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1Summary of “A Universally-Truthful Approximation Scheme for Multi-unit Auction” Berthold Vöcking, Author Thatchaphol Saranurak, Presenter I. I NTRODUCTION mechanism running in polynomial time with an approximation ratio better than 2. They showed the impossibility, if an The paper considers multi-unit auction problem. In this output allocation of mechanism is restricted with the certainauction, a set of m identical items has to be allocated to constraint. However, the main result of this paper refutesn bidders. Each bidder i bids his valuation function vi : this question and shows that there is randomized PTAS for{0, ..., m} → R≥0 , indicating how much i is willing to pay for universally truthful mechanism.each amount of items. We consider only the set V of valuation To get such result, there are two majors tools used cruciallyfunctions which are non-decreasing and vi (0) = 0. in this papers : ∆-perturbed maximizer and consensus function After everyone bids, we get an allocation s = with drop-outs.(s1 , s2 , ..., sn ) ∈ {0, ..., m}n indicating how many items each nbidder gets. The set of feasible allocation A = {s | i=1 si ≤ II. ∆- PERTURBED MAXIMIZERm} is set of allocations allocating at most m items to bidders.Besides, vi depends on only his amount of items si , not For convenience of deﬁning ∆-perturbed maximizer, let uss−i , that is vi (s) = vi (si ). The objective of this problem consider an instance of multiple-choice knapsack problem.is to ﬁnd allocation s ∈ A maximizing the social welfare Suppose there are n classes of objects and m objects in eachv(s) = i vi (s). class. Each object k has weight wk and proﬁt pk . We want an A mechanism for multi-unit auctions is a pair (f, p) where algorithm for selecting exactly one object from each class, withf : V n → A is a social choice function gathering valu- summation of weight of objects at most m, and maximizingation functions and then outputs allocation for bidders, and summation of proﬁt. Observe that we can maximize socialp = (p1 , p2 , ..., pn ), pi : V n → R is payment scheme. It is welfare using such algorithm, by deﬁning a tuple (i, j) be anassumed that valuation functions are not given explicitly, but in object for allocating j items to bidder i and set w(i,j) = j andthe form of black box that can be queried by mechanism. Thus, p(i,j) = vi (j). The allocation from knapsack would be feasiblesize of instance of problem, excluding valuation functions, are because number of items allocated is Σk wk ≤ m, and indeedthe number of bits representing m, as items are identical, and social welfare Σi vi = Σk pk is maximized. Nevertheless,the number of bidders : poly(n, log m). Therefore polynomial polynomial-time constraint is not satisﬁed yet as number oftime mechanism in this context should run in poly(n, log m) objects is n × m which is not poly(n, log m), and also generaltime. knapsack is NP-hard problem. Deﬁning utility of bidder i to be his valuation subtracted To deal with this issue, we introduce perturbed valuationby his payment vi (f (vi , v−i )) − pi (vi , v−i ), a deterministic function.mechanism is truthful if, for each bidder i, his utility is Deﬁnition 1 (Perturbed valuation function). Let ∆ > 0. Foralways maximized when he bids the true vi . For randomized 1 ≤ i ≤ n, 0 ≤ j ≤ m, let xj be a random variable chosen imechanism, which is actually a probability distribution over independently, uniformly from [0, 1]. For k ∈ {1, ..., m}, q(k)deterministic mechanisms, there are 2 kinds of truthfulness. is number of factor 2 of k, e.g. q(96) = q(25 × 3) = 5. Hence,Firstly, a mechanism is universally truthful if each of these q(k) ≤ log m . Let q(0) = log m + 1 as an exception.deterministic mechanisms in the distribution is truthful. In thiscase, randomness is used only for guaranteeing other aspects vi (j) = v(j) + (2q(j) + xj )∆ ie.g. social welfare. Secondly, the weaker one, a mechanismis truthful in expectation if, given that bidders do not know Now, by setting proﬁt of object p(i,j) = vi (j) insteadrandom bits of mechanism, the expected bidder’s utility is of vi (j), we deﬁne ∆-perturbed maximizer be the algorithmmaximized by telling the truth. that solves such knapsack instance, and so maximizes Σi vi . It is known that there is deterministic mechanism for this We claim we can achieve the expected running time ofauction running in polynomial-time and has approximation poly(n, log m, P/∆), where P is the second largest numberratio of 2, and if the mechanism follows Robert’s characteri- of maximum bid vi (m) of all bidders. As we will see later,zation, i.e., afﬁne maximizer with VCG-based payments, then ∆ will be set to make P/∆ be poly(n, log m) for some ﬁxedthis bound is tight [3]. For truthful in expectation mechanism, > 0. Thus, ∆-perturbed maximizer run in poly(n, log m).Dobzinski and Dughmi [2] presented that there is FPTAS, and, Brieﬂy, the term q(·) allows us to consider onlymoreover, also questioned the existence of universally truthful poly(n, log m, P/∆) objects and the term x enables us to
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2solve knapsack in expected time polynomial in number of Informally, the reason why we can use randomness toobjects. We illustrate some part of the proof and informal bound the running time of knapsack problem, which isexplanation as follows. NP-hard, to be polynomial, is because the “hard instances” of knapsack problem are “isolated” in some sence, as illustrated in the left ﬁgure below. By perturbing the these instancesA. Term q with random variable, the expected running time is an average To show how we can focus only small portion of objects, running time of hard instances with many easy instanceswe have to deﬁne breakpoints of valuation function. Let surrounding them. Consequently, the running time might beVi = (vi (0), vi (1), ..., vi (m)) denotes the non-decreasing decreased dramatically, as in the right ﬁgure below.sequence of bids of bidder i. We partition Vi into subsequencesViq = (vi (k))|q(k)=q for 0 ≤ q ≤ log m + 1. To illustrate,Viq = (vi (1 · 2q ), vi (3 · 2q ), vi (5 · 2q ), ...) with an exception log m +1Vi = (vi (0)). The q-breakpoints of bidder i aresmallest indices k in Viq such that vi (k) is at least 0, ∆, 2∆, ...The breakpoints of bidder i are the union of all q-breakpoints.Lemma 2. Any allocation s maximizing v (s) = Σi vi (s)satisﬁes that, for each bidder i, si is a breakpoint of Vi . Proof: Suppose that si is not a breakpoint. Then si > 0because 0 is ( log m +1)-breakpoint. Let q = q(si ), then, forsome k, si = (2k + 1)2q . We deﬁne si = si − 2q = (2k)2q =k2q+1 and deﬁne q = q(si ). We get that q ≥ q + 1. Next, consider Viq and si is in some interval of ∆, si ∈ C. Some properties of ∆-perturbed maximizer[c∆, (c + 1)∆) for some c. There must be a predecessor of si Lemma 3. The ∆-perturbed maximizer selects an allocationin such interval, otherwise si in a breakpoint. We also know maximizing the social welfare, which has additive error v (s)−that the predecessor is (2k − 1)2q = si − 2q+1 . This gives v(s) ≤ (2 log m + 3)∆n.vi (si ) < vi (si − 2q+1 ) + ∆ < vi (si ) + ∆. Thus, we get Proof: Consider vi (s) = vi (s) + (2q(s) + xj )∆. Since, i q ≤ log m + 1 and xj ≤ 1, so vi (s) − vi (s) ≤ (2 log m + vi (si ) = vi (si ) + (2q + xj )∆ i i n n 3)∆. As a result, v (s) − v(s) = i=0 vi (s) − i=0 v i (s) ≤ < (vi (si ) + ∆) + (2q − 2 + 1)∆ (2 log m + 3)∆n. = vi (si ) + 2q ∆ Lemma 4. Bidder i get nothing si = 0, if his max bid vi (m) < ≤ vi (si ) ∆. Now, consider allocation s with s−i = s−i and si = si −2q Proof: Since, for any amount of items j > 0, vi (j) ≤as deﬁned. This allocation is feasible because we allocate less vj (m) < ∆, and, by deﬁnition, q(j) + 1 ≤ q(0).items but v (s) < v (s ) which yields a contradiction. vi (0) = (2q(0) + x0 )∆ i As a result, we may input only objects which are break-points into knapsack solver. To count number of all break- ≥ (2q(j) + 2)∆points, let bidder i∗ = arcmaxi vi (m) be the one who bids > vi (j) + (2q(j) + 1)∆the highest max bid. Considering bidder i = i∗ , there are ≥ vi (j)at most P/∆ q-breakpoints for each q. With a certain trick Because vi (0) > vi (j), ∆-perturbed maximizer selects si =omitted here, we can ﬁrst ﬁnd allocation, by solving knapsack, 0.for all bidders excluding i∗ , and then supplement such allo-cation to maximize v (s) in polynomial-time. Thus, there are III. C ONSENSUS FUNCTION WITH DROP - OUTSpoly(n, log m, P/∆) objects inputted to ∆-perturbed maxi-mizer. Let > 0, and τ is a random variable chosen uniformly from [0, 1]. We want a function l with following properties 1) For a ∈ R, it fails to compute with low probability :B. Term x P r[lτ (a) = ⊥] = . The term x is chosen independently uniformly at random 2) If it can be computed lτ (a) = ⊥, then lτ (a) ≤ a. Also,from [0, 1], which ﬁts into a framework of smoothed analysis we want a “gap” between them to be small : a−lτ (a) ≤for knapsack problem [1]. Particularly, the term x makes the 1/ − 1.expected number of Pareto-optimal subsets be polynomial in 3) If l(a1 ) < l(a2 ), then it guarantees a certain separationnumber of objects. Furthermore, the dynamic programming a1 < l(a2 ) − 1.framework of Nemhauser and Ullmann [4] can solve knapsack To deﬁne lτ (a), we introduce xτ (i) = (i+τ ) 1 for each i ∈ Z.in time polynomial in number of Pareto-optimal subsets. Thus, The interval between each xτ (·) is 1/ . Now let k be thethe expected running time is polynomial. largest integer such that xτ (k) ≤ a as in the ﬁgure.
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3 = () ( + 1) . =⊥ 1 ∆i -perturbed maximizer. However, there are actually only two different allocations to be combined. To see this, let 1/ i∗ = argmaxi vi (m) be the one who bids the highest max bid. For bidder i = i∗ , v (−i) = vmax are all the same. Since max these bidders use the parameter ∆i = Li /N = Lτ (vmax )/N Let d be the difference between a and xτ (k + 1). Now, in ∆-perturbed maximizer, they compute the same allocationlτ (a) = xτ (k) if d 1, otherwise it fails to compute s(i) . Moreover, in the case that Li = ⊥, they all also drop outlτ (a) = ⊥. By this deﬁnition, this satisﬁes the second property together.because, if it can be computed, so lτ (a) = xτ (k) ≤ a. Also, Thus, we can call ∆-perturbed maximizer for only twosince lτ (a) is at the left border of the interval and, as d 1, a times, once for any i = i∗ , which yields s(i) , and once, for i∗ , ∗cannot be closer than 1 to the right border, so the gap between to combine the i∗ -th element from s(i ) into s(i) . Note that ∗lτ (a) and a is at most 1/ − 1. both s(i) and s(i ) are feasible allocations because they are result from knapsack solver. But we have to show feasibility (2 ) 1 of the combined allocation s. 1 2 n Lemma 5. s is feasible, or in the other word i=0 si ≤ m. 1 1/ Proof: There are 3 cases. First, when consensus function Now, suppose there are two number a1 and a2 such that can be computed for both i∗ and i = i∗ and also agree the same ∗lτ (a1 ) lτ (a2 ). Then a1 and a2 must be in different intervals. value : Li∗ = Li = ⊥. Thus, s(i ) = s(i) so there is nothing to ∗ ∗And because a1 cannot be closer than 1 to the right border of be combined. Hence, s(i ) = s(i) = s. And since s(i ) and s(i)its interval, so a1 lτ (a2 ) − 1, satisfying the third property. are feasible, so is s. Second case is when one of them fail to n compute : Li∗ = ⊥ or Li = ⊥. If both fail, then i=0 si = 0 ( + 1) and is trivially feasible. If only one of them fails, then s is 1 the combination of feasible allocation and empty allocation, which means the number of items allocated never increase, so 1/ s is feasible. The last case is when they both can be computed but do not For the ﬁrst property, because τ is the random variable of agree the same value : Li∗ = Li = ⊥. This implies Li∗ Liinterval of size 1, we can also view xτ (k + 1) = ((k + 1) + because we can rewrite them as Lτ (v(−i∗ ) ) and Lτ (vmax ),τ ) 1 as a random variable of interval of size 1/ , picked from max respectively, and v(−i∗ ) ≤ vmax by deﬁnition. Moreover, by(a, a + 1 ]. And it fails to compute if it turns out to be in max 1 the third property of consensus function, if Lτ (v(−i∗ ) ) Li ,(a, a + 1]. Hence, P r[lτ (a) = ⊥] = 1/ = . max then v(−i∗ ) Li /N = ∆i . This means the max bids of In fact, in the mechanism, we use the multiplicative maxvariation of this consensus function. Particularly, we deﬁne all bidders except i∗ are less than ∆i , and ∆i -perturbedLτ (a) = lτ (logN a) where N is some constant to be deﬁned. maximizer will give them nothing. So only bidder i∗ may getWe can verify that Lτ has similar properties as before. some items, implying i si = si∗ ≤ m. Next, we show how much social welfare we can achieve. 1) For a ∈ R, P r[Lτ (a) = ⊥] = . Let opt denote the optimal welfare for the given valuations. 2) If Lτ (a) = ⊥, then Lτ (a) ≤ a. The multiplicative gap is a/Lτ (a) ≤ N 1/ −1 . Lemma 6. The expected social welfare of the computed 3) If Lτ (a1 ) Lτ (a2 ), then a1 Lτ (a2 )/N . allocation s is at least (1 − 4 )opt. Proof: To begin with, observe that the probability that IV. T HE MECHANISM Li∗ and Li both can be computed is 1 − 2 , because by the We will introduce the payment scheme p later in the proof of ﬁrst property of consensus function, the probability that it failstruthfulness of mechanism. For now, the algorithm of the social to compute is , for both Li∗ and Li . In the following, we willchoice function f works as follows. Let N = 2(log m+3)n/ . prove that, if Li∗ and Li can be computed, approximation ratioPick τ uniformly from [0, 1] and ﬁx it. For each bidder i, we is at least 1−2 , which altogether gives the approximation ratioget the number si of the items allocated to bidder i by the at least (1 − 2 )2 ≥ (1 − 4 ) and, thus, proves the lemma.following way. We ﬁrst analyze only the social welfare of allocation ∗ • Compute Li = Lτ (v (−i) ), where v (−i) = maxj=i vj (m) s(i ) calculated by ∆i∗ -perturbed maximizer. By property of max max ∗ is the largest max bid excluding i’s. ∆i∗ -perturbed maximizer, the additive error is v (s(i ) ) − ∗ • If Li = ⊥, then si = 0. Bidder i drops out and gets v(s(i ) ) ≤ 2(log m + 3)n∆i∗ = 2(log m + 3)nLi∗ /N = Li∗ , nothing. by deﬁnition of N . As we also know that Li∗ = Lτ (v(−i∗ ) ) ≤ max • If Li = ⊥, then call ∆i -perturbed maximizer with ∆i = v(−i∗ ) ≤ vmax ≤ opt, this gives max Li /N . This call gives us the allocation s(i) . We set ith ∗ ∗ (i) element of s(i) to si . That is si = si . v(s(i ) ) ≥ v (s(i ) ) − Li∗ ∗To sum up, the result allocation is a combination of allo- ≥ v (s(i ) ) − optcations, where bidder i gets ith element of allocation from ≥ opt − opt = (1 − )opt
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4 The last inequality holds because, suppose that s∗ is the also ∆i , as ∆i = Li /N = Lτ (v (−i) )/N . Furthermore, maxoptimal allocation maximizing v(·). Since, by deﬁnition, we what ∆i -perturbed maximizer maximizes is exactly his utilityonly add positive values to valuation function to get the vi (s) − pi . Two conditions are satisﬁed and this concludes the ∗perturbed one, so v (s∗ ) ≥ v(s∗ ), and because s(i ) maximizes proof. ∗v (·), so v (s(i ) ) ≥ v (s∗ ) ≥ v(s∗ ) = opt. One may observe that now the payment pi = −( 2q(si ) + Now, to actually analyze the social welfare of the combined xsi )∆i + j=i vj (sj ) ) is negative, which is not practical. iallocation s, there are 2 cases when both Li∗ and Li can be We can ﬁx this problem by making bidder i pay additionalcomputed. First, when Li∗ = Li , we know that there is no payment hi chosen according to Clarke’s pivot rule. To clarify, ∗combination in this case, and thus s = s(i ) , which means the let ∆i : V n → A be a function mapping valuation functions tosocial welfare is at least (1 − )opt. the allocation selected by ∆i -perturbed maximizer. We deﬁne Next, when Li∗ = Li , by the analysis in last lemma, in this hi (v−i ) = v (∆i (0, v−i )) be the “perturbed” social welfare ofcase, there is only bidder i∗ who gets the items. Hence, the the allocation from ∆i -perturbed maximizer, given that bidder (i∗ )social welfare is vi∗ (si∗ ). Also, the others’ max bids are all i bids zero. We can verify that 0 ≤ pi + hi ≤ vi .less than ∆i , which means v(−i∗ ) ∆i = Li /N . Thus, the max Lemma 9. For each bidder i, the payment is non-negativemechanism achieves social welfare and does not exceed the amount that the bidder is willing to (i∗ ) ∗ ∗ pay : 0 ≤ pi + hi ≤ vi . vi∗ (si∗ ) = v(s(i ) ) − vi (s(i ) ) i=i∗ Proof: Suppose the allocation s maximizes v (·) and the ≥ (1 − )opt − (n − 1)v(−i∗ ) allocation s maximizes v (∆i (0, v−i )) = hi . We ﬁrst show max (1 − )opt − (n − 1)Li /N that pi + hi ≥ 0. By deﬁnition of hi , bidder i bid zero, so he gets nothing si = 0. Thus, hi = v (s ) = (2q(0) + (1 − )opt − Li x0 )∆i + j=i vj (s ). Also by deﬁnition, pi = −( (2q(si ) + i ≥ (1 − 2 )opt xsi )∆i + j=i vj (s) ). Therefore, all we need to show is i 2q(0) + x0 ≥ 2q(si ) + xsi and j=i vj (s ) ≥ j=i vj (s). i i For convenience in proving truthfulness, we refer to direct The former inequality is because if si = 0, then they are equal.characterization of truthful mechanism. If si = 0, then 2q(0) + x0 ≥ 2q(si ) + 2 2q(si ) + xsi by i i deﬁnition of q(0). The latter one is because we can observeProposition 7. A mechanism is truthful if and only if it satisﬁes that s is chosen to maximizes the term j=i vj (·).these conditions for every i and every v−i : Next, we show that pi + hi ≤ vi , that is vi − pi ≥ hi . Since, 1) Payment pi does not depend on vi . It may, though, vi − pi = v (s), hi = v (s ), and s maximizes v (·), therefore depend on the allocation s and v−i . That is we can v (s) v (s ). (i) deﬁne the payment like this pi = qs (v−i ). Finally, we need to show that ∆ is set so that, for ﬁxed , 2) Social choice function f maximizes the utility. That is P/∆ = poly(n, log m) to conclude that ∆i -perturbed maxi- (i) mizer runs in polynomial time, and so does the mechanism. By f (v) = argmaxs ( vi (s) − qs (v−i ) ). To informally explain correctness of the proposition, sup- our deﬁnitions, P is the second largest max bid, so P = v(−i∗ ) . maxpose that, for bidder i, the mechanism selects allocation s Also, by the gap of consensus function, a/Lτ (a) ≤ N 1/ −1when he tell the truth, and selects s when he lies. Since, for any a, therefore, v(−i∗ ) /Lτ (v(−i∗ ) ) ≤ N 1/ −1 . This gives max maxs = f (v) = argmaxs ( vi (s)−qs ), so the utility when he tells 1/ −1the truth vi (s) − qs is never less than when he lies vi (s ) − qs . P = v(−i∗ ) ≤ N Li∗ = N 1/ ∆i∗ maxThus, the mechanism is truthful. Since N = 2(log m + 3)n/ and ∆i∗ ≤ ∆i , so P/∆ ≤Lemma 8. The described mechanism is universally truthful. P/∆i∗ = N 1/ = poly(n, log m) for ﬁxed . Proof: To prove universal truthfulness, we need to show Theorem 10. There exists a randomized polynomial timethat, after we ﬁx all the random variables, τ and xj , the i approximation scheme for multi-unit auction that is universallymechanism is still truthful. Considering bidder i, there are truthful.2 cases. First, when Li = ⊥, since Li = Lτ (v (−i) ) which maxdepends on only v−i . So bidder i cannot do anything, andhas no incentive to lie. Next, when Li = ⊥, his allocationis selected by ∆i -perturbed maximizer. Since ∆i -perturbed R EFERENCESmaximizer maximizes v (s), we can set his payment pi to [1] R ENÉ B EIER AND B ERTHOLD VÖCKING . R ANDOM KNAPSACK INsatisfy conditions in direct characterization by this way. EXPECTED POLYNOMIAL TIME . J. Comput. Syst. Sci., 69(3):306–329, 2004. [2] S HAHAR D OBZINSKI AND S HADDIN D UGHMI . O N THE POWER OF v (s) = vi (si ) + (2q(si ) + xsi )∆i + i vj (sj ) RANDOMIZATION IN ALGORITHMIC MECHANISM DESIGN . CoRR, j=i ABS /0904.4193, 2009. [3] S HAHAR D OBZINSKI AND N OAM N ISAN . M ECHANISMS FOR MULTI - −pi UNIT AUCTIONS . J. Artif. Intell. Res. (JAIR), 37:85–98, 2010. We can see that the payment pi does not depend on vi . [4] H ANS K ELLERER , U LRICH P FERSCHY, AND DAVID P ISINGER . Knap- sack problems. S PRINGER , 2004.Because q(·), xsi , v−i trivially do not depend on vi , and i
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