» MAKE: NOISE — Discuss this article

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• My solution to the first one
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  My solution is this, it may be a little out of bounds though. The guy who can move at 1 min per bridge has the 10 min guy jump on his back. He holds the flash light and they cross. 1 min gone, then the 1 min guy goes back across, 2 min gone, repeat with the 5 min guy on the back, 4 min gone, repeat with the 2 min guy walking, 6 min gone.

  Posted by Athrawn17 on February 08, 2006 at 19:08:44 Pacific Time

• My solution to the first one
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  Yeah, that solution is totally out of bounds. One solution is

  (1,2)->
  <-(2)<br />(5,10)->
  <-(1)<br />(1,2)->

  Posted by mike_m on February 08, 2006 at 21:17:10 Pacific Time

• My solution to the first one
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  yup.
  
  me, too.
  
  That didn't take long once you realized its optimal to send the slow pokes at the same time.

  Posted by DaveVonNatick on February 10, 2006 at 05:38:26 Pacific Time

• Hint to the second puzzle
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  This might be a spoiler for some, so if you want to solve the problem completely on your own, don't read this. If you don't know where to start, this might help you a bit.
  
  The solution becomes much easier to see if you simplify the problem to one pirate, then try it with two pirates, then three, and so on. Keep in mind that only 50% of the pirates need to agree to the plan.
  
  You should end up with a horribly skewed distribution of the gold (one pirate gets a LOT more than the others). My only concern is that if these pirates are all so logical, why would they agree to such a plan in the first place?

  Posted by sschultz on February 10, 2006 at 07:29:22 Pacific Time

• Hint to the second puzzle
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  I was thinking along those lines, too.
  
  If there were two pirates left, the senior would decide to take all and with 50% of the vote he could do this without being beheaded.
  
  With 3 pirates... The most senior could offer the most junior just one coin for his vote... since that's better than the alternative.
  
  But it gets a bit fuzzy for me at 4 pirates. Suggestions?

  Posted by DaveVonNatick on February 21, 2006 at 15:15:34 Pacific Time

• Hint to the second puzzle
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  I am having trouble closing a recursive solution because I am unable to close on what 'rational' behavior entails. In the 3 pirate case, if #1 offered #2 one coin and #3 none, would #2 vote for or against? This is better than (99,0,1), which would otherwise pass, but worse than (X,100,0) if he betrays #1. Is rationality decided on by more than the mere disparity of results from a personal Y/N vote?

  Posted by Jeff-S on March 15, 2006 at 08:13:24 Pacific Time

• Hint to the second puzzle
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  For sake of argument, lets call the pirates P5, P4, P3, P2, and P1, to reflect their years of seniority. Lets also name the round in which a pirate proposes a distribution after the pirate; therefore round P5 is the round in which pirate P5 proposes a distribution. Finally, lets describe the distribution as a set, starting with the proposing pirate and working downwards, for example (P5,P4,P3,P2,P1). Working backwards:
  
  Round P1: this is obvious—P1 only needs 1 vote to win, therefore he proposes (100). The vote is 1 for (P1) and 0 against; P1 wins.
  
  Round P2: P2 only needs 1 vote to win (50%), therefore he proposes (100,0). The vote is 1 for (P2) and 1 against (P1); P2 wins.
  
  Round P3: P3 needs 2 votes to win (67%). P3 knows that if he loses, P2 will get it all (see Round P2); therefore, there is nothing that P3 can propose that P2 would vote for. However, P3 also knows that if he loses P1 will get nothing in Round P2. Therefore, P1 will be an ally with as little as 1 piece. P3 proposes (99,0,1). The vote is 2 for (P3, P1) and 1 against (P2); P3 wins.
  
  Round P4: P4 needs 2 votes to win (50%). P4 knows that if he loses, P3 will get 99; therefore, P3 will not be an ally for anything less than 100. However, P4 also knows that if he loses, P2 will get nothing (see Round P3). Therefore, P2 will be an ally with as little as 1 piece. P4 proposes (99,0,1,0). The vote is 2 for (P4, P2) and 2 against (P3, P1); P4 wins.
  
  Round P5: P5 needs 3 votes to win (60%). P5 knows that if he loses, P4 will get 99; therefore, P4 will not be an ally for anything less than 100. However, P5 also knows that if he loses, P3 and P1 will get nothing (see Round P4). Therefore, P3 and P1 will be allies with as little as 1 piece. P5 proposes (98,0,1,0,1). The vote is 3 for (P5, P3, P1) and 2 against (P4, P2); P5 wins.
  
  Without actually working it, my guess is round P15 will look like (93,0,1,0,1,0,1,0,1,0,1,0,1,0,1).
  

  Posted by AndrewBrooman on March 31, 2006 at 12:51:05 Pacific Time

• Hint to the second puzzle
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  But at round 15, P14 would get nothing under the proposed P15 plan. So the rational thing for P14 to do is to make a deal with P13 to scuttle the deal P15 is proposing by offering P13 2 gold coins from the stash that P14 will win if P13 helps P14 vote down P15.
  
  Then you get into prisoners' dilemmas all the way until P1 gets it all. So P15, if he wants to keep his life, better just give it all to P1 from the start.

  Posted by NeilHunt on May 19, 2006 at 07:46:08 Pacific Time