SudoCue Users

Posted: **Sun Nov 18, 2007 11:21 pm**

Let's use this code from another topic.

.-----------------------.-----------------------.-----------------------.
|7 6 1 |8 3 2 |4 9 5 |
|4 9 8 |1 6 5 |2 37 37 |
|3 5 2 |4 9 7 |6 1 8 |
:-----------------------+-----------------------+-----------------------:
|9 38 67 |63 4 1 |378 5 2 |
|58 4 35 |2 7 9 |1 368 36 |
|2 1 67 |36 5 8 |37 4 9 |
:-----------------------+-----------------------+-----------------------:
|1 37 9 |57 8 6 |35 2 4 |
|568 2 35 |57 1 4 |9 3678 367 |
|56 78 4 |9 2 3 |58 67 1 |
'-----------------------'-----------------------'-----------------------'

Let's suppose that number 5 is in R8C1 or R9C1 (they are in line) then R5C1=8, R4C7=8, R8C7=5. Then we can eliminate 5 from R9C1.

Posted: **Sun Nov 18, 2007 11:55 pm**

Hi Novajlija, welcome to the site.

Let's suppose that number 5 is in R8C1 or R9C1 (they are in line) then R5C1=8, R4C7=8, R8C7=5. Then we can eliminate 5 from R9C1.

The last node in your chain should be R9C7=5.

The chain is a classic Forcing Chain, but you do not need to consider R8C1=5, just R9C1.

The grid:

Code: Select all

.------------------.------------------.------------------.
| 7     6     1    | 8     3     2    | 4     9     5    |
| 4     9     8    | 1     6     5    | 2     37    37   |
| 3     5     2    | 4     9     7    | 6     1     8    |
&#58;------------------+------------------+------------------&#58;
| 9     38    67   | 36    4     1    |*378   5     2    |
|*58    4     35   | 2     7     9    | 1    *368   36   |
| 2     1     67   | 36    5     8    | 37    4     9    |
&#58;------------------+------------------+------------------&#58;
| 1     37    9    | 57    8     6    | 35    2     4    |
| 568   2     35   | 57    1     4    | 9     3678  367  |
|-56    78    4    | 9     2     3    |*58    67    1    |
'------------------'------------------'------------------'

This is a AIC that does the same elimination:

(5=8)r5c1-(8)r5c8=(8)r4c7-(8=5)r9c7 => r9c1<>5

It looks a bit like a W-Wing with the same pair at both ends.

Ruud

Posted: **Mon Nov 19, 2007 2:15 am**

Hi Ruud!

I'm sorry, it's just oversight, it is R9C7=5.

And about rule: for me is easier to consider 2 or 3 cells in line (in the same box), at the same time, and see what is result.

I need help because I'm not sure is this right way.

Novajlija = Fresh Blood

Posted: **Wed Nov 21, 2007 3:21 am**

The chain is a classic Forcing Chain, but you do not need to consider R8C1=5, just R9C1.

Ruud's suggestion is good one since it simplifies the logical chain.

Also, consider the converse by assuming that r9c7=8, giving
(8=7)r9c2=(3)r7c2=(5)r8c3=>r9c1<>5

SudoCue Users

Is this valid?

Is this valid?