Difference between revisions of "Alice"

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m (Reasoning (Part 3))
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==Table==
 
==Table==
# D = 1,2
 
# A = 3,4
 
# N = 4
 
# I = 1
 
# E = 1, (9 or 10)
 
# S = 1
 
# P = 2
 
# R = 3
 
# Y = 1
 
 
==Reasoning (NEW)==
 
# 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. 
 
# 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.
 
# Thus E=1.  And S,I = 1 as well.  D+A+N+D=10. 
 
# 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!
 
 
# 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!
 
# 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.
 
# 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. 
 
# 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. 
 
# 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.
 
# 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.
 
# 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.
 
# If E=3... Since I!=2 we must have I=3 and S=3.  But that is a contradiction.
 
# 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. 
 
# 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.
 
# 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.
 
# 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.
 
# 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.
 
# 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)==
 
# 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.
 
#S=6 => GAMI=13 => I goes with right => I=6 or 4.  I!=6 => I=4 => GAM = 9 => G=2 => impossible. 
 
# 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.
 
# S = 5 => GAMI=16.  Nonsequence.  I=5 or 3.  => GAM=11 or 13, nonsequences.  Impossible.
 
# 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. 
 
# SEARING=30.  N!=1,2,3.  N=4,5,6,7,8.  N=8=>SEAR=6 nonsequence.  Thus R=6 or 8, both impossible. N=7 => SEAR=9 nonsequence.  Thus R=5 or 7.  7 impossible and 5 => SEA=4 impossible. 
 
# N=4 => I=4 or 3.  Both impossible.
 
# N=5 => I=5 or 4.  I!=4 so I=5 and G=5.  SEAR=15, NS.  so R=5 => too many 5s. 
 
# N=6 => SEAR=12.  Nonsequence so R=6 or 4.  4 impossible so R=6 => SEA=6.
 
# REVISED=34.  E=1,2,3,5,7,8,9.  E=9=>REVI=7 NS => I=9,7 impossible.  E=8=>REVI=10.  If NS then I=6,8 both impossible.  So REVI=10 S => R=1 => too many 1s.  E=7 => REVI=13 NS.  I=7 or 5 => REV=6 S or 8 NS => REV=6 => R=1 =>too many 1s. 
 
# E=1,2,3,5.  E=5 => REVI=19 NS => I=5,3.  I!=3 => I=5 => REV=14 NS.  Impossible. 
 
 
 
{| border=1
 
{| border=1
 
| 1 || D || S || I || E
 
| 1 || D || S || I || E
Line 59: Line 13:
 
| 4 || A || N || B || T
 
| 4 || A || N || B || T
 
|-
 
|-
| 5 || D ||  
+
| 5 || D || I || R ||
 +
|-
 +
| 6 || R || O || W || N
 +
|-
 +
| 7 || A || M || E || G
 +
|-
 +
| 8 || A
 +
|-
 +
| 9 || S
 +
|-
 +
| 10 || P || R || T || E
 
|-
 
|-
| 6 || R || O || G || N
+
| J || I || R || L || E
 
|-
 
|-
| 7 || A
+
| Q || N || O || T || V
 
|-
 
|-
| 8 || M
+
| K || C || U || R || Y
 
|}
 
|}
  
==Reasoning==
 
  
# 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.
+
Another 70 pointer is 10JQK101010 = '''trouper'''
# 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. 
 
# DANDIES=13.  Speculate:
 
## Assume that we have a 4-sequence.
 
### Assume that the sequence is first.  Then D1=1,A1=2,N1=3,D2=4,I1=1,E2=2,S1=1.
 
### Apply this information to PRAISED=13. 
 
#### 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).
 
#### Then D=D1.  Then the last three must be 1 => E2=1.  A contradiction.
 
### 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. 
 
## Pop.  We must have a 3 sequence.
 
### Assume that b=1.  Then 4*x+3*b+3 = 13.  4*x=7.  Impossible.
 
### Assume that b=2.
 
#### 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. 
 
#### Assume that the sequence is first.  Then D1=2,A1=3,N1=4, D2=1, I1=1, E2=1,S1=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.
 
# 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.
 

Latest revision as of 08:01, 23 August 2006

Puzzle_Boat

[1]

Table

1 D S I E
2 P Y E D
3 R A E S
4 A N B T
5 D I R
6 R O W N
7 A M E G
8 A
9 S
10 P R T E
J I R L E
Q N O T V
K C U R Y


Another 70 pointer is 10JQK101010 = trouper