# Cutting the Cake Problem

The traditional method for two children to divide a piece of cake fairly between them is "you cut, I choose". What process accomplishes this is intuitively clear. Is there a similar process which works for $$3$$ children?

Suppose that three children $$(1),(2)$$ and $$(3)$$ have associated measures (ways of evaluating the pieces of cake) $$\alpha_{1},\alpha_{2}$$ and $$\alpha_{3}$$. Assume for the simplicity of the notation that the sum of total pieces of the cake is equal to $$1$$.

Question : Does the cake have three subsets $$C_{1},C_{2}$$ and $$C_{3}$$ that are pairwise disjoint and their union is equal to $$C$$, and that have moreover the property that $$\alpha(C_{1}) \geq \frac{1}{3}, \quad \alpha(C_{2} )\geq \frac{1}{3} , \quad \alpha(C_{3}) \geq \frac{1}{3}?$$

Solution: Let one of the three children, say $$(1)$$, divide the cake into what he regards as three equal pieces $$C_{1},C_{2}$$ and $$C_{3}$$: In other words $$\alpha(C_{1})= \alpha(C_2)=\alpha(C_{3})=\frac{1}{3}$$. Let each of $$(2)$$ and $$(3)$$ claim dibs on one of those three pieces, the one he prefers the most, the one he would be quite satisfied to have. It is possible that $$(2)$$ and $$(3)$$ will claim different pieces, and it is possible that they will claim the same one, but no matter, atleast one of the three pieces, $$C_{1},C_{2},C_{3}$$ will be left unclaimed. Let $$(1)$$ remove that piece (one of those pieces) and eat it. So far everyone is satisfied. Child (1) got what he believes to be exactly a third, and in the judgement of the children $$(2)$$ and $$(3)$$ the best piece (the claimed piece) is still available. The problem reduces therefore, to dividing whats fairly left, - and that's the classical problem of dividing a piece of cake between two claimants. Let that problem be solved by "you cut, i choose" and everybody is happy.

Solution The Proposed solution is wrong; Whats wrong with it?

• You might want to check out the book "Cake Cutting Algorithms" as well: amazon.com/Cake-Cutting-Algorithms-Jack-Robertson/dp/1568810768 Aug 15, 2010 at 7:58
• – JRN
Jan 16, 2014 at 14:46
• Dec 14, 2014 at 14:38
• I think Numberphile did a video on something like this Oct 3, 2019 at 22:00

After giving to (1) one of unclaimed pieces, less than 2/3 of cake can be left (in measure of (2) or (3)), so one of them will be unsatisfied.

Edit: The right solution is the following.
1. Let (1) divide cake into three equal pieces.
2. Ask (2) and (3) "What is the best one?". If their answers are different, we are done: each one of (2) and (3) gets what he (or she) wants, and (1) takes unclaimed piece. Otherwise go to step 3.
3. Ask (2) and (3) "Is there any other piece, you would satisfied with?". If at least one of the answers is "Yes" we are done similar to step 2. Otherwise go to step 4.
4. (1) takes any of two unclaimed pieces.
5. Thanks to the answer "No" at the step 3 we know, that at least 2/3 of the cake left in both measures (the one of (2) and the one of (3)). Therefore we can apply "you cut, I choose" method to the leaving 2 pieces.

Edit: This problem can be solved with arbitrary number of children using the algorithm for finding maximum matching in a bipartite graph. Suppose, number of children is n. Cutting the cake algorithm is the following.
1. Let (1) divide cake into three equal pieces.
2. Ask others the following: "Please, list all the pieces, you will be satisfied with."
3. Consider bipartite graph with 2n-1 vertices: n pieces of cake and all children except (1). Children is connected with a cake iff it will be satisfied with it. 4. Find maximum matching in this graph. Lets call vertex "free" iff it is not matched in this matching with another one.
5. If there are no free children, give pieces of the cake according to the matching (and free piece of cake is for (1)). In this case we are done. Otherwise go to step 6.
6. In the algorithm of maximum matching we are able to find its certificate, i.e. pair of sets (U,V) with U containing all free children and some non-free, and V containing some pieces of cake, such that 1) amount of elements in V is equal to amount of non-free elements in U; 2) every vertex in U has edges only to vertexes form V 3) there are no edges from chosen matching, connecting vertex from V to some vertex outside U. In our case it means that every piece of cake outside V is less than 1/n in all measures of children from U.
7. Give pieces of cake to children outside U (except (1) ) according to chosen matching.
8. Give any free piece of cake outside V to (1).
9. Note, that according to 6, all pieces of cake given at 7 and 8 are unimportant for elements of U. Therefore there is enough cake left for them. Divide it according to this algorithm between elements of U (note, that (1) is not in U, and, therefore size of U is less than n).

There is a simple and elegant solution for N children, suggested by Banach and Knaster (Steinhaus, 1948):

"The partners being ranged A, B, C, ... N, A cuts from the cake an arbitrary part. B has now the right, but is not obliged, to diminish the slice cut off. Whatever he does, C has the right (without obligation) to diminish still the already diminished (or not diminished) slice, and so on up to N. The rule obliges the "last diminisher" to take as his part the slice he was the last to touch. This partner being thus disposed of, the remaining n-i persons start the same game with the remainder of the cake. After the number of participants has been reduced to two, they apply the classical rule for halving the remainder.

The rules are obvious; as to the methods, they are the following ones: If in any stage of the procedure you have to cut a slice, do it in a way to give to it a value of 1/n-th of the whole cake in your subjective es- timation; if you are asked to diminish a slice already cut off by another person, diminish it to the size you consider to be 1/n-th of the whole in value but abstain from action if it has already this value or a smaller one in your estimation. It is easy to prove that the methods explained here secure to every partner at least a part equal in value to the 1/n-th of the whole. The greed, the ignorance, and the envy of other partners can not deprive him of the part due to him in his estimation; he has only to keep to the methods described above. Even a conspiracy of all other partners with the only aim to wrong him, even against their own interests, could not damage him.

The hypotheses underlying the procedure described are: (1) the ideal shares due to the partners are not contested by any of them; (2) the object is continuous, so as to make it possible to cut off it a slice of value pV, V being the value of the whole and p any fraction between 0 and 1; the same applies to every part which can be cut off the object; (3) the sum of values of parts is equal to the value of the whole; the same applies to every part. The postulates (2) and (3) have to be valid for every individual estimation."

There is a result due to Dubins and Spanier (JSTOR required) that says that if $(M,\mathcal{M})$ is a measurable space and $\alpha_1,\alpha_2,\ldots,\alpha_n$ are atomless probability measures on $(M,\mathcal{M})$ and $c_1,c_2,\ldots,c_m$ are finitely many nonnegative numbers such that $\sum_{j=1}^m c_j=1$, then there exists a partition $\Pi$ of $M$ into measurable sets such that all the probability measures agree on $\Pi$ and have exactly measures $c_1,\ldots,c_m$. The result has been independently rediscovered by Aumann in the context of game theory, (Lemma 4.1).

One can even show that there exists a sub-$\sigma$-algebra of $\mathcal{M}$ on which all measures agree and are atomless.

Two things:

1. You are assuming that (1) cut the cake into three pieces of equal value to him, and then took one piece. Now there are two pieces left for (2) and (3) to choose between, and you claimed that this problem could be solved by the original "you cut, I choose" method. But, since the cake is already cut, they cannot use that method.

2. (1) cut the cake into three pieces of equal value to himself and then took a piece for himself, but this piece may have been more valuable to (2) or (3) than any of the other pieces, so they would not be satisfied.

I have an idea of how to do it, but its not complete:

Let (1) take three pieces of string to mark off where he intends to cut the cake into three slices of what he believes are equally valued. Then there are two cases for people (2) and (3). If (2) and (3)'s optimal slices are different, then they should just take those slices and (1) can take the last one- everyone is satisfied.

If both (2) and (3) want the same slice, then remove the string and start again with (2) setting the string on the cake this time instead.

By doing this eventually you reduce the problem to the case where when any person cuts the cake into equal slices, the other two people will want the same piece.

• Your premise (#2) of ... this piece may have been more valuable to (2) or (3) than any of the other pieces is wrong. (1) only takes a piece after (2) and (3) choose their favorite piece, which was not the piece that (1) takes. Jan 17, 2014 at 1:17

Person 1 cuts the piece of cake in half and person 2 chooses. Now person 1 cuts their portion into thirds and hands the knife to person 2 who cuts their portion into thirds. Then person #3 chooses a piece from person 1 and a piece from person 2. Everyone, therefore, has at least 1/3 of the piece of cake.

Fiktor is correct. In proportional division, you want to achieve the result, that each of the n player will end up with at lest 1/n cake in her own valuation. The procedure above does not achieve this.

Counter example:

Total valuation of cake

Value P1 P2 P3
3  5  8


So each player should end up with at least

Value P1 P2 P3 1 5/3 8/3

Player 1 makes first cut, valuation after first cut:

Value P1 P2 P3
C1    1   3  1
C2    1   2  4
C3    1   1  3


P3 always chooses C2, P2 chooses C1. So P1 is chooses C3. So it remains:

   Value P2 P3
C1    3   1
C2    2   4


Say P3 now makes the second cut, so that C1 and C2 have same valuation:

   Value P2 P3
C1    >3   5/2
C2    <2   5/2


so P2 chooses C1. Now P3 is left with a piece that she values at 5/2 which is less than the proportional 8/3.

• "archive" or "achieve"? Aug 14, 2010 at 19:05
1. Person (1) cuts the cake in 3 pieces
2. Person (2) and (3) choose one of the pieces
3. if they choose 2 diffrent pieces everything is fine ,if not go to 4
4. Person (1) can choose one of the 2 unchoosen pieces
5. Now there are 2 persons and 2 pieces left and you can reapply the i cut you choose system on each of them
6. So Person (2) cuts one of the 2 left pieces and person (3) chooses one of the new pieces
7. Then Person (3) cuts the last piece and person (2) can choose one of the new pieces

So in the end everyone has cut a piece and everyone has one desired piece.

Or to shorten the last 3 steps:
Person (2) can try to make the 2 left pieces equal by cutting of a small piece of piece  and putting it to piece  and let person (3) choose one of the pieces.

From a game-theory point of view, the reasoning is that the cutting method should ensure that whoever makes the cuts is trying in his best interest to cut fairly.

Hence, the adaptation for three children, in practical terms is that kid #1 makes on single cut, and lets kid number 2 chose a piece for cutting. Then Kid #3 picks his piece, followed by kid #2 and kid #1 in this order.

That way, if you are doing the first cut, you want to have one of the pieces with exactly half the size of the other. If the small piece is too small, then kid #1 (yourself) will likely end up with it. If it's too large, then kid #3 will have it.

Then kid #2 will cut what should be two-thirds of a cake, and let kid #3 chose. Hence it falls to the traditional method. If he decides to cut the small piece, then kid #3 will get the large one.

Kid #3 just choses as he pleases, he doesn't need a strategy.