝑔 Ԧ 𝑦 + Ԧ 𝑓𝑗 = 𝑗

• Ԧ

𝑓𝑗 has 1 at 𝑗𝑢ℎ coordinate, 0 elsewhere

• In words, 𝑗 cannot prevent elimination by adding
• r removing his edge to 𝑤𝑜

𝑤1 𝑤2 𝑤3 𝑤𝑜 𝑤𝑜−1 𝑤4 𝑤1 𝑤2 𝑤3 𝑤𝑜 𝑤𝑜−1 𝑤4

No Finite Approximation 

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• Proof (𝑙 = 𝑜 − 1 continued):

➢ 𝑔: 0,1 𝑜−1\{0} → {1, … , 𝑜 − 1} ➢ 𝑔 Ԧ

𝑦 = 𝑗 ⇔ 𝑔 Ԧ 𝑦 + Ԧ 𝑓𝑗 = 𝑗

• Ԧ

𝑓𝑗 has 1 only in 𝑗𝑢ℎ coordinate

➢ Pairing implies…

• The number of strings on which 𝑔 outputs 𝑗 is

even, for every 𝑗.

• Thus, total number of strings in the domain

must be even too.

• But total number of strings is 2𝑜−1 − 1 (odd)

➢ So impartiality must be violated for some

pair of Ԧ 𝑦 and Ԧ 𝑦 + Ԧ 𝑓𝑗

𝑤1 𝑤2 𝑤3 𝑤𝑜 𝑤𝑜−1 𝑤4 𝑤1 𝑤2 𝑤3 𝑤𝑜 𝑤𝑜−1 𝑤4

Back to Impartial Selection

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• Question: So what can we do to select impartially?
• Answer: Randomization!

➢ Impartiality now requires that the probability of an agent

being selected be independent of his outgoing edges.

• Examples: Randomized Impartial Mechanisms

➢ Choose 𝑙 nodes uniformly at random

• Sadly, this still has arbitrarily bad approximation.
• Imagine having 𝑙 special nodes with indegree 𝑜 − 1, and all other

nodes having indegree 0.

• Mechanism achieves

Τ 𝑙 𝑜 ∗ 𝑃𝑄𝑈 ⇒ approximation = 𝑜/𝑙

• Good when 𝑙 is comparable to 𝑜, but bad when 𝑙 is small.

Random Partition

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• Idea:

➢ What if we partition 𝑊 into 𝑊

1 and 𝑊 2, and select 𝑙 nodes

from 𝑊

1 based only on edges coming to them from 𝑊 2?

• Mechanism:

➢ Assign each node to 𝑊

1 or 𝑊 2 i.i.d. with probability ½

➢ Choose 𝑊

𝑗 ∈ 𝑊 1, 𝑊 2 at random

➢ Choose 𝑙 nodes from 𝑊

𝑗 that have most incoming edges

from nodes in 𝑊

3−𝑗

Random Partition

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• Analysis:

➢ We want to approximate 𝐽 = # edges incoming to nodes in 𝑃𝑄𝑈.

• Let 𝑃𝑄𝑈

1 = 𝑃𝑄𝑈 ∩ 𝑊 1, and 𝑃𝑄𝑈2 = 𝑃𝑄𝑈 ∩ 𝑊 2.

• Let 𝐽1 = # edges incoming to 𝑃𝑄𝑈

1 from 𝑊 2.

• Let 𝐽2 = # edges incoming to 𝑃𝑄𝑈2 from 𝑊

1. ➢ Note that 𝐹 𝐽1 + 𝐽2 = 𝐽/2. (WHY?) ➢ With probability ½, mechanism picks 𝑙 nodes from 𝑊

1 that have

most incoming edges from 𝑊

2 (thus at least 𝐽1 incoming edges).

• Because they’re at least as good as 𝑃𝑄𝑈

1. ➢ With probability ½, mechanism picks 𝑙 nodes from 𝑊

2 that have

most incoming edges from 𝑊

1 (thus at least 𝐽2 incoming edges).

➢ The expected total incoming edges is at least

• 𝐹

1 2 ⋅ 𝐽1 + 1 2 ⋅ 𝐽2 = 1 2 ⋅ 𝐹 𝐽1 + 𝐽2 = 1 2 ⋅ 𝐽 2 = 𝐽 4

Random Partition

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• Generalization

➢ Divide into ℓ parts, and pick 𝑙/ℓ nodes from each part

based on incoming edges from all other parts.

• Theorem [Alon et al. 2011]:

➢ ℓ = 2 gives a 4-approximation. ➢ For 𝑙 ≥ 2, ℓ~𝑙1/3 gives 1 + 𝑃

1 𝑙1/3 approximation.

Better Approximations

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• Alon et al. [2011] conjectured that for randomized

impartial 1-selection…

➢ (For which their mechanism is a 4-approximation) ➢ It should be possible to achieve a 2-approximation. ➢ Recently proved by Fischer & Klimm [2014] ➢ Permutation mechanism:

• Select a random permutation (𝜌1, 𝜌2, … , 𝜌𝑜) of the vertices.
• Start by selecting 𝑧 = 𝜌1 as the “current answer”.
• At any iteration 𝑢, let 𝑧 ∈ {𝜌1, … , 𝜌𝑢} be the current answer.
• From {𝜌1, … , 𝜌𝑢}\{𝑧}, if there are more edges to 𝜌𝑢+1 than to 𝑧,

change the current answer to 𝑧 = 𝜌𝑢+1.

Better Approximations

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• 2-approximation is tight.

➢ In an 𝑜-node graph, fix 𝑣 and 𝑤, and suppose no other

nodes have any incoming/outgoing edges.

➢ Three cases: only 𝑣 → 𝑤 edge, only 𝑤 → 𝑣, or both.

• The best impartial mechanism selects 𝑣 and 𝑤 with probability ½

in every case, and achieves 2-approximation.

• But this is because 𝑜 − 2 nodes are not voting!

➢ What if every node must have an outgoing edge? ➢ Fischer & Klimm [2014]:

• Permutation mechanism gives between Τ

12 7 and Τ 3 2

approximation.

• No mechanism gives better than 4/3 approximation.

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PageRank Axiomatization

PageRank

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• An extension of the impartial selection problem

➢ Instead of selecting 𝑙 nodes, we want to rank all nodes

• The PageRank Problem: Given a directed graph,

rank all nodes by their “importance”.

➢ Think of the web graph, where nodes are webpages, and

a directed (𝑣, 𝑤) edge means 𝑣 has a link to 𝑤.

• Questions:

➢ What properties do we want from such a rule? ➢ What rule satisfies these properties?

PageRank

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• Here is the PageRank Algorithm:

➢ Start from any node in the graph. ➢ At each iteration, choose an outgoing edge of the current

node, uniformly at random among all its outgoing edges.

➢ Move to the neighbor node on that edge. ➢ In the limit of 𝑈 → ∞ iterations, measure the fraction of

time the “random walk” visits each node.

➢ Rank the nodes by these “stationary probabilities”.

• Google uses (a version of) this algorithm

➢ It’s seems a reasonable algorithm. ➢ What nice axioms might it satisfy?

PageRank

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• In a formal sense…

➢ Let 𝑞𝑗 = stationary probability of visiting 𝑗. ➢ Let 𝑂(𝑗) = set of nodes that have an edge to 𝑗. ➢ Then, 𝑞𝑗 = σ𝑘

Τ 𝑞𝑘 𝑝𝑣𝑢𝑒𝑓𝑕(𝑘) ⇒ 𝑜 equations, 𝑜 variables!

• Another way to do this:

➢ Let 𝐵 be a matrix with 𝐵𝑗,𝑘 =

Τ 1 𝑝𝑣𝑢𝑒𝑓𝑕(𝑗) for every 𝑗, 𝑘 ∈ 𝐹.

➢ Then, we are searching for a solution 𝑤 such that 𝐵𝑤 = 𝑤. ➢ One method: start from any 𝑤0, and compute lim

𝑙→∞ 𝐵𝑙𝑤0

• Note: 𝐵𝑙 can be computed using log 𝑙 matrix multiplications!

Axioms

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• Axiom 1 (Isomorphism)

➢ Permuting node names permutes the final

ranking.

• Axiom 2 (Vote by Committee)

➢ Voting through intermediate fake nodes

cannot change the ranking.

• Axiom 3 (Self Edge)

➢ 𝑤 adding a self edge cannot change the

• rdering of the other nodes.
• Axiom 4 (Collapsing)

➢ Merging identically voting nodes cannot change the

• rdering of the other nodes.
• Axiom 5 (Proxy)

➢ If 𝑙 nodes with equal score vote for 𝑙 other nodes

through a proxy, it should be no different than a direct 1-1 voting.

𝑏 𝑏

PageRank

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• Theorem [Altman and Tennenholtz, 2005]:

An algorithm satisfies these five axioms if and only if it is PageRank.

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Facility Location

Apprx Mechanism Design

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• 1. Define the problem: agents, outcomes, values
• 2. Fix an objective function (e.g., maximizing sum of

values)

• 3. Check if the objective function is maximized

through a strategyproof mechanism

• 4. If not, find the strategyproof mechanism that

provides the best worst-case approximation ratio

• f the objective function

Facility Location

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• Set of agents 𝑂
• Each agent 𝑗 has a true location 𝑦𝑗 ∈ ℝ
• Mechanism 𝑔

➢ Takes as input reports ෤

𝑦 = (෤ 𝑦1, ෤ 𝑦2, … , ෤ 𝑦𝑜)

➢ Returns a location 𝑧 ∈ ℝ for the new facility

• Cost to agent 𝑗 : 𝑑𝑗 𝑧 = 𝑧 − 𝑦𝑗
• Social cost 𝐷 𝑧 = σ𝑗 𝑑𝑗 𝑧 = σ𝑗 𝑧 − 𝑦𝑗

Facility Location

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• Social cost 𝐷 𝑧 = σ𝑗 𝑑𝑗 𝑧 = σ𝑗 𝑧 − 𝑦𝑗
• Q: Ignoring incentives, what choice of 𝑧 would

minimize the social cost?

• A: The median location med(𝑦1, … , 𝑦𝑜)

➢ 𝑜 is odd → the unique “(n+1)/2”th smallest value ➢ 𝑜 is even → “n/2”th or “(n/2)+1”st smallest value ➢ Why?

Facility Location

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• Social cost 𝐷 𝑧 = σ𝑗 𝑑𝑗 𝑧 = σ𝑗 𝑧 − 𝑦𝑗
• Median is optimal (i.e., 1-approximation)
• What about incentives?

➢ Median is also strategyproof (SP)! ➢ Irrespective of the reports of other agents, agent 𝑗 is best

• ff reporting 𝑦𝑗

Median is SP

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No manipulation can help

Max Cost

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• A different objective function 𝐷 𝑧 = max

𝑗

𝑧 − 𝑦𝑗

• Q: Again ignoring incentives, what value of 𝑧

minimizes the maximum cost?

• A: The midpoint of the leftmost (min

𝑗

𝑦𝑗) and the rightmost (max

𝑗

𝑦𝑗) locations

• Q: Is this optimal rule strategyproof?
• A: No!

Max Cost

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• 𝐷 𝑧 = max𝑗 𝑧 − 𝑦𝑗
• We want to use a strategyproof mechanism.
• Question: What is the approximation ratio of

median for maximum cost?

• 1. ∈ 1,2
• 2. ∈ 2,3
• 3. ∈ 3,4
• 4. ∈ 4, ∞

Max Cost

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• Answer: 2-approximation
• Other SP mechanisms that are 2-approximation

➢ Leftmost: Choose the leftmost reported location ➢ Rightmost: Choose the rightmost reported location ➢ Dictatorship: Choose the location reported by agent 1 ➢ …

Max Cost

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• Theorem [Procaccia & Tennenholtz, ‘09]

No deterministic SP mechanism has approximation ratio < 2 for maximum cost.

• Proof:

Max Cost + Randomized

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• The Left-Right-Middle (LRM) Mechanism

➢ Choose min

𝑗

𝑦𝑗 with probability ¼

➢ Choose max

𝑗

𝑦𝑗 with probability ¼

➢ Choose (min

𝑗

𝑦𝑗 + max

𝑗

𝑦𝑗)/2 with probability ½

• Question: What is the approximation ratio of LRM

for maximum cost?

• At most

(1/4)∗2𝐷+(1/4)∗2𝐷+(1/2)∗𝐷 𝐷

=

3 2

Max Cost + Randomized

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• Theorem [Procaccia & Tennenholtz, ‘09]:

The LRM mechanism is strategyproof.

• Proof:

1/4 1/4 1/2 1/4 1/4 1/2 2𝜀 𝜀

Max Cost + Randomized

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• Exercise for you!

Try showing that no randomized SP mechanism can achieve approximation ratio < 3/2.