Hidden Pair in a Chain

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Hidden Pair in a Chain

Postby daj95376 » Fri Aug 14, 2015 5:56 pm

Assuming that basic techniques have already been applied.

Code: Select all
 +-----------------------------------------------+
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   3W  .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   /   3X  /   |   /   /   /   |  37Y 37Z  /   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 +-----------------------------------------------+

 Chain Segment:   ... = 3r3c2 - 3r8c2 = hp(37)r8c78 - ???

It follows that <7> is locked in r8c78 and must have been eliminated elsewhere in [r8] and [b9]. It can't contribute any eliminations from the Hidden Pair.

There is an SL for <3> in [r8], and a simpler chain segment -- ... = 3r3c2 - 3r8c2 = 3r8c78 - 3[b9] -- exists for eliminations of <3> in [b9].

Thus, I'm left with the only realistic eliminations (in the chain) associated with the Hidden Pair being the candidates Y and Z.

Agree/disagree ?

_
daj95376
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Re: Hidden Pair in a Chain

Postby ronk » Fri Aug 14, 2015 10:23 pm

daj95376 wrote:Thus, I'm left with the only realistic eliminations (in the chain) associated with the Hidden Pair being the candidates Y and Z.

Agree/disagree ?

Agreed, in your scenario no more and no less than a conventional hidden pair.
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Re: Hidden Pair in a Chain

Postby eleven » Sun Aug 16, 2015 12:25 pm

Moreover to be able to continue the chain you need a candidate x in the 2 cells with a strong inference
.. = 3r3c2 - 3r8c2 = (hp37-x)r8c78 = ..
This candidate always must build a hidden pair x7 in the other direction
.. =(hpx7-3)r8c78=3r8c2 .. or
.. =(hpx7-hp37)r8c78=3r8c2 ..
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