McEs, A Hacker Life: PUZZLE: Group Strategy

Monday, May 07, 2007

PUZZLE: Group Strategy

Coming back from Montreal yesterday, Peyman told me the following puzzle which is probably the coolest of its kind I've ever heard. Quite the like that you try hard proving to be wrong first. Anyway:

(I first tried to write it down as prisoners and all, but gave up. Here it goes bare:)
There are 2k men numbered 1 to 2k. And a room with 2k boxes inside (also numbered 1 to 2k), and each man's number is written on a paper and placed in one of the boxes, such that every box has a number in it and the distribution is uniform.

Now there is a game (trial, whatever): each man goes into the room, opens and sees the number inside k of the boxes of his choice, and comes out, altering nothing in the room, and not communicating with anyone during the experience. Before the game starts the men have a chance to develop a strategy, but other than that they can't communicate at all. Anyway, after everyone has visited the room, if all of the men can tell in which box their number is placed, they win, otherwise they lose.

Show that the men can develop a strategy with a expected winning probability that is not less than a positive constant regardless of k.

Labels: puzzle

¶ 7:46 PM

Comments:

Hi!
My solution would be the following.

The man are numbered 1 to 2k so they decided each looks at exactly one box, the one he is numbered.
Then after each one looked they line up from left to right standing in the position of the number they saw in the box. E.g. The one who saw the number 1 will be at the leftmost, than the one who saw number two lines up next etc.

The only flaw is that this means that they don't communicate verbal but through another channel.

Fabian

# posted by

Anonymous : May 08, 2007 2:09 AM

if all of the men can tell in which box their number is placed, they win, otherwise they lose.
How they tell that? Is it ordered and do other mens inform when man number x tells in which box his number is placed?

# posted by

Anonymous : May 08, 2007 5:00 AM

If they really cannot communicate in any way (nothing non-verbally, no change in the distribution of the pieces of paper allowed, no nothing) every person is presented with exactly the same situation. Hard to image that there can exist any kind of strategy in that case. The best solution would be to open half the boxes - with chances of winning course depending on k.

So maybe the are allowed to modify the papers, or their distribution?

Or maybe I just don't get it. :-)

# posted by

Anonymous : May 08, 2007 8:03 AM

Is 'k' a variable here or does it replace '1000' as I read this I thought you were saying 1 - 2000, not 1-2(variable k).

# posted by

skvidal : May 08, 2007 8:33 AM

k is a variable. At the end each reports to a judge or something. No communication, really. And there exists a solution. As I said, this is the kind of puzzle that you try hard to prove wrong, but it isn't.

# posted by

behdad : May 08, 2007 11:10 AM

If everyone looks in all boxes and thus finds the correct box for themselves then the probability of success is 1 - independent of k.

(However note that the number of trials is not independent of k)

ps. I've had a bit to drink so may be missing something obvious!

# posted by

Adam : May 09, 2007 9:00 PM

You missed the fact that there are 2k men/boxes, but each man can look into only k boxes, that is, only half of the boxes.

# posted by

behdad : May 09, 2007 11:57 PM

When reporting the number of the box containing their own number, can they choose the order in which to report to the judge (or whatever) or are they called in a fixed order one by one, and at the first fail the game is lost? Can other prisoners see prisoner number n saying something "I know mine, I'll go now"?
That's a _very_ important detail of the game rule set :)

# posted by

Anonymous : May 10, 2007 3:47 AM

kitty, "no communication" includes that too. Each one reports to the judge without others seeing/hearing/whatever.

# posted by

behdad : May 10, 2007 11:35 AM

I think I got it after reasoning in terms of lists... but it is horribly counterintuitive, especially because my approximate calculation gives a _huge_ minimal chance of success for k->oo (about 30%)
Nice one, anyway, I'll be talking about it to a few friends of mine :)

# posted by

Anonymous : May 11, 2007 12:23 PM

About 30% sounds right. The strategy itself is very elegant, but getting a good bound is a bit harder.

I didn't want to give the 30% away at first. It's very surprising though when you find it out.

# posted by

behdad : May 11, 2007 12:59 PM

SOLUTION

# posted by

behdad : June 02, 2007 7:01 PM

About Me

Twitter Updates