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Journal Chacham's Journal: Puzzle: 88 Hats 10

While looking through some Oracle Blog i hit a link to one of "essays" at JSoftware. The one i have been looking at lately is 88 Hats:

88 people stand in a circle, each having a hat with a number from 0 to 87 written on it. Everyone can see the numbers on other people's hats but can not see his own number. They simultaneously write a number on a piece of paper and give it to the judge. If at least one of them wrote a number that is on his own hat then everyone wins, otherwise everyone loses. What strategy should they use to guarantee victory?

        (Numbers on the hats do not have to be all different. People can not exchange any information during the procedure but can agree on some strategy beforehand.)

88 was a bit too large for me, so i decided to start easy, and see if a simple logical plan could be deduced (without reviewing the solution provided).

One hat is very easy. Just pick the number.

Two hats is slightly more difficult, but not much. Gurer ner gjb tebhcf bs ahzoref, bar jurer gurl ner gur fnzr, bar jrer gurl ner qvssrerag (rnpu tebhc unf gjb zrzoref). Gurersber, gurl fubhyq ynory gurzfryirf N naq O. N fubhyq pubbfr gur fnzr nf O, naq O fubhyq pubbfr gur bccbfvgr bs N. Gung jvyy pbire obgu pnfrf.

I'm working on three, on and off. I'm guessing the strategy should be somewhat similar to two, but it is obviously different being each person has to deal with multiple others. Four and on, i would think, should then be similar.

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Puzzle: 88 Hats

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  • Before hand everyone picks a number, say 20. Then, they look around the room. If they see someone with a 20, they go and stand behind them. Within a short amount of time, someone will notice that just about everyone is standing behind them. So they write the agreed upon number. If you want to increase the odds that the number is not one that's skipped, then pick 4. For number a everyone stand to the North of the person. Number B to the West. Number C to the East, and Number D to the South. Odds are
    • by Chacham ( 981 )
      What if everyone has the same number? Or only two numbers are distributed? There will be noone to stand behind.
      • In that case, then once the people notice that there's no one to stand behind, ie, that there's only a handful of numbers showing, then the default plan is for everyone to pick one of the numbers they see and write it down. Odds are that one will hit it right.

        That or video tape everyone trying to stand behind someone who isn't there and sell at a performance art piece entitled, "Brownian Motion." ;-)

  • Is each number only used once? Does each hat has a different number? If so it's easy, find hat 0, hat 1, hat 2 .... iteratively until you find the missing number, all 88 players should be able to know there own number by removing all the others.
  • What strategy should they use to guarantee victory?
    tell the guy next to them what the number on his hat is ;)
    • by huckda ( 398277 )
      lemme try that again as I read the last line of the puzzle ;) (damn the no information exchange!)

      agree to the strategy that if two people have the same number on their hats that they will be given a greater distance between them and their neighbors

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