# Mediated envy-free and efficient cake cutting with n=2?

Is there an algorithm in literature to compute an efficient (pareto optimal) and envy-free cake cutting when there are only $n=2$ players and a mediator?

-
Will Mediator work for free? – Ilya Nikokoshev Dec 24 '09 at 13:22
I am not sure what you mean by "Mediator" - what exactly is his job? In any case, you may find some information on PEEF (Pareto-Efficient-Envy-Free) division here: en.wikipedia.org/wiki/… – Erel Segal-Halevi Sep 4 '15 at 5:17

Huh? I cut, you choose. Why do we need a Mediator?

-
I suspect that "envy-free" might rule this out, though I don't know what assumptions are going into this, or what "envy-free" means in a technical sense. Perhaps if I'm bad at cutting, then I will envy you for getting the bigger piece? The use of an undefined technical term like this from economics makes the question sound like a homework question. – Kenny Easwaran Oct 22 '09 at 14:40
>Perhaps if I'm bad at cutting, then I will envy you for getting the bigger piece? And what if you're bad at choosing? Then it's hopeless! I think your objection is silly. – TonyK Oct 22 '09 at 14:54
That didn't work very well :-( I meant to quote Kenny, and add further commentary,but comments don't parse like answers do... – TonyK Oct 22 '09 at 14:55
Cut and choose is not pareto optimal, which is: there could be a better assignment that makes some of the players better off without making anybody worse off. This is the case, because in the cut and choose method, the person that cuts is guaranteed to have 1/2 of it view of the cake, whereas the person the chooses get at least one half. – unluckyjoe Oct 23 '09 at 0:36
@TonyK: Suppose each player assigns a different measure to the cake, and player $i$ seeks to maximize $\mu_i(A_i)$, where $\mu_i$ is that player's measure and $A_i$ is the piece that player ends up with. Then it is certainly not true that any improvement for the cutter must be worse for the chooser. – Steven Landsburg Oct 23 '15 at 4:06

Take a look at: