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I found a puzzle like this online, so I thought I'd try making my own.

There are 4 dice, each with 6 different letters on them, such that 24 of the 26 letters of the English alphabet are represented (Q and Z are missing). Lets say I rolled these dice a number of times and was able to make the following words with the letters by re-arranging the dice in any order:

  1. CLAP
  2. WOKE
  3. CASH
  4. JUNK
  5. BITE
  6. GIFT
  7. LEWD
  8. GERM
  9. FOXY
  10. RUST
  11. LIMP
  12. BUOY
  13. JAKE

What letters are on the faces of each die?

Some notes: I know for a fact that this is solvable using logic, and I also know for a fact that there are multiple solutions. I'll accept any solution that is provably correct and accompanied by some explanation of your process.

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5,6 mean BE are found on the same pair of dice as GF. 8 means E is not on the same die as G. EF and BG go together.

FX=BU (9,12) But we know F is not with B, so F is with U, B with X. EFU, BGX so far

AE=UN (4,13) E with U means A with N EFU, BGX, AN

Looking at those letters which cannot be on the same die as E, we now have ABDGIJKLMNORSTWXY which leaves CEFHPUV. P is the only one of these present in (11), so P is correct. (1) then rules out C, so we have EFUHPV, BGX, AN

Removing the fully known die from the words:

CLA
WOK
CAS
JNK
BIT
GIT
LWD
GRM
OXY
RST
LIM
BOY
JAK

It is clear that L=S. G is not on the same die as I or M (GIT, GRM), so it must be with L (LIM).

EFUHPV, BGXLS, AN.

K is the only letter which appears everywhere none of the other 5 (BGXLS) are present, therefore it must be the final letter from that die. EFUHPV, BGXLSK, AN.

Removing the newly completed die and eliminating duplicates:

CA
WO
JN
IT
WD
RM
OY
RT
IM
JA

We have C=J, A=N, W=Y, O=D, T=M, R=I.

The third and fourth dies may each have either CJ, AN; either WY, OD; either TM, RI.

So one possible assignment is EFUHPV, BGXLSK, CJWYTM, ANODRI

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  • $\begingroup$ Nice! I like your method too, I'll have to see if I can use it to make a better puzzle. $\endgroup$ – VictorHenry Jun 9 '15 at 15:57
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So a possible set of answers is: 1st die: B,G,L,S,X,K. 2nd die: E,U,F,P,H,V. These two (1st and 2nd dice) should be a unique solution. In the other two dice, just the first two letters are unique. And the last for can be exchanged between the last two dice. 3rd dice: R,I,J,C,O,D. 4th dice: M,T,A,W,Y,N.

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I find, there are 4 possibilities......

! EFHPUV and BGKLSX are unique. Remained 2 dice can be 1. ANMTWY & CJIRDO 2. ANMTDO & CJIRWY 3. ANIRWY & CJMTDO 4. ANIRDO & CJMTWY.

Explanation:-

! I see the frequency of E is 5, so I start with E.. After analysing I get EFHPU might be in one dice(as there is 0 frequency for V and all remained alphabets was likeable to be in any other dice, Do Its clear that V is with 'EFHPU')

! Now we see (1) and (3) roll, LS and C is in different dice... Then see (5)&(6) BG is in same dice.. Following same logic, see there are 3 groups BGX, MT and IR must be in different dice. But MT and IR can't be in 'LS' dice... So BGX is in. We see WY and DO must be in different group... So from (2) roll K is in dice 'LSBGX'......

! finally we have 2 unique dice and 3 pairs 1. AN & CJ 2. MT & IR 3. WY & DO must be in different group, So gives 4 possibilities. !

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  • $\begingroup$ I don't see any explanation for EFHPU other than 'analysing'. Could you elaborate? $\endgroup$ – frodoskywalker Jun 9 '15 at 16:05

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