I have (rather recently) dabbled in game theory. I need it to design an algorithm to share chores. Obviously this is a kind of cake-cutting problem. So far, I have fought my way through An Introduction to Game Theory by Martin J. Osborne, but I'm still feel far from comfortable with it. I have a solid foundation in calculus, know how to deal with ODEs and PDEs (but I try to avoid them if I can. :)) And yes, I'm not a mathematician, I"m an engineer.


The 'cake' needs to be split among $k\geq 2$ players. The twist is that valuation of the players is not finitely additive, it has a maximum i.e. there is an amount of cake that they will find more valuable than a larger amount (kind of being afraid of overeating $-$ or being on a diet).


Does anybody know of a starting point to how tackle this? (Efficient or equitable solutions would be the most interesting.) All the resources I found, treat only the finitely additive case. I would also be grateful for any freely downloadable material.

EDIT: I'm looking for efficient and/or equitable ways of splitting the cake, regardless of the protocol to achieve it. If there also is a protocol for that, the better for me. :)

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Count Zero
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  • Just because it's game theory doesn't mean that you should keep us guessing what the question is :-) Are you looking for a particular protocol for cutting the cake? Or do you have a particular protocol in mind and are looking for optimal/equilibrium strategies for the eaters under that protocol? Or are you just looking for efficient or equitable divisions of the cake, independent of protocols for achieving them? – joriki Oct 14 '12 at 14:22
  • @joriki: I edited the question. Hope that ends the guessing game. `:)` – Count Zero Oct 14 '12 at 14:28
  • You haven't specified a valuation for the eaters -- are you just generally interested in methods for finding efficient and/or equitable divisions that work if the valuations aren't additive, or do you want to find an efficient and/or equitable division for a particular valuation? – joriki Oct 14 '12 at 14:36
  • @joriki: The valuation is random, but known, subject to this non-additivity condition (i.e. has 1 maximum). – Count Zero Oct 14 '12 at 14:38
  • You may have better luck describing it as a "compensation" problem, rather than cake cutting. For discrete problems (finitely many chores) the math involved is very engineery, some Lagrange multipliers and stuff. I believe I've read Econ papers where special sorts of non-additive utility functions were considered. – Jack Schmidt Oct 14 '12 at 15:11
  • @JackSchmidt: That's very likely: there are compensation schemes, where extra time is not paid proportionally to discourage employees from doing more than a desired limit (e.g. avoid overproduction). Do you have any links? That could be an answer and I'd certainly upvote it! – Count Zero Oct 14 '12 at 15:17
  • @CountZero: done. Definitely don't accept my answer, as it just (1) recommends a (cheap, good) book and an easy technique, (2) confirms what you already know, and (3) gives you one more search term "unimodal" to help out. – Jack Schmidt Oct 14 '12 at 15:51

4 Answers4


TLDR: Borgers book is \$35 and awesome for the additive case of compensation. From this you can Lagrange multipliers to solve nice variations. Add "unimodal" to your search to find articles where the marginal utility goes up for a bit and then back down.

I found cake-cutting to have some additional complications that can be solved in a manageable way by simply paying people a little to make sure things are equal. In other words, we have a cake where people have different (usually finitely additive) utility functions and those difference produce the opportunity for extra-good in the division, but we also have a pile of money where everyone's utility function agrees and is plain old "dx", the easiest of all.

Compensation problems with finitely many goods and additive utility functions are very easy to understand, and are worked out in a very nice textbook way in Borger, 2010 (\$35 print, over \$960 online form the publisher, impressive compensation scheme). The techniques used do not change too much with different utility functions if they are not too weird (for instance, unimodal, continuous function in your case, or monotone, continuous in the case I read about), since you are just optimizing differentiable functions of a single variable, and so the maximums occur either at corners or at solutions to Lagrange multiplier constraints. In the additive case, all the utility functions are linear, so the calculus disappears, and the maximums occur at (at least one of the) corners.

I did not find an easily digestible article discussing your case (which is probably very similar to the example you give: adjust compensation to encourage neither under-production nor over-production; "unimodal" should help the search). The cases I read about were "auctions" where the risk involved means utility functions are "sub-linear", so that people are less willing to risk large amounts of money, so the marginal utility of the cash decreases (but never goes negative as I think it does in your case).

  • Börgers, Christoph. Mathematics of social choice: Voting, compensation, and division. Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2010. xii+245 pp. ISBN: 978-0-898716-95-5 MR2574481 DOI:10.1137/1.9780898717624
Jack Schmidt
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The utility functions you describe are called "single-peaked". Here are two references:

✔ Resource-monotonic solutions to the problem of fair division when preferences are single-peaked, Social Choice and Welfare, Vol. 11, No. 3. (1994), pp. 205-223, doi:10.1007/bf00193807, by William Thomson

✔ Population-monotonic solutions to the problem of fair division when preferences are single-peaked, Economic Theory, Vol. 5, No. 2. (1995), pp. 229-246, doi:10.1007/bf01215201, by William Thomson

Note, however, that these references deal with a homogenous resource (i.e. each person cares only about the amount of resource he gets, not about which part he gets).

For the case of a heterogenous resource, with non-additive valuation functions, see these references.

Erel Segal-Halevi
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You; might find the book Cake-Cutting Algorithms by Jack Roberson and William Webb, A.K.Peters, Natick, 1998 of value.

Joseph Malkevitch
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The terms equitable or efficient need to be defined. And the action choices need to be specified. Suppose there are no other action choices and the players have to eat the share of cake they get. Let the utility/valuation of player $i$ be single-peaked/unimodal function $u_i(x_i)$ of his share $x_i$.

Suppose equitable goal focuses on the lowest utility players: allocation $a$ with utilities $(a_1,a_2,...,a_n)$ in increasing order is better than allocation $b$ with utilities $(b_1,b_2,...,b_n)$ in increasing order if there is a $k$ such that $a_k>b_k$ and $a_i=b_i$ for $i<k$. Here is an approach for most equitable allocation: Determine utility maximizing share $x_i$ for each player. If the sum of shares exceeds 1, some shares need to be reduced. If the sum of shares is less than 1, some shares need to be increased. Increase of decrease the shares of the players with highest utility while keeping their utility equal. For example, increase or decrease their shares until, sum of shares reaches 1 or if their utility reaches the level of the next highest utility. In the latter case, start adjusting shares of the new set of players with highest utilities. The algorithm can be made efficient.

Suppose efficient goal focuses on the sum of utilities: allocation $a$ with utilities $(a_1,a_2,...,a_n)$ in increasing order is better than allocation $b$ with utilities $(b_1,b_2,...,b_n)$ if $\sum_i{a_i}>\sum_i{b_i}$. The idea here is to equate marginal utilities.

First some notation. Assume utility functions are differentiable and concave. Define inverse of marginal utility functions $r_i(s)$ as $\mathrm{d}u_i(x_i)/dx_i\mid_{x_i=r_i(s)}=s$. If there are multiple solutions to the above equation choose highest for positive $s$ and lowest for negative $s$ (to be closest to peak). If $\mathrm{d}u_i(x_i)/dx_i<s\forall{x_i}$, choose $r_i(s)=0$.

Again start with utility maximizing share $x_i$ for each player. If the sum of shares exceeds 1 determine minimum $s>0$ such that $\sum_i{r_i(s)}=1$. If the sum of shares is less than 1 determine maximum $s<0$ such that $\sum_i{r_i(s)}=1$. The solution is $x_i=r_i(s)$. Kinks in utility function and noncaoncavity can be accomodated with little more work.

To be relevant economically, this problem needs more structure. Some thoughts: (1) Free disposal assumption (players can costlessly dispose of piece of cake not needed) would ensure non-decreasing utility functions. (2) There is no reason all players should get same weight. For example, if a person is always unhappy (utility function is below the utility functions of other players), equitable allocation to bring him up to others doesn't seem appropriate. (3) If players could trade (had enough cash or any other commodity that they value identically), you could do any allocation and the players would trade to end up with the efficient allocation. Marginal utility $s$ would be the market price.

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