2
$\begingroup$

Given that each letter in the English alphabet has a position: $$a = 1, b = 2, ..., z = 26$$ Can you place 16 different letters such that:

  • Each row, column and diagonal forms a 4-letter valid English word. You could use dictionary.com.
  • The sum of all letter positions is maximized. That is, using able counts as $a+b+l+e = 1+2+12+5 = 20$.

"And what if such a grid doesn't exist?" — @Bass

For every letter that is used twice or more, all words containing the letter won't count in the total. That is, a beer won't count any point alongside with all other letters in words in the grid that have an e.

Improve this question
$\endgroup$
4
  • $\begingroup$ How certain are you that it's possible to create a(n unavoidably double) word square from 16 distinct letters at all, not to even mention the diagonals? $\endgroup$
    – Bass
    Commented Feb 19, 2022 at 12:48
  • $\begingroup$ @Bass, actually, I am not certain at all. I have improved the puzzle thanks to your relevant remark :) $\endgroup$
    – JKHA
    Commented Feb 19, 2022 at 12:57
  • $\begingroup$ Can the diagonals be read in either direction? $\endgroup$
    – hexomino
    Commented Feb 20, 2022 at 21:05
  • $\begingroup$ @hexomino, not precised in the puzzle so yes! $\endgroup$
    – JKHA
    Commented Feb 20, 2022 at 22:27

1 Answer 1

Reset to default
1
$\begingroup$

The best I've found so far has a one letter repeat

 T Y M P
 H A I L
 U N C O
 S K E T

With the caveat that

"tymp" does not appear at dictionary.com but does appear on Merriam-Webster. I hope this is acceptable.

I'm not sure how to score this (maybe OP can advise).

Improve this answer
$\endgroup$
1
  • $\begingroup$ I believe your score is 27, the "double letter negates all points from words using said letter" rule really reduces said score $\endgroup$
    – Auribouros
    Commented Mar 28, 2022 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.