1. Ron Lavi Presented by Yoni Moses

2.  Introduction ◦ Combining computational efficiency with game theoretic needs  Monotonicity Conditions ◦ Cyclic Monotonicity ◦ Weak Monotonicity  An Example – Machine Scheduling Problem

3.  Approximation for Combinatorial Auctions ◦ Fractional allocation ◦ Integral allocation  Impossibility results

4.  m items (Ω) are allocated to n players  i is the value given by player i to a bundle S (a subset of Ω)  Valuations are ◦ Monotone: ◦ Normalized:  Goal: Find allocation such that is maximized.

5.  Problem: a general valuation’s size is exponential is n and m.  Possible representations: ◦ Bidding languages model ◦ access model  But polynomial algorithms that use these representations only obtain an approximation. VCG requires the exact optimum!

6.  Given: an algorithm for CA that outputs a c- approximation.  Construct: A randomized c-approx. mechanism that is truthful in expectation  Plan: ◦ First, solve for the fractional domain ◦ Next, move back to the original domain, using randomization

7.  Solve using Linear Programming  Allocation x gives player i a fraction of subset S.  The value is  Constraints: ◦ A player receives at most one integral subset ◦ An item cannot be over-allocated  Goal: ◦ maximize the sum of values

9.  The algorithm’s time complexity is polynomial. ◦ We can assume the bidding languages model, where the LP has size polynomial in the size of the bid (for example: k-minded players) ◦ We can assume general valuations with query- access, and the LP is solvable with a poly. num of demand queries ◦ The number of non-zero coordinates is poly. because we obtain x in polynomial-time  Solution is optimal => We can use VCG! ◦ but it’s a solution for the fractional domain…

10.  Definition: Algorithm A “verifies a c- integrality-gap” for the LP program CA-P if it receives real numbers and outputs an integral point which is feasible for CA-P and

11.  Suppose A verifies a c-integrality-gap for CA-P (in poly. time), and x is any feasible point of CA-P.  Then x/c can be decomposed to a convex combination of integral feasible points (in poly. time)

13.  Individual rationality (non-negative utility) is satisfied, regardless of the randomized choice:  VCG is individually rational:  Thus, by definition: for any l

14.  Lemma: The decomposition-based mechanism is truthful in expectation, and obtains a c-approx. to the social welfare  Proof:  The expected social welfare is.  Since x* is the optimal (fractional) allocation, the c- approx. is obtained.  Truthfulness: First, we show that the expected price equals the fractional price over c:

15.  Now, fix the other players’ valuation.  x* is the fractional optimum obtained when player i declares. z* is the frac. optimum obtained when i declares..  Since VCG fractional prices are truthful:  Divide this formula by c. Using the previous formula and by definition of the decomposition, we get:

16.  The left hand side is the expected utility for declaring.  The right hand side is the expected utility for declaring.  Thus, the lemma follows.

17.  This analysis is for one-shot mechanisms, where a player declares his valuation up-front ◦ for example: the bidding languages model.  For an iterative mechanism such as the query-access model, the solution is weakened to ex-post Nash ◦ If all other players are truthful, player i will maximize her expected utility by being truthful.

18.  How de we decompose x/c into ?  We use a new LP called P and its dual D.  notation: E is the set of nonzero fractions in the allocation. primaldual

19.  Constraints 1.11 of P describe the decomposition.  If the optimum satisfies, we’re almost done. ◦ But P has exponentially many variables!  We’ll use the dual D. Its number of variables is poly. ◦ Of course, D’s constraints are analogous to P’s variables => D has exponentially many constraints.  We can still solve D in polynomial time, using the ellipsoid method and our verifier A as a separation oracle.

20.  Claim: If w, z is feasible for D: If not, A can be used to find a violated constraint in poly. time  Proof:  Suppose.  Let A receive w as input. Its output is an integral allocation.  Since A is a c-approx. to the fractional optimum:  Due to the violated inequality of the claim:  Thus constraint 1.12 is violated for :

21.  Claim: The optimum of D is 1, and the decomposition is polynomial-time computable.  Proof: is feasible, hence the optimum is at least 1.  By the previous claim, it is at most 1.  To solve P, we first solve D with this separation oracle:  Given w,z, if, return the  separating hyperplane.  Otherwise, find the violated constraint (which implies the separating hyperplane)

22.  Due to the oracle, the ellipsoid method uses a poly. number of constraints  Thus, there is an equivalent program with only these constraints.  Its dual is a program equivalent to P, but with a poly. number of variables. ◦ Solving that gives us the decomposition.

23.  We still need an algorithm for verifying a c-integrality-gap…  Claim: We’re given A’, a c-approx. for general CA. ◦ The approximation is with respect to the fractional optimum.  Using A’,we can obtain A, a c-integrality-gap verifier for CA-P, with a poly. time overhead on top of A’.  Proof:  Given (the weights in A’s input), we need to build from them a valid valuation that can be used as input for A’. ◦ We can’t assume that w is non-negative and monotone.  Define for non-negativity  Next, Define for monotonicity.

24.  is valid and can be represented with size |E|.  Let  A’ gives c-approx. So such that ◦ Remember that  But in order to construct a verifier, we need this formula to hold for (w instead of ). ◦ Now we only consider coordinates in E ◦ Some coordinates in w (but not in ) can be negative  To fix the first problem, define :  For any (i,S) such that, set:  All other coordinates of are set to 0

25.  By construction,  To fix the second problem, define :   Clearly,  So now we have, which is feasible for CA-P such that

26.  Now we know how to build a verifier using a c-approx. for CA.  We still have to find an algorithm that approximates the fractional optimum.  The following greedy algorithm will give us a approx. to the fractional optimum (proof is skipped).  Input:  Iteration:  Let  Set.  Remove from E all (i’,S’) with i’=i or  If E isn’t empty, reiterate.

27.  The decomposition-based mechanism with Greedy as the integrality-gap verifier is individually rational and truthful- in-expectation and obtains an approximation of to the social welfare.

28.  The notion of truthfulness-in-expectation is inferior to truthfulness ◦ It assumes to players are only interested in their expected utility. But Don’t they care about the variance as well?  Stronger notion: universal truthfulness. Players maximize their utility for every coin toss ◦ Still, “deterministic truthfulness” is better. ◦ In classic algorithms, the law of large numbers can be used to approach the expected performance. But in mechanism design, we cannot repeat the execution because it affects the strategic properties.  Conclusion: deterministic mechanisms are still a better choice.

29.  Notations:  is the domain of values  The social choice function is onto A (domain of alternatives)  Definition: f is an “affine maximizer” if there exist weights such that for all :  Of course, we might prefer other function forms. For example, due to computational complexity, revenue maximization, etc. ◦ But what other forms are implementable:

30.  Theorem:  Suppose and. Then f is dominant-strategy implementable iff it is an affine maximizer. ◦ In other words, if we have unrestricted value domain and nontrivial alternative domain, we have to use an affine maximizer. ◦ Note that any affine maximizer is implementable (can be shown by generalizing VCG arguments).  We will prove one side of a weaker theorem.  Definition: f is neutral if for all, if an alternative x exists such that for all i and, then f (v)=x ◦ In a neutral affine maximizer, all constants will be zero.

31.  Theorem:  Suppose and. Then if f is dominant- strategy implementable and neutral, it must be an affine maximizer.  The proof will require two monotonicity conditions: ◦ Positive Association of Differences (PAD) ◦ Generalized-WMON

32.  Definition: f satisfies PAD if the following holds for any : f(v)=x. for any and any i, Claim: Any implementable function f, on any domain, satisfies PAD. Proof: Let. In other words, players up to i declare according to v’. The rest declare according to v. f(v’) = x

33.  Now, suppose that for some and,.  For every alternative we have. In addition:  Reminder: f satisfies W-MON if for every player i, every and every with,.  W-MON implies that. By induction,. Which means f(v’)=x.

34.  In W-MON, we fix a player and fix the other players’ declarations. ◦ We can generalize W-MON by dropping this.  Definition: f satisfies Generalized-WMON if for every with f(v)=x and f(v’)=y there exists a player i such that  Another way of looking at it: if f(v)=x and then.

35.  Claim: If the domain is unrestricted and f is implementable then f satisfies Generalized-WMON  Proof:  Fix any v, v’. Suppose that f(v’) = x and v’(y) – v(y) > v’(x) – v(x). Assume by contradiction that f(v) = y.  Fix a vector such that v’(x) – v’(y) = v(x)- v(y) -.  Define v’’:  Using PAD, the transition v->v’’ implies f(v’’)=y and the transition v’->v’’ implies f(v’’)=x. contradiction.

36.  Define:  Note that P(x,y) is not empty (assuming that v exists such that f(v) = x)  Also, if then for any,  Explanation: take v with f(v)=x and v(x)-v(y)=.  Construct v’ by increasing v(x) by and setting the other coordinates as in v. By PAD, f(v’)=x and v’(x) – v’(y) =

37.  Proof (i):  Suppose by contradiction that.  There exists We assumed that, we know that a v’ exists such that v’(x)-v’(y) = and f(v’)=x  Due to our assumption,. This contradicts Generalized-WMON

38.  Proof (ii): For any take some and fix some.  Also, fix some v such that for all.  By the above argument,  Since, it follows that f(v)=y.  Thus, as needed.

39.  Proof:  For any, fix some.  Choose any v such that for all  By Generalized-WMON, f(v)=x.  And by adding the 2 equations, we get:

40.  The proof of the thorem follows…  Based on separation lemma