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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
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.
2.
Shapley-Shubik Index (SS)
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 < w1 ≤ … ≤ wn
• q : the quota of a voting game
• G(q) : a game with quota q
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
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
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.
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 < w1 ≤ …
≤ wn and any quota q1 , q2, for i != n, SSi (q1) − SSi
(q2) can be at most 1 / (n−i+1) and this bound is tight.
For i = n, SSi (q1) − SSi(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 < 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.
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 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)).
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)
(or BFi (q1) < BFj (q1)).
There is also a quota q2, which holds that SSi (q2) =
SSj (q2) (or BFi (q2) = BFj (q2)).
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 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 w1, we
say that w1 is essential for w if for all 1 ≤ t ≤ n,
∑t−1i=1 wi ≥ wt − w1.
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 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}; (w1, … , wn, w1); 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 1/2n−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 http://en.wikipedia.org/wiki/PP_(complexity)
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 Quotaf problem 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 PowerComparef 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
fi(G1) > fj(G2).
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
PowerComparef problem is PP-complete.
• So the result immediately implies that Quotaf 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
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⌋) ・2i−n
▫
For i = n, the difference of BF ≤ 1 − 1/2n−1
Conclusion
2.
How can the choice of quota affect the relative power of
players?
▫
If wi < wj , 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?