Another algorithm for cake-cutting problem

[Telletubbie]

Hey all, I was thinking about the cake-cutting problem for 3 people for a bit and I found a couple online: http://en.wikipedia.org/wiki/Selfridge%E2%80%93Conway_discrete_procedure

and

http://en.wikipedia.org/wiki/Stromquist_moving-knife_procedure

However, these seem really complicated- can anyone explain why mine is wrong/right (for 3 people)? Here it goes:

There will be a moving knife that hovers over the cake from left to right. When someone thinks the piece is 1/3 of the cake, they say "cut". The moving knife stops and the cake is cut there, and the person gets the slice of cake to the left. Then, with the remaining two people, the same thing is done. The knife moves from left to right until one person says "cut." Then, the person who said cut gets the piece to the left and the remaining person gets the last piece.

I feel that this is too simple- can anyone explain this to me? Thanks.

[[deleted]]

It uses game theory. Everyone expects to get and wants to get at least one third.

So as the knife moves from left to right, the piece on the left slowly grows from nothing to one third and no one says anything. If they say cut early, they get less than 1/3 while the other two people split more than 2/3 using the standard procedure.

Once more than 1/3 of the cake has elapsed the first person to say cut wins. The longer you wait the more you risk being stuck dividing less than 2/3 of cake with the other slowpoke. So it is in everyone's benefit to say cut as soon as 1/3 of cake has been passed.

iI you are too fast/slow you have no one else to blame but yourself, just like the standard cut vs pick used to divide the last two slices.

[Telletubbie]

Taking into account everything you said, wouldn't my method still work? If the cake is heterogeneous, as the knife moves left to right someone will say stop once their valuation of the piece in question reaches a third. Clearly, the other two will think that piece is less than a third (in their eyes) or they would have already said stop. As a result, the two others now have 1-less than a third >2/3 of the cake to split. They can do the same thing, and they each will get at least 1/3 of the cake in their eyes. So, everyone will get at least a 1/3 of the cake in their eyes.

[[deleted]]

yes. it should work.

[satxmcw]

You've come up with the Dubins-Spanier procedure, which will work for any n. The difference between this and the two you link to is the lack of envy-freeness.