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Help us code the next solve method

Click Solve It to get as far as possible, then read about the solve method below suggested by Mark Summerville.

The next challenge of course is.. can anyone code the Javascript to do it? Please submit any code under the GNU General Public License, or we can't use it. Don't modify any of the existing solve methods, but steal freely from the object methods provided.

Please send the code as a new Javascript source document. For debugging, we recommend Mozilla although please test using MS Explorer so that everyone can use it.

If you manage to do it, you've got to email us!

Other solve methods

Other solve rules, which can be seen at John Perry's website.

Solve Method E Explanation - by Mark Summerville

I've been looking at the Start grid on the website that does not get solved by your current algorithm (The Times 'Very Hard' one I believe). I have come up with the following algorithm, it may need some refining and I don't know how I'd go about programming it.

The general idea is to combine 'cyclical groups'. A cyclical group is a set of cells that are in the same block, row or column. A cyclical group also has no 'slack' in it. ie any additional information will allow us to complete all the cells in that group.

When the solver has got as far as it can with the puzzle I started to look for these groups. There are the following groups:

1. Column 1 A&H are 17
2. Row H 1&4 are 17
3. Column 4 CD&H are 37, 13, 17
4. Top Middle Block C4 C5 & A6 are 37 36 67

There are many more, but these are the key ones. In each case being able to rule out a number will allow us to fill in the vales of all the rest in the cyclical group.

The other key thing about these 4 groups is that they overlap ie 1H, 4H and 4C are in two cyclical groups each. This 'links' the cyclical groups so that it becomes one large cyclical group. ie knowing any cell will define all the others.

If 1A is a 7 then using the princliple of cyclical groups then 6A is also a 7. (Similarly if 6A is a 7 then 1A is a 7). That means that neither can be a 7. When we've established that 1A is a 1 and 6A is a 6 then the existing solving rules can be used to complete this puzzle.

If you know of any other su doku that are not solvable using the existing rules then please let me know as I may be able to use them to refine this new rule. [Please email us and we will pass on the message to Mark]

Please note that we have disabled Solve Method F on this page because it now cracks this sudoku!