Denote the cells in a row, column or square a through i. Now you can write the Sudoku requirement of having 1 though 9 once as 10^a+10^b+10^c+10^d+10^e+10^f+10^g+10^h+10^i = 1111111110, if you can find a way of enforcing the numbers being integers, which is left as an exercise to the reader. (^ represents exponentiation, of course.) If you can solve these equations for each row, column and square simultaneously, you'll solve the Sudoku.
I considered making a linear version by requiring that the sum is the sum of 1 though 9, i.e. a+b+c+d+e+g+g+h+i = 45, but here's a counter example to prove it won't work. We start out with a top-left block of a valid solution:
123
456
789
Now we move some numbers around:
213
366
789
The row, column and block sums are unchanged, but the solution has become unvalid due to the double 6. QED. |
