Not wanting to think about it much, I tried just successive substitution from a “random” starting point and that gave me the right answer. So I’ve tried a bunch more starting points and so far they’ve all converged to the answer. Is that just luck or is there a reason to think this should just converge? I’d think there might be cycles where it would get stuck. And then I wondered if there was a pattern based on the number of digits. I tried with 6 digits and immediately hit a cycle with that one.
Place an X or O in each empty cell of this grid so that four consecutive X’s or O’s do not appear horizontally, vertically, or diagonally. The solution is unique.
In each row, use the left-hand clue to fill in a five-letter word in the unshaded squares from left to right. Then write a letter in the shaded square to make a six-letter word that fits the right-hand clue. If you do it correctly, the shaded squares will spell out a word when read from top to bottom.
Add a letter to the first set of letters below and form a common word by anagramming the six given letters plus the one you added. Enter the word in the squares on the right. For each succeeding row, add the last letter of the previous word you wrote to the new set of letters before anagramming. The last letter of your fifth word should be the letter you added to the first word. Can you complete the daisy chain of anagrams?
Form four common five-letter words reading across. Use ten letters exactly twice, including the seven different letters given plus three others that you must determine. Since B, F, and M already appear twice, they can’t be used again. What are the words?
