Look at Row 5 Column 9 (R5C9). It must be either 1 or 4. If we pretend that it does equal 4 then we find out quickly that this can't be true. Follow the logical steps below:

Let R5C9 = 4

Then R4C8=1

Then R7C8=9

Then R7C2=1

Then R5C2=4 .... but R5C2 can't equal 4 since we let R5C9 equal 4.

Therefore R5C9 can't be 4, and so R5C9=1

I think this might be known as a "jellyfish" if you want to Google more information.]]>

Check out R1C3. If that's a 6, then there are 4 squares in a rectangle (R2C3, R2C5, R3C3, and R3C5) that are all either 1 or 4; this would lead to multiple solutions, which would mean the puzzle is invalid. So R1C3 must be 4.

Question for others: Is the assumption that there are no multiple solutions valid? It should be true of any well-created puzzle, and I use often it in situations like the one here, but it just kinda feels wrong.

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Some solvers like to avoid using uniqueness assumptions in their solution, but I don't see the point in that since by definition, a valid sudoku must have a single unique solution. Why limit the number of solving techniques in your toolbox?]]>

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If you keep solving the puzzle with this assumption, i.e. that R1C3 is 6, you will come to a point where you'll realize that, in fact, no solution exists, i.e. putting a 6 in R1C3 is wrong. So, no, this will not lead to multiple solutions, but rather to no solution at all.

[quote=rdwells]

Question for others: Is the assumption that there are no multiple solutions valid? It should be true of any well-created puzzle, and I use often it in situations like the one here, but it just kinda feels wrong.

[/quote]

This assumption is valid, however, this approach is still wrong.

[b]Now about solving this particular puzzle, notice that R7C2 and R8C9 will always have the same value (either a 1 or a 9) that is enforced by R7C8, however, that value cannot be 1 because that would prevent both R5C2 and R5C9 from having a 1. Therefore, the value in both R7C2 and R8C9 must be a 9.[/b]]]>

[IMG]http://i.imgur.com/aRs1L69.png?1[/IMG]

And here, the puzzle is finally solved by making another uniqueness deduction. This is a BUG+1 grid which allows us to immediately conclude r8c2=1, and the puzzle is reduced to singles.]]>