YAPP – Part II

In previous post “YAPP (Yet Another Prisoners Problem)” we considered the problem of 100 prisoners each opening 50 out of 100 boxes, surviving iff all 100 prisoners found their unique id in one of the 50 boxes.

Remarkably there was a strategy enabling the prisoners to survive with probability > 30%. Even more remarkably we established that the survival probability continues to stay over 30% as the number of prisoners and boxes tend to infinity.

In this post I wanted to consider a variation of the problem, similar spirit, yet seemingly more difficult than the previous.

Let’s state the problem I have in mind (in it’s full generality):

Problem

n condemned prisoners awaiting execution are offered a chance to live (for unknown reasons).
The prison administration has labelled each prisoner with a unique number id from 1 to n.

Each of the prisoners are led alone into a room with 2n closed boxes each containing a number from 1 to n in some random arrangement. The multiplicity of each id number is exactly 2 (that is, there are exactly two boxes containing the same id).

Each prisoner is allowed to open n of the 2n boxes.

The deal is: If all n prisoners manage to find their number id in some box after opening n boxes then they may all live. Otherwise they will all die.

The prisoners may communicate a strategy with eachother before they enter the room, but not after.
They are not allowed to manipulate the boxes nor leave messages to following prisoners.
All boxes are closed when a prisoner enters the room.

The question is: What strategy should the prisoners choose to maximize their chance of surviving?

 
Let’s first make a few easy remarks:

  • The independent probability of all prisoners surviving is now (\frac{3}{4})^n, as each prisoners individual chance of success is \frac{3}{4}. To see this, each person chooses n out of 2n boxes, so the probability that the first of the two instances of id k is not among the n boxes is \frac{1}{2}. The same goes for the second instance of id k. Therefore the probability that they are both not among the n chosen boxes is \frac{1}{4} i.e the probability that at least one is chosen is \frac{3}{4}. Selections are independent so total survival probability among n prisoners is hence (\frac{3}{4})^n.
  • When we compare the problems, we compare the settings where we have n prisoners over n boxes in the original problem vs n prisoners over 2n boxes in this problem.
  • The previous strategy no longer works as we are no longer dealing with classical permutations. Because of the double multiplicity of each number, paths may self-intersect prematurely (below picture illustrates this phenomenon).

new boxes

A few Questions:

  1. Is this problem more or less hard than the previous problem? After all, we have twice as many boxes, but on the other hand each prisoner has the opportunity to find their number in two boxes instead of one and is allowed to open twice as many boxes.
  2. Can we salvage anything from the previous strategy, i.e “can we fix the argument?”
  3. Is there a correspondence between the two problems? (can we translate a configuration (or a group of configurations) in one problem to a configuration in the second problem?)
  4. What is the total number of configurations (permutations) in this problem?
  5. If we can find an alternative argument, how would we go about proving optimality? (can we adapt/reuse the proof by Curtin and Warshauer(see previous post)?)

I should probably mention at this point that I have unfortunately not managed to find an argument that works as well as the argument for the previous problem (please feel free to comment if you have any nice ideas). Nevertheless I have a partial remediation of the previous strategy that is somewhat better than the default independent strategy. I will begin discussing this strategy in what remains of this post.

Let’s first look at the easiest question – Question 4.

The number of ways of permuting n distinct numbers is as familiar n!

In this problem we are dealing with a so called multiset of n distinct numbers, each of multiplicity 2.

In general a multiset {a_1,...,a_n} with multiplicities m_1,...,m_k can be permuted in \left( \begin{array}{c} n \\ m_1,...,m_k \end{array} \right) = \frac{n!}{m_1!....m_k!} ways. To see this just note that there are n! ways of permuting {a_1,...,a_n} as distinct objects. In each rearrangement we can permute the objects with multiplicity m_i in m_i! ways among eachother, each leading to the same rearrangement. Therefore there are only \frac{1}{m_i!} as many true rearrangements. The same argument inductively for each i=1,...,k establishes the formula.

In this problem we therefore have  \left( \begin{array}{c} 2n \\ 2,...,2 \end{array} \right) = \frac{(2n)!}{2^n} different permutations.

As we might have expected this space is vastly larger than the space of n! permutations in the previous problem.

Let’s get to the proposed strategy.

As we saw, the obvious issue with pointer following in the new problem setting is that paths no longer partition into disjoint cycles. The paths have a tendency to curl back into themselves leading to degenerate cycles.

To straighten out the curls we can impose a new rule:
If we encounter number k for the second time, then visit box number n+k.

NewStrategy

What are the implications of this rule?

We certainly won’t visit each box more than once, since each number has multiplicity 2.
As the rule implies, box k is visited the first time number k is encountered, then box n+k the second time and thereafter we will never see number k again, so both boxes are henceforth off limits.
Because of this fact the path will eventually lead to the discovery of every number id (of course we will stop once we have discovered the number id we are looking for).

This is somewhat good news, we have a procedure for exploiting structure in the permutation.
We are interested in the proportion of permutations s.t every path taken by the prisoners have length at most n under the rules of the strategy.
The next natural question is if we can quantify how well this strategy ought to work?

In the previous problem we counted the number of permutations N_{k,n} containing a k-cycle (k > \frac{n}{2}) by indirectly expanding the following recursion:

N_{i,j} = \begin{cases} jN_{i-1,j-1}, & if\;\; i > 1 \\ (j-1)! & if \;\;i=1 \end{cases}

In this problem we have more choice.
Let M_{i,j,k} denote the number of permutations including a cycle of length i where j denotes #(remaining numbers with multiplicity 1) and k denotes #(remaining numbers with multiplicity 2). At each stage, we can either choose a number with multiplicity 2 (i.e a new number) or we choose a number of multiplicity 1 (i.e a number already discovered). Finally once we have our cycle of length i we can permute the remaining multiset at will.

This leads to the following recursion:

M_{i,j,k} = \begin{cases} jM_{i-1,j-1,k} + kM_{i-1,j+1,k-1}, & if\;\; i > 1 \\ \left( \begin{array}{c} j+(k-1) \\ 1,...,1,2,...,2 \end{array} \right) & if \;\;i=1 \end{cases}

We are interested in M_{r,0,n} \;\; (r > n).
Unfortunately unwinding this recursion into a closed form expression looks too difficult (don’t hesitate to let me know if you can think of a nice combinatorial argument for counting k-cycles).
Things are also made further complicated by the fact that cycles of length greater than n need not be unique occurrences in any given permutation, making the probability of failure difficult to evaluate.

Resorting to Monte-Carlo methods we obtain the following simulated survival curve.

SurvivalCurve

Unfortunately it seems the combinatorial explosion of long cycles take over fairly quickly, rendering the strategy far from ideal.

Below you can try running a few simulations yourself, using the Flash implementation hacked together for this purpose.
(Click here if you see a blank space area below)

 

I’ll admit defeat for now and come back to this problem again some time in the future.

 

Exercise 1 (How many words? – Easy)
 
How many words (not necessarily in any language dictionary) can you form of the string “PRISONERSPROBLEM”?
 
(Hint: Think of multisets)

 

Exercise 2 (An Optimal Strategy – (Possibly) Hard)
 
Find an optimal solution to the second Prisoners Problem (or even a good quantified strategy for that matter).
 
Advertisements
This entry was posted in Probability, Uncategorized and tagged , , , , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s