Write a letter in each square such that every group of four contiguous squares is a four-letter existing English word (i.e. it can be found in the Merriam-Webster dictionary).



  1. The first column (words A,B,C and D) can contain any words.
  2. The letters in the colored squares have to be copied in the next column following the arrows with the same color.
  3. The letters in the white squares must not be not copied.
  4. When copying the letters on the next column, the colors of the squares on the destination word does not play any role.
  5. When following an arrow the copied letters must keep the same order they had in the original word.
  6. Letters copied from two different words can be alternated in the destination word.

Example: Let A1, A2, A3, A4 be the four letters of the word A and B1, B2, B3, B4 be the four letters of the word B. The first word of the second column is composed of the letters A1, A4, B1, and B3. Some possible orders can be A1 A4 B1 B3 or A1 B1 A4 B3 or B1 B3 A1 A4 or B1 A1 A4 B3 but not A4 A1 B1 B3 because swapping A1 and A4 breaks the rule 5.

I think there are multiple solutions of this puzzle, I created it based on another puzzle I found on this site that had slightly different rules.


EDIT 2020-03-23: simplified a little bit.

Find exactly the solution I have in mind. Here are some more constraints to achieve that:

  • The letters in the white squares of the first column are (in alphabetical order): C, E, F, I.
  • The letters in the white squares of the second column are (in alphabetical order): E, K, N, U.
  • The letters in the white squares of the third column are (in alphabetical order): C, D, H, U.
  • The last word is "LAST".

Note that with these constraints you know all the letters to be placed on the diagram. Fun fact: the letters in the white squares of the first three columns form an anagram of the sentence "check if undue".

I hope you enjoy this puzzle. If someone solves the harder version I will accept their answer. If no one solves the harder version by 2020-03-31 at 23:59 UTC I will accept any other answer of my choice.

  • $\begingroup$ This seems to have a lot of solutions without the "harder version" restrictions, and it seems to be basically impossible to find what that restriction... $\endgroup$
    – Deusovi
    Mar 22 '20 at 17:37
  • $\begingroup$ @Deusovi funnily enough you get the last word right. It was not meant to be a computer puzzle, now I'm thinking of adding the [computer-puzzle] tag. $\endgroup$
    – melfnt
    Mar 22 '20 at 17:59
  • 1
    $\begingroup$ Yeah, I guessed the last word because of the hint. But the rest of the puzzle is very hard to do -- maybe if there were clues to the starting words or some of the middle words, it would be more doable. $\endgroup$
    – Deusovi
    Mar 22 '20 at 18:09
  • $\begingroup$ @Deusovi I have added more constraints to simplify the harder version. Maybe you want to try to solve it now. $\endgroup$
    – melfnt
    Mar 23 '20 at 18:37

The harder version: ...................


  • $\begingroup$ Congrats! Was it too simple to find this solution? $\endgroup$
    – melfnt
    Mar 23 '20 at 19:19
  • 1
    $\begingroup$ @melfnt - I didn't even notice this question until you had made the latest revisions to it. With those revisions, yes, it wasn't too difficult. I started at the end and worked backwards. $\endgroup$ Mar 23 '20 at 19:28
  • $\begingroup$ It's still a fun problem to work on, though. $\endgroup$ Mar 23 '20 at 19:30

Solutions to this (the 'base' puzzle) are relatively easy to find. Here's one solution of very many:

enter image description here

(Colors altered from the original because of colorblindness reasons, and to make reading the diagram easier.)

The harder puzzle seems to be only reasonably attackable through computer search, though... even with a few guesses at the last word, making sure the white squares anagram to the right thing doesn't seem to be very human-doable to me.

  • 1
    $\begingroup$ Your answer is particularly interesting because it contains a reference to another prominent PSE user (namely @DEEM). $\endgroup$
    – trolley813
    Mar 22 '20 at 21:21

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