Difference between revisions of "Alice"

Revision as of 17:06, 22 August 2006

Puzzle_Boat

[1]

Table

1. D = 1,2
2. A = 3,4
3. N = 4
4. I = 1
5. E = 1, (9 or 10)
6. S = 1
7. P = 2
8. R = 3
9. Y = 1

Reasoning (NEW)

1. TROUBLE=67. Clearly, E>=7. Thus E in DANDIES is different from the E in TROUBLE. In fact, E=1,2,3. E cannot be 4 since then I+E+S = 12. Assume E=2. Then I+E+S=6=>D+A+N+D=7. If the second D=2, then D+A+N=5, which is impossible (since its a sequence). Likewise, no sequence makes D+A+N+D=7.
2. Assume E=3. Then I+E+S=9 => D+A+N+D= 4. Then D,A,N,D=1. Applied to PRAISED, S=E=D=1, contradicting E=3.
3. Thus E=1. And S,I = 1 as well. D+A+N+D=10.
   1. If D=1 (going with the triplet), then D+A+N=9.

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.
      1. Assume that b=1. Then 4*x+3*b+3 = 13. 4*x=7. Impossible.
      2. Assume that b=2.
         1. Assume that the sequence is second. Then I1=2,E2=3,S1=4,D1=1,A1=1,N1=1,D2=1. But when applied to PRAISED, E2=1 a contradiction.
         2. Assume that the sequence is first. Then D1=2,A1=3,N1=4, D2=1, I1=1, E2=1,S1=1.
            1. Apply to PRAISED. No matter what, the triplet must be last then. If D=D1, then E2=2 and we have a contradiction. Thus, D=D2, E2=1.
4. Popping since the above is correct. PRAI = 10. If this is a sequence, then P=1, R=2,A=3, I=4. But we already have 4 1's, so this is not the case. Therefore, the sequence is PRA and I is part of a quartet. P+P+1+P+2 = 9 => 3*P = 6 => P=2, R=3,A=4.