Solve the following crypt arithmetic problem using constraint satisfaction.

The crypt arithmetic problem is given as: TWO + TWO = FOUR.

The easiest place to start is F. F can only be 1. This can be explained as follows: when the letters are at the highest possible value (987+987) = 1974, so

Now we move to O. O cannot be 1 as it is taken by F, so can be 2,3,4,5,6,7,8,9,0.

If O=2, then R is 4. W cannot be assigned a number, as if W is 5 and more, we will have a carry-over into T and 11 (12-1) cannot be divisible by 2, so then no number will be assigned to T, so 5 or more cannot be correct. W cannot be 1 because of F. W cannot be 2 because of O. If W was 3, then U=6 but T=6 (12 divided by 2), so it cannot be 3. W cannot be 4 because R is 4. So, O isn't 2.

If O was 3, then R=6. If R= 6 then T cannot be assigned a number, since if there is a carryover of 1, 12 divided by 2 is 6, but R is 6. So, O isn't 3.

If O is 4, then R=8.

Now we have

If there was a carryover into the TTO column, then 13 divided by 2 is not a whole number. So T has to be 7.

Now,

Now, we find W and U.

W cannot be o, 1 and 4 from just looking, so we have 2, 3, 5, 6, 7, 8, 9

If W was 2, then U= 4 but O is 4.

If W was 3, then U= 6 and that is the answer.

This is one solution. There are multiple solutions possible for this question. So, your solution may not match my solution.