JC Van Hay wrote:As an example of another possible path to the solution of puzzle 10, as well as for any other puzzle containing a Loop, ...

Using here only "basics" (elimination of all the candidates not in the solutions of each unit and each digit), out of the 16 remaining possible combinations of the bilocals 69B19+24B37,

15 lead to a contradiction and 1 gives the solution : [9r2c1+6r3c2]+[4r1c8+2r2c7]+[6r8c7+9r9c8]+[2r7c2+4r8c1]

I implemented that process and had a surprise showing an error in the manual solution pointed above.

Here is the problem, with a possible misunderstanding of what JC Van Hay is doing

With the following scenario (always in puzzle 10)

9r1c2 6r2c3 4r2c9 2r3c8 2r7c2 6r7c8 4r8c1 9r8c9 and the set of rules I am using, I am locked here

- Code: Select all
`2 9 137 |14 345 34 |58 78 6 `

37 5 6 |29 8 29 |37 1 4

8 13 4 |16 3567 367 |9 2 35

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59 7 28 |3 469 1 |26 49 58

35 46 13 |8 2 469 |15 49 7

19 46 28 |7 469 5 |26 3 18

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137 2 9 |5 37 8 |4 6 13

4 8 37 |26 1 26 |37 5 9

6 13 5 |49 3479 3479 |18 78 2

No problem to continue with 37 r2c17 r8c37 establishing 7r2c1 7r8c3, and showing quickly the contradiction, but we are far beyond the basic rules. I already used derived pairs, hidden pairs and derived XW to come to that point.

Question to JC Van Hay:

What do you do with that scenario (how do you conclude it is false)