Alice

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Revision as of 16:29, 22 August 2006 by Jcone (talk | contribs) (Reasoning)

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Puzzle_Boat

[1]

Table

  1. D1 = 1
  2. A1 = 2
  3. N1 = 3
  4. D2 = 4
  5. I1 = 1
  6. E1 = 1
  7. S1 = 1

Reasoning

  1. TROUBLE=67. For a 4 sequence, b,b+1,b+2,b+3,x,x,x; b=10 => x = 9; b=9 impossible; b=8 => x = 10. For a 3 sequence, b,b+1,b+2,x,x,x,x; b=10 impossible; b = 9 impossible; b = 8 => x=10. E1 = 9 or 10.
  2. If a sequence of 3, then the smallest the base of the sequence, b, must be b+(b+1)+(b+2) +1+1+1+1 <= 13. Thus, 3*b+7 <= 13. b <= 2. If it's a 4-sequence, then we have b+(b+1)+(b+2)+(b+3)+1+1+1 <= 13. 4*b+9 <= 13. b <= 1.
  3. DANDIES=13. Speculate:
    1. Assume that we have a 4-sequence.
      1. Assume that the sequence is first. Then D1=1,A1=2,N1=3,D2=4,I1=1,E2=2,S1=1.
      2. Apply this information to PRAISED=13.
        1. If D=D2, then the sequence must end there, and we must have 1+1+1+1+2+3+4. Giving us P1=1,R1=1,A2=1,S2=2,E2=3. A contradiction (because of E).
        2. Then D=D1. Then the last three must be 1 => E2=1. A contradiction.
      3. Pop. Assume that the sequence is second. Then D1=1,A1=1,N1=1,D2=1,I1=2,E2=3,S1=4. Again, applying to praised yields the same contradiction that E2=1.
    2. Pop. We must have a 3 sequence.