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[MATH] gets the marked piece up to [MATH] ’s mark. [MATH] gets her second favorite piece. [MATH] gets her favorite piece among the two remaining. |
[MATH] gets the remaining piece. Clearly [MATH] now dominates [MATH] since the initial allocation was envy-free, [MATH] is the cutter and [MATH] got the unique partial piece. It remains to show that the resulting allocation is still envy-free. First, [MATH] as the cutter gets a piece of equal value to her favorite piec... |
Given all the above ingredients, we are ready to prove our main theorem. Proof of Theorem We first argue about the correctness of the Main Protocol . It suffices to prove that phases one and two terminate with the desired domination structures. Then it is straightforward that the third phase results in an envy-free all... |
Phase one: Lemma guarantees that once the for loop is completed in phase one, we have a domination graph where node [MATH] has out-degree at least one. |
We first consider the case where at least 2 agents got an insignificant piece in lines . Then Lemma guarantees that after line agent [MATH] dominates all those agents. If lines |
are executed, Lemma guarantees that some agent becomes dominated by both agents [MATH] and [MATH] and phase one is successfully completed. Otherwise, lines are executed and the algorithm terminates returning a complete envy-free allocation. The latter holds because when agent [MATH] dominates all other agents, we only ... |
Core property , however, takes care of the envy-freeness after line because so far we have only executed Core with no excluded agents from competition (5 times). |
The remaining case is when the same agent got the insignificant piece in all 4 iterations in lines , i.e., when the condition in line is true. Among the 4 Core allocations produced in lines , let [MATH] be an allocation satisfying the conditions of Lemma . The execution of Correction with input [MATH] gives an allocati... |
where the insignificant piece with respect to [MATH] is given to a different non-cutter, due to Correction property Since the insignificant piece has gone now to a different agent, Lemma implies that running Core one more time with agent [MATH] as the cutter results in agent [MATH] dominating a new agent in addition to... |
[MATH] for all agents [MATH] , i.e., [MATH] is envy-free. Core property and Lemma guarantee that the allocation at the end of phase one is envy-free as well. |
Phase two: Having obtained a partial allocation where one agent, say [MATH] , is dominated by two others, say [MATH] and [MATH] , we execute Core twice on the current residue with the remaining agent [MATH] as the cutter (line of the main protocol). First, suppose that there is at least one agent excluded from competit... |
We conclude that, if line is reached, there was no domination between [MATH] and [MATH] before each of the last two Core executions. That is, these executions excluded no one from competition. Therefore, we can combine Core property (which guarantees there is only one partial piece) with the fact that the allocation so... |
A possible complication now is that the second Core execution might result again in [MATH] getting the partial piece. Let [MATH] be the overall partial allocation so far, i.e., at line We execute Correction on the Core allocation [MATH] (out of the two last ones) where [MATH] is smaller. By Core property , we have that... |
Finally, we argue about the overall partial allocation being envy-free. Clearly, [MATH] and [MATH] are non-envious. By the choice of [MATH] , we have [MATH] , where [MATH] thus [MATH] is not envious. On the other hand, [MATH] is not envious since by Core property she was indifferent between her piece and [MATH] ’s piec... |
Counting Queries: In the worst case, at most [MATH] calls to Core are required followed by a call to Cut and Choose . Theorem directly gives an upper bound [MATH] cut and [MATH] evaluation queries. For a more detailed argument matching the statement of the theorem, see the last part of the next section. |
The Core Protocol An important building block of the whole algorithm is the Core protocol, used for allocating part of the current residue every time it is called. We begin with a high-level idea of how Core works. It takes as input an agent, specified as the cutter, the current residue, and the current partial allocat... |
Core first asks the cutter to divide the residue into four equally valued contiguous pieces. The cutter is going to be the last one to receive one of these four pieces. Regarding the remaining three agents, each of them will either be immediately allocated her favorite piece or will be asked to place a mark on certain ... |
As seen in the pseudocode description of Core , there are two possible types of marks that can be placed; [MATH] -marks and [MATH] -marks. The type of mark that the agents will be asked to place depends mainly on the conflicts that arise for the favorite and second favorite pieces of each agent. The conditions that det... |
To describe the protocol, we need to formalize conflicts between agents for certain pieces. Agent [MATH] cuts the current residue [MATH] in four equal-valued pieces (according to her). |
Let [MATH] be the set of agents who may compete for pieces. if there exists [MATH] who has no competition in [MATH] for her favorite piece |
then [MATH] is allocated her favorite piece and is removed from [MATH] if every agent in [MATH] has a different favorite piece then |
Everyone gets her favorite piece and the algorithm terminates. for every agent [MATH] do if (1) [MATH] has no competition for her second favorite piece [MATH] or |
(2) [MATH] has exactly one competitor [MATH] for [MATH] [MATH] also considers [MATH] as her second favorite, and [MATH] [MATH] each have exactly one competitor for their favorite piece |
then 10 [MATH] makes a [MATH] -mark. 11 else 12 [MATH] makes a [MATH] -mark. 13 14 Allocate the pieces according to a rightmost rule: |
15 if an agent has the rightmost mark in two pieces then 16 Out of the two partial pieces, considered until the second rightmost mark (which always exists by Lemma below), she is allocated the one she prefers. |
17 The other partial piece is given to the agent who made the second rightmost mark on it. 18 else 19 Each partial piece is allocated—until the second rightmost mark—to the agent who made the rightmost mark on that piece. |
20 if any non-cutters were not given a piece yet then 21 Giving priority to any remaining agents in [MATH] (but in an otherwise arbitrary order), they choose their favorite unallocated complete piece. |
The cutter is given the remaining unallocated complete piece. Algorithm 2 [MATH] Definition 6 During an execution of Core , let [MATH] be a set of pieces and [MATH] be a subset of non-cutters. We say that an agent [MATH] has competition for a piece [MATH] , if (1) |
[MATH] is not dominated by everyone in [MATH] , and (2) there exists [MATH] such that [MATH] is [MATH] ’s favorite or second favorite piece in [MATH] . We call [MATH] competitor of [MATH] for the piece [MATH] |
Definition helps us identify whether we need to perform a [MATH] -mark or [MATH] -mark on the available pieces. Furthermore, in some cases where we know that certain domination patterns appear, it is convenient to prevent some agents from competing for any piece. Hence, Core also takes as an input a subset [MATH] of ag... |
The main result for Core , which is crucial for the entire algorithm to work, is the next theorem. Theorem The Core protocol satisfies Core properties and , and makes at most [MATH] cut queries and [MATH] evaluation queries. |
The proof of Theorem is based on a series of lemmas regarding the properties of Core We start by establishing the following key lemma. |
Lemma 4 Let [MATH] , be the four pieces created by the cutter in the initial step of Core . After all markings have taken place, at most two pieces have marks and each marked piece has at least two marks. |
Proof. We prove the statement for the case where no agent is excluded from competition. Note that when someone is excluded from competition, the lemma is either straightforward (at least two agents excluded) or reduces to case (1.) below. |
To facilitate the proof we use a convenient matrix notation to describe instances. E.g., Each row corresponds to a non-cutter, call them [MATH] . Each column corresponds to a piece: [MATH] [MATH] [MATH] and [MATH] . The number in cell [MATH] indicates the rank of piece [MATH] for agent [MATH] . A blank cell [MATH] mean... |
1. Line is executed and some agent [MATH] is removed. If the remaining agents have a different favorite piece, then the lemma is vacuously true. If the remaining agents have the same favorite piece, then one of conditions or will hold, and thus they each make a [MATH] -mark on their favorite piece. |
2. Line is not executed and all agents have a different favorite piece; the lemma is clearly true since there are no marked pieces in this case. |
3. Line is not executed and exactly two agents have the same favorite piece. The instance is isomorphic to: If agent [MATH] has no competition for her favorite piece, she takes it and leaves. The remaining agents make exactly one 2-mark each, on the first piece. |
If agent [MATH] has competition with exactly one agent for her favorite piece: In this case, [MATH] makes a 3-mark, marking [MATH] and [MATH] , and [MATH] makes a [MATH] -mark, marking [MATH] . Therefore, [MATH] will definitely have at least [MATH] marks. [MATH] can make a [MATH] -mark or a [MATH] -mark, depending on w... |
If agent [MATH] has competition with exactly two agents for her favorite piece: Agents [MATH] and [MATH] make [MATH] -marks, on [MATH] and [MATH] , and thus at least two pieces are marked with [MATH] marks each. It remains to show that no other piece is marked. If [MATH] ’s second favorite piece is [MATH] or [MATH] she... |
4. Line is not executed and all agents have the same favorite piece. If they also have the same second favorite piece, then they all make [MATH] -marks on the same two pieces. If they have different second favorite pieces, they all make [MATH] -marks on [MATH] . Otherwise, the instance is isomorphic to: |
In this case, [MATH] and [MATH] make a [MATH] -mark on [MATH] and [MATH] , while [MATH] makes a [MATH] -mark on [MATH] : two pieces marked with at least [MATH] marks each. ∎ |
An almost immediate corollary is the following: Corollary 1 All pieces allocated in lines and of the Core protocol are unmarked, and therefore they are allocated as complete pieces. |
Proof. If there existed two marked pieces and the same agent had the rightmost mark in both, then they will both be allocated in line . By Lemma , there are no other marked pieces, hence the pieces that have remained in lines and are unmarked. Otherwise, the else part in line is executed (no agent has the rightmost mar... |
Given Lemma , and since only marked pieces are allocated partially, it follows that the cutter and at least one other agent receive complete pieces, each of which the cutter values as [MATH] of the input residue, thus establishing Core property See |
Towards proving the remaining Core properties, we first establish the following two lemmata. Note that the lemmata still hold when some agents are excluded from competition. |
Lemma 5 If all non-cutters have placed their marks, as dictated by the Core protocol, and some agent [MATH] has marked a piece [MATH] her value for [MATH] up to her mark is equal to her value for her favorite unmarked piece. |
Proof. Assume agent [MATH] has made an [MATH] -mark. We need to prove that her [MATH] -th favorite piece has remained unmarked after all agents have placed their marks. Suppose this is not the case. If [MATH] had made a [MATH] -mark and someone marked her second favorite piece, it means she had competition for it. Thus... |
Lemma 6 If agent [MATH] has made an [MATH] -mark, [MATH] , she receives a piece with value at least equal to the value she has for her [MATH] -th favorite piece out of those that were still unallocated when she made her marks. |
Proof. First, we prove the statement for the case where [MATH] gets a partial piece, i.e., [MATH] gets her piece in line or line . If [MATH] makes a [MATH] -mark on a piece [MATH] and she is allocated a part of [MATH] in either step, she either gets the part of [MATH] up to her mark (when she is the “other agent” of li... |
Next, we prove it for the case where [MATH] is allocated a complete piece in line , i.e., she did not have the rightmost mark on her favorite piece, nor was she the “other agent” of line . Then, by Lemma [MATH] ’s [MATH] -th favorite piece has remained unmarked and will be allocated completely. It remains to show that ... |
To continue with the analysis, we need to understand what could possibly cause agents to experience envy during the execution of the protocol. For this, suppose that an agent [MATH] has made a mark on a piece. If an agent [MATH] is allocated this piece strictly beyond [MATH] ’s mark, Lemma is not enough to ensure that ... |
Lemma 7 If agent [MATH] marked a piece [MATH] and she does not have the rightmost mark in two pieces, then no agent [MATH] will be allocated [MATH] strictly beyond [MATH] ’s mark. |
Proof. Since [MATH] has a mark on it, by Corollary , it could not have been allocated in line or If [MATH] has the rightmost mark on [MATH] , and since [MATH] does not have the rightmost mark in another piece, it is her who gets (a part of) [MATH] in line If [MATH] does not have the rightmost mark on [MATH] , then [MAT... |
We are now ready to prove Core properties and See Proof. Envy-freeness for an agent [MATH] who either is the cutter, or was allocated her favorite piece completely is straightforward: If [MATH] is the cutter, she considers all pieces to be equal and is allocated one of them completely, while the other agents are each a... |
Let [MATH] be a non-cutter who did not get her complete favorite piece. This means she was asked to make an [MATH] -mark, which, by definition, means that pieces left unmarked by [MATH] have value at most her value for her [MATH] -th favorite piece. Let [MATH] be this value. By Lemma [MATH] gets a piece of value [MATH]... |
See Proof. Agent [MATH] is allocated her favorite piece completely in Line , since she is dominated by [MATH] and [MATH] and has no competition, thus satisfying (1). Towards (2), in case the remaining non-cutters, [MATH] and [MATH] , have no competition for their favorite piece out of those remaining, each is allocated... |
Counting Queries Recall the discussion in Section about the residue being a finite union of intervals and not a single interval. It is important that our algorithm should know, at any time during its execution, the values of all the agents for all the intervals that make up the residue. Otherwise, our queries cannot be... |
When we ask agents to place a mark on a piece, this corresponds to 1 cut query, since we can simulate this action with a cut query. Thus Core requires [MATH] evaluation and [MATH] cut queries. We assume that in the beginning of the current execution all agents know the values of all the intervals that make up the resid... |
The cutter is asked to cut the residue into four pieces in line . ( [MATH] cut queries) The ordering each non-cutter has for the pieces is necessary for lines , therefore all three non-cutters are queried for their value for each of the four pieces. ( [MATH] evaluation queries) |
According to the conditions in lines and , each of the non-cutters remaining (at most three) will make either a 2-mark or a 3-mark. Worst case, three agents remain and all make a 3-mark. ( [MATH] cut queries) |
We ask evaluation queries so that every agent learns the value of each marked piece up to the second rightmost mark. Worst case, there are two marked pieces and for each we asked everyone but the agent that placed the second rightmost mark. ( [MATH] evaluation queries) |
Note that the protocol now has all the information needed to check the condition on line , find the values of the partial pieces, calculate gain, and know the values of all the intervals that make up the new residue. |
It should be noted that in a few cases we know we have less queries. In particular, the calls to Core in lines and of the Main Protocol , are guaranteed to produce at most one partial piece. Therefore, in these 3 calls we only need [MATH] cut queries and [MATH] evaluation queries for everyone to learn the value of the ... |
Moreover, on the second execution of line , agents [MATH] and [MATH] (as they are called in the Main Protocol ) never need to know the value of the marked piece up to the second rightmost mark. |
In particular, the total number of queries of the Main Protocol is [MATH] cut and [MATH] evaluation queries. The Correction Protocol |
Correction takes as input an allocation [MATH] , produced by a single execution of Core . It outputs a redistribution of the pieces [MATH] |
such that the insignificant piece goes to a different agent. Some of the pieces are marked (the ones partially allocated by Core ), while others are unmarked (the pieces allocated completely). Note that each of the marked pieces of [MATH] still has exactly two marks, the rightmost and the second rightmost marks of the ... |
The redistribution should satisfy further properties, so that both the envy-freeness of the overall partial allocation and certain dominations are preserved. Towards bounding the “local” envy that the redistribution may cause, the notion of gain (see Section ) is crucial. |
In the pseudocode description below, we refer to the cutter in allocation [MATH] as [MATH] , the non-cutter who holds the insignificant piece as [MATH] the non-cutter who gets the insignificant piece after executing Correction as [MATH] , and the remaining non-cutter as [MATH] |
Let [MATH] [MATH] be the agents having the two marks on the insignificant piece, and suppose [MATH] was given this piece in allocation [MATH] |
The insignificant piece is allocated to [MATH] if there is no other partial piece then Agents choose their favorite piece in the order [MATH] [MATH] [MATH] |
else Find the rightmost mark not made by [MATH] on the other partial piece. Let [MATH] be the agent who made it. Agent [MATH] is allocated the partial piece. |
The last non-cutter chooses her favorite among the two complete pieces. The cutter is allocated the remaining (complete) piece. 10 |
Algorithm 3 [MATH] The main result about Correction is the next theorem. The remaining of this section is dedicated to its proof. |
Theorem The Correction protocol satisfies Correction properties and , and makes no queries. We stick to the notation we used in the description of Correction . That is, given an allocation [MATH] [MATH] is the cutter, [MATH] is the non-cutter who holds the insignificant piece, [MATH] is the non-cutter who gets the insi... |
Correction property trivially holds, since in line of Correction the insignificant piece is allocated to agent [MATH] See See Proof. |
If [MATH] is a non-cutter who received her favorite unmarked piece in Core , she has made no marks. Therefore, since by definition [MATH] and [MATH] have made marks on the insignificant piece and [MATH] is the cutter, |
[MATH] is the only non-cutter agent who could have formerly been allocated her favorite piece. We need to show that in this context [MATH] will be the first to choose one of the complete pieces in Correction If we are in the case of line , this follows immediately. Otherwise, since [MATH] has made no marks, she is not ... |
See For the proof of Correction Property we assume that [MATH] is produced by a call to Core where no agent was excluded from competition. Note that this is always the case when Correction is used in our main protocol (see the proof of Theorem ). We need Lemma below, but first it is helpful to establish some tie-breaki... |
Lemma 8 Any non-cutter [MATH] who was not allocated her favorite whole piece when [MATH] was produced, receives in [MATH] value at least equal to that of her favorite piece in [MATH] among those formerly allocated to agents in [MATH] |
Proof. Let [MATH] . In most cases we will show the stronger—but easier to prove—property that [MATH] ’s new allocation has value to her at least equal to that of her second favorite piece among all four pieces in [MATH] |
We start by proving this for the case when Correction allocates a partial piece [MATH] to [MATH] Note that this may happen only if [MATH] has a mark on [MATH] (lines ). Thus [MATH] ’s value for [MATH] up to her mark is equal to her value for her favorite unmarked piece, and therefore there is only one piece which [MATH... |
Next, we move to the case where [MATH] is allocated an unmarked piece [MATH] by Correction , distinguishing among the two sub-cases where that could happen: (a) in line , and (b) in line , where there are one and two marked pieces respectively. |
For sub-case (a), we claim that it suffices to show that [MATH] is allocated value at least equal to that of her favorite unmarked piece out of those formerly allocated to agents in [MATH] . To see that, note that the fact that there is only one marked piece, immediately implies that she is allocated value at least equ... |
[MATH] ’s favorite unmarked piece out of those formerly allocated to agents in [MATH] is still unallocated and she can take it. For sub-case (b), first notice that if [MATH] does not have a mark in both pieces, then her favorite unmarked piece (i.e., the one she chooses in line ) is either her overall favorite or secon... |
Proof of Correction Property Towards proving Correction property , recall Definition . Given a partial allocation [MATH] and a suballocation [MATH] of [MATH] , let [MATH] be an agent who dominates agents in [MATH] . If [MATH] is agent [MATH] ’s piece in [MATH] and [MATH] is [MATH] ’s favorite piece out of those allocat... |
Now assume [MATH] was produced by Core (and recall that no agent was excluded from competition). By Correction property , if [MATH] was a non-cutter in [MATH] who was formerly allocated her favorite complete piece, then |
[MATH] and hence [MATH] Similarly, if [MATH] was the cutter in [MATH] , then by line of Correction we have [MATH] Therefore the difficulty lies in proving that the property holds for non-cutters who did not get their favorite whole piece when [MATH] was produced. Suppose that after the permutation output by Correction ... |
# Source: arxiv 1807.00322 # Title: Colimits of monoids # Sections: all # Downloaded: 2026-03-03T02:41:22.140707+00:00 Colimits of monoids |
Abstract If [MATH] is a cocomplete monoidal category in which tensoring from both sides preserves coequalizers, then the category [MATH] of monoids over [MATH] is cocomplete. The same holds if [MATH] has regular factorizations and tensoring only preserves regular epimorphisms. As an application a lifting theorem for an... |
MSC 2000 : Primary 18D10, Secondary 18A30 Keywords: Monoids in monoidal categories, coequalizers, (regularly) monadic functors Introduction |
While limits in categories of monoids in [MATH] , the category of monoids over a monoidal category [MATH] , are well know, since the forgetful functor [MATH] creates limits (see e.g. |
), not much seems to be known about the existence of colimits in [MATH] . Assuming that [MATH] is cocomplete and [MATH] preserves colimits, it is shown in |
that certain pushouts exist in [MATH] , while deals with the more restrictive situation that [MATH] is locally finitely presentable and [MATH] preserves directed colimits: Then [MATH] is locally presentable and, hence, cocomplete. |
In this note we will consider the case where [MATH] is cocomplete and [MATH] is monadic. Then, by a well known result on monadic functors (see e.g. , 4.3.4] ), the category [MATH] will be cocomplete provided that it has coequalizers. |
An application of this criterion becomes possible by a result on the existence of coequalizers in [MATH] in the preprint . Unfortunately, neither the publication |
with a similar title nor the actual arXiv-version contains this result such that it is nearly impossible to become aware of it. Since the author of this note is convinced that this result on the existence of coequalizers in [MATH] should become part of the “official” literature and the authors of that preprint have no ... |
, its proof is included in Section below. We add a short remark concerning what might have been the motivation for the construction described in that paper. A similar comment is in place with respect to the interesting application concerning left adjoints of functors between categories of monoids induced by a monoidal ... |
, our proof uses crucial elements of that proof. We make clear, however, that what appears as an ad hoc construction in in fact is based on general principles for the lifting of adjunctions. |
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