» MAKE: NOISE — Discuss this article

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• Poison Pills
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  All the solutions look so complicated too me, but not necessary. You don't need to indentify the poison pill, just isolate it into a pile. (and eat from another )
  
  Divide the 12 into 3 groups of 4. Put 2 groups on the scale.
  
  If they balance, the poison pill is in the unused pile, eat any on the scale.
  
  If they don't, remove the heavier pile, place the new one on the scale. If the new pile is also the heaviest, the lighter pile contains a lighter poison pill.
  
  If they balance, the heavier poison pill is in the first pile removed.
  
  If the new pile weighed is lighter then someone lied to you and more than one pill different.

  Posted by idearat on May 19, 2008 at 17:21:25 Pacific Time

• crystal orbs
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  Dont get it -what if the first floor it breaks on is the 3rd. How would you determine that starting on 14?

  Posted by ubay on April 09, 2008 at 15:06:38 Pacific Time

• Poison Pills
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  Number the pills 1 through 12.
  
  First, weigh 5,6,7 and 8 on the left against 9,10,11 and 12 on the right.
  
  If the left side is heavier, record the number "+9"; if the right side is heavier record the number "-9". If they balance, record "0"
  
  Second, weigh 2,3,4 and 11 on the left against 5,6,7 and 12 on the right.
  
  If the left side is heavier, record the number "+3"; if the right side is heavier, record the number "-3". If they balance, record "0".
  
  Third, weigh 1,4,7 and 11 on the left side against 2,5,8 and 10 on the right side.
  
  If the left side is heavier, record the number ""+1"; if the right side is heavier, record the number "-1". If they balance, record "0".
  
  That completes the three allowed weighings.
  
  Now, add the three numbers you have recorded. The sum will represent the number of the odd pill, with the sign indicating whether it is heavy ("+") or light ("-"). However, if the final sum is 9, 10, 11 or 12, reverse the sign for the final result.
  
  
  
  

  Posted by davidhy on March 15, 2008 at 17:08:27 Pacific Time

• Crystal Orbs -Correction
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  Sorry, messy markup. Let me try again with the table and formula.
  
  

  
  
    o 0   1   2   3   4   5   6   7 
  
  d_|______________________________
  
  0 | 1   1   1   1   1   1   1   1
  
  1 | 1   2   3   4   5   6   7   8 
  
  2 | 1   2   4   7   11  16  22  29
  
  3 | 1   2   4   8   15  26  42  64
  
  4 | 1   2   4   8   16  31  57  99
  
  5 | 1   2   4   8   16  32  63  120
  
  
  
  

  
  (etc). Note that with unlimited orbs you
  can distinguish 2^d floors.
  
  
  
  To get a formula for this you can use generating functions (or other combinatorial methods) and conclude that in the n^th row, the entry is the sum of the first n terms in row d of Pascal's triangle (with the left-hand "1" indexed by 0 in the traditional fashion.) Thus in particular the 2-orb row has entries
  
  
  1 + n + n(n-1)/2 = (n² + n +2)/2
  
  
  and the three-orb row has entries
  
  1 + n + n(n-1)/2 + n(n-1)(n-2)/6
  
  
  
  
  

  Posted by Robert Dawson on March 05, 2008 at 05:56:24 Pacific Time

• Crystal Orbs
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  To prove that 14 is the minimum, work by recursion. For o orbs, and d drops, let F(o,d) be the maximum number of floors that can be distinguished. If you drop an orb at a certain floor and it breaks, you know the correct floor (first one at which the orb breaks) is there or lower, and you have one fewer orb and one fewer drop to continue with. If it doesn't, you know the correct floor is higher and you have the original number of orbs but one fewer drop. If you plan correctly the two new searches (whose lengths we assume we already know - that's the recursion) do not overlap or have a gap between them. Thus:
  
  
  F(o,d) = F(o-1,d-1) + F(o,d-1)
  
  
  To start the recursion, note that with no drops we can distinguish between one state.
  
  
  We can now work out the terms of this recursion in a table:
  
  
  o d0 1 2 3 4 5 6 7<br/>
  
  
  0 1 1 1 1 1 1 1 1
  
  1 1 2 3 4 5 6 7 8
  
  2 1 2 4 7 11 16 22 29
  
  3 1 2 4 8 15 26 42 64
  
  4 1 2 4 8 16 31 57 99
  
  5 1 2 4 8 16 32 63 120
  
  
  (etc). Note that with unlimited orbs you
  can distinguish 2^d floors.
  
  
  
  To get a formula for this you can use generating functions (or other combinatorial methods) and conclude that in the n^th row, the entry is the sum of the first n terms in row d of Pascal's triangle (with the left-hand "1" indexed by 0 in the traditional fashion.) Thus in particular the 2-orb row has entries
  
  
  1 + n + n(n-1)/2 = (n<sup>2 + n +2)/2
  
  
  and the three-orb row has entries
  
  1 + n + n(n-1)/2 + n(n-1)(n-2)/6</sup>
  
  
  
  
  

  Posted by Robert Dawson on March 05, 2008 at 05:49:11 Pacific Time

• Radar Date - wrong revised
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  OK, then::
  
  12/9/1921
  
  Not only does it treat zero's consistently, it's later than my last attempt.

  Posted by jaa_baba on February 26, 2008 at 08:08:51 Pacific Time

• pills
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  you can also split the pills into two piles of 6. Take the lighter pile after the first weighing, and split into two piles of 3. Weigh them, and keep the remaining pile of 3. Pick any two pills from the pile of 3. If they weigh the same, the remaining pill is the safe one. If they don't weigh the same, eat the lighter one. This is the same binary search that you're using, but is a little easier to follow than the initial split into three piles.

  Posted by thowland on February 25, 2008 at 20:23:44 Pacific Time

• Radar Date - wrong wrong wrong
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  9/09/1909 not a palindrome?

  Posted by jaa_baba on February 25, 2008 at 20:05:49 Pacific Time

• 14?!?
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  My officemate and I spent about 20 or 30 minutes on this and confidently declared the answer to be 19. (We also noted that it could be 18 if it could be guaranteed that SOME floor would break the orb.)
  
  Then I saw the 14, gasped and closed the window.
  
  We reworked our solution and got the same thing you did. However our previous confidence made us question how it could be *proven* that this is a minimum. You've got a demonstration, but that's hardly a proof.
  
  (We did at least establish, with a lot of handwaving, that 14 was the minimum for this scheme type.)

  Posted by drysdam on February 25, 2008 at 09:58:38 Pacific Time