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Summary of “A Universally-Truthful ApproximationScheme for Multi-unit Auction”

Berthold Vöcking, AuthorThatchaphol Saranurak, Presenter

I. INTRODUCTION

The paper considers multi-unit auction problem. In thisauction, a set of m identical items has to be allocated ton bidders. Each bidder i bids his valuation function vi :{0, ...,m} → R≥0, indicating how much i is willing to pay foreach amount of items. We consider only the set V of valuationfunctions which are non-decreasing and vi(0) = 0.

After everyone bids, we get an allocation s =(s1, s2, ..., sn) ∈ {0, ...,m}n indicating how many items eachbidder gets. The set of feasible allocation A = {s |

∑ni=1 si ≤

m} is set of allocations allocating at most m items to bidders.Besides, vi depends on only his amount of items si, nots−i, that is vi(s) = vi(si). The objective of this problemis to find allocation s ∈ A maximizing the social welfarev(s) =

∑i vi(s).

A mechanism for multi-unit auctions is a pair (f, p) wheref : V n → A is a social choice function gathering valu-ation functions and then outputs allocation for bidders, andp = (p1, p2, ..., pn), pi : V n → R is payment scheme. It isassumed that valuation functions are not given explicitly, but inthe form of black box that can be queried by mechanism. Thus,size of instance of problem, excluding valuation functions, arethe number of bits representing m, as items are identical, andthe number of bidders : poly(n, logm). Therefore polynomialtime mechanism in this context should run in poly(n, logm)time.

Defining utility of bidder i to be his valuation subtractedby his payment vi(f(vi, v−i)) − pi(vi, v−i), a deterministicmechanism is truthful if, for each bidder i, his utility isalways maximized when he bids the true vi. For randomizedmechanism, which is actually a probability distribution overdeterministic mechanisms, there are 2 kinds of truthfulness.Firstly, a mechanism is universally truthful if each of thesedeterministic mechanisms in the distribution is truthful. In thiscase, randomness is used only for guaranteeing other aspectse.g. social welfare. Secondly, the weaker one, a mechanismis truthful in expectation if, given that bidders do not knowrandom bits of mechanism, the expected bidder’s utility ismaximized by telling the truth.

It is known that there is deterministic mechanism for thisauction running in polynomial-time and has approximationratio of 2, and if the mechanism follows Robert’s characteri-zation, i.e., affine maximizer with VCG-based payments, thenthis bound is tight [3]. For truthful in expectation mechanism,Dobzinski and Dughmi [2] presented that there is FPTAS, and,moreover, also questioned the existence of universally truthful

mechanism running in polynomial time with an approximationratio better than 2. They showed the impossibility, if anoutput allocation of mechanism is restricted with the certainconstraint. However, the main result of this paper refutesthis question and shows that there is randomized PTAS foruniversally truthful mechanism.

To get such result, there are two majors tools used cruciallyin this papers : ∆-perturbed maximizer and consensus functionwith drop-outs.

II. ∆-PERTURBED MAXIMIZER

For convenience of defining ∆-perturbed maximizer, let usconsider an instance of multiple-choice knapsack problem.Suppose there are n classes of objects and m objects in eachclass. Each object k has weight wk and profit pk. We want analgorithm for selecting exactly one object from each class, withsummation of weight of objects at most m, and maximizingsummation of profit. Observe that we can maximize socialwelfare using such algorithm, by defining a tuple (i, j) be anobject for allocating j items to bidder i and set w(i,j) = j andp(i,j) = vi(j). The allocation from knapsack would be feasiblebecause number of items allocated is Σkwk ≤ m, and indeedsocial welfare Σivi = Σkpk is maximized. Nevertheless,polynomial-time constraint is not satisfied yet as number ofobjects is n×m which is not poly(n, logm), and also generalknapsack is NP-hard problem.

To deal with this issue, we introduce perturbed valuationfunction.

Definition 1 (Perturbed valuation function). Let ∆ > 0. For1 ≤ i ≤ n, 0 ≤ j ≤ m, let xj

i be a random variable chosenindependently, uniformly from [0, 1]. For k ∈ {1, ...,m}, q(k)is number of factor 2 of k, e.g. q(96) = q(25×3) = 5. Hence,q(k) ≤ blogmc. Let q(0) = blogmc+ 1 as an exception.

v′i(j) = v(j) + (2q(j) + xji )∆

Now, by setting profit of object p(i,j) = v′i(j) insteadof vi(j), we define ∆-perturbed maximizer be the algorithmthat solves such knapsack instance, and so maximizes Σiv

′i.

We claim we can achieve the expected running time ofpoly(n, logm, P/∆), where P is the second largest numberof maximum bid vi(m) of all bidders. As we will see later,∆ will be set to make P/∆ be poly(n, logm) for some fixedε > 0. Thus, ∆-perturbed maximizer run in poly(n, logm).

Briefly, the term q(·) allows us to consider onlypoly(n, logm, P/∆) objects and the term x enables us to

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solve knapsack in expected time polynomial in number ofobjects. We illustrate some part of the proof and informalexplanation as follows.

A. Term q

To show how we can focus only small portion of objects,we have to define breakpoints of valuation function. LetVi = (vi(0), vi(1), ..., vi(m)) denotes the non-decreasingsequence of bids of bidder i. We partition Vi into subsequencesV qi = (vi(k))|q(k)=q for 0 ≤ q ≤ blogmc + 1. To illustrate,

V qi = (vi(1 · 2q), vi(3 · 2q), vi(5 · 2q), ...) with an exception

Vblog mc+1i = (vi(0)). The q-breakpoints of bidder i are

smallest indices k in V qi 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) = Σiv′i(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 (blogmc+1)-breakpoint. Let q = q(si), then, forsome k, si = (2k+1)2q . We define s′i = si − 2q = (2k)2q =k2q+1 and define q′ = q(si). We get that q′ ≥ q + 1.

Next, consider V qi and si is in some interval of ∆, si ∈

[c∆, (c+1)∆) for some c. There must be a predecessor of siin such interval, otherwise si in a breakpoint. We also knowthat the predecessor is (2k − 1)2q = si − 2q+1. This givesvi(si) < vi(si − 2q+1) + ∆ < vi(s

′i) + ∆. Thus, we get

v′i(si) = vi(si) + (2q + xji )∆

< (vi(s′i) + ∆) + (2q′ − 2 + 1)∆

= vi(s′i) + 2q′∆

≤ v′i(s′i)

Now, consider allocation s′ with s′−i = s−i and s′i = si−2q

as defined. This allocation is feasible because we allocate lessitems but v′(s) < v′(s′) which yields a contradiction.

As a result, we may input only objects which are break-points into knapsack solver. To count number of all break-points, let bidder i∗ = arcmaxivi(m) be the one who bidsthe highest max bid. Considering bidder i 6= i∗, there areat most P/∆ q-breakpoints for each q. With a certain trickomitted here, we can first find allocation, by solving knapsack,for all bidders excluding i∗, and then supplement such allo-cation to maximize v′(s) in polynomial-time. Thus, there arepoly(n, logm, P/∆) objects inputted to ∆-perturbed maxi-mizer.

B. Term x

The term x is chosen independently uniformly at randomfrom [0, 1], which fits into a framework of smoothed analysisfor knapsack problem [1]. Particularly, the term x makes theexpected number of Pareto-optimal subsets be polynomial innumber of objects. Furthermore, the dynamic programmingframework of Nemhauser and Ullmann [4] can solve knapsackin time polynomial in number of Pareto-optimal subsets. Thus,the expected running time is polynomial.

Informally, the reason why we can use randomness tobound the running time of knapsack problem, which isNP-hard, to be polynomial, is because the “hard instances” ofknapsack problem are “isolated” in some sence, as illustratedin the left figure below. By perturbing the these instanceswith random variable, the expected running time is an averagerunning time of hard instances with many easy instancessurrounding them. Consequently, the running time might bedecreased dramatically, as in the right figure below.

C. Some properties of ∆-perturbed maximizerLemma 3. The ∆-perturbed maximizer selects an allocationmaximizing the social welfare, which has additive error v′(s)−v(s) ≤ (2 blogmc+ 3)∆n.

Proof: Consider v′i(s) = vi(s) + (2q(s) + xji )∆. Since,

q ≤ blogmc+1 and xji ≤ 1, so v′i(s)− vi(s) ≤ (2 blogmc+

3)∆. As a result, v′(s)− v(s) =∑n

i=0 v′i(s)−

∑ni=0 vi(s) ≤

(2 blogmc+ 3)∆n.

Lemma 4. Bidder i get nothing si = 0, if his max bid vi(m) <∆.

Proof: Since, for any amount of items j > 0, vi(j) ≤vj(m) < ∆, and, by definition, q(j) + 1 ≤ q(0).

v′i(0) = (2q(0) + x0i )∆

≥ (2q(j) + 2)∆

> vi(j) + (2q(j) + 1)∆

≥ v′i(j)

Because v′i(0) > v′i(j), ∆-perturbed maximizer selects si =0.

III. CONSENSUS FUNCTION WITH DROP-OUTS

Let ε > 0, and τ is a random variable chosen uniformlyfrom [0, 1]. We want a function l with following properties

1) For a ∈ R, it fails to compute with low probability :Pr[lτ (a) = ⊥] = ε.

2) If it can be computed lτ (a) 6= ⊥, then lτ (a) ≤ a. Also,we want a “gap” between them to be small : a−lτ (a) ≤1/ε− 1.

3) If l(a1) < l(a2), then it guarantees a certain separationa1 < l(a2)− 1.

To define lτ (a), we introduce xτ (i) = (i+τ) 1ε for each i ∈ Z.The interval between each xτ (·) is 1/ε. Now let k be thelargest integer such that xτ (k) ≤ a as in the figure.

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1/𝜖

𝑙 𝑎 = 𝑥𝜏(𝑘) 𝑎

1

𝑑

𝑥𝜏(𝑘 + 1) 𝑙 . =⊥

Let d be the difference between a and xτ (k + 1). Now,lτ (a) = xτ (k) if d > 1, otherwise it fails to computelτ (a) = ⊥. By this definition, this satisfies the second propertybecause, if it can be computed, so lτ (a) = xτ (k) ≤ a. Also,since lτ (a) is at the left border of the interval and, as d > 1, acannot be closer than 1 to the right border, so the gap betweenlτ (a) and a is at most 1/ε− 1.

1/𝜖

𝑎1 1 𝑙𝜏(𝑎2)

𝑎2

1

Now, suppose there are two number a1 and a2 such thatlτ (a1) < lτ (a2). Then a1 and a2 must be in different intervals.And because a1 cannot be closer than 1 to the right border ofits interval, so a1 < lτ (a2)− 1, satisfying the third property.

1/𝜖

𝑎 1 𝑥𝜏(𝑘 + 1)

For the first property, because τ is the random variable ofinterval of size 1, we can also view xτ (k + 1) = ((k + 1) +τ) 1ε as a random variable of interval of size 1/ε, picked from(a, a + 1

ε ]. And it fails to compute if it turns out to be in(a, a+ 1]. Hence, Pr[lτ (a) = ⊥] = 1

1/ε = ε.In fact, in the mechanism, we use the multiplicative

variation of this consensus function. Particularly, we defineLτ (a) = lτ (logNa) where N is some constant to be defined.We can verify that Lτ has similar properties as before.

1) For a ∈ R, Pr[Lτ (a) = ⊥] = ε.2) If Lτ (a) 6= ⊥, then Lτ (a) ≤ a. The multiplicative gap

is a/Lτ (a) ≤ N1/ε−1.3) If Lτ (a1) < Lτ (a2), then a1 < Lτ (a2)/N .

IV. THE MECHANISM

We will introduce the payment scheme p later in the proof oftruthfulness of mechanism. For now, the algorithm of the socialchoice function f works as follows. Let N = 2(logm+3)n/ε.Pick τ uniformly from [0, 1] and fix it. For each bidder i, weget the number si of the items allocated to bidder i by thefollowing way.

• Compute Li = Lτ (v (−i)max

), where v (−i)max

= maxj 6=ivj(m)

is the largest max bid excluding i’s.• If Li = ⊥, then si = 0. Bidder i drops out and gets

nothing.• If Li 6= ⊥, then call ∆i-perturbed maximizer with ∆i =

Li/N . This call gives us the allocation s(i). We set ith

element of s(i) to si. That is si = s(i)i .

To sum up, the result allocation is a combination of allo-cations, where bidder i gets ith element of allocation from

∆i-perturbed maximizer. However, there are actually onlytwo different allocations to be combined. To see this, leti∗ = argmaxivi(m) be the one who bids the highest maxbid. For bidder i 6= i∗, v (−i)

max= vmax are all the same. Since

these bidders use the parameter ∆i = Li/N = Lτ (vmax)/Nin ∆-perturbed maximizer, they compute the same allocations(i). Moreover, in the case that Li = ⊥, they all also drop outtogether.

Thus, we can call ∆-perturbed maximizer for only twotimes, once for any i 6= i∗, which yields s(i), and once, for i∗,to combine the i∗-th element from s(i

∗) into s(i). Note thatboth s(i) and s(i

∗) are feasible allocations because they areresult from knapsack solver. But we have to show feasibilityof the combined allocation s.

Lemma 5. s is feasible, or in the other word∑n

i=0 si ≤ m.

Proof: There are 3 cases. First, when consensus functioncan be computed for both i∗and i 6= i∗and also agree the samevalue : Li∗ = Li 6= ⊥. Thus, s(i

∗) = s(i) so there is nothing tobe combined. Hence, s(i

∗) = s(i) = s. And since s(i∗) and s(i)

are feasible, so is s. Second case is when one of them fail tocompute : Li∗ = ⊥ or Li = ⊥. If both fail, then

∑ni=0 si = 0

and is trivially feasible. If only one of them fails, then s isthe combination of feasible allocation and empty allocation,which means the number of items allocated never increase, sos is feasible.

The last case is when they both can be computed but do notagree the same value : Li∗ 6= Li 6= ⊥. This implies Li∗ < Li

because we can rewrite them as Lτ (v(−i∗)max

) and Lτ (vmax),respectively, and v(−i∗)

max≤ vmax by definition. Moreover, by

the third property of consensus function, if Lτ (v(−i∗)max

) < Li,then v(−i∗)

max< Li/N = ∆i. This means the max bids of

all bidders except i∗ are less than ∆i, and ∆i-perturbedmaximizer will give them nothing. So only bidder i∗ may getsome items, implying

∑i si = si∗ ≤ m.

Next, we show how much social welfare we can achieve.Let opt denote the optimal welfare for the given valuations.

Lemma 6. The expected social welfare of the computedallocation s is at least (1− 4ε)opt.

Proof: To begin with, observe that the probability thatLi∗ and Li both can be computed is 1 − 2ε, because by thefirst property of consensus function, the probability that it failsto compute is ε, for both Li∗ and Li. In the following, we willprove that, if Li∗ and Li can be computed, approximation ratiois at least 1−2ε, which altogether gives the approximation ratioat least (1− 2ε)2 ≥ (1− 4ε) and, thus, proves the lemma.

We first analyze only the social welfare of allocations(i

∗) calculated by ∆i∗ -perturbed maximizer. By property of∆i∗ -perturbed maximizer, the additive error is v′(s(i

∗)) −v(s(i

∗)) ≤ 2(log m+3)n∆i∗ = 2(log m+3)nLi∗/N = εLi∗ ,by definition of N . As we also know that Li∗ = Lτ (v(−i∗)

max) ≤

v(−i∗)max

≤ vmax ≤ opt, this gives

v(s(i∗)) ≥ v′(s(i

∗))− εLi∗

≥ v′(s(i∗))− εopt

≥ opt− εopt = (1− ε)opt

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The last inequality holds because, suppose that s∗ is theoptimal allocation maximizing v(·). Since, by definition, weonly add positive values to valuation function to get theperturbed one, so v′(s∗) ≥ v(s∗), and because s(i

∗) maximizesv′(·), so v′(s(i

∗)) ≥ v′(s∗) ≥ v(s∗) = opt.Now, to actually analyze the social welfare of the combined

allocation s, there are 2 cases when both Li∗ and Li can becomputed. First, when Li∗ = Li, we know that there is nocombination in this case, and thus s = s(i

∗), which means thesocial welfare is at least (1− ε)opt.

Next, when Li∗ 6= Li, by the analysis in last lemma, in thiscase, there is only bidder i∗ who gets the items. Hence, thesocial welfare is vi∗(s

(i∗)i∗ ). Also, the others’ max bids are all

less than ∆i, which means v(−i∗)max

< ∆i = Li/N . Thus, themechanism achieves social welfare

vi∗(s(i∗)i∗ ) = v(s(i

∗))−∑i 6=i∗

vi(s(i∗))

≥ (1− ε)opt− (n− 1)v(−i∗)max

> (1− ε)opt− (n− 1)Li/N

> (1− ε)opt− εLi

≥ (1− 2ε)opt

For convenience in proving truthfulness, we refer to directcharacterization of truthful mechanism.

Proposition 7. A mechanism is truthful if and only if it satisfiesthese conditions for every i and every v−i :

1) Payment pi does not depend on vi. It may, though,depend on the allocation s and v−i. That is we candefine the payment like this pi = q

(i)s (v−i).

2) Social choice function f maximizes the utility. That isf(v) = argmaxs( vi(s)− q

(i)s (v−i) ).

To informally explain correctness of the proposition, sup-pose that, for bidder i, the mechanism selects allocation swhen he tell the truth, and selects s′ when he lies. Since,s = f(v) = argmaxs( vi(s)−qs ), so the utility when he tellsthe truth vi(s)−qs is never less than when he lies vi(s′)−qs′ .Thus, the mechanism is truthful.

Lemma 8. The described mechanism is universally truthful.

Proof: To prove universal truthfulness, we need to showthat, after we fix all the random variables, τ and xj

i , themechanism is still truthful. Considering bidder i, there are2 cases. First, when Li = ⊥, since Li = Lτ (v (−i)

max) which

depends on only v−i. So bidder i cannot do anything, andhas no incentive to lie. Next, when Li 6= ⊥, his allocationis selected by ∆i-perturbed maximizer. Since ∆i-perturbedmaximizer maximizes v′(s), we can set his payment pi tosatisfy conditions in direct characterization by this way.

v′(s) = vi(si) + (2q(si) + xsii )∆i +

∑j 6=i

v′j(sj)︸ ︷︷ ︸−pi

We can see that the payment pi does not depend on vi.Because q(·), xsi

i , v−i trivially do not depend on vi, and

also ∆i, as ∆i = Li/N = Lτ (v (−i)max

)/N . Furthermore,what ∆i-perturbed maximizer maximizes is exactly his utilityvi(s)−pi. Two conditions are satisfied and this concludes theproof.

One may observe that now the payment pi = −( 2q(si) +xsii )∆i +

∑j 6=i v

′j(sj) ) is negative, which is not practical.

We can fix this problem by making bidder i pay additionalpayment hi chosen according to Clarke’s pivot rule. To clarify,let ∆i : V

n → A be a function mapping valuation functions tothe allocation selected by ∆i-perturbed maximizer. We definehi(v−i) = v′(∆i(0, v−i)) be the “perturbed” social welfare ofthe allocation from ∆i-perturbed maximizer, given that bidderi bids zero. We can verify that 0 ≤ pi + hi ≤ vi.

Lemma 9. For each bidder i, the payment is non-negativeand does not exceed the amount that the bidder is willing topay : 0 ≤ pi + hi ≤ vi.

Proof: Suppose the allocation s maximizes v′(·) and theallocation s′ maximizes v′(∆i(0, v−i)) = hi. We first showthat pi + hi ≥ 0. By definition of hi, bidder i bid zero,so he gets nothing s′i = 0. Thus, hi = v′(s′) = (2q(0) +x0i )∆i +

∑j 6=i v

′j(s

′). Also by definition, pi = −( (2q(si) +xsii )∆i +

∑j 6=i v

′j(s) ). Therefore, all we need to show is

2q(0) + x0i ≥ 2q(si) + xsi

i and∑

j 6=i v′j(s

′) ≥∑

j 6=i v′j(s).

The former inequality is because if si = 0, then they are equal.If si 6= 0, then 2q(0) + x0

i ≥ 2q(si) + 2 > 2q(si) + xsii by

definition of q(0). The latter one is because we can observethat s′ is chosen to maximizes the term

∑j 6=i v

′j(·).

Next, we show that pi+hi ≤ vi, that is vi−pi ≥ hi. Since,vi − pi = v′(s), hi = v′(s′), and s maximizes v′(·), thereforev′(s) > v′(s′).

Finally, we need to show that ∆ is set so that, for fixed ε,P/∆ = poly(n, logm) to conclude that ∆i-perturbed maxi-mizer runs in polynomial time, and so does the mechanism. Byour definitions, P is the second largest max bid, so P = v(−i∗)

max.

Also, by the gap of consensus function, a/Lτ (a) ≤ N1/ε−1

for any a, therefore, v(−i∗)max

/Lτ (v(−i∗)max

) ≤ N1/ε−1. This gives

P = v(−i∗)max

≤ N1/ε−1Li∗ = N1/ε∆i∗

Since N = 2(logm + 3)n/ε and ∆i∗ ≤ ∆i, so P/∆ ≤P/∆i∗ = N1/ε = poly(n, logm) for fixed ε.

Theorem 10. There exists a randomized polynomial timeapproximation scheme for multi-unit auction that is universallytruthful.

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[3] SHAHAR DOBZINSKI AND NOAM NISAN. MECHANISMS FOR MULTI-UNIT AUCTIONS. J. Artif. Intell. Res. (JAIR), 37:85–98, 2010.

[4] HANS KELLERER, ULRICH PFERSCHY, AND DAVID PISINGER. Knap-sack problems. SPRINGER, 2004.