## Looking for the flaw in the logic

Everything about Sudoku that doesn't fit in one of the other sections

### Looking for the flaw in the logic

Wikipedia's been updated, mathematics discussion page, at the bottom. I wouldn't have added it if the method didn't seem solid, but I need to know for sure.

I'm polling for anyone who can disprove the logic, preferable WITHOUT a band generator, as they appear unreliable and far more complex than the problem calls for. (If my solution is correct, band generators are probably WHY no one else has found it yet.)

I've got over 613 sextillion legal combinations, which is over the current accepted method by a factor of nearly 100. As a result, I'm prepared for some very negative feedback. Doesn't bother me. All I need to know is can anyone quantifiably DISPROVE it?
Weaselgrip

Posts: 2
Joined: 06 March 2006

On Wikipedia, Weaselgrip wrote:We begin with 3 blocks (1, 5, & 9) that can legally have their total allotted combinations without affecting each other, since they share no axes. Each of these blocks has 9! possibilities (9*8*7*6… or 362,880). Thus, our calculation begins with (9!^3).
Well there's your first mistake. You're assuming that each choice of B1/5/9 leads to an equal number of completions of the whole grid. To see that this doesn't work, try applying the same logic to the 2x2 case (2x2 boxes each of 2x2 cells, in a grid of overall size 4x4 cells), bearing in mind that the utterly uncontroversial answer in that case is 288 grid completions.

Whatever method you come up with for the 3x3 (ordinary sudoku) case, it has to produce the same result as on Wikipedia or it's just plain wrong. The 6.67e21 figure has been verified time and again using a wide range of techniques by different authors.
Red Ed

Posts: 633
Joined: 06 June 2005

I have managed to disprove it. On several fronts, but the first was along these lines. Thanks.

Got a lot of work to do.
Weaselgrip

Posts: 2
Joined: 06 March 2006