Alice

From CSWiki
Revision as of 18:34, 22 August 2006 by Jcone (talk | contribs) (Reasoning (Part 3))

Jump to: navigation, search

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.
  4. Apply to SPRAYED = 16. Again, the E cannot be TROUBLE's E. So here, E=1 => Y=1, D=1. Contradiction!. Clearly each letter can appear more than twice. Doh!
  1. First point still holds. E=1 or 3. If E=3 then I+S+E=9, and D+A+N+D=4. => D,A,N,D=1. There should be NO MORE 1s. Now to PRAISED. Again, by the same reasoning, the E in PRAISED must be 1 or 3. Since we've run out of 1s, E there must be 3, implying that P+R+A+I=4. But this means that P,R,A=1, a contradiction!
  2. E=1 and S,I=1. D+A+N+D=10. One of the Ds must be equal to 1. D,S,I,E =1. There should be NO MORE 1s on the board. Move to PRAISED. By the same logic, E=1 or 3. Since there can be no more 1s, we must have that S=E=D=1. P+R+A+I=10 which is fine since I=1 and P,R,A=2,3,4. By before, D,A,N=2,3,4 or A,N,D=2,3,4. Either way, we now have all 1s accounted for and 2 of 2,3,4 each accounted for.
  3. For SPRAYED=16, E<=4. E cannot be 5 since Y+E+D=15. If E=4 then Y+E+D=12 and S+P+R+A = 4. Since they would all have to be 1s, we would have a contradiction. If E=1 then Y=1, a contradiction.
  4. If E=3 then Y+E+D=9. Then S+P+R+A=7. They cannot form a sequence, thus A must go with Y,E,D. If Y,E,D is a triplet, then S+P+R=4, which is impossible. If Y,E,D is a sequence, then A=1 which is also impoosible.
  5. So E=2. Then Y+E+D=6 and S+P+R+A = 10. Y,E,D must form a triplet lest Y=1. If we let A=2, then S+P+R=8. But 1+2+3 = 6 and 2+3+4=9. So we must have a 4 sequence. S,P,R,A=1,2,3,4. So we'd have Y,E,D=2. From before, P=2. All 2s are accounted for.
  6. Back to DANDIES. D,A,N=2,3,4 or A,N,D=2,3,4. But A cannot be 2, so D=2, A=3, N=4.
  7. ANDRIES. E=1,2,3,4,5. E=5 => A+N+D+R=6 => A+N+R >= 5. But since A,N,R>=3, A+N+R>=9. If E=4, then A+N+D+R=9. A+N+D+R cannot be a sequence since it's >= 10. A+N+D+R cannot be 4 of a kind. Thus, R must go with the second half. R!=2. So then R=4. Then A+N+D=5 which is impossible.
  8. If E=3... Since I!=2 we must have I=3 and S=3. But that is a contradiction.
  9. If E=2 and since I!=2 we must have I=1,E=2,S=3. A+N+D+R=15. 4*b+6=15 => 4*b=9. Impossible.
  10. E=1. A+N+D+R=18. Since R!=1, this must be a sequence. Only option is A=3,N=4,D=5,R=6.
  11. EYEBROW=28. O=1,2,3,4,5,6,7,8. Cannot be 8 lest B=1. Cannot be 1 or 2. So O=3,4,5,6,7. If O=7 then E+Y+E+B=7. Impossible. If O=3 then since R!=2, R=3 and W=3. E+Y+E+B=19. Since we've run out of 3s B!=3. So it is a sequence which is impossible.
  12. If O=4 then R and W cannot both be 4. If R=3 and W=5 then B != 2. Then the sequence is E+Y+E+B=16. Impossible.
  13. If O=5 then E+Y+E+B=13. If R=W=5 then B!=5, making an impossible sequence. If R=4, W=6 then B=3. This gives E+Y+E=10. Impossible.
  14. O=6 then E+Y+E+B=10. if B is 6 then E+Y+E=4, impossible. Thus B is 4. E,Y,E=1,2,3. Either R,W=5,7 or R,W=6,6.


Reasoning (Part 3)

  1. GAMIEST=31. S!=2. S!=9. S=1,3,4,5,6,7,8. S=8 => G+A+M+I=7. Impossible (< 10). S=7 => GAMI=10 => G=1. Impossible.
  2. S=6 => GAMI=13 => I goes with right => I=6 or 4. I!=6 => I=4 => GAM = 9 => G=2 => impossible.
  3. S=1,3,4,5. S=1 => E,T=1 => impossible. S=4 => GAMI=19.

E,T != 4,4 so E,T=3,5. I!=2 so GAMI sequence. Impossible.

  1. S = 5 => GAMI=16. Nonsequence. I=5 or 3. => GAM=11 or 13, nonsequences. Impossible.
  2. S=3 => E,T=3,3 or E,T=2,4. 3,3 is impossible. So E,T=2,4. GAMI=22. G,A,M,I=4,5,6,7 impossible because too many 4s. Thus, non-sequence so I=1, and GAM=21. G,A,M=6,7,8.
1 D S I E
2 P Y E D
3 R A E S
4 A N B T
5 D
6 R O G
7 A
8 M

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.