QJohnson wrote:On very tough ones, I often have to guess at least once. I just always guess where a cell can have only one of two values so that the guess confirms one of them. Is this what they mean here in other messages by a "forcing chain?"
No, it means something very different. There are various types of forcing chain -- here is an example of one in the present puzzle that has what are called "strong" links, and eliminates the '8' from r6c7:
r6c7-4-r3c7-1-r3c9-5-r8c9-8-r8c7-8-r6c7
where A-x-B means
1) cells A,B are in the same unit (row/column/box), *and*
2) candidate digit x occurs in both cells A,B, *and*
3) x occurs nowhere else in that unit (this is what makes the link "strong", because it implies that x must occur in exactly one of the two cells).
The chain refers to the stage of solution when the candidate grid is
- Code: Select all
7 156 456 | 2 458 146 | 3 458 9
19 3 459 | 14589 7 149 | 2 458 6
269 269 8 | 3 45 69 |[14] 7 [15]
----------------------+----------------------+-------------------
3 2678 267 | 1478 9 1247 | 5 1246 128
2689 4 25679 | 1578 258 127 | 1689 3 128
289 2589 1 | 458 6 3 |[489] 249 7
----------------------+----------------------+-------------------
12689 12689 269 | 469 24 5 | 7 1269 3
4 2679 3 | 679 1 279 |[689] 2569 [258]
5 12679 2679 | 679 3 8 | 169 1269 4
The reason for calling this a "forcing" chain, is that a pattern such as A-p-B-q-C-r-D-x-E-x-A "forces" the conclusion A<>x, as can be seen by breaking the chain into two segments at any chosen intermediate cell (say we pick the C cell to get segments A-p-B-q-C and C-r-D-x-E-x-A). Then ...
from A-p-B-q-C, going right-to-left, we get
C<>q => B=q => B<>p => A=p => A<>x
but from C-r-D-x-E-x-A, going left-to-right, we get
C=q => C<>r => D=r => D<>x => E=x => A<>x
Since it must be the case that either C<>q or C=q, and both cases lead to A<>x, it follows that A<>x.
I think it's important to note that it is
not necessary to use any of the above logic in order to find the chain or to apply its "candidate elimination rule". This amounts to applying an established theorem without having to prove it every time.
(
<rubylips' solver> solves this puzzle without Nishio, by finding lots of chains of different varieties.)