Plant-and-Steal: Truthful Fair Allocations via Predictions

Introduction

Allocating items among self interested agents in a "fair" way is an age-old problem, with many applications such as splitting inheritance and allocating courses to students. As a starting point, consider the case of two agents. When the items are divisible, the famous cut-and-choose procedure achieves fairness in two senses. Firstly, no agent wants to switch their allocation with the other; i.e., there is no envy among the agents. Secondly, each agent gets a bundle of items which they value at least as much as their value for all the items divided by 2; that is, each one gets their "fair share". When moving to the case of indivisible goods, which is relevant to scenarios such as splitting inheritance and allocating courses, things get trickier. For instance, if there's a single item, the agent that does not receive that item does not get an envy-free allocation, nor do they get their "fair share" according to the previous definitions. Therefore, it is clear that some fairness needs to be sacrificed in this case.

The study of fair allocations with indivisible goods has been a fruitful research direction, with many meaningful notions of fairness studied (see survey by Amanatidis et al. [10]). In this paper, we focus on the notion of the Maximin Share, or MMS, introduced by Budish [18]. For two agents, this notion captures the value an agent will ensure if we implement the cut-and-choose procedure. That is, assume Alice splits the items into two bundles, and then Bob takes one of them (adversarially), and Alice gets the second one. The MMS captures exactly how much value Alice can guarantee for herself. Generalizing the notion for n agents is pretty straightforward -the MMS is the minimum value Alice can guarantee for herself when she partitions the items into n bundles, assuming n -1 bundles are taken adversarially.

We study the case where agents have additive valuations over goods. 1 For the case of two agents, the allocation produced by the cut-and-choose procedure guarantees each of the agents their MMS value. For more than two agents, the existence of such an allocation is not longer guaranteed. Kurokawa et al. [30] show an instance of three agents, where in every allocation, at least one of the agents does not get their MMS value. Since allocating all the agents their MMS value is not always feasible, various papers studied the existence of approximately optimal allocation. An allocation is an α-approximate MMS allocation for α > 1 if every agents gets at least an 1/α fraction of their MMS value. Feige et al. [22] introduce an instance where one cannot find an α-approximate allocation for α < 40 39 . On the other hand, [30] show there always exists 3 2 -approximation. The 3 2 factor was gradually improved [16,24,23,8,4,3,5], where the state-of-the-art algorithm achieves an approximation of 959/720 < 4/3 [3]. Adding incentives to the mix further complicates matters. Amanatidis et al. [7] study the case of two additive agents, and m items, where the algorithm (or mechanism) does not know the values of the agents. Thus, the algorithm's designer is faced with the task of devising an allocation rule such that (i) agents will maximize their allocated value by bidding truthfully, and (ii) the resulting allocation is an α-approximate MMS allocation for an α close to 1 as possible. [7] show that no incentive-compatible algorithm can approximate the MMS to a factor better than ⌊ m 2 ⌋, and this is matched by the following trivial mechanism -the first agent picks their favorite item, and the second agent gets the rest.

For 2 < n < m,foot_1 a trivial truthful algorithm that lets the first n -1 agents pick a single item in some order and gives the last agent the rest achieves an ⌊ m-n+2 2 ⌋-approximation, and no better mechanism is known. It is conjectured that one cannot drop the dependence in m for n > 2. We are left with a stark disparity. On the one hand, assuming agents' values are public information, approximate solutions are known to exist. On the other hand, when considering private values, it seems that only trivial approximations are possible. The goal of this paper is to bridge these two regimes using predictions.

We study the problem of truthful allocations that approximate the MMS, taking a learning-augmented point of view. In the learning-augmented framework, the algorithm designer aims to tackle some intrinsic hardness of the problem at hand, which might arise due to computational constraints, space constraints, input arriving piecemeal online, or incentive constraints, among others. To help the designer overcome these constraints, the algorithm is given some side information which is a function of the input, or a prediction, in order to improve the algorithm's performance. The hope is that if the prediction is accurate, then the performance is greatly improved over the performance without the prediction (termed consistency). On the other end, if the prediction is inaccurate then the performance of the algorithm is comparable to the performance of the best algorithm that is not given access to predictions (termed robustness). The learning-augmented framework has proven useful in bypassing impossibilities that arise due to incentive issues [14,1,25,15,40,33,13].

When designing a learning-augmented mechanism, one should think of realistic predictions. For instance, predicting the entire valuation profile of all agents seems to be a strong assumption. A more plausible assumption is to have some ordinal ranking over the items of the agents. Indeed, it seems unlikely that the algorithm can accurately predict Alice's value for a car, but it is plausible that the algorithm can guess that Alice values the car more than she values the table. Ideally, the algorithm's performance should remain robust if the predicted ordering is almost perfect, with only a few pairs of items whose real ordering is swapped in the prediction. Another desired property is to make the prediction as space-efficient as possible, as previous results [20,31,32] show that succinct predictions are crucial for learning parameters from few samples and for incorporating a PAC-learnable component in the learning-augmented framework. allocation using the predictions. It then applies a recursive plant-and-steal procedure to determine the final allocation.

Experiments. Finally, In Section 6, we demonstrate how several components in our design come into play when experimenting with synthetic data. We run different variants of mechanism on two player instances, and show that when predictions are accurate, then only using predictions is nearly optimal, if predictions are noisy, then the stealing component ensures robustness, and our Plant-and-Steal framework achieves best-of-both-worlds guarantees.

We summarize the bounds we obtain in Table 1. Further Related Work. In addition to the the studies mentioned above, we give a comprehensive review of further related work in Appendix B.

Preliminaries

In the setting we study, there is a set N of n agents and a set M of m indivisible items. Each agent has a private additive valuation over the items, unknown to the mechanism designer, where the value of agent i for item j is v ij (also denoted as v i (j)). For a bundle S ⊆ M of items, v i (S) = j∈S v ij .

The fairness notion we focus on is the following. Definition 2.1 (Maximin Share). The Maximin Share (MMS) of agent i with valuation v i and n agents is

that is, if i were to partition the items into n bundles, and then n -1 of those bundles are taken adversarially, what is the value i can guarantee for themselves. When clear from the context, we omit n and use µ i to denote the MMS of i with n agents.

We are interested in mechanisms that produce approximately optimal allocations, as defined next. Definition 2.2 ((γ, k)-approximate MMS Allocation). An allocation X = (X 1 , . . . , X n ) is (γ, k)approximate MMS allocation for γ > 1 and a natural number k if for every agent i,

we say the allocation is a γ-approximate MMS allocation.

We study mechanism that get some prediction on the input. Definition 2.3 (Learning-Augmented Mechanism). A learning-augmented mechanism takes agents' reports r = (r 1 , . . . , r n ) and predictions p in some prediction space P, and outputs a partition of the items

where agent i gets X i (r, p).

For learning-augmented mechanisms, truthfulness should hold for any possible prediction p. Definition 2.4. A learning-augmented mechanism is truthful if for every agent i and every possible report of other agents r -i and every possible prediction p,

for every r i .

We next define the consistency and robustness measures according to which we measure the performance of our mechanisms. Definition 2.5 (α-consistency). Consider a prediction function f P which takes a valuation profile and outputs a prediction in prediction space P. A learning-augmented mechanism is α-consistent for α > 1 and prediction function f P if for every valuation profile v and every prediction p = f P (v), X(v, p) is an α-approximate MMS allocation. Definition 2.6 ((β, k)-robust). A learning-augmented mechanism is (β, k)-robust for β > 1 and natural number k if for every valuation profile v and every prediction p, X(v, p) is an (β, k)approximate MMS allocation. If k = n, we say the mechanism is β-robust.

For ease of presentation, for valuation v i , report r i and prediction p i , we use v ℓ i , r ℓ i , p ℓ i to denote both the ℓ th highest good according to the valuation/report/prediction and its value. Note that, we may use v ℓ i for ℓ > m, in this case, v ℓ i = 0. For ℓ = 1, i.e., the highest good we use v * i , r * i , p * i .

Ordering Predictions and Kendall tau Distance. Most of our mechanisms use predictions which take the form of an ordering over agents' items. That is, f P (v) outputs a vector of orderings p = (p 1 , . . . , p n ), where p ℓ i is the ℓ th highest valued item of i in M according to p. Accordingly, for agent i, let v ℓ i be the ℓ th highest valued item according to v. For two items j ̸ = j ′ , We use j ≻ pi j ′ to denote that j is higher ranked than j ′ according to p.

When studying imprecise predictions, we want to quantify the degree to which the prediction is inaccurate. For this, we use the following measure. For an agent i, we define our noise level with respect to the Kendall tau distance (also known as bubble-sort distance) between v and p. Definition 2.7 (Kendall tau distance). The Kendall tau distance counts the number of pairwise disagreements between two orders. For i's valuation v i and predicted preference order p i , we define

That is, the number of pairs of items where the prediction got their relative ordering wrong. We also denote

We note that the Kendall tau distance between v i and p i , K d (v i , p i ), can go from 0 to m 2 .

3 Plant-and-Steal Framework

In this section, we present the framework which is used to devise learning-augmented mechanisms for two agents. The ideas presented here also inspire the more complex learning-augmented mechanism for n > 2 agents. Missing proofs of this section appear in Appendix C. Our framework, which we term Plant-and-Steal is given the set of goods, an allocation procedure A, the prediction p and reports v. The framework operates as follows:

1. It first applies A on the predictions p to divide the set of goods into two bundles A 1 , A 2 .

The procedure A should be an allocation procedure with good MMS guarantees. We use different allocation procedures depending on the type of prediction given and on the consistency-robustness tradeoffs we are aiming for. 2. Planting phase: For each agent i, it picks i's favorite item in set A i according to prediction, and "plants" this item in the bundle A j of the other agent j ̸ = i. Let T 1 , T 2 denote the sets that result in this planting phase.

Stealing phase:

To obtain the final allocation, each agent i now "steals" back their favorite item from set T j of agent j ̸ = i according to reports. Notice this is the first and only place where we use agents' reports.

This procedure is trivially truthful because the only step where we use agents' reports is the one where they pick exactly one item to steal back from T j , and this T j only depends on predictions, and not reports (Lemma 3.1). To obtain robustness, we notice that each agent gets one of their two favorite items according to their true valuations (Lemma 3.2). This implies a robustness of m -1.

We show that for balanced allocations, we get improved robustness guarantees (Lemma 3.4).

For S ⊆ M , and agent i, let v * i (S) (p * i (S),r * i (S)) be the max valued item in S according to v i (p i , r i ). for g ∈ M and S ⊆ M , denote S + g := S ∪ {g} and S -g = S \ {g}. The Plant-and-Steal framework is presented in Mechanism 1.

We show that for any allocation function A and predictions p given to the framework, the resulting mechanism is truthful. Lemma 3.1 (Truthfulness Lemma). For any allocation procedure A, Plant-and-Steal mechanism using A is truthful.

Since the framework is truthful, from now on, we assume that r = v. Next, we show that the Plant-and-Steal mechanism ensures that for each agent, an item is allocated with a value that is at least as good as their second-best option according to their value. Lemma 3.2. Consider the allocation (X 1 , X 2 ) returned by Plant-and-Steal with some allocation procedure A. For any agent i, then

We next claim that if i gets one of their two favorite items and any k -1 additional items, i's value is an m -k-approximation to µ i . Lemma 3.3. For any agent i, let S ⊆ M be a subset of the items of size

We immediately get the following. Lemma 3.4 (Robustness Lemma). Let A be an allocation rule guaranteeing min{|A 1 |, |A 2 |} ≥ k, then when Plant-and-Steal uses A, the resulting mechanism is (m -k)-robust.

Proof. By Lemma 3.2, we are guaranteed that each agent gets one of their two favorite items according to their report. Combining with the condition on A and Lemma 3.3, the proof is finished.

Ordering Predictions

In this section, we consider the case of two agents, where the predictions (and in fact, also the reports) given to the mechanism are preference orders of agents over items. Our mechanisms makes use of the Plant-and-Steal framework instantiated by Round-Robin based allocation procedures. We first present the round-robin allocation procedures we'll use, and give their approximation guarantees when the input is accurate. Next, we prove the robustness and consistency guarantees. Finally, we quantify the accuracy of the predictions using the Kendall tau distance, and obtain fine-grained approximation results, where the approximation smoothly degrades in the accuracy.

Amanatidis et al. [6] studied mechanisms where the preference orders of the agents over items are public (while valuations are private). They showed that no truthful mechanism can achieve a better approximation than 5/4 in this setting. This implies that when the predictions are preference orders, no learning-augmented mechanism can obtain consistency better than 5/4, no matter if the robustness is bounded or not. Proposition 4.1 (Corollary of Amanatidis et al. [6]). For any ϵ > 0, no mechanism that is given preference orders as predictions can obtain consistency 5/4 -ϵ.

Round-Robin Allocation Procedures. The two allocation procedures we use to instantiate the Plant-and-Steal framework take as input preference orders of agents over items:

• Balanced-Round-Robin: the agents take turns, and at each turn, an agent takes their highest ranked remaining item. This results in a balanced allocation. • 1-2-Round-Robin: the agents take turns, where we compensate the second agent, who might not get their favorite item, to take two items each turn.

In this section, we only prove consistency-robustness guarantees when Balanced-Round-Robin is used as the allocation procedure. In Appendix E we show different tradeoffs when 1-2-Round-Robin is used.

ALGORITHM 2: Balanced-Round-Robin Input :Preference orders of agents over items v = (v1, v2).

Consider the allocation procedure depicted in Algorithm 2. In order to implement the two allocation procedures, we only needs to receive preference orders over items. Let A i = (a 1 i , . . . , a |Ai| i

) be agent i's allocation by the algorithm, where a k i is the k'th choice of agent i. We observe the following. Observation 4.1. The output (A 1 , A 2 ) of the Balanced-Round-Robin procedure, satisfies:

; that is, in round k an agent gets one of their top 2k items. Amanatidis et al. [8] show that first allocating large items to agents, and then using a Round-Robin to allocate the remaining items to the remaining agents, gives a 2-approximation to the MMS. We observe that for two agents, Round-Robin as is, without the initial step, achieves this approximation guarantee. The proof of the following Lemma is deferred to Appendix D. Lemma 4.1. Let (A 1 , A 2 ) be the allocation of Balanced-Round-Robin. For every agent i,

We next use the allocation procedure to instantiate the Plant-and-Steal framework.

Round-Robin-Based Mechanism. The mechanism we analyze, B-RR-Plant-and-Steal, results from instantiating Plant-and-Steal with Balanced-Round-Robin as A.

We first show that if the predictions correspond to the preference orders of the real valuations, then B-RR-Plant-and-Steal outputs the same allocation as Balanced-Round-Robin. Lemma 4.2. When predictions correspond to actual values, B-RR-Plant-and-Steal outputs the same allocation as Balanced-Round-Robin.

We are now ready to prove the performance guarantees of our mechanisms. Proof. By Lemma 3.1, the mechanism is truthful. By Observation 4.1, each agent receives at least ⌊m/2⌋ items; combining with Lemma 3.4, we get that the mechanism is ⌈ m 2 ⌉-robust. Finally, if predictions correspond to valuations, by Lemma 4.1 and Lemma 4.2, the allocation is a 2-approximation to the MMS. Thus, the mechanism is 2-consistent.

We note that by Amanatidis et al. [7], our robustness guarantee matches the optimal obtainable approximation by any truthful mechanism (up to the rounding).

Noisy Predictions

We now analyze Mechanism B-RR-Plant-and-Steal's performance under varying levels of noise. Consider the case where the Kendall tau distance between v and p is at most d. Our main theorem in this section shows that combining the Plant-and-Steal framework with a Round-Robin allocation procedure obtains O( √ d)-approximation to the MMS when the Kendall tau distance is d. Missing proofs of this section appear in Appendix F.

To prove the approximation ratio, we relate the value that agent i obtains from the allocation, v i (X i ), to their maximin share, µ i , by considering the worst possible set of items that agent i might receive under the Round-Robin procedure when acting on their true preferences. Specifically, we define this worst-case set as R i = {v 2j i } j∈{1,...,⌊m/2⌋} . In Eq. ( 2) of Lemma 4.1, we prove that v i (R i ) ≥ µ i /2. Therefore, obtaining an allocation that is a factor of c times v i (R i ) ensures a factor of c/2 of the MMS value.

We further simplify the analysis by applying the zero-one principle 3 . The zero-one principle basically let's us reduce the analysis to instances where the values are either 0's or 1's. For threshold τ ≥ 0, let h τ (q) = 1 if q ≥ τ and 0 otherwise, and let v τ i (S) = j∈S h τ (v i (j)). By the zero-one principle, for two sets S, T ⊆ M , in order to show that v i (S) approximates v i (T ), it is enough to show that v τ i (S) approximates v τ i (T ) for every threshold τ ≥ 0. Lemma 5.1. For c > 1 and for any two sets S, T ⊆ M , if for every threshold τ ≥ 0,

Thus, we will show that when the Kendall tau distance is d, for every threshold τ ≥ 0,

Recall that A i is the set of items assigned to i after running the Round-Robin procedure on the predictions p. We first show that for Kendall tau distance d, the additive approximation

Lemma 5.2. If the Kendall tau distance between p and v is at most d, then for any threshold τ ≥ 0, we have that

We note that although v τ i (A i ) gives an additive approximation to v τ i (R i ), it can still be the case that the Kendall tau distance is constant, yet v i (A i ) does not give any multiplicative approximation to µ i . 4 Therefore, we must use the fact that agent i gets to "steal" an item according to their true valuation in the Plant-and-Steal procedure in order to get our approximation guarantee. By combining these results, we prove the following theorem concerning the approximation ratio of B-RR-Plant-and-Steal's performance under varying levels of noise, d.

Experimental Results

In this section 6 , we give experiments which illustrate the role of different components of our framework for two players under various noise levels of the predictions. The predictions we use for our experiments are the predicted values of the items. The noise we introduce permutes the vectors of values to match the instance's Kendall tau distance, and uses the permuted vector as prediction. We show that our framework is almost optimal for small amounts of noise while still showing resilience for higher noise levels. Moreover, we study the performance of variants which only use specific components of our framework.

When using predictions, our initial allocation procedure is a cut-and-choose procedure, implemented as follows:

• We use the first player's prediction to implement a bag-filling algorithm which sorts the items by values, and then partitions the items into two sets using a greedy procedure that assigns each item to the set with current lowest value. • We use the second player's prediction to allocate the agent the set with the higher predicted value of the two.

This allocation ensures that the second agent obtains their MMS value according to the prediction. In the data we generates, we observe that in a sampled valuation, the two sets chosen by the bag-filling algorithm gives the two sets the same value, up to 0.5%, which ensures that the lowest valued set obtains a 1.026-approximation to the MMS.

We inspect the following mechanisms:

1. Random: a mechanism that ignores reports and predictions and randomly partitions the items into two sets of size m/2. 2. Random-Steal: a mechanism that ignores predictions, randomly partitions the items into two sets of size m/2, and then implements the stealing phase where each player takes their favorite item from the other player's set according to reports. 3. Partition: a mechanism that ignores reports, and partitions the items according to predictions, using the cut-and-choose procedure described above. 4. Partition-Steal: a mechanism that partitions the items according to predictions, using the cut-and-choose procedure described above, and then implements the stealing phase where each player takes their favorite item from the other player's set according to reports. 5. Partition-Plant-Steal: a mechanism that implements the Plant-and-Steal framework.

partitions the items according to predictions, using the cut-and-choose procedure described above, "plants" each player's favorite item according to predictions, and then "steals" each player's favorite item from the other player's set according to reports.

Experiments. We consider two-player scenarios with m = 100 items. For each distance measure, we generate 1000 valuation profiles. For each pair of valuation profiles and corresponding Kendall tau distance, we generate 100 predictions based on the distance. We then assess the performance of the mechanisms described earlier on these instances. We examine two distinct cases regarding the relationship between the players' preference orders: the Correlated case, where both players have identical preference orders, although their valuation magnitudes differ, and the Uncorrelated case, where the preference orders of the players are generated independently and chosen uniformly at random. Further details on the procedures used to generate the valuations and predictions are provided in Appendix A.

Benchmark. We plot the percentage of these instances where both players get at least (1 -ϵ) of their MMS value for ϵ = 0.1, 0.05, 0.02.

Results. The results are shown in Figure 1. We first examine the performance of the two mechanisms that do not use predictions, Random and Random-Steal. Scenarios with correlated values perform significantly worse, as there is a non-negligible probability of an unbalanced partition of the relatively few high and medium valued items in a random partition. For ϵ values of 0.02, 0.05, 0.1, the Random strategy success rate is 11%, 25%, and 43%, respectively, under correlated preferences, compared

1 5 10 40 160 640 2,560 0 0.2 0.4 0.6 0.8 1 Correlated ϵ = 0.02 1 5 10 40 160 640 2,560 0 0.2 0.4 0.6 0.8 1 ϵ = 0.05 1 5 10 40 160 640 2,560 0 0.2 0.4 0.6 0.8 1 ϵ = 0.1 1 5 10 40 160 640 2,560 0 0.2 0.4 0.6 0.8 1 Kendall tau distance Uncorrelated 1 5 10 40 160 640 2,560 0 0.2 0.4 0.6 0.8 1 Kendall tau distance 1 5 10 40 160 640 2,560 0 0.2 0.4 0.6 0.8 1 Kendall tau distance to 33%, 43%, and 60% under uncorrelated preferences. Moreover, adding the stealing component significantly improves the success rate only in the uncorrelated case, as Random-Steal achieves success rates of 66%, 75%, and 87%. In the correlated case, as each player has a highly valuable item stolen, their obtained value is not expected to increase.

In the mechanisms that use predictions, Partition, Partition-Steal and Partition-Plant-Steal, the performance degrades as a function of noise, as expected. When comparing the performance of Partition, which only relies on the prediction component of our framework, and Random-Steal, which only relies on the stealing component of our framework, we notice that in the uncorrelated case, for small amount of noise guarantee a higher success rate, while as the noise increases, the stealing component becomes more instrumental to the performance. This is in tact with the theoretical results, where using the prediction is crucial to achieve the consistency guarantees, which take place when the prediction is accurate, while stealing is important to achieve robustness guarantees in case the prediction is inaccurate. As described above, in the case where the valuations are correlated, stealing is not expected to help. Interestingly, on fully noisy input, even Random outperforms Partition as Partition might partition the items into unequally-sized sets, which performs worse than the equally-sized sets Random outputs.

Our experiments show that Partition-Plant-Steal performs as well as the Partition strategy for small amounts of noise and outperforms it on uncorrelated instances for large amounts of noise. Moreover, for any amount of noise, it outperforms Random-Steal and converges to it for a fully noisy input. This illustrates the "best of both worlds" tradeoff obtained by our framework.

Finally, when comparing the Partition-Plant-Steal strategy to the Partition-Steal strategy, we observe that Partition-Plant-Steal outperforms Partition-Steal in the correlated case with a small amount of noise (worst-case scenario) for ϵ = 0.02, as planting guarantees your favorite items would not be taken. In other scenarios, Partition-Steal outperforms Partition-Plant-Steal because "planting" removes a valuable item from the player's set that might be taken otherwise, especially in the uncorrelated case.

A Experimental Supplement

Generating valuations. To generate interesting valuations for the players, we use a multi-step function to generate item values, since if the values are close together, any balanced partition obtains good MMS guarantees, without considering reports and predictions. Specifically, we consider a four-step (High/Med/Low/Extra-Low) random valuation function, where an item has a High valuation with probability 8/m, a Medium valuation with probability 1/4, a Low valuation with probability 1/2 and an Extra-Low valuation with the remaining probability. A High valuation is drawn from U [1000, 2000], a Medium valuation is drawn from U [400, 800], a Low valuations is drawn from U [100, 200] and an Extra-Low valuation is sampled from U [1,2]. Figure 2 shows the value distribution generated by this process for two players. We generate values for m = 100 items.

We generate valuations satisfying one of the two types of relations between players' preferences:

• Correlated: the two preference orders are identical (but not the values).

• Uncorrelated: Both preference orders are chosen independently and uniformly at random. Generating predictions. To generate predictions, we take valuations and permute elements randomly to create noise. We generate predictions under varying noise levels according to the Kendall tau distance between the valuations and the predictions. We very the Kendall tau distance between 1 to 2560, where 2560 corresponds to the expected noise level of a random permutation of 100 items. To randomly choose a permutation of a certain noise level, we start with the ordered permutation and then choose two indices j < k u.a.r. and swap items r and r + 1 for r ∈ {j, . . . , k -1} if it increases the Kendall Tau distance by one. We repeat this process until the distance of the resulting permutation equals the desired value.

B Related Work

The notion of the maximin share allocation was introduced by Budish [18] as an ordinal notion, and extended to the notion we adopt by Bouveret and Lemaître [17]. Using machine learning advice in algorithm design was used in theory [21,38] and practice [29]. The learning-augmented framework of studying consistency-robustness tradeoffs was introduced by Lykouris and Vassilvitskii [34]. [35,39] studied the performance of algorithms using imprecise predictions. Fair division with incentives. The two closest papers to ours are Amanatidis et al. [6,7]. In [6], they initiate the study of truthful mechanisms for approximating the MMS value for agents with additive valuations. They show that no truthful mechanism can get an approximation better than 1/2 for the MMS in the case of 2 agents and 4 items. They give the best known approximation guarantee for n agents and m items of ⌊ m-n+2 2 ⌋. Finally they consider the public ranking model, where the ranking over items is public information. Using this, they are able to obtain a n+1 2 -approximation algorithm. One can view this as an algorithm that is given a prediction over the input, but does not provide robustness guarantees. [7] Fully characterize truthful mechanism for 2 agents with additive valuations. They use this characterization to provide a strong lower bound of ⌊ m 2 ⌋ for any truthful mechanism.

[12] design truthful mechanisms for dichotomous submodular valuations that maximize welfare, along with desirable fairness properties such as EFX and NSW. For additive binary valuations, they also maximize the MMS in a truthful manner. [26] bypass the impossibilities imposed by [7,37] for truthful fair allocations with indivisible and divisible goods by considering Bayesian Incentive Compatible mechanisms with symmetric priors. They are able to obtain EF-1 allocations for indivisible goods and proportional allocations for indivisible goods.

Finally, [9] study the Nash equilibrium for simple mechanisms for agents with additive valuations. They show that for every number of agents, the Pure Nash equilibrium of the Round-Robin procedure produces an EF-1 allocation. For two agents, they show that the Pure Nash equilibrium of Plaut and Roughgarden [36] cut-and-choose procedure produces an EFX and MMS allocation.

1-out-of-k. As stated above, the MMS value of an agent is defined by the highest value an agent can guarantee for themselves when partitioning the items into n different bundles, where n is the number of agents, and then getting the lowest valued bundle. Thus, an agent get a value larger than the worst one-out-of-n bundles that define the MMS.

Noticing that finding an allocation that satisfies the MMS value of each agent is a demanding task (which was shown to be infeasible in some cases by Kurokawa et al. [30]), Budish [18] relaxed the notion and defined the 1-out-of-n + 1 MMS to be the worst bundle out of the bundles that define the MMS when partitioning the items using an additional bundle. [18] showed it is possible to achieve this benchmark when adding a small number of access goods. There has been an effort to find the smallest k for which an allocation that guarantees a 1-out-of-k MMS for each agent exists. [2] were able to show the existence for k = 2n -2, [27,28] achieved k = ⌈ 3n 2 ⌉, and recently, [5] showed the smallest up-to-date k = ⌈ 4n 3 ⌉. In our n-agent mechanism, our robustness guarantee approximates this relaxed benchmark for k = ⌈ 3n 2 ⌉.

Learning-Augmented Mechanisms. Agrawal et al. [1] and Xu and Lu [40] first explored the learning-augmented framework in a mechanism design setting, where [1] studied the facility location problem while [40] applied the framework to several settings such as revenue-maximization, path auctions, scheduling and two-facility games. [14] give nearly optimal consistency-robustness tradeoffs to the strategyproof scheduling with unrelated machines. [25] use predictions to design mechanisms with improved Price of Anarchy bounds. [33,19] study revenue maximization auctions with predictions, and [13] devise bicriteria mechanisms.

C Deferred proofs from Section 3

Proof of Lemma 3.1. We show that agent 1 is better off reporting their true valuation, a symmetric argument holds for agent 2. First, notice that sets T 1 and T 2 are determined using predictions, ignoring the reports. Next, notice that the item j2 is chosen only using agent 2's report. Therefore, the only way agent 1 can affect their allocation is by choosing which item in T 2 is allocated to them. agent 1 gets their favorite item in T 2 according to their report. Therefore, it is clear that the agent maximize their utility by reporting their true value.

Proof of Lemma 3.2. Consider some agent i. We claim for every partition of the items into two non-empty sets, T 1 , T 2 , i is always guaranteed to have one of their two favorite items according to their true valuation v i in X i . This is because either (1) i has one of their two favorite items in T ℓ , ℓ ̸ = i, and i gets their favorite item from T ℓ ; or (2) i's two favorite items are in T i , and in this case, i gets all items from T i but one, so i is guaranteed one of them.

By definition, µ i = min j∈{1,2} v i (S j ). We have,

where the before last inequality is since if S j ⊆ S ′ for some j, then v 2 i + v i (S ′ ) ≥ µ i ; therefore S 1 \ S ′ and S 2 \ S ′ are two disjoint non empty subsets and

hence the maximum number of elements in one of these subsets is m -k.

D Deferred proofs from Section 4

Proof of Lemma 4.1. By Observation 4.1, we have

Since i's favorite item must be absent from some set of the sets defining the MMS value,

By Equations ( 1),(2), we have:

Proof of Lemma 4.2. Let j 1 be the first item assigned in Balanced-Round-Robin to agent 1. By definition, j 1 is agent 1's favorite item in M according to p 1 . Clearly, in Plant-and-Steal, j 1 is also agent 1's favorite item in A 1 ⊆ M according to p 1 . Hence, ĵ1 = j 1 . By the definition of Plant-and-Steal, j 1 ∈ T 2 . Since we assume the prediction corresponds to agent 1's actual value, j 1 is also agent 1's favorite item in T 2 ⊆ M , which implies j1 = j 1 .

Similarly Let j 2 be the first item assigned in Balanced-Round-Robin to agent 2. By definition,

Therefore, j 2 is also agent 2's favorite item in A 2 according to p 2 . Hence, ĵ2 = j 2 . Since we established that ĵ1 = j 1 , we have that T 1 ⊆ M \ {j 1 } and j 2 ∈ T 1 . Since we assume the prediction corresponds to agent 1's actual value, j 2 is also agent 1's favorite item in T 1 , implying j2 = j 2 . We get that X 1 = A 1 and X 2 = A 2 as required.

E 1-2-RR-Plant-and-Steal Mechanism: A 3/2-consistent, ⌊ 2m /3⌋-robust Mechanism

In this section, we show that using a modified round-robin allocation procedure to instantiate the Plant-and-Steal framework give an improved 3/2 consistency guarantee, while maintaining O(m) robustness. Consider the Balanced-Round-Robinallocation procedure depicted in Algorithm 2. One can show that the agent that picks first actually gets a value at least as large as their MMS, while for the second agent this analysis is indeed tight. 7 In order to compensate agent 2, 1-2-Round-Robin lets this agent pick two items each round. See Algorithm 3 for details.

Let a k i be agent i's kth choice in 1-2-Round-Robin, we observe the following. Observation E.1. The output (A 1 , A 2 ) of the 1-2-Round-Robin procedure, satisfies:

Amanatidis et al. [6] show that 1-2-Round-Robin guarantees each agent 2/3 of their MMS. We provide the proof for completeness. Lemma E.1 (Amanatidis et al. [6]). Let (A 1 , A 2 ) be the allocation of 1-2-Round-Robin. For every agent i, v i (A i ) ≥ 2µi /3.

Proof. We first prove the approximation for player 1 (the first player to be allocated). First, observe that v 1 (M ) ≥ 2µ 1 . Let I 1 = {v 3k-2 1 : k = 1, . . . , ⌈m/3⌉} be the worst possible allocation agent 1 might get in the 1-2-Round-Robin allocation. Notice that v 1 (I

The mechanism we consider in this section, 1-2-RR-Plant-and-Steal, results from instantiating Plant-and-Steal with 1-2-Round-Robin as A.

The next lemma states that if the predictions correspond to the preference orders of the real valuations, then 1-2-RR-Plant-and-Steal outputs the same allocation as 1-2-Round-Robin. The proof is omitted as it is identical to the proof of 4.2. Lemma E.2. When predictions correspond to actual values, 1-2-RR-Plant-and-Stealoutputs the same allocation as 1-2-Round-Robin.

We next show that in 1-2-RR-Plant-and-Steal we are able to achieve a better consistency, while slightly weakening the robustness guarantee. Theorem E.1. Mechanism 1-2-RR-Plant-and-Steal is truthful, 3/2-consistent and ⌊ 2m 3 ⌋-robust.

Proof. By Lemma 3.1, the mechanism is truthful. By Observation E.1, each agent receives at least ⌈m/3⌉ items; combining with Lemma 3.4, we get that the mechanism is ⌊ 2m 3 ⌋-robust. Finally, if predictions correspond to valuations, by Lemma E.1 and Lemma E.2, the allocation is 3/2approximation to the MMS. Thus, the mechanism is 2/3-consistent.

F Deferred proofs from Section 5

Proof of Lemma 5.1. Let S = {s 1 , . . . , s k } (|S| = k) and T = {t 1 , . . . , t ℓ } (|T | = ℓ). We have the following.

where we use the identity ∞ 0 h τ (q)dτ = q.

Proof of Lemma 5.2. Let ⌊ m 2 ⌋ ≤ m i ≤ ⌈ m 2 ⌉ be the number of items agent i gets by Mechanism B-RR-Plant-and-Steal. Let A i = {a 1 i , a 2 i , . . . , a mi i } be the items assigned to agent i in the Round-Robin according to the predicted orderings p, where a ℓ i is the item allocated to i in the ℓ th round of Round-Robin. First, by Observation 4.1, we have:

For a fixed τ ≥ 0, let L τ = v τ i (R i ) be the number of values larger than threshold τ in R i . We show that if the Kendall tau distance is at most d, then it must be the case that

Note that h τ (v k i ) = 1 for k ≤ 2 • L τ since R i gets every second item by the sorted values of agent i. This implies that if

there are strictly more than ⌈ √ d⌉ items whose rank according to the true valuation is at most 2 • L τ , and their rank according to the prediction is at least 2 • L τ + 1. We show that this implies that the Kendall tau distance is larger than d, yielding a contradiction. Formally, let

..,2•Lτ } be the set of items whose rank is at most 2 • L τ according to the real values but not according to the predictions, and let

..,m} be the set of items whose rank is strictly larger than 2 • L τ according to the real values but not according to the predictions. By the above,

d⌉, and for each pair

1. j rank according to v i is at most 2 • L τ and j ′ rank according to v i is at least 2 • L τ + 1;

2. j ′ rank according to p i is at most 2 • L τ and j rank according to p i is at least

That is, j and j ′ are ordered oppositely in the ordering according to p i and v i . Since there are |G 1 | • |G 2 | > d such pairs, we get that the Kendall tau distance is strictly greater than d, a contradiction.

Proof of Theorem 5.1. Recall that, in Eq. ( 2) of Lemma 4.1, we establish that:

In addition, we apply the zero-one principle to show that Lemma 5.1 holds for the sets X i and R i with c = √ d + 3. The proof follows by combining these two results.

Notice that |A i \ X i | ≤ 2, because in the "stealing" phase, agent i might not take the "planted" item from A i back, and the other agent might take one item from A i . 8 Moreover, by Lemma 3.2, either v 1 i or v 2 i are in X i . Therefore, for every threshold τ ≥ 0,

where the inequality follows Lemma 5.2.

i ) = 0, and Lemma 5.1 holds with c = 0. Therefore, the interesting case is when h τ (v 2 i ) = 1. Consider the ratio

On the other hand, by Eq. ( 6), setting

We get that Lemma 5.1 holds for X i and R i with c = √ d + 3. Thus,

), where the last inequality follows Eq. ( 5).

G.1 No Mechanism with < 6/5 Consistency and Bounded Robustness

In [7] they define the following family of mechanisms. Definition G.1 (Singleton Picking-Exchange Mechanisms [7]). A mechanism X is a singleton picking-exchange mechanism if for each i ∈ {1, 2}, there is exactly one of two sets: either N i ⊆ M , or E i = {ℓ i } for a single item ℓ i ∈ M . If N i is non-empty, then the mechanism lets player j ̸ = i pick item ℓ ∈ N i that maximizes v j (ℓ), and i gets N i \ {ℓ}. If both E 1 , E 2 are non-empty, then the agents exchange the two items

empty and there will be no exchange.

[7] showed the following. Lemma G.1. In order for a mechanism to be truthful and have a bounded approximation, it has to be a singleton picking-exchange mechanism

We make use of this characterization in our impossibility. Theorem G.1. For any ϵ > 0, there is no truthful a mechanism with consistency 6/5 -ϵ and bounded robustness.

Proof. Consider the case where p 1 = p 2 = (1/2, 1/2, 1/3, 1/3, 1/3). Notice that for the predictions, µ 1 = µ 2 = 1. We show that for any singleton-picking-exchange mechanism, no agent obtains both large items (of value 1/2). Consider agent 1 (the argument is symmetric for agent 2). If N 1 is non-empty, then if both large items are in N 1 , surely 1 will only get one of them. If both large items are in N 2 , then agent 2 will surely pick one of them, and agent 1 will only get one of them. If one large item is in N 1 and the other is in N 2 , each agent i will pick the large item in N i . If agent 2 has a large item in E 2 , then since N 1 is non-empty, E 1 is empty and agent 2 will keep the large item. Now consider the case where E 1 is non-empty. In this case, E 1 contains one item, and N 1 is empty. Since E 2 can contain at most one item, and there are more than 2 items, in this case, E 2 = ∅, and |N 2 | = 4. Therefore, N 2 contains at least one large item. Since agent 2 will always pick the large item, agent one only gets one large item. We conclude that for any singleton picking-exchange mechanism, the large items are split among the agents. Since there are 3 small items, there must be an agent that gets at most one small item, and this agent has an overall value of at most 1/2 + 1/3 = 5/6, while the MMS is 1. Thus the claim follows.

G.2 4-Consistent, (m -1)-Robust Mechanism Using a 3 log m + 1-Space Prediction Let us formally define a mechanism that uses a space-s prediction Definition G.2. A learning-augmented mechanism is a space-s mechanism if the prediction space P can be represented by the elements of {0, 1} s .

We first give a simple mechanism that only requires 3 log m + 1 bits of information about the valuations v 1 and v 2 . It will only need to know an index j 0 in [m] together with a bit b to produce an approximately-optimal allocation, and an additional 2 log n bits to implement the planting phase. The mechanism will utilize the Plant-and-Steal framework in conjunction with the well-known bag-filling allocation procedure:

We see that, in order to predict the behaviour of the mechanism above, one only needs to predict accurately the index j 0 on which the mechanism terminates, as well as a bit b ∈ {1, 2} that encodes

) and terminate whether the algorithm terminates due to the condition v1([j]) v1([m]) ≥ 1 2 being satisfied or due to the condition v2([j]) v2([m]) ≥ 1 2 being satisfied. This can be encoded using log m + 1 bits. We also see that the The Plant-and-Steal framework when used with the Bag-Filling allocation procedure gives a truthful 4-consistent and a m -1-robustfoot_8 allocation mechanism. The truthfulness and robustness follow immediately from Lemmas 3.1 and 3.4 respectively. The 4-consistency holds for the following reason. It is a well-known fact (see i.e. [10]) that the partition (A 1 , A 2 ) given by the bag-filling algorithm satisfies v 1 (A 1 ) ≥ µ 1 /2 and v 2 (A 2 ) ≥ µ 2 /2. By inspecting the Plant-and-Steal framework (Algorithm 1), we see that both agent 1 and agent 2 will either (i) retain their most preferred item in A 1 and A 2 respectively or (ii) Lose this item, but obtain an item that they prefer even more. Overall, this implies that in the worst case the difference v 1 (A 1 ) -v 1 (X 1 ) will equal to the value of the second-most favorite item of Agent 1 in A 1 . This implies that v 1 (X 1 )

G.3 2 + ϵ-Consistent, ⌈ m 2 ⌉-Robust Mechanism Using a O(log m/ϵ)-Space Prediction We now show that a better consistency of 2 + ϵ can be achieved at the cost predicting O(log m/ϵ) bits of information about the valuations v 1 and v 2 . We will also obtain a better robustness of ⌈ m 2 ⌉. To do this, we will use the Plant-and-Steal framework in conjunction with the Cut-and-Balance allocation procedure. We first explain how the mechanisms above can be implemented by only

• if |S2| > 1 additionally satisfying v1(S ′ ) ≤ v1(S1 \ { ĵ, ĵ′ })/2, for some ĵ ∈ arg maxj∈S 1 v1(j) and ĵ′ ∈ arg max ℓ∈S 1 \ ĵ v1( ĵ) obtaining O(log m/ϵ) bits of information about the valuations v 1 and v 2 . This follows from the following proposition, the proof of which is given in Appendix G.4.

Additionally, there exist integers α 3 , β 3 , α 4 and β 4 such that the set

, where ĵ ∈ arg max j∈S1 v 1 (j) and ĵ′ ∈ arg max j∈S1\ ĵ v 1 (j).

The main ideas for proving Proposition G.1 are: (i) using the sets L 1 and L 2 to handle elements x whose value v(x) is large, and separate the remaining items into the set S (ii) Showing that the remaining items can be separated into well-behaved subsets of the form S [α i , β i ].

The proposition above implies that the sets S 1 , S 2 and S ′ can be represented exactly via sets L 1 and L 2 , together with the indices {α 1 , • • • , α 4 , β 1 , • • • β 4 }. We will also need to know the index i 2 ∈ {1, 2}. Since the sets L 1 and L 2 have a size of O(1/ϵ), all this information amounts to O(log m/ϵ) bits as claimed.

The following proposition implies the truthfulness, the robustness and the consistency of the mechanism that combines the Cut-and-Balance allocation procedure with the Plant-and-Steal framework. Theorem G.2. The Plant-and-Steal framework, when used with Cut-and-Balance allocation procedure, gives a truthful, 2 + ϵ-consistent and a ⌈m/2⌉-robust allocation mechanism.

Proof. Truthfulness follows from Lemma 3.1. Since the sets A 1 and A 2 both have size at most ⌈m/2⌉, the robustness follows via Lemma 3.4.

The proof of (2 + ϵ)-consistency is deferred to Appendix G.5. The main challenge for showing the bound on consistency is the fact that both the Cut-and-Balance allocation procedure and the Plant-and-Steal framework can reduce the consistency by a factor of 2. Naively, one would expect the overall consistency to be close to 4, given that each stage can lose a factor of 2 in consistency. However, our insight is that for the instances, on which the Cut-and-Balance allocation procedure loses a factor of 2 in consistency, the Plant-and-Steal framework will have consistency close to 1, and vice versa. This allows us to prove a tighter bound of 2 + ϵ on the consistency of our overall algorithm.

Proposition G.2. There exists a partition

• For every x in L 1 and y in S 1 we have v 1 (x) > v 1 (y). Analogously, for every x in L 2 and y in S 2 we have v 1 (x) > v 1 (y)

• There are ĵ, ĵ′ ∈ L 1 satisfying ĵ ∈ arg max ℓ∈S1 v 1 (ℓ) and ĵ′ ∈ arg max ℓ∈S1\ ĵ v 1 (ℓ),

We do this by inspecting two types of items, large items, with value greater than ϵµ 1 /4, and small items items with value at most ϵµ 1 /4. We first show that there are O(1/ϵ) large items, therefore, separating these items into two bundles require at most O(1/ϵ) intervals. Moreover, we can find a separation of the larges items into two sets, L 1 , L 2 , and a single index j ∈ [m] such that all small items to the left of j (including) together with L 1 form S 1 , and all items to the right of j (excluding) together with L 2 form S 2 , such that S 1 , S 2 satisfy the approximation requirement. It is easy to see that this increases the number of intervals by at most 1.

We start by showing there are not too many large items. Lemma G.2. There are at most ⌈ 8 ϵ ⌉ items with value strictly greater than ϵµ 1 for agent 1.

Proof. Let items with value greater than ϵµ 1 /4 be the large items. Suppose there are at least ⌈

8 ϵ ⌉ + 1 large items. If ⌈ 8 ϵ ⌉ is even, consider a partition (S 1 , S 2 ) such that each S i gets at least ⌈ 8 ϵ ⌉/2 large items and the rest are allocated arbitrarily. If ⌈ 8 ϵ ⌉ is odd, consider the allocation in which each S i gets (⌈ 8 ϵ ⌉ + 1)/2 large items and the rest are allocated arbitrarily. In either case, each S i gets at least ⌈ 8 ϵ ⌉/2 ≥ 4 ϵ large items. Thus, min{v 1 (S 1 ), v 1 (S 2 )} > ϵµ 1 /4 • 4 ϵ = µ 1 , a contradiction.

We are now ready to prove Proposition G.2.

Proof of Proposition G.2. Consider the set of large items, L = {j ∈ [m] : v 1 (j) > ϵµ 1 /4}, and let S = M \ L be the set of small items.

We give a constructive proof which finds both sets L 1 , L 2 and an index j satisfying the condition stated in the lemma. Let

We use the following procedure to find j.

1. Let j ℓ = 0 and j r = m.

we can rename both bundles)

. By the definition of L 1 , L 2 , it must be the case that v 1 (L 1 ) ≥ v 1 (T 1 ∩ L). Thus, Since T 1 \ (T 1 ∩ L) ⊆ S, it must be that

On the other hand, if

, by condition 2b of the above procedure, it must be the case that when j ℓ was equal m -1,

We get that

where the second inequality follows since

Case 2: 0 < j < m. In this case, since both j ℓ and j r were moved, there were some values of j ℓ and j r such that v 1 (S ℓ ) ≤ v 1 (S r ) and some values such that v 1 (S ℓ ) > v 1 (S r ). Assume initially that v 1 (S ℓ ) ≤ v 1 (S r ). Since at each step of the procedure, the the lower-valued bundle can increase by at most ϵµ 1 /4, when the first item is added to S ℓ such that v 1 (S ℓ ) > v 1 (S r ), it must be the case that v 1 (S ℓ ) ≤ v ( S r ) + ϵµ 1 /4. It is easy to see that the invariant where |v 1 (S ℓ ) -v 1 (S r )| ≤ ϵµ 1 /4 is kept throughout the run of the procedure. Therefore, this also holds for the final S 1 and S 2 . Thus, we can use the same reasoning of Eq. ( 7) to conclude that min{v 1 (S 1 ),

Thus, the sets S 1 and S 2 have a form

and have the form required. If |S 1 | < |S 2 | we can swap our definitions for the sets S 1 and S 2 , thus ensuring that |S 1 | > |S 2 |. Due to our definitions of L 1 and L 2 we have for every x in L 1 and y in S 1 we have v 1 (x) > v 1 (y). Analogously, for every x in L 2 and y in S 2 we have v 1 (x) > v 1 (y). We can ensure that There are ĵ, ĵ′ ∈ L 1 satisfying ĵ ∈ arg max ℓ∈S1 v 1 (ℓ) and ĵ′ ∈ arg max ℓ∈S1\ ĵ v 1 (ℓ), by adding such values from S ∩ [α 1 , β 1 ] to L 1 (we see that after this all other properties still hold). Overall, we see that |L 1 | + |L 2 | ≤ ⌈ 8 ϵ ⌉ + 2, as required. Now, we proceed to proving the second half of Proposition G.1. We will need the following lemma. Lemma G.3. Let k 1 and k 2 be positive integers satisfying k 1 > k 2 , and let f be a function mapping [k 1 ] to non-negative real numbers. Then, there exist a pair of integers α, β, α ′ and β

Proof. We prove the lemma using the probabilistic method. Let j be a uniformly random integer in [k 1 ], and choose α, β, α ′ and β ′ such that

We see that indeed a set chosen as above can be represented as a union of two intervals. Now, since j is chosen uniformly at random form [k 1 ], we see that for every element i in [k 1 ] we have

Thus via linearity of expectation we have:

Thus, since f (i) is non-negative for all values of i, we see that for some specific choice of j it has to be the case that

which finishes the proof. Now, we apply the lemma above. If m < 4⌈ t ϵ ⌉ + 2 we can satisfy Proposition G.2 by:

2. Define L 2 := S 2 , put the smallest ⌊m/2⌋ -|S 2 elements of S 1 into S, and define L 1 to contain the rest of elements in S 1 .

Define α

Overall, this allocates S ′ to be the bottom ⌊m/2⌋ elements of S 1 . We see that this suffices to guarantee the properties that S ′ needs to satisfy in Proposition G.1.

Therefore, henceforth we can assume that m > 4⌈

First of all, since every element in L 1 is larger than any element in S ∩ [α 1 , β 1 ]), we see that this is also true in average

Then, applying Lemma G.3 to the set S ∩ [α 1 , β 1 ]) we see that there exist disjoint subsets

Combing Equations 8 and 9, we see that taking

)/2 and the other requirements of Proposition G.1.

Since by Proposition G.2, the set S does not contain the two largest elements ĵ and ĵ′ of S 1 , as well as the fact that every element in L 1 is at least as large as any element in S ∩ [α 1 , β 1 ]), we see that every every element in

). this implies that:

Then, we can again apply applying Lemma G.3 to the set S ∩ [α 1 , β 1 ]) we see that there exist disjoint subsets

Combing Equations 10 and 11, we see that taking

as required in Proposition G.1. Note that this also implies that v 1 (S ′ ) ≤ v 1 (S 1 })/2 since ĵ, ĵ′ have the top two largest values of v 1 in S 1 . Overall, this finishes the proof of Proposition G.1.

G.5 Proof of (2 + ϵ)-consistency.

It remains to show that the algorithm is 2 + ϵ-consistent. We will be referencing the variables ĵ1 , ĵ2 , j1 , j2 , T 1 and T 2 within the Plant-And-Steal framework (Algorithm 1).

We first reason about agent 2. First, notice that since agent 2 has a higher value for Si2 ,

Since the mechanism had a chance to pick item ĵ2 from T 1 as j2 , it must be the case that v 2 ( j2 ) ≥ v 2 ( ĵ2 ) (and possibly j2 = ĵ2 ). If j1 = ĵ1 , then T 2 \ j1 = Si2 \ ĵ2 , and

Otherwise, j1 ∈ Si2 , and

Since ĵ2 is the item with the highest value for agent 2 in Si2 , v 2 ( j2 ) ≥ v 2 ( ĵ2 ) ≥ v 2 (k 1 ). Combining with Eq. ( 12), we get that

In this case, the same exact arguments used for agent 2 can be harnessed to show that v 1 (X 1 ) ≥ (1 -ϵ/4)µ 1 /2 ≥ µ 1 /(2 + ϵ). Thus, it is left to consider the case where i 1 = 1.

Consider the (S 1 , S 2 ) partition that is set in the first step of Cut-and-Balance-and-Choose. Since v 1 (S ′ ) ≤ v 1 (S 1 )/2, we have

If j2 = ĵ2 , we have that

where the first inequality follows since v 1 ( j1 ) ≥ v 1 ( ĵ1 ).

Note also that if

Therefore, we assume j2 ̸ = ĵ2 and |S 2 | > 1, and let ĵ′

, where the first inequality is since, v 1 j1 = max j∈T2 v 1j ≥ v 1 ĵ1 . The second inequality is since v 1 ( ĵ2 ) ≥ 0, the third inequality is by ĵ′ 1 definition since j2 ∈ S1 \ ĵ1 by our assumption that k 2 ̸ = ĵ2 . Finally, we have have

where the last inequality is since |S 2 | > 1. Since we handles the case |S 2 | = 1 earlier, we can here assume

2+ϵ , finishing the proof.

H.2 Implementation Details

As discussed, in order to utilize the Round-Robin mechanism, we first allocate a single item to each agent with a high predicted value.

ALGORITHM 7: Allocate-Large Input :Set of agents N , set of items M , reports rN , predictions pN Output :A partial allocation • i∈B Xi, updated sets of agents and items N, M , respectively

Before describing the tentative allocation mechanism, we first give a procedure, Allocate-Best, which performs a single round of Round-Robin according to a specific order, and preferences (either predictions or reports), denote o.

The tentative allocation mechanism repeatedly invokes Allocate-Best according to given predictions, until all items are tentatively allocated. As previously mentioned, the first round of the tentative allocation is performed according to the given order, and in all subsequent rounds, the order is reversed (recall that reversing the order enhances the robustness guarantees).

The final phase in the mechanism is a recursive plant and steal algorithm. The input to this algorithm is an ordered set of agents N , along with their predictions, reports, and a tentative allocation for each agent. At each recursive invocation, the algorithm splits the set of agents into two (almost) equal-size ordered sets N 0 and N 1 . Then the mechanism "plants" for the i th agent in each set N b their highest (according to predictions) valued item in their tentative allocation in the tentative set of the i th agent in N ¬b . Then we perform one round of Round-Robin, where the items available to the agents of set N b are those tentatively allocated to the agents of N ¬b (after the planting phase), and the allocations are determined according to agents reports. See Figure 3 for an illustration of a single round of plant and steal. The algorithm then recurses on each of the sets N 0 and N 1 , until all sets are of size 1. At this point, the single agent in the set is further allocated its remaining tentatively allocated items, and the process terminates. Let i 0 , i 1 denote the i th agent in N0, N1 respectively.

Given the above implementation details, it remains to prove Theorem H.1 regarding truthfulness, consistency and robustness of the mechanism. The proof is given in Appendix H.3.

H.3 Missing Details from Section H

In this section we prove Theorem H.1, which we now recall. Theorem H.1. The Learning-Augmented-MMS-for-n-Agents Mechanism (Mechanism 6) is truthful, 2-consistent and (µ n i /α)-robust for n = ⌈3n/2⌉ and α = max{m -n -1, 1}.

First, we give a simple observation regarding Algorithm 7.

which in particular contains j. Hence, the first item in A 1 is allocated into X 1 . We now assume the claim holds for k = 1 and ℓ -1 and prove it for ℓ. Assume without loss of generality that ℓ is odd so that i ℓ ∈ N 0 .

In step ℓ of the planting phase, the mechanism plants ℓ 0 's (the proof for ℓ 1 is identical) first (according to value p ℓ 0 ) item in A ℓ 0 . Then, during the tentative allocation phase, agent ℓ 0 is the ℓ th to choose among the items in A N1 minus the items that were allocated to the ℓ -1 agents that were before her in the tentative Round-Robin. By the induction hypothesis, every agent preceding her chose the item the mechanism planted for them previously in that round. Therefore, the item j that the mechanism planted for agent ℓ 0 is still available. Moreover, let M ℓ-1 denote the set of items after ℓ -1 rounds of the tentative Round-Robin in Algorithm 9. Further let A ℓ-1 N1 denote the set of items after ℓ -1 rounds of the Allocate-Best algorithm invoked in the stealing phase with the set N 0 , i.e., A ℓ-1 N1 = A N1 \ j∈N0,j<ℓ {X j }. Since the order in which the agents plant and steal in each round of the recursion is equivalent to the order in which the corresponding tentative allocation round was performed, it holds that A ℓ-1 N1 ⊂ M ℓ-1 . Since j = p * ℓ 0 (M ℓ-1 ), and p ℓ 0 = r ℓ 0 , it holds that r * ℓ 0 (A ℓ-1 N1 ) equals j. Therefore ℓ 0 will choose j to X ℓ 0 as claimed. Proving the claim for a general k is almost identical. At the planting phase of the k th round, the mechanism plants for every agent ℓ 0 ∈ N k 0 their k th item of A i in A N k 1 and vice versa. A similar argument to the one above, shows that this item will remain available until its their turn to choose an item for allocation, as by the recursion hypothesis, all agents preceding i in the Round-Robin will select the items the mechanism planted for them. Hence, the k th item in A ℓ 0 will be allocated to X ℓ 0 . Finally, once the set agent i belongs to becomes a singleton, by our halting condition, X i ← X i ∪ A i , so together with the previous argument, we get that for every ℓ, X i = A i as needed.

We continue to prove that the mechanism is robust. Since when m < n, µ n i = 0 for every agent i, and each agent trivially gets their MMS value, we assume from now on that m ≥ n and show the mechanism achieves (m -n -1)-robustness for µ n i . We first prove in Lemma H.3 that for each agent i, v i (X i ) ≥ v ⌈3n/2⌉ i , and then prove in Lemma H.5 that the value of this item is not too small compared to µ ⌈3n/2⌉ i . Lemma H.3. For every agent i, v i (X i ) ≥ v ⌈3n/2⌉ i . Proof. We first prove the claim for agents that were allocated a value during the invocation of Algorithm 7. By the definition of the algorithm and its truthfulness when agent i is allocated an item, at most n -1 items were previously allocated to other agents. Hence, she can always choose her n th highest valued item. Therefore, we have

We continue to prove the claim for the set of agents with no large predicted values. Consider the ℓ th agent in N , i ℓ , and consider the following coloring process. Initially, color all items in M black. We will then color all items i ℓ was able to choose from green, and items allocated before she had the chance to choose from gray (note that these colors are unrelated to the ones in the figure). Note that an item turns green when it belongs to the tentative allocation of opposite set to i ℓ 's and has not been taken by agents preceding her in the allocation order. We claim that by the time no black items remain, at most ⌈3n/2⌉ -1 have turned gray, implying that at some point during the recursion, i ℓ could have chosen their ⌈ 3n 2 ⌉ th highest valued item (according to r i ℓ ). We let N k denote the set of agents to which i ℓ belongs to at depth k of the recursion, starting with

We further let b k ∈ {0, 1} denote the index of the set to which i ℓ belongs to: i ℓ ∈ N k b k . We will separately bound the number of items turned gray due to agents in N k b k and N k ¬b k . In the first iteration, for k = 1, let A 1

denote the tentative sets allocated to the agents of N 1 0 and N 1 1 after the planting phase (i.e., at the beginning of the stealing phase). The number of items that turn gray due to agents in N 1 b 1 is G 1 b 1 = ⌈ℓ/2⌉ -1, since i ℓ has access to all items in A 1 N ¬b 1 excluding the ⌈ℓ/2⌉ -1 items that were allocated to the agents in her set preceding her in the ordering. (The rest of the items in A 1 N ¬b 1 turn green.)

Turning to G 1 ¬b 1 , each agent in the opposite set to hers, N 1 ¬b 1 , is allocated a single item (from A 1

) before continuing to the next round of the recursion. Therefore, G 1 ¬b 1 = |N 1 ¬b 1 | (and no item turns green).

The recursion then continues with N 2 = N 1 b 1 and in reversed order (due to the order being reversed). Therefore, at the beginning of the second iteration, i ℓ is in location |N 1 b 1 | -⌈ℓ/2⌉ in N 2 . After the partition phase, i ℓ is in set N 2 b 2 and in location ⌈

⌉ -1 due to agents in her set preceding here in the ordering. Also, G 1 ¬b 1 = |N 1 ¬b 1 | due to allocations to agents in the opposite set to hers.

From now on, the order is preserved, so for every k ≥ 3,

We continue by bounding

and if N k is odd, then either N k b k is odd and N k ¬b k is even or vice versa. In the first case, G k ¬b k = ⌈N k /2⌉ and we recurse with N k b k which is of size ⌊N K /2⌋. In the second case, G k ¬b k = ⌊N K /2⌋ and we recurse with N k b k of size ⌈N k /2⌉

. Hence, we have the following recursion formula. For even ℓ, T (ℓ) = ℓ/2 + T (ℓ/2), and for odd ℓ, either (a) T (ℓ) = ⌈ℓ/2⌉ + T (⌊ℓ/2⌋) or (b) T (ℓ) = ⌊ℓ/2⌋ + T (⌈ℓ/2⌉). In Claim H.4 below, we prove that for such a function, if it also holds that T (1) = 0 and T (2) = 1, then T (ℓ) ≤ ℓ -1. Therefore, we get that

We continue to bound

where L is the number of indices k for which the fraction X/2 k is rounded up. Observe that for every X, L can be bounded above by ⌈log X⌉ as L exactly equals the number of 1 bits in the binary representation of X. Hence, the overall number of items that turn gray can be bounded as follows:

Therefore, the number of items that turn gray by the end of the recursion is at most ⌈3n/2⌉ -1, and so i ℓ get their ⌈3n/2⌉ highest valued item v ⌈3n/2⌉ i ℓ .

We now prove the claim regarding the cost of the recursion that was used in the previous lemma. Proof. We prove the claim by induction on n. By T (1) = 0 and T (2) = 1 so the induction basis holds. We now assume correctness for all values smaller than n and prove for n.

If n is even then Finally, we prove that the highest valued item allocated to each agent i is not too small compared to their MMS. Lemma H.5. Consider an MMS for agent i, and let j * be the highest valued item of i in her allocation.

Proof. Consider an MMS allocation of M for k = ⌈3n/2⌉, and let A i be the set such that v i (A i ) = µ k i . By the assumption on j * , its value is higher then the highest valued item in

as at least k -1 items must be allocated to the k -1 additional agents, it holds that v i (j * ) ≥ µ 3n/2 i /(m -⌈3n/2⌉ -1).

Proof of Theorem H.1. The theorem follows by Lemmas H.1, H.2, H.3, and H.5.

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