Question

I have three special four-sided dice. They have one letter on each side. When I roll them together I get three random letters which I try to rearrange into a word. In my eight goes so far I have made the words: CAT, SON, POD, RIG, PEG, TAP, DIN, APE What are the letters on each dice?

Answer 1

Explanation:

4 letters per die.

the hidden meaning is that each die has a different set of 4 letters, right ?

CAT and TAP tell us that one die has C and P.

TAP and APE tell us that one die has E and T.

the third one has A.

APE tells us that the vowels are on at least 2 different dice.

PEG tells us that the die with A also has G.

so, CAT, PEG, TAP, APE are fully used.

we need to focus on the remaining words SON, POD, RIG, DIN. we start with POD and RIG, as we have at least 1 letter already fixed to a die.

die 1 : C P R N

die 2 : E T O I

die 3 : A G D S

POD means O on die 2 and D on die 3 or vice versa.

RIG means R in die 1 and I on die 2 or vice versa.

that means die 2 is "full". no other letter can be in that die.

let's assume D is on 3 (so, O on die 2) and I on die 2 (so, R on die 1), then DIN means N is on die 1 or die 2.

as die 2 is already full, N must be on die 1.

that makes die 1 "full".

and we are left with SON, which fits with the other assumptions for O and N, leaving S as last letter for die 3.

see list above. this is definitely a valid solution.

let's see if there are others.

we flip the first assumption and put D on die 2 and O on die 3.

die 1 : C P I S

die 2 : E T D R

die 3 : A G O N

but now, as we cannot put R or I of RIG on die 3 (because G is already there), we have to put R on die 2 (otherwise D and I of DIN would be on the same die), and we are left with S or N of SON to put on die 3. but O is in this scenario already on die 3. so, this assumption leads us to an impossible scenario.

that means O has to be on die 2 and D on die 3.

so, let's flip the second assumption and put I of RIG on die 1 and R on die 2.

die 1 : C P I S

die 2 : E T O R

die 3 : A G D N

as the O is on die 2, die 2 is "full" abs we would have to put either I or N of DIN on die 3. which is in conflict with the D that's already there.

again, an impossible scenario.

that means R has to be in die 1 and I on die 2.

again, die 2 is "full".

and for SON we can now only put N on die 1, because on die 3 it would be in conflict with the D.

so, indeed, our first scenario is the only possible solution.

Answer 2

1. ADGS
2. CNPR
3. EIOT

Explanation:

You want to know the letters on each of three 4-sided dice, given that on different rolls they can spell the words CAT, SON, POD, RIG, PEG, TAP, DIN, APE.

Initial Assignment

We can arbitrarily assign the letters CAT to dice 2, 1, 3. This forces assignment of other letters so that we have on the three dice ...

1. AG
2. CP
3. TE

Compatibility

Looking at the remaining letters, we can draw a compatibility table as at the bottom of the attachment. Letters are not compatible if they show up in the same word. They are not compatible with a die if letters from that die are in the same word.

This shows that letters DNOS are not excluded from being assigned to die 1. However, the only compatible pair of letters in that group is DS, making ...

die 1 = ADGS

This leaves letters INR not excluded from die 2. However, the only compatible pair in this group is NR, making ...

die 2 = CNPR

The remaining letters IO are compatible with each other, making ...

die 3 = EIOT