The New Sudoku Players' Forum

[pjb]

My perspective on my solution is that the 3 at r2c5 must be either true or false - no argument here I hope.
A. If true then the 3 at r8c5 is false.
B. If false, the the simple AIC infers that the 3 at r8c9 is true, which would cause the 3 at r8c5 again to be false.
As it happens, proposition A is the correct alternative. It follows then that the AIC's intermediates are false, but this is not a problem. It could have been the other way round after all.
Cheers, Phil

[eleven]

...you and Phil are both right; you've just expressed the answer differently...
... Bat's observations and Phil's solution are not in conflict....

Yup.

One method uses a pincer attack to determine that r8c5<>3 if r2c5=3 and r8c5<>3 if r2c5<>3.
One method uses logic to determine that r2c5=3.

Both is logic, the difference is just how you express it - and how you find it.

Code: Select all

     *------------------------------------------------------------*
     | 168   2     18    |  9     178   147   | 3     5     46    |
     | 1368  5     7     |  348   138   2     | 9     18    46    |
     | 138   4     9     |  38    6     5     | 7     2     18    |
     *-------------------+--------------------+-------------------|
     | 5     18    6     |  7     2     9     | 4     3     18    |
     | 89    189   4     |  36    5     36    | 2     18    7     |
     | 2     7     3     |  1     4     8     | 6     9     5     |
     *-------------------+--------------------+-------------------|
     | 148   6     18    | #2348  9    *134   | 5     7    #23    |
     | 4789  89    2     |  5    #378   347   | 1     6    *39    |
     | 179   3     5     |  26    17    167   | 8     4     29    |
     *------------------------------------------------------------*

All you have to spot is, that 3r7c5 implies both 3r7c9 (3-links) and 2r7c9 (8 and 2 links), which is a contradiction.

Or you might notice the 2 strong links for 2 and 8 with a common cell (r7c4), which gives you 2r7c9=8r8c5.
Then 2r7c9->3r8c9 (or 3r7c46) shows, that r8c5 cannot be 3 (this is leren's L3 wing).
Or you take 8r8c5->9r8c2->9r9c9->2r7c9 => r7c9=2.
You could also show that r8c5 cannot be 7: 2r7c9->2r9c4->36r59c6->147r78c6,r9c5.

The notation is secondary.

[bat999]

...I have difficulty understanding some solutions.

OK, I've come to terms with it.

I know about the two states (true/false).
Toggle the states and eliminate candidates that would be targeted by both states.
In our case it means candidates that can see both ends of the chain.
I don't have a problem with any of that.

But sometimes I see a situation where if we used a particular state it would break the rules.
It would create a house without any of a particular candidate or a house with more than one of a particular candidate.
This information tells me which candidate to keep, rather than which targets to consider eliminating.
But that's not part of the plan.

We set out to perform a pincer attack.
Don't be distracted by the contradiction, stick with the original plan.
One state will eliminate the targets, the other state won't happen anyway (whether or not it breaks the rules).