Figure 5 - uploaded by Sebastian Wild
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When considering cut lengths L (i) /j, no j larger than k /i is relevant. The sketch shows a cutting with L (i) and j = k /i. Note how we have k maximal pieces for sure (dark); there may be many more (light).
Source publication
Segal-Halevi, Hassidim, and Aumann (AAMAS, 2015) propose the problem of
cutting sticks so that at least k sticks have equal length and no other stick
is longer. This allows for an envy-free allocation of sticks to k players, one
each. The resulting number of sticks should also be minimal.
We analyze the structure of this problem and devise a linear...
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Citations
... We thank Chao Xu for pointing us towards the work by Cheng and Eppstein [CE14] and noting that the problem of envy-free stick-division [RW15b] is related to proportional apportionment as discussed there. He also observed that our approach for cutting sticksthe core ideas of which turned out to carry over to this article -could be improved to run in linear time. ...
Proportional apportionment is the problem of assigning seats to parties according to their relative share of votes. Divisor methods are the de-facto standard solution, used in many countries.
In recent literature, there are two algorithms that implement divisor methods: one by Cheng and Eppstein (ISAAC, 2014) has worst-case optimal running time but is complex, while the other (Pukelsheim, 2014) is relatively simple and fast in practice but does not offer worst-case guarantees.
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... In fact, it is equivalent to the following envy-free stick division problem: given m sticks of different lengths, make a minimal number of cuts such that there are at least k pieces with equal lengths and no other piece is longer. Reitzig and Wild [12] devise an algorithm that solves the envy-free stick division problem in time O(m + min (k, m) log min (k, m)). For our purposes, it is sufficient that Equalize(k) can be done in bounded time. ...
We consider the classic problem of envy-free division of a heterogeneous good (aka the cake) among multiple agents. It is well known that if each agent must receive a contiguous piece then there is no finite protocol for the problem, whenever there are 3 or more agents. This impossibility result, however, assumes that the entire cake must be allocated. In this paper we study the problem in a setting where the protocol may leave some of the cake un-allocated, as long as each agent obtains at least some positive value (according to its valuation). We prove that this version of the problem is solvable in a bounded time. For the case of 3 agents we provide a finite and bounded-time protocol that guarantees each agent a share with value at least 1/3, which is the most that can be guaranteed.
We consider the classic problem of envy-free division of a heterogeneous good ("cake") among several agents. It is known that, when the allotted pieces must be connected, the problem cannot be solved by a finite algorithm for three or more agents. The impossibility result, however, assumes that the entire cake must be allocated. In this article, we replace the entire-allocation requirement with a weaker partial-proportionality requirement: the piece given to each agent must be worth for it at least a certain positive fraction of the entire cake value. We prove that this version of the problem is solvable in bounded time even when the pieces must be connected. We present simple, bounded-time envy-free cake-cutting algorithms for (1) giving each of n agents a connected piece with a positive value; (2) giving each of three agents a connected piece worth at least 1/3; (3) giving each of four agents a connected piece worth at least 1/7; (4) giving each of four agents a disconnected piece worth at least 1/4; and (5) giving each of n agents a disconnected piece worth at least (1 - ε)/n for any positive ε.
