Manipulating the Quota in Weighted Voting Games (M. Zuckerman, P. Faliszewski, Y. Bachrach, and E. Elkind) ‏ Presented by: Sen Li Software Technologies Applied Research group ECE

Introduction

A weighted voting game is described by: 1. A set of players 2. A list of player’s weights 3. A quota A coalition of the players is said to be winning if the total weight of its members meets or exceeds the quota.

Introduction An important issue in weighted voting is how to measure the power of each voter. i.e. its ability to affect the final outcome. This question is critical when the agents have to decide how to distribute the payoffs. ▫Because a natural approach is to pay each agent according to his contribution.

Introduction Intuitively, we might think that a player’s voting power is always directly proportional to its weight. However, this is NOT true. For example: We have 4 voters: { A, B, C, D} with weights:{10, 5, 2, 1 } Who has the strongest power? A? If the quota is 10, then A does have the strongest power. But if the quota is 18, then A, B, C, D all have veto power. So they have equivalent powers.

Introduction From last example, we can see that: By modifying the quota, central authority can change a player’s voting power.

Shapley-Shubik Index & Banzhaf Index We have showed that an agent’s power is not always directly proportional to its weight. So, we need some other ways to measure an agent’s power. Voting powers are traditionally identified by its power index. There are two famous indices: 1.Shapley-Shubik Index (SS) ‏ 2.Banzhaf Index (BF) ‏

Introduction This presentation wants to answer 4 Questions: 1.By manipulating the quota, how much can the central authority change a player’s voting power? 2.How can the choice of quota affect the relative power of players? 3.Is there an efficient algorithm to determine if there is a quota making a player dummy? i.e., reduces its power to 0. 4.Is there an efficient algorithm to check which of two quotas makes a player more powerful?

Terminology I : a set of players, |I| = n w : a weight vector, 0 < w 1 ≤ … ≤ w n q : the quota of a voting game G(q) : a game with quota q

Terminology SS i (q) : the value of Shapley Index for player i in the game with quota q BF i (q) : the value of Banzhaf Index for player i in the game with quota q Dummy : A player with zero voting power

First question: How much can centre change a player’s voting power?

How much can centre change a player’s power? There are two ways to quantify the “How much”: ▫ The worst case ratio between a player’s power index values for two different quotas. ▫ The worst case difference between a player’s power index values for two different quotas.

How much can centre change a player’s power? It is more natural to use ratio in general case. Unfortunately, we can not use ratio in here. Because a players’ power index value might be 0.

How much can centre change a player’s power? Theorem 1. Given a set of players I, there exists a weight vector w, and quota q1, q2, such that weight for i != n, we have SS i (q1) = BF i (q1) = 0, while SS (q2) != 0, BF i (q2) != 0. On the other hand, for any w such that 0 < w 1 ≤ … ≤ w n and any q1, q2 ≤ w(I), we have SS n (q1) / SS n (q2) ≤ n, BF n (q1) / BF n (q2) ≤ 2 n−1, and these bounds are tight.

How much can centre change a player’s power? Therefore, at least in some weighted voting games, the center can change the agent’s power index to 0. That is, we can not use worst case ratio. We can only use worst case difference to quantify.

How much can centre change a player’s power? Theorem 2. For a set of players I, any weight vector w, 0 < w 1 ≤ … ≤ w n and any quota q1, q2, for i != n, SS i (q1) − SS i (q2) can be at most 1 / (n−i+1) and this bound is tight. For i = n, SS i (q1) − SS i (q2) can be at most 1 − 1/n, and this bound is tight.

How much can centre change a player’s power? Theorem 3. For a set of players I, any weight vector w, 0 < w 1 ≤ … ≤ w n and any quota q1, q2, for i != n, BF i (q1) − BF i (q2) can be at most (n-i choose ⌊ (n−i)/2 ⌋ ) ・ 2 i−n and this bound is tight. For i = n, we have BF i (q1) − BF i (q2) can be at most 1 − 1/2 n−1 and this bound is tight.

How much can centre change a player’s power? What do we learn from the first section: ▫ We can NOT use worst case ratio to quantify the “How much”. ▫ But we can use worst case difference. ▫ There exists upper-bounds for worst case differences.

Second question: How can the choice of quota affect the relative power of players?

Changing the quota could affect the power of two players Let’s rephrase the question: Could changing the quota be affecting the relative power of two players i and j ? For example, suppose that w i < w j, but the center prefers player i to j. Obviously, SS i (q) ≤ SS j (q) (or BF i (q) ≤ BF j (q)). So, the best that the center can do, hopefully, is to find a quota that satisfies SS i (q) = SS j (q) (or BF i (q) = BF j (q)).

Changing the quota could affect the power of two players Theorem 4. Consider a set of players I, and a vector of weights w. For each two players i and j with w i < w j. There is a quota q1, which holds that SS i (q1) < SS j (q1) (or BF i (q1) < BF j (q1)). There is also a quota q2, which holds that SS i (q2) = SS j (q2) (or BF i (q2) = BF j (q2)).

Changing the quota could affect the power of two players The pervious theorem tells us: If w i < w j, the center can always find a quota q such that: power index of i = power index of j.

Changing the quota could affect the power of two players Also, the center may want to find a quota that ensures all players have different power index. The quota that can satisfy this constraint is called a separating quota. However, it is NOT always possible to find such quota. Due to the time issue, we ignore the proof here.

Third question: Is there an efficient algorithm to determine if there is a quota making a player dummy?

An algorithm to determine if a player can be dummy The answer is: Yes, such algorithm exists.

An algorithm to determine if a player can be dummy Before specifying the algorithm, we need to define a term first. Definition 9. Given a weight vector w and a weight w 1, we say that w 1 is essential for w if for all 1 ≤ t ≤ n, ∑ t−1 i=1 w i ≥ w t − w 1.

An algorithm to determine if a player can be dummy Theorem 10. Let w be a vector of weights. A weight w 1 is essential for w if and only if there is no quota q, such that n + 1 is a dummy in a game G(q) = [{1, …, n, n + 1}; (w 1, …, w n, w 1); q]. That is, a player can never be dummy if and only if its weight is essential.

An algorithm to determine if a player can be dummy The previous theorem yields a simple algorithm for testing whether there exists a quota making a given agent dummy. This can be done using O(n) time. Moreover, we can now check what quota minimizes the power index of a given player.

An algorithm to determine if a player can be dummy Theorem 11. There exists a polynomial time algorithm that finds the value of the quota which minimizes the BF of a given player. Proof: Use the algorithm described before to check if there is a quota that makes the agent dummy, and if so, return this quota. Otherwise, return quota q = min{w1, …,wn}. Under q, the BF of our agent is 1/2 n−1, since the only coalition it contributes to is the empty set.

Fourth question: Is there an efficient algorithm to check which of two quotas makes a player more powerful?

An algorithm to determine which of the two values of quota makes a player more powerful In the previous section, we showed that when the center can choose any quota, minimizing a player’s power index becomes easy. However, deciding which of two given quotas favors a player is hard!!!

An algorithm to determine which of the two values of quota makes a player more powerful How hard can it be? It is PP-hard, which is believed to be considerably stronger than NP-hard: Any PP-hard is NP-hard, but not vice versa.

An algorithm to determine which of the two values of quota makes a player more powerful PP stands for probabilistic polynomial time. Formally, we say that a language L belongs to PP if there exists an NP machine N such that: x ∈ L if and only if the probability that N accepts x is at least 1/2. The paper does not talk too much detail about PP. If you are interested, you can get more information from

An algorithm to determine which of the two values of quota makes a player more powerful Then we gonna show why our problem is PP-hard. Let’s first define our problem: determine which of two quotas makes a player more powerful.

An algorithm to determine which of the two values of quota makes a player more powerful Definition 12. Let f be either SS or BF. Let Quota f problem be: Given I, w, two quota q1 and q2, and an index i ∈ I. Let G 1 = [ I; w; q1], G2 = [ I; w; q2]. The task is to decide whether f i (G1) > f i (G2). Definition 13. Let f be either SS or BF. Let PowerCompare f problem be: Given two weighted voting games, G1 and G2, a player i in G1, and a player j in G2, does it hold that f i (G1) > f j (G2).

An algorithm to determine which of the two values of quota makes a player more powerful Quota f is a special case of PowerCompare f. In 2008, two researchers have showed that PowerCompare f problem is PP-complete. So the result immediately implies that Quota f is also in PP.

An algorithm to determine which of the two values of quota makes a player more powerful Discussion: PP-hardness sounds scary and can be a barrier to manipulation by the central authority. However, PP-hardness does not necessarily imply that the problem is hard on average. Proving that manipulating the quota is hard in this sense is still an open problem.

An algorithm to determine which of the two values of quota makes a player more powerful Moreover, it is known that both SS and BF are easy to compute if the weights are polynomially bounded (i.e. given in unary). (Matsui and Matsui 2000) ‏ To solve the previous PP-hard problem, we can compute a player's power index for both quotas, and choose the one that gives us a better outcome. Therefore, computational complexity alone does NOT provide proper protection from this manipulation.

Conclusion

As we said at the beginning of the presentation, this talk wants to answer 4 questions.

Conclusion 1.How much can the central authority change a player’s voting power by manipulating the quota? ▫ For i != n, the difference of SS ≤ 1 / (n−i+1) ▫ For i = n, the difference of SS ≤ 1 − 1/n ▫ For i != n, the difference of BF ≤ (n-i choose ⌊ (n−i)/2 ⌋ ) ・ 2 i−n ▫ For i = n, the difference of BF ≤ 1 − 1/2 n−1

Conclusion 2.How can the choice of quota affect the relative power of players? ▫If w i < w j, center can always find a quota q satisfies that: power index of i = power index of j. ▫ The center may NOT be able to find a quota that ensures each player has different power index.

Conclusion 3.Is there an efficient algorithm to determine if there is a value of quota making a player dummy? Yes, such algorithm exists.

Conclusion 4.Is there an efficient algorithm to check which of two quotas makes a player more powerful? No, the algorithm is PP-hard.

Conclusion What can we learn from this paper? 1.Central authority can manipulate quota to change players’ powers, but this change is bounded. 2.There might not exist a quota to ensure that all players have different power index. 3.We have an efficient algorithm to determine if there is a value of quota making a player dummy. 4.Checking which of two quotas makes a player more powerful is PP-hard.

Further Questions We know manipulations through quota control are possible, what measures can be taken against such manipulations? Are there other payoff division schemes that are more resistant to such manipulations?