Earlier today I set you this puzzle by one of the world’s leading logicians about a group of five power-hungry philosophers. Here it is again with the solution.
The philosopher’s ruling council
In the ancient land of Philosophia there is a ruling council of five philosophers, linearly ranked by power and prestige, with various accompanying benefits accruing in the order of this rank. Philosopher 1 is the recognized philosopher king, the most powerful, and then philosopher 2 and so on.
It is time to pick a new council, and according to the long agreed-upon procedure, the lowest-ranked philosopher proposes a new council and ranking. The newly proposed council can in principle include any citizen at all from Philosophia—candidates are not limited to the current council members, although strangely, it usually happens that the new council is constituted by previous council members. Given the new proposal, the council votes. If a majority approve, then this is the new council and ranking; otherwise, the lowest-ranked philosopher is kicked off the council and the next lowest-ranked philosopher makes a proposal. This process continues until a new council and ranking is approved.
As mentioned above, these philosophers are a selfish crowd, all hell-bent on becoming the new council’s philosopher king. Each member prefers being on the new council above all other things, and will never vote in favor of a council that they are not on. Secondly, being on the proposed council, they would prefer to have as high a rank as possible (that is, a low number—being philosopher 1 is the best, 2 is second best, and so forth). Third, given that they will be on the council with a certain rank, they prefer that the council is as small as possible, so as not to have to share power unnecessarily (but superior rank on a larger council is preferred).
Philosopher 5, the lowest-ranked philosopher starts off, proposing a council and a ranking.
Can you suggest a proposal that guarantees philosopher 5 becomes philosopher king?
Solution
As I mentioned in the original post, you solve the puzzle by working backwards. In order to know what philosopher 5 will propose, we seem to need to know already what philosopher 4 will propose, since that will be the alternative if we vote philosopher 5’s proposal down.
Let us start with the easiest case, where the council has only one philosopher, the king. In this case, he will be the lowest ranked philosopher, and he can propose that the new ruling council is the same as the old ruling council, having only himself as a member. This is clearly the most desired situation, according to the philosopher values we mentioned, and so he will vote in favor and the plan will be adopted.
If there are two philosophers currently on the council, then philosopher 2 must propose a plan that will pass unanimously, since that is the only way to have a majority of two. But philosopher 1 will not approve of any plan except the plan of him being on the council alone, since that is what he can attain if philosopher 2’s plan is rejected. And we said that philosopher 2 will only vote to approve a plan in which she is on the council. So philosopher 2’s proposal, whatever it is, will be rejected and we will end up eventually with the council of just philosopher 1.
If there are three philosophers on the council, then philosopher 3, needing two votes, will propose a council of two, consisting of himself as king and philosopher 2 as the next member. This will be approved by philosophers 2 and 3, since they both prefer it to the result of the two-philosopher case. And furthermore, this is the best possible arrangement for philosopher 3, since he is made king on a council of size two.
If there are four philosophers currently, then philosopher 4 needs three votes. She will not get the vote of philosopher 3, who would be very well off if her plan were to be rejected, but she can get approval for a three-member council consisting of philosopher 2, herself, and philosopher 1, in that order. Each of these three will vote to approve, since they are better off with this than the alternative, and this is furthermore the best possible for philosopher 4, because she needs a three-member council, but must make philosopher 2 better off, and so philosopher 2 must become king (or should we say queen), and then philosopher 4 next, which is optimal, and then philosopher 1 will be third, which is better than he would get under the competing proposal of philosopher 3.
Consider now the case of five philosophers on the council. Philosopher 5 needs three votes and therefore will propose a council of size three, and indeed he can propose the council: himself, philosopher 1, philosopher 3, in that order. Philosopher 5 likes this very well, becoming king, and philosopher 1 prefers this to the preceding philosopher 4 plan, since he will have rank two instead of rank three. Philosopher 3 also likes this plan, since he wasn’t on philosopher 4’s proposed council at all.
Let us summarize the proposals as follows:
1 philosopher:proposed council 1
2: proposal rejected
3: 3 2
4: 2 4 1
5: 5 1 3
If you want to read further discussion of this puzzle, it can be found on Joel David Hamkins’ Substack Infinitely More. (Some of the content is free, some is for paid subscribers).
Hamkins, who wrote the puzzle, is the O’Hara Professor of Logic at the University of Notre Dame and was previously Professor of Logic at the University of Oxford. Infinitely More is an amazing resource for anyone who enjoys the intersection of maths and philosophy. It includes many essays, as well as serialisations of his books and puzzles.
Hamkins is very active online: he is in the top 0.01 per cent of contributors to MathOverflow.
I hope you enjoyed today’s puzzle. I’ll be back in two weeks.
I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
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