A standard 9x9 Sudoku uses the digits 1 to 9.

You are only allowed two distinct primes to represent 1 to 9.

Find out the minimum number of characters (digits + signs) needed to construct a Sudoku puzzle with a unique solution.

Allowed signs are plus, minus, division, multiplication, factorial, and exponentiation.


2 Answers 2


The minimal Sudoku contains 17 digits: 2 each of seven of the nine digits, 3 of another digit, and none of the last one. (By symmetry, it doesn't matter which two digits are selected to appear 3 and 0 times in the Sudoku.)

If we select our two primes as

2 and 3, then we can represent the digits as $$3-2,2,3,2+2,2+3,3!,?,2\wedge3,3\times3.$$ (We omit 7 because we only need to represent all but one of the digits for the minimal Sudoku.)

So we choose the minimal Sudoku with

seventeen digits filled, two each of $1,2,4,5,6,8,9$ and three of $3$.

How many symbols do we need to use then?

Each of $1,4,5,8,9$ requires three symbols, $2$ and $3$ require only one, and $6$ requires two. So the total is $$5(2\times3)+(2\times1)+(2\times2)+(3\times1)=30+2+4+3=39.$$

  • $\begingroup$ Excellent !!!........ $\endgroup$
    – Uvc
    May 27, 2019 at 17:09

I got it down to

37 symbols

Using all the same methods as @RandAlThor, but choosing a

17 clue sudoku with a slightly more off-balance clue distribution (1,3,4,2,1,2,0,2,2):

enter image description here

For double checking purposes, here's the sudoku as solved by https://sudokusolver.net/

enter image description here

For further improvement, it's very likely that some digit(s), particularly those that occur twice, can be replaced by adding both a 2 and a 3, giving a sudoku that has more clues but fewer symbols. Since I cannot figure out a non-boring way to do the search for those, I'll leave it to someone else.

EDIT @Oray helpfully pointed out in the comments that factorials were allowed, and OP says we can cheat with the exponentiation symbol too so here's a version with

29 symbols:

enter image description here

  • $\begingroup$ Very elegant...!!! $\endgroup$
    – Uvc
    May 27, 2019 at 19:01
  • $\begingroup$ 3! instad of 3+3 makes it 35 :) $\endgroup$
    – Oray
    May 27, 2019 at 19:22
  • $\begingroup$ @Oray Oh, right, we were allowed to use that. Oops :-) $\endgroup$
    – Bass
    May 27, 2019 at 19:46
  • $\begingroup$ We can cheat a little bit..use the power notation and omitting “^” carrot symbol..8 can be written with 2 characters..bringing it down further to 33..just a tad less than double the minimum of 17. $\endgroup$
    – Uvc
    May 27, 2019 at 19:47
  • $\begingroup$ Given these new tricks, the latest version is probably pretty close to a hard minimum: the only 3-symbol clues occur only once each, so they cannot be replaced with anything, and replacing 2-symbol clues seems unlikely to bring the number of symbols down very much. $\endgroup$
    – Bass
    May 27, 2019 at 20:20

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