ArchieB wrote:As for the method you use to arrive at the answer, it's basically trial-and-error, i.e. "9 in r7c9 doesn't work so r7c9=5".

You're miss-stating the point of the proof which tells us *nothing* about what r7c9 is. You're describing a proof by contradiction, the contrapositive of forcing chains.

The procedure to finding forcing chains varies with the specifics of the puzzle. Here's what I did in this one:

- Code: Select all
` . . . | . . . | . . . `

. . . | . . . |24 49 29

. . . | . . . | . . .

----------+-----------+----------

. . . | . . . | . . .

. . . | . . . |45 14 15

. . . | . . . | . . .

----------+-----------+----------

89 . . |68 . . |56 . 59

. . . | . . . | . . .

89 . . |68 . . |26 19 129

I see that r2c789 is a triplet that could ONLY be 249 or 492 -- exactly two possible states.

Similarly, r5c789 is a triplet that could ONLY be 415 or 541.

They each must have a 4 column 7 or 8. This combines all six cells into a single group that can have exactly two states:

249

415

OR

492

541

Since r7c9 must be either 5 or 9, it eliminates the first state which must have BOTH a 5 and a 9 in the 9th column.

From there I just follow the shortest paths to make the forcing chains given.

If you agree that if each of the four cells in r12c12 had two candidates like this:

- Code: Select all
`[12][13]`

[23][34]

... that we could deduce without T&E that r2c2=4 ...

... then you must also agree that given these cells in rows 3, 5 and 7, columns 7, 8 and 9:

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`[12][13][35]`

[23][--][--]

[34][--][14]

... that we could deduce without T&E that r7c9=1 ...

... and if you do, you must accept this as well:

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`[12][13][35]`

[24][--][--]

[45][--][15]

... which is essentially the same set of forcing chains that are in the puzzle.

If you're not convinced, please see

this post and see if you can draw the line that distinguishes T&E from other logic.