A Social Welfare Otimal Sequential Allocation Procedure Thomas Kalinowsi Universität Rostoc, Germany Nina Narodytsa and Toby Walsh NICTA and UNSW, Australia May 2, 201 Abstract We consider a simle sequential allocation rocedure for sharing indivisible items between agents in which agents tae turns to ic items. Suosing additive utilities and indeendence between the agents, we show that the exected utility of each agent is comutable in olynomial time. Using this result, we rove that the exected utilitarian social welfare is maximized when agents tae alternate turns. We also argue that this mechanism remains otimal when agents behave strategically. 1 Introduction There exist a variety of mechanisms to share indivisible goods between agents without side ayments [2,, 4, 5, ]. One of the simlest is simly to let the agents tae turns to ic an item. This mechanism is arameterized by a olicy, the order in which agents tae turns. In the alternating olicy, agents tae turns in a fixed order, whilst in the balanced alternating olicy, the order of agents reverses each round. Bouveret and Lang [1] study a simle model of this mechanism in which agents have strict references over items, and utilities are additive. They conjecture that comuting the exected social welfare for a given olicy is NP-hard suosing all reference orderings are equally liely. Based on simulation for u to 12 items, they conjecture that the alternating olicy maximizes the exected utilitarian social welfare for Borda utilities, and rove it does so asymtotically. We close both conjectures. Surrisingly, we rove that the exected utility of each agent can be comuted in olynomial time for any olicy and utility function. Using this result, we rove that the alternating olicy maximizes the exected utilitarian social welfare for any number of items and any linear utility function including Borda. Our results rovides some justification for a mechanism in use in school laygrounds around the world. 2 Notation Each of the n agents has a total reference order over the items. A rofile is an n-tule of such orders. Agents have the same utility function. An item raned in th osition has a 1
utility g. For Borda utilities, g = + 1. The utility of a set of items is the sum of their individual utilities. Preference orders are indeendent and drawn uniformly at random from all! ossibilities full indeendence. Agents tae turns to ic items according to a olicy, a sequence π = π 1... π {1, 2,..., n}. At the -th ste, agent π chooses one item from the remaining set. Without loss of generality, we suose π 1 = 1. For rofile R, u R i, π denotes the utility gained by agent i suosing every agent always chooses the item remaining that they ran highest. We write u i π for the exectation of u R i, π over all ossible rofiles. We tae an utilitarian standoint, measuring social welfare by the sum of the utilities: sw R π = n i=1 u Ri, π. By linearity of exectation, swπ = n i=1 u iπ. To hel comute exectations, we need a sequence given by 1 = 2 = 1 and = 1/2 2j + 1 2j for and = / for all. Asymtotically, we have = 2 1 π + O. Comuting the Exected Social Welfare.1 Recursions Bouveret and Lang [1] conjectured that it is NP-hard to comute the exected social welfare of a given olicy as such a calculation deends on the suer-exonential number of ossible rofiles. However, as we show now, the exected utility of each agent can be comuted in just On 2 time for an arbitrary utility function, and On time for Borda utilities. We begin with this last case, and then extend the results to the general case. Let P n denote the set of all olicies of length for n agents. For 2, we define an oerator P n P 1 n maing π π, by deleting the the first entry. More recisely, For examle, π = 1211 and π = 211. π i = π i+1 for i [ 1]. Lemma 1. For Borda scoring, n 2 agents, 2 items and π P n with π 1 = 1, we have: u 1 π = + u 1 π, u i π = + 1 u i π, i 1 and these values can be comuted in On time. Proof. Agent 1 ics her first item, giving her a utility of. After that, from her ersective, it s the standard game on 1 items with olicy π, so she exects to get an utility of u 1 π. This roves the first equation. For the other agents, it is more involved. Let i {2,..., n} be a fixed agent. For q {1,..., }, let a i q, π denote the robability that under olicy π agent i gets the item with utility q. Note that this robability does not deend on the 2
utility function but only on the raning: it is the robability that agent i gets the item of ran q + 1 in her reference order. By the definition of exectation, u i π = a i q, πq. 1 q=1 There are three ossible outcomes of the first move of agent 1 with resect to the item that has utility q for agent i. With robability q 1/, agent 1 has iced an item with utility less than q for agent i, with robability q/, agent 1 has iced an item with utility more than q, and with robability 1/ it was the item of utility equal to q. In the first case there are only q 2 items of utility less than q left, hence the robability for agent i to get the item of utility q is a i q 1, π. In the second case there are still q 1 items of value less than q, hence the robability to get the item of utility q is aq, π. In the third case, the robability to get the item of utility is zero, and together we obtain Substituting this into 1 yields u i π = q=1 aq, π = q 1 a iq 1, π + q a iq, π. 2 [ q 1 a iq 1, π + q ] a iq, π q = q=1 In the first sum we substitute q = q 1 and this yields q 1q a i q 1, π + q=1 qq a i q, π u i π = 1 q =0 q a iq, π q + 1 + q=1 q a i q, π q The first term in the first sum and the last term in the second sum are equal to zero, so they can be omitted and we obtain u 2 π = 1 q =1 q a iq, π q + 1 + q=1 1 q=1 q 1 [ ] q q = a i q, π q + 1 + q a i q, π q = + 1 The time comlexity follows immediately from the recursions. 1 q=1 a i q, π q = + 1 u 2 π. Examle 1. Consider two agents with Borda utilities and the olicies π 1 = 121212 and π 2 = 111222. We comute exected utilities and exected social welfare for each of them using Lemma 1. Table 1 shows results u to two decimal laces. Note that exected values comuted in all examles in the aer coincide with the results obtained by the brute-force search algorithm from [1].
π 1 = 121212 π 2 = 111222 π u 1 π u 2 π swπ π u 1 π u 2 π swπ 1 2 0.00 1.00 1.00 2 0.00 1.00 1.00 2 12 2.00 1.50.50 22 0.00.00.00 212 2.7 4.50 7.17 222 0.00.00.00 4 1212.7 5. 12.0 1222 4.00 7.50 11.50 5 21212 8.00 10. 18. 11222 9.00 9.00 18.00 121212 14.00 12.40 2.40 111222 15.00 10.50 25.50 Table 1: Exected utilities and exected utilitarian social welfare comutation for π 1 = 121212 and π 2 = 111222 Due to the linearity of Borda scoring the robabilities a i q, π in the roof of Lemma 1 cancel, and this will allow us to solve recursions exlicitly and to rove our main result about the otimal olicy for Borda scoring in Section 4. In the general case, we can still comute the exected utilities ui, π, and thus swπ, but we need the robabilities a i q, π from the roof of Lemma 1: a i q, π is the robability that under olicy π, agent i gets the item raned at osition q + 1 in her reference order. Comuting these robabilities using 2 adds a factor of to the runtime. Lemma 2. For n 2 agents, 2 items, a olicy π P n and an arbitrary scoring function g, the exected utility for agent i is u i π = a i q, πgq q=1 and can be comuted in On 2 time. Lemma 2 allows us to resolve an oen question from [1]. Corollary 1. For n agents and an arbitrary scoring utility function g, the exected utility of each agent, as well as the exected utilitarian social welfare can be comuted in olynomial time..2 Secial olicies For some secial olicies, the recursions in Lemma 1 can be solved exlicitly. A articularly interesting olicy is the strictly alternating one denoted AltPolicy: π = 12... n12... n12... n.... The following roosition can be roved by induction using the recursions from Lemma 1. 4
Proosition 1. For n = 2 agents, let π be the strictly alternating olicy of length starting with 1. The exected utilities and utilitarian social welfare for Borda scoring are swπ = 1 [ ] 2 2 + 1 2 1 + 1 + +1 = + 1 2 1 π + O +1, 2 1 + 1 u 1 π, u 2 π = +1 if is even, +1 + 1 +1, 2 1 if is odd. Examle 2. Consider the olicy π 1 from Examle 1. Using Proosition 1, we get: u 1 π = +1 = 14, u 2 π = 2 1 + 1 +1 = 11.7 + 2.19/ = 12.4 and swπ = 2.4. These values coincide with results in Table 1. For n > 2 agents, we first solve the recursions for the exected utility of agent i when the number of items is i 1 mod n. Using these values, we aroximate the exected utility for the remaining combinations of and i. As the olicy is determined by the number of items we simlify notation by letting u i be the exected utility of agent i for the allocation of items. Then u 11 = 1, u 21 = u 1 = = u n1 = 0 and u 1 = + u n, 1, u i = + 1 u i 1, 1 i = 2,..., n for 2. Decouling these recursions we get for the first agent u 1 = for n, u 1 = + n + 1 u 1, n for > n. and this allows us to write down the exected utility exactly for one residue class modulo n er agent. Lemma. For i {1, 2,..., n}, if i 1 mod n then the exected utility of agent i equals i + 1 + 1 u i = n + 1 Proof. We start with i = 1 and rove by induction that u 1 = +1 n+1 for all 0 mod n. The induction starts at = n with u 1 = n = + 1/n + 1. For > n with 0 mod n, we have by induction u 1 = + n + 1 u n n + 1 1, n = + n + 1 n + 1 = + 1 n + 1. Now we roceed by induction on i. For i 2 our recursion yields u i = + 1 u i 1, 1 = + 1 1 i 1 + 1 1 + 1 n + 1 = i + 1 + 1. n + 1 For the remaining residue classes mod n we rovide asymtotic statements. 5
Proosition 2. For n > 2 agents, let π be the strictly alternating olicy of length starting with 1. The exected utilities and utilitarian social welfare for Borda scoring are, for all i {1,..., n}, u i π = 2 n2 + O, swπ = n + 1 n + 1 + O. Proof. For i 1 mod n this follows from Lemma. Otherwise let be the unique element of {1,..., n 1} such that i 1 mod n. The following estimates rove the claim. From u i, u i u i,+n it follows that i + 1 + 1 n + 1 u i + n i + 1 + n + 1, n + 1 hence 2 n + 1 2 u i 2 + 2 + n. n + 1 For a fixed number n of agents, we write the number of items as = n + r with 0 r < n. We call a olicy π = π 1... π balanced if {π in+1,..., π in+n } = {1, 2,..., n} for all i {0,..., 1} and {π n+1,..., π n+r } = r. For any balanced olicy, the exected utility of any agent lies between that of agents 1 and n in the alternating olicy. Thus we have the following Corollary. Corollary 2. When items are allocated to n agents with Borda scoring, then for any balanced olicy, every agent has exected utility 2 /n + 1 + O. As in [1], we define asymtotic otimality of a sequence of olicies π =1,2, where π is a olicy for items by lim swπ max π P swπ = 1. We will show that every balanced olicy is asymtotically otimal. To this end we analyze the best reference mechanism BestPref which allocates each item to the agent who refers it most breaing ties at random. Proosition. For two agents, and Borda utilities, the exected utilitarian social welfare of BestPref is +14 1. Proof. Let S be the set of all ermutations of {1, 2,..., }. Then the exected utilitarian social welfare is 1 max{i, a i } = 1 max{i, a i }.!! a S i=1 i=1 a S We slit the inner sum into two sums: one for all ermutations a = a 1,..., a with a i i, and one for the remaining ermutations. Hence, we get 1 i + a i.! i=1 a S : a i i a S : a i >i
We comute the value of each inner sum searately. In the first sum each term equals i, so we have to determine the number of terms. For a i there are i ossible values 1, 2,..., i, and for a fixed value of a i there are 1! ermutations of the remaining values. So the first sum has 1!i terms of value i, hence it equals 1!i 2. The second sum contains for each j {i + 1,..., } exactly 1! terms of value a i = j, hence it equals 1! j=i+1 j = 1! + 1/2 ii + 1/2. So the exected utilitarian social welfare is 1 1! [i 2 + 12! + 1 12 ] ii + 1 i=1 = 1 [ + 1 + i 2 i ] + 14 1 =. 2 i=1 Proosition 4. For n > 2 agents, and Borda utilities, the exected utilitarian social welfare of BestPref is n2 + + O1. n+1 2 Proof. The exected utilitarian social welfare for this rocedure is the exected value of the random variable X = q=1 max 1 i n α iq where the n vectors α i = α i1, α i2,..., α i are random ermutations of the set {1, 2,..., } that are drawn indeendent and uniform from the set of all! ermutations. The interretation is that we fix an order of the items, and a iq is the value of item q for agent i according to her random reference order. We decomose X as a sum of random variables X q = max 1 i n α iq and calculate the exected value of these. For the robability that X q taes value j we can write n n n 1 j 1 1 P X q = j = = + j 1 n n j 1 =1 n n j j 1 =. The exected value of X q is n n j j 1 E X q = j = 1 j [j n j 1 n ] n [ ] [ = 1 1 ] j n+1 j + 1j n = 1 1 1 j n+1 j n+1 j n n n j=0 j=0 j=0 = 1 [ n+1 1 n n + 1 1n+1 1 ] 2 1n + O n 1 = 1 [ n n n + 1 n+1 + 1 ] 2 n + O n 1 = n n + 1 + 1 2 + O1/. 7
Now summation over q yields the result. Clearly, BestPref reresents an uer bound on the exected utilitarian social welfare for any allocation mechanism. Hence Corollary 2 imlies that all balanced olicies are asymtotically otimal. Corollary. Every sequence π =1,2, of balanced olicies is asymtotically otimal. This is Proosition 5 of Bouveret and Lang [1]. However, the roof in [1] is incorrect as we exlain in the next subsection.. Analysis of the roof of the asymtotic otimality of balanced olicies For i = 1,..., let the sequence of stages i 1n + 1, i 1n + 2,..., in be called the i-th round, i.e. for a balanced olicy, in every round each agent ics exactly one item. Bouveret and Lang start with the observation that in the first round the first agent gets utility and the second one 2 1/ = 1+o1. They say that the third agent gets Θ which is too wea for what they want to derive: If the third agent would get /2 this would be Θ but not + O 1 which is what the claim next. The roof of this exectation of + O 1 is already nontrivial and could be done as follows. The j-th agent of the first round gets her most referred item with robability 1 j 1 / j 1 = j+1 and her second most referred item with robability 2 j 2 / j 1 = j 1 j+1, so her exected utility from 1 the first round is at least j + 1 + j 1 j + 1 = j 12 = + O 1. Then Bouveret and Lang continue by stating that in the second round the starting agent gets her second referred item with robability 1 n 1. This is only true if the second round 1 starts with the same agent as the first round. Otherwise one has to tae into account the robability, that the first agent of the second round too her second favourite item already in the first round because her first choice was not available. For instance if the starting agent of the second round was second in the first round then her robability to ic her second referred item in the second round equals 2 n 2 = 1 n. 2 n 2 They argue that the starting agent of round two gets utility at least 1 n 1 1 = 1 1 + O 1, and this last equality is clearly wrong. In order to show that every agent from the second round gets utility at least 1 + O 1 it would be necessary to consider not only the robability that the agent gets her second most referred item in the second round, but also robabilities for other items just lie for the first round we had to tae into account the most referred and the second most referred item. It seems ossible that this can be done maybe just the third referred item is sufficient, but it is in no way obvious how to do it. 8
It seems to be very difficult to generalize this from the second round to the following rounds. Their claim is that in round i every agent gets utility i + 1 + O 1. It might be that this is true although highly non-obvious for bounded i, but Bouveret and Lang use this statement for all i u to which tends to infinity with. Even if the i + 1 + O 1 utility for round i would be correct, their calculation of the total utility [ + O 1 + + + 1 + O 1 ] of any agent is still wrong. It should be: i + 1 + O 1 = i=1 1 2 + O1 = 2n 12 2n 2 4 Otimality of the Alternating Policy + 2n + O1. We now consider the roblem of finding the olicy that maximizes the exected utilitarian social welfare for Borda utilities. Bouveret and Lang [1] stated that this is an oen question, and conjectured that this roblem is NP-hard. We close this roblem, by roving that AltPolicy is in fact the otimal olicy for any given with two agents. Theorem 1. The exected utilitarian social welfare is maximized by the alternating olicy for two agents suosing Borda utilities and the full indeendence assumtion. Note that by linearity of exectation this imlies otimality of the alternating olicy for every linear scoring function g = α + β with α, β R, α 0. In articular, the result also holds for quasi-indifferent scoring where g = N + + 1 for sufficiently large N. 121 L 1eft follow * 1 R 1ight deviate from * 1 12 11 L2 R2 L2.5 112 L R L R L R 7.17 7 7 1212 1121 1221 1112 1211 1122 L4 12.29 R4 L4 11.8 R4 L4 12.25 R4 L4 11.5 R4 L4 11.75 R4 L4 12 R4 12121 11212 12212 11121 12112 11221 12221 11112 12122 11211 12211 11122 18.2 18.42 18.5 17.5 18.45 18.5 18. 17 18.5 17.5 18.4 18 122 R2 L 111 R 1222 1111 L4 R4 L4 R4 11.5 10 12111 11222 12222 11111 17. 18 17 15 Figure 1: The olicy tree of deth 5. In the following let π always be the alternating olicy of length. We also recall that due to symmetry we can only consider olicies that starts with 1, e.g. olicy 212 is equivalent to 121. To rove Theorem 1 we need to rove that for any olicy π of length the exected utilitarian social welfare is smaller or equal to the exected utilitarian social welfare of π. That is, dsw π = swπ swπ 0. We roceed in two stes. First, we describe dsw π recursively, by reresenting the olicy π in terms of its deviations from π. Second, given the recursive descrition of dsw π, we rove by induction that this difference is never ositive Proosition 5. The roof is not trivial as the natural inductive aroach to derive dsw π 0 from dsw π 0 does not go through. Hence, we will rove a stronger result in Theorem 2 that imlies Proosition 5 and Theorem 1. 9
Recursive definition. To obtain a recursive definition of dsw π, we observe that any olicy π can be written in terms of its deviations from AltPolicy olicy π. We exlain this idea using the following examle. Consider a olicy π = 1121. There are two ways to extend π with a refix to obtain olicies of length 5: π = 11121 and π = 21121 which is equivalent to π = 12212. We say that π = 12212 follows AltPolicy in extending π as its refix is 12 which coincides with the alternation ste. We say that π = 11121 deviates from AltPolicy in extending π as its refix is 11 which does not corresond to the alternation ste. Next we define a notion of the olicy tree, which is a balanced binary tree, that reresents all ossible olicies in terms of deviations from π. The main urose of this notion is to exlain intuitions behind our derivations and roofs. We start with the olicy 1, which is the root of the tree. We exand a olicy to the left by a refix of length one. We can follow the strictly alternation olicy by exanding 1 with refix 2. This gives olicy 21 which is equivalent to 12 due to symmetry. Alternatively, we can deviate from AltPolicy by exanding 1 with refix 1. This gives olicy 11. This way we obtain all olicies of length 2. We can continue exanding the tree from 12 and 11 following the same rocedure and eeing in mind that we brea symmetries by remembering only olices that start with 1. The following examle show all olices of length at most 5. By convention, given a olicy π in a node of the tree we say that we follow AltPolicy on the left branch and deviate from AltPolicy on the right branch. Examle. Figure 1 shows a tree which reresents all olicies of length at most 5. A number below each olicy shows the value of the exected utilitarian social welfare for this olicy. As can be seen from the tree, AltPolicy is the otimum olicy for all. Consider, for examle, π = 12212. We can obtain this by deviations from π5 shown as the dashed ath: 1 L1 12 L2 121 R 1121 L4 12212. Next we give a formal recursive definition of dsw π. We recall that from Lemma 1 the recursions for AltPolicy u 1 π, u 2 π = + u 2 π 1, + 1 u 1π 1 For any π P, 2 we obtain a similar recursion that deends on whether π follows or deviates from π in extension of π at each ste. In the first case, the refix of π is 12 and in the second case the refix is 11. So we have + u 2 π, +1 u 1 π if π = 12..., Then u 1 π, u 2 π = u 1 π u 1 π = u 2 π u 2 π = + u 1 π, +1 u 2 π if π = 11.... { u 2 π u 2 π 1 if π = 12..., u 1 π u 2 π 1 if π = 11..., u1 π u 1 π 1 if π = 12..., u2 π u 1 π 1 if π = 11.... { +1 +1 10
We introduce notations to simlify the exlanation. Using Proosition 1 we define δ = u 1 π u 2 π = 1 [ + 1 + 1+1 +1 ], and we define the sets A = { u 1 π u 1 π, u 2 π u 2 π : π P }. Note that for an element a, b A corresonding to a olicy π P we have a + b = swπ swπ = dsw π. Hence, π has a higher exected utilitarian social welfare than π if and only if a + b > 0. The recursions above rovide a descrition of the sets A. We have A 1 = {0, 0} because π1 is the only olicy of length 1, and for 2 the set A consists of the elements b, +1 a and a + δ 1, +1 b δ 1 where a, b runs over A 1. Theorem 1 is equivalent to the following statement. Proosition 5. Let A 1 = {0, 0} and { A = b, + 1 } a for 2 where a, b A 1, δ = 1 { a + δ, + 1 } b δ Figure 2 shows the sets A, = 1,..., 4 in the olicy tree. + 1. Then a + b 0 for all a, b A. A 1 A 2 0,0 0,0 L 1 / f 1 R 1 / g 1 1, - 2 A L 2 / f 2 R 2 / g 2 0,0 1,-2 2 L 2 / f 2 R 2 / g 2 -, 4 2, -8 2 L / f R / g A 4 0,0 11,-55 24 L / f R / g L / f R / g L / f R / g 7-25, -2, 5 8 8 4,-15 8-8 15, 10-45, 1, -5 8 8 8 Figure 2: The sets A, = 1,..., 4 in the olicy tree. Proving otimality. We might try to rove Proosition 5 inductively by deriving a + b 0 for the oint a, b A corresonding to olicy π from a + b 0 for a, b A 1 corresonding to olicy π. Unfortunately, the induction hyothesis is too wea as the following examle shows. Examle 4. Assume a, b = 12, 11.9 A 10 corresonding to some olicy π P 10. Let a, b A 11 be obtained from 12, 11.9 by deviating from π 11. With δ 11 = 2.74 we obtain a + b = 9.257 + 9.92 > 0. Thus a, b satisfies Proosition 5 while a, b violates it. To remedy this roblem we would lie to strengthen the roosition, for examle by roving a + b fa, b for all a, b A where f is some function with fx, y 0 for 11
Figure : The set A 10 and a more detailed view of the region around the origin. all x, y. The difficulty of finding such a function is indicated by Figure showing the set A 10. Different marers distinguish the oints arising from A 9 by following π from those deviating from π. The ey idea of our roof is to strengthen Proosition 5 in another direction. We describe this strengthening first and then outline the induction argument. Consider a olicy π that is reresented by a node n π at level in the olicy tree. Instead of requiring the inequality a + b 0 only for the oint a, b A that corresonds to olicy π, we also require it for i all olicies that lay on the ath that follow only the right branches from n π and ii all olices that lay on the ath that starts from n π by following the left branch once and then only follow the right branches. To formalize this idea, for 1 we define functions f, g : R 2 R 2 by f x, y = y, +2x and g +1 x, y = x + δ +1, +2y δ +1 +1. Note that A +1 = f A g A for all 1, as f encodes the case when we follow the left branch and g the right branch. Figure 2 illustrates this corresondence. We also consider iterated comositions of these functions. For every 1 let G 0 denote the identity on R 2, i.e. G 0 x, y = x, y, and for m 1 let G m denote the function G m = g +m 1 g +m 2 g. Alying G m to the oint a, b A corresonding to π P gives the oint a, b A +m which corresonds to the olicy π P +m that is obtained from π by following the right branch m times. For all 1 and m 1, we define the function F m = G +1,m 1 f. F m corresonds to starting in level, following the first left branch and then m 1 right branches. For x, y A, G m x, y A +m for m 0 and F m x, y A +m for m 1. Proosition 5 is a consequence of the following theorem. Theorem 2. Let A 1 = {0, 0} and { A +1 = f A g A = y, + 2 } + 1 x { x + δ +1, + 2 } + 1 y δ +1 for 1 where x, y A, δ = 1 + 1. Then for every 1 and every x, y A the following statements are true. 12
1. For all m 0, if x, y = G m x, y then x + y 0. 2. For all m 1, if x, y = F m x, y then x + y 0. In articular, the first statement with m = 0 imlies Proosition 5 and hence Theorem 1. We first give a high-level overview of the roof, and resent the details in Section 5. We start with a few technical lemmas to derive an exlicit descrition of functions x, y = F m x, y and x, y = G m x, y. This gives us exlicit exressions for the sums x + y and x + y in terms of x and y. Then we roceed through the induction roof. We summarize the induction ste here. Suose the statements of the theorem are already roved for all sets A l with l <. Let x, y be an arbitrary element of A. Suose x, y = f 1 x, ỹ for some x, ỹ A 1 the case x, y = g 1 x, ỹ is similar. Figure 4 shows x, ỹ and x, y that is obtained from x, ỹ by following the left branch. By the induction hyothesis, x + y 0 whenever x, y {G 1,m x, ỹ, F 1,m x, ỹ}. The corresonding nodes are highlighted in gray in Figure 4. To comlete the induction ste we need to show x + y 0 for x, y 0,0 f -1 x,y ~ ~ g -1 x,y to bound x+ y ~ ~ ~ ~ G -1m-1 x,y Figure 4: Schematic reresentation of the roof of Theorem 2. {G m x, y, F m x, y}. The corresonding nodes are indicated by dashed circles. The result for x, y = G m x, y gray and dashed follows immediately as x, y = G m x, y = F 1,m+1 x, ỹ. For x, y = F m x, y we first exress x + y in terms of x and ỹ. Then, by induction x +y 0 for x, y = G 1,m 1 x, ỹ. Inverting the reresentation of x +y in terms of x and ỹ we derive a bound x + ỹ cm 0, deending on m, and this stronger bound is used to rove x + y 0 for x, y = F m x, y. The extension of Theorem 2 to n agents is not straightforward. Firstly, it requires deriving exact recursions for the exected utility for an arbitrary. This is not trivial, as Proosition 2 only rovides asymtotics. An easier extension might be to other utility functions. The alternating olicy is not otimal for all scoring functions. For examle, it is not otimal for the -aroval scoring function which has gi = 1 for i and 0 otherwise. However, we conjecture that AltPolicy is otimal for all convex scoring functions which includes lexicograhical scoring. 1
5 Proof of Theorem 2 5.1 Technical lemmas We start with the observation that +1 = { if is even, +1 if is odd. In the following lemma we describe the functions G m and F m exlicitly. Lemma 4. For 1, m 0, and x, y = G m x, y, and x, y = F m x, y, we have m m x = x + δ +j, y y = + m + 1 + 1 δ +j. + j For 1, m 1, and x, y = F m x, y, we have m x = y + δ +j, y = + m + 1 j=2 m x + 1 j=2 δ +j + j Proof. The exression for x follows immediately from the definitions. For y we roceed by induction on m. The start for m = 0 is trivial: y = y. So assume m 1 and let x, ỹ = G,m 1 x, y. Then x, y = g +m 1 x, ỹ, and using the induction hyothesis we obtain y = + m + 1 + m ỹ δ +m = + m + 1 + m m 1 y + m + 1 δ +j δ +m + j m y = + m + 1 + 1 Finally, the exressions for x and y follow from x, y = G +1,m 1 y, +2 +1 x.. δ +j + j In the next two lemmas we calculate how the functions G m and F m affect the coordinate sum x, y. Lemma 5. Let 1, m 0, and x, y = G m x, y. 1. If is odd, then 2. If is even, then x + y = x + y + x + y = x + y + 4 m mm + 1 y 1 m+1/2 + 1 +2j 1. 5 m mm + 1 y + m + 1 +1 1 m/2 +2j.. 14
Proof. With Lemma 5 we obtain m x + y = x + δ +j + + m + 1 = x + y + m + 1 y + m = x + y + m + 1 y 1 = x + y + m + 1 y 1 m y + 1 1 + m 1 + j m m + 1 j + j m m + 1 j 1 = x + y + m mm + 1 y 1 + 1 Using 4 it is easy to chec that for odd and for even, δ +j + j δ +j + j + 1 +j +j m 1 +j m + 1 j +j m 1 +j m + 1 j +j. m 1 +j m + 1 j +j = m+1/2 m 1 +j m + 1 j +j = m + and this concludes the roof. Lemma. Let 1, m 1, and x, y = F m x, y. 1. If is odd, then x + y = x + y + 2. If is even, then m m 1m x + 1 x + y = x + y + Proof. We roceed exactly as in the roof of Lemma 5. +2j 1, m/2 +2j, + m 1 +2 1 m 1/2 +2j+1. 7 m m 1m x 1 m/2 + 1 +2j. 8 In the roof of our main result we need some rough bounds on the numbers. The following wea estimates will be sufficient for the induction ste in the roof of Theorem 2 below. 15
Lemma 7. For 2, m 0 we have +2 m+1/2 mm + 1, 2 +1 +2 m/2 +1 m2 2. Proof. For m = 0 both of these inequalities are trivially true. So assume m 1. For i = 1,..., m + 1/2, using +2i 2 1/2 we have +2i 2 +2i 1 + 2i 2 + 2i 1 +2i 2 1 2, and summation over i yields +2 m+1/2 m + 1 4 mm + 1. 2 The second inequality is also trivial for m = 1. For m 2 and i = 1,..., m/2, using +1+2i 2 1/2 we have +1+2i 2 +1+2i and summation over i yields + 1 + 2i 2 1 + 1 + 2i 1 +1+2i 2 1 2, +1 +2 m/2 +1 m 4 m2 2. 5.2 The induction argument We roceed by induction on. For = 1 we only have to consider x, y = 0, 0. For x, y = G 1m 0, 0, it follows from 5 that x + y mm + 1 = 1 m+1/2 +2j 1 0. For x, y = F 1m 0, 0, it follows from 7 and = 1/2 that x + y m 1m = + m 1 1 m 1/2 +2j+1 m 12 = 1 m+1/2 +2j 1 0. We now assume that > 1 and the statements of the theorem are already roved for all sets A l with l <. Let x, y be an arbitrary element of A. We distinguish two cases. 1
Case 1: x, y = f 1 x, ỹ for some x, ỹ A 1. If x, y = G m x, y = F 1,m+1 x, ỹ then x + y 0 follows immediately from the induction hyothesis alied to x, ỹ. So suose x, y = F m x, y = F m ỹ, + 1 x. We need to consider the two arities of searately. If is odd, then it follows from Lemma that x + y = ỹ + + 1 x + m m 1m + 1ỹ + m 1 +2 1 m 1/2 +2j+1. 9 By induction ỹ + +1 x 0, and using +1 1/2 we conclude that x + y 0 is immediate if ỹ m 12 +1. Hence we may assume ỹ > m 12 +1. Let x, y = G m m 1,m 1 x, ỹ. By Lemma 5, x + y = x + ỹ + m 1 m 1m ỹ and by induction x + y 0. Hence + m 1 1 m 1/2 +2j 1, x + ỹ m 1m + 1 m 1/2 +2j 1 m 1 ỹ m 1, and substituting into 9 yields together with > +2, x + y < 1 m x + + 1 m 1 ỹ + 1 +1 +2 m 1/2 +1 = x + ỹ m + 1ỹ + 1 +1 +2 m 1/2 +1. With x + ỹ 0 and ỹ > m 12 +1 m this imlies x + y < 1 m 1 2 +1 +2 m 1/2 +1, and finally, x + y < 0 by Lemma 7. If is even, then it follows from Lemma that x + y = ỹ + + 1 x + m m 1m 1 m/2 + 1ỹ +2j. 10 17
By induction ỹ + +1 x 0, so x + y 0 is immediate if ỹ m 1+1. Hence we may assume ỹ > m 1+1. Let x, y = G 1,m 1 x, ỹ. By Lemma 5, x + y = x + ỹ + m 1 m 1m ỹ 1 m/2 +2j 2, and by induction x + y 0. Hence x + ỹ m 1m + 1 m/2 +2j 2 m 1 ỹ, and substituting into 10 yields x + y < 1 m x + + 1 m 1 ỹ + 1 +2 m/2 With x + ỹ 0 and ỹ > m 1+1 this imlies x + y < 1 and finally, x + y < 0 by Lemma 7. = x + ỹ mm 1 +2 m/2, Case 2: x, y = g 1 x, ỹ for some x, ỹ A 1. m + 1ỹ + 1 +2 m/2. If x, y = G m x, y = G 1,m+1 x, ỹ then x + y 0 follows immediately from the induction hyothesis alied to x, ỹ. So suose x, y = F m x, y = F m x + δ, + 1 ỹ δ. Again we discuss odd and even searately. If is odd then it follows from Lemma that x + y = x + δ + + 1 ỹ δ + m + 1 x + δ m 1m + m 1 +2 1 m 1/2 +2j+1. 11 By induction x + δ + +1ỹ δ 0, and using +2 1/2 we conclude that x + y 0 is immediate if m x + δ / + 1 m 1 2 /. So with δ = 1 we may assume x > + 1m 12 m 18. 12
Substituting δ = 1 into 11, rearranging terms, and using +2 = / + 1 we obtain x +y = x+ỹ+ 1 ỹ+ m 2m 1 x m 1 m 1/2 + 1 +2j+1 m + 1 + + 1. 1 For m = 1 we rearrange terms and use x + ỹ 0 and 12 to obtain x + y = x + ỹ + 1 ỹ + 1 + 1 x 1 + 1 + + 1 = + 1 x + ỹ x + 1 1 + 1 + For m 2 let x, y = F 1,m 1 x, ỹ. By Lemma, + 1 1 + 1 x + x + y = x + ỹ + m 1 m 2m 1 x 1 m 1/2 +2j 1, and by induction x + y 0. So < 0. x + ỹ m 2m 1 + 1 m 1/2 +2j 1 m 1 x, and substituting into 1 yields x + y x + ỹ m + 1 x m + 1 + + 1 + 1 +1 +2 m 1/2 +1. With x + ỹ 0 and 12 we obtain x + y m + 1m 12 m + 1 2m + 1 + + 1 +1 +2 m 1/2 +1 = m 12 and finally x + y < 0 by Lemma 7. If is even then it follows from Lemma that + 1 + 1 m + 1 + 1 +1 +2 m 1/2 +1 < 1 m 1 2 +1 +2 m 1/2 +1, x + y = x + δ + + 1 ỹ δ + m + 1 x + δ m 1m 1 m/2 +2j. 14 19
By induction x + δ + +1ỹ δ 0, and we conclude that x + y 0 is immediate if m x + δ / + 1 m 1m/. So with δ = 1 + we may assume x > + 1m 1 Substituting δ = 1 + into 14 and rearranging terms we obtain x + y = x + ỹ + ỹ + m x m 2m 1 + 1 +. 15 + m 1 1 m/2 For m = 1 we rearrange terms and use x + ỹ 0 and 15 to obtain +2j m1 + + 1. 1 x + y = x + ỹ + 1 ỹ + 1 + 1 x 1 + + 1 = + 1 x + ỹ x + 1 1 + + 1 1 x + + + 1 For m 2, let x, y = F 1,m 1 x, ỹ. By Lemma, x + y = x + ỹ + m 1 m 2m 1 x and by induction x + y 0. So + m 2 +1 1 m 2/2 +2j, < 0. x + ỹ m 2m 1 + 1 m 2/2 +2j m 1 x m 2 +1, and substituting this into 1, taing into account +1 =, yields x + y x + ỹ With x + ỹ 0 and 15 we obtain x + y < 1 and finally x + y < 0 by Lemma 7. Strategic Behaviour m x + + + 1 + 1 +2 m/2. +2 m/2 m 1m So far, we have suosed agents sincerely ic the most valuable item left. However, agents can sometimes imrove their utility by icing less valuable items. To understand such strategic behaviour, we view this as a finite reeated game with erfect information. [9] 20
roves that we can comute the subgame erfect Nash equilibrium for the alternating olicy with two agents by simly reversing the olicy and the references and laying the game bacwards. More recently, [7] rove this holds for any olicy with two agents. We will exloit such reversal symmetry. We say that a olicy π is reversal symmetric if and only the reversal of π, after interchanging the agents if necessary, equals π. The olicies 1212 and 1221 are reversal symmetric, but 1121 is not. The next result follows quicly by exandingand rearranging exressions for the exected utilitarian social welfare using the fact that we can comute strategic lay by simly reversing the olicy and rofile and suosing truthful behaviour. Theorem. For two agents and any utility function, any reversal symmetric olicy that maximizes the exected utilitarian social welfare for truthful behaviour also maximizes the exected utilitarian social welfare for strategic behaviour. As the alternating olicy is reversal symmetric, it follows that the alternating olicy is also otimal for strategic behaviour. Unfortunately, the generalisation of these results to more than two agents is comlex. Indeed, for an unbounded number of agents, comuting the subgame erfect Nash equilibrium becomes PSPACE-hard [8]. 7 Conclusions Suosing additive utilities, and full indeendence between agents, we have shown that we can comute the exected utility of a sequential allocation rocedure in olynomial time for any utility function. Using this result, we have roven that the exected utilitarian social welfare for Borda utilities is maximized by the alternating olicy in which two agents ic items in a fixed order. We have argued that this mechanism remains otimal when agents behave strategically. There remain oen several imortant questions. For examle, is the alternating olicy otimal for more than two agents? What haens with non-additive utilities? References [1] S. Bouveret and J. Lang. A general elicitation-free rotocol for allocating indivisible goods. In: Proceedings of the 22nd International Joint Conference on Artificial Intelligence IJCAI 2011. Ed. by T. Walsh. 2011,. 7 78. [2] S.J. Brams, P.H. Edelman, and P.C. Fishburn. Fair Division of Indivisible Items. English. In: Theory and Decision 55.2 200,. 147 180. doi: 10.102/B:THEO.000 0024421.85722.0a. [] S.J. Brams and P.C. Fishburn. Fair division of indivisible items between two eole with identical references: Envy-freeness, Pareto-otimality, and equity. In: Social Choice and Welfare 17.2 2000,. 247 27. doi: 10.1007/s00550050019. [4] S.J. Brams and T.R. Kalan. Dividing the Indivisible. In: Journal of Theoretical Politics 1.2 2004,. 14 17. doi: 10.1177/095129804041118. 21
[5] S. Brams, D. Kilgour, and C. Klamler. The undercut rocedure: an algorithm for the envy-free division of indivisible items. In: Social Choice and Welfare 9.2-2012,. 15 1. doi: 10.1007/s0055-011-0599-1. [] D.K. Herreiner and C. Pue. A simle rocedure for finding equitable allocations of indivisible goods. In: Social Choice and Welfare 19.2 2002,. 415 40. doi: 10.1007/s00550100119. [7] T. Kalinowsi, N. Narodytsa, T. Walsh, and L. Xia. Elicitation-free Protocols for Allocating Indivisible Goods. Fourth International Worsho on Comutational Social Choice. 2012. [8] T. Kalinowsi, N. Narodytsa, T. Walsh, and L. Xia. Strategic Behavior when Allocating Indivisible Goods Sequentially. In: Twenty-Seventh AAAI Conference on Artificial Intelligence AAAI-1. 201. [9] D.A. Kohler and R. Chandrasearan. A class of sequential games. In: Oerations Research 19.2 1971,. 270 277. doi: 10.1287/ore.19.2.270. 22