"yielding chain"

Advanced methods and approaches for solving Sudoku puzzles

"yielding chain"

Postby StrmCkr » Thu Sep 21, 2006 10:09 am

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Last edited by StrmCkr on Sat Dec 13, 2014 5:58 am, edited 2 times in total.
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Re: "yielding chain"

Postby re'born » Thu Sep 21, 2006 11:19 am

StrmCkr wrote:
Case Study 1:

A| 5 3479 479 |
B| 379 6 2 |
C| 1 3489 489 |

in this specific quad i have noticed that all unsolved squares are connected by the number 9.

this number complicates the normally exposed strong\weak links of all the cells.

I start by removing that number, and can then see the underling chain of events and hidden links between the cells.


A| 5 347 47 |
B| 37 6 2 |
C| 1 348 48

this exposes several events.

In cells A2, A3 a (47) pair are hidden pairs.
C2, C3 are hidden pairs (48)
and B1 is a singled 3 thanks to removal by the hidden pair.



I'm already lost at this point and did not procede further. In what sense do you mean that A2 and A3 form a hidden pair? In the box, this makes no sense as 4 and 7 appear elsewhere. In the row it does not make sense either as looking at the entire row again yields additional 4's and a 7. For the same reasons, C2 and C3 are not hidden pairs. Are you using an alternate definition of hidden pair?

Of course, all of this assumes that there is some logical reason for removing the 9 from every cell in the box and expecting to obtain results. Please clarify.
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Postby StrmCkr » Sat Sep 23, 2006 7:18 am

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Postby Myth Jellies » Sat Sep 23, 2006 8:46 am

Worse.

As far as I can tell, nothing you have presented can legitimately invalidate the following possibility...
Code: Select all
5   3   4
7   6   2
1   8   9

...in your first case, and yet somehow you have solved for 3 in B1. While I can see the logic of temporarily removing a pervasive digit from a group to better see locked sets and almost locked sets, nothing you have presented substantiates many of the reductions you are making.
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