Manipulating the Quota in Weighted Voting Games (M. Zuckerman, P. Faliszewski, Y. Bachrach, and E. Elkind) ‏

1. 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

2. Introduction

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

4. 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.

5. 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.

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

7. 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: • Shapley-Shubik Index (SS)‏ • Banzhaf Index (BF)‏

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

10. Terminology • I : a set of players, |I| = n • w : a weight vector, 0 < w1 ≤ … ≤ wn • q : the quotaof a voting game • G(q) : a game with quota q

11. Terminology • SSi (q) : the value of Shapley Index for player i in the game with quota q • BFi (q) : the value of Banzhaf Index for player i in the game with quota q • Dummy : A player with zero voting power

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

13. 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.

14. 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.

15. 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 SSi (q1) = BFi (q1) = 0, while SS(q2) != 0, BFi (q2) != 0. On the other hand, for any w such that 0 < w1 ≤ … ≤ wn and any q1, q2 ≤ w(I), we have SSn (q1) / SSn (q2) ≤ n, BFn (q1) / BFn (q2) ≤ 2n−1, and these bounds are tight.

16. 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.

17. How much can centre change a player’s power? • Theorem 2. For a set of players I, any weight vector w, 0 < w1 ≤ … ≤ wn and any quota q1 , q2, for i != n, SSi (q1) − SSi (q2) can be at most1 / (n−i+1) and this bound is tight. For i = n, SSi (q1) − SSi(q2) can be at most1 − 1/n, and this bound is tight.

18. How much can centre change a player’s power? • Theorem 3. For a set of players I, any weight vector w, 0 < w1 ≤ … ≤ wn and any quota q1 , q2, for i != n, BFi(q1) − BFi(q2) can be at most(n-i choose ⌊(n−i)/2⌋)・2i−n and this bound is tight. For i = n, we have BFi(q1) − BFi(q2) can be at most 1 − 1/2n−1 and this bound is tight.

19. 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.

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

21. 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 wi < wj , but the center prefers player i to j. • Obviously, SSi (q) ≤ SSj(q) (or BFi (q) ≤ BFj (q)). • So, the best that the center can do, hopefully, is to find a quota that satisfies SSi (q) = SSj (q) (or BFi (q) = BFj (q)).

22. 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 wi < wj. There is a quota q1, which holds that SSi (q1) < SSj (q1) (orBFi (q1) < BFj (q1)). There is also a quota q2, which holds that SSi (q2) = SSj (q2) (orBFi (q2) = BFj (q2)).

23. Changing the quota could affect the power of two players • The pervious theorem tells us: If wi < wj, the center can always find a quota qsuch that: power index of i = power index of j.

24. 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.

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

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

27. 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 vectorw and a weightw1, we say that w1 is essential for w if for all 1 ≤ t ≤ n, ∑t−1i=1 wi ≥ wt − w1.

28. An algorithm to determine if a player can be dummy • Theorem 10. Let w be a vector of weights. A weight w1 is essential for wif and only if there is no quota q, such that n + 1 is a dummy in a game G(q) = [{1, … , n, n + 1}; (w1, … , wn, w1); q] . • That is, a player can never be dummy if and only if its weight is essential.

29. 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.

30. 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 1/2n−1, since the only coalition it contributes to is the empty set.

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

32. 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!!!

33. 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.

34. 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 http://en.wikipedia.org/wiki/PP_(complexity)

35. 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.

36. 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 Quotafproblem be: Given I, w, two quota q1 and q2, and an index i ∈ I. Let G1 = [ I; w; q1], G2 = [ I; w; q2]. The task is to decide whether fi (G1) > fi (G2). • Definition 13. Let f be either SS or BF. Let PowerComparefproblem be: Given two weighted voting games, G1 and G2, a player i in G1, and a player j in G2, does it hold that fi(G1) > fj(G2).

37. An algorithm to determine which of the two values of quota makes a player more powerful • Quotaf is a special case of PowerComparef . • In 2008, two researchers have showed that PowerComparefproblem is PP-complete. • So the result immediately implies that Quotaf is also in PP.

38. 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.

39. 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.

40. Conclusion

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

42. 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⌋)・2i−n • For i = n, the difference of BF≤ 1 − 1/2n−1

43. Conclusion • 2. How can the choice of quota affect the relative power of players? • If wi < wj, center can always find a quota qsatisfies 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.

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

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

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

47. 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?