reallyjoel wrote:This is a pretty obvious one, but it's not mentioned on

http://www.angusj.com/sudoku/hints.php, so i thought I'd throw it out there.

If you have three cells with candidates 1 and 2, and 2 of them have candidates 3 also, and candidate 3 is ONLY in those two cells in that box/row, then you can eliminate all other cells 1's and 2's, because wether the 3 goes in one or the other, the other one will make a pair with the third cell.

I can only guess this is very common knowledge.

/Joel

Really realyjoel, you *are* describing a specific case of naked triples. A through D below are all basically the same structure. In each case, the first three cells must be 1, 2 and 3 in some order, so the 4th through 9th cells must not be 1, 2 or 3. You've described B -- but they're all basically the same thing:

- Code: Select all
` Before - - - - - - - - - - After`

A: (123) (123) (123) (1234) => (123) (123) (123) (---4)

B: (123) (123) (12-) (124) => (123) (123) (12-) (--4)

C: (123) (1-3) (12-) (1234) => (123) (1-3) (12-) (---4)

D: (-23) (1-3) (12-) (1234) => (-23) (1-3) (12-) (---4)

Even this line below qualifies as a triplet, though it includes a pair within which could be processed first:

- Code: Select all
`(-23) (-23) (1-3) (1234) => (-23) (-23) (1-3) (---4) => (-23) (-23) (1--) (---4) `

or

(-23) (-23) (1-3) (1234) => (-23) (-23) (1--) (12-4) => (-23) (-23) (1--) (---4)

This idea extends to quads, quints, etc. N cells with N values:

- Code: Select all
`E: (12345)(12345)(12345)(12345)(12345)(123456)`

F: (1---5)(-23--)(-2-4-)(---45)(12--5)(123456)

In both E and F, the first 5 cells must be 1 through 5 in some order, so the 6th cell is 6. Read Argusj again.