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URs and pseudo cells

 
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Marty R.



Joined: 12 Feb 2006
Posts: 5179
Location: Rochester, NY, USA

PostPosted: Sat Oct 09, 2010 4:46 pm    Post subject: URs and pseudo cells Reply with quote

I've run into this situation a number of times and don't know what to do. Below is row 9 from the Free Press 10-8 puzzle:

Code:

+-----------+--------+---------------+
| .   .   . | . .  . | .   .     .   |
| .   .   . | . .  . | .   .     .   |
| .   .   . | . .  . | .   .     .   |
+-----------+--------+---------------+
| .   .   . | . .  . | .   .     .   |
| .   .   . | . .  . | .   .     .   |
| .   .   . | . .  . | .   .     .   |
+-----------+--------+---------------+
| .   .   . | . .  . | .   .     .   |
| .   .   . | . .  . | .   .     .   |
| 289 289 . | . 12 . | 168 12368 368 |
+-----------+--------+---------------+

Play this puzzle online at the Daily Sudoku site

I'm looking at the implications of a 36 DP. R9c89 have extra candidates 128. These combine with r9c125 to form a 1289 quad, forcing r9c7=6. Proceeding from there I arrive at an invalidity.

What eliminations can now be made?
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sat Oct 09, 2010 5:06 pm    Post subject: Reply with quote

After basics:
Code:
+----------------------+----------------------+----------------------+
| 59     19     56     | 8      7      4      | 2      136    136    |
| 238    238    268    | 9      5      1      | 68     7      4      |
| 48     148    7      | 26     3      26     | 5      18     9      |
+----------------------+----------------------+----------------------+
| 345    6      1      | 345    8      39     | 7      49     2      |
| 234578 2348   258    | 12345  12     2379   | 1468   14689  168    |
| 2478   248    9      | 124    6      27     | 3      5      18     |
+----------------------+----------------------+----------------------+
| 6      7      3      | 12     4      8      | 9      12     5      |
| 1      5      28     | 36     9      36     | 48     248    7      |
| 289    289    4      | 7      12     5      | 168    12368  1368   |
+----------------------+----------------------+----------------------+

After basics and an X-wing on 1:
Code:
+----------------------+----------------------+----------------------+
| 59     19     56     | 8      7      4      | 2      136    136    |
| 238    238    268    | 9      5      1      | 68     7      4      |
| 48     148    7      | 26     3      26     | 5      18     9      |
+----------------------+----------------------+----------------------+
| 345    6      1      | 345    8      39     | 7      49     2      |
| 234578 2348   258    | 2345   12     2379   | 1468   4689   68     |
| 2478   248    9      | 124    6      27     | 3      5      18     |
+----------------------+----------------------+----------------------+
| 6      7      3      | 12     4      8      | 9      12     5      |
| 1      5      28     | 36     9      36     | 48     248    7      |
| 289    289    4      | 7      12     5      | 168    2368   368    |
+----------------------+----------------------+----------------------+

Marty,

I see it as either one of R1C89 is 1, and / or there is a pseudo-triple 289 in R9C12(8+9), which forces R9C5=1 and R9C7=6. (R9C89 are a pseudocell 89.)

Looking further, the condition in R1 causes 8 in R3C8, while the condition in R9 causes 8 in R2C7, so the "and" is not correct.

It turns out that R9C7 is not 6 solves the puzzle. (There is a UR along the way.)

Keith
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Sat Oct 09, 2010 6:02 pm    Post subject: Reply with quote

[Withdrawn: off topic!]

Last edited by daj95376 on Sat Oct 09, 2010 6:49 pm; edited 2 times in total
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Marty R.



Joined: 12 Feb 2006
Posts: 5179
Location: Rochester, NY, USA

PostPosted: Sat Oct 09, 2010 6:20 pm    Post subject: Reply with quote

Quote:
I see it as either one of R1C89 is 1, and / or there is a pseudo-triple 289 in R9C12(8+9), which forces R9C5=1 and R9C7=6. (R9C89 are a pseudocell 89.)

Looking further, the condition in R1 causes 8 in R3C8, while the condition in R9 causes 8 in R2C7, so the "and" is not correct.

It turns out that R9C7 is not 6 solves the puzzle. (There is a UR along the way.)

Thank you.

Is it possible to dismiss the logic in the first two paragraphs above? In other words, if the 1289 quad forces a 6 and that leads to an invalidity, can we eliminate that 6 right then and there?

What I'm really trying to do is determine if there is a general "rule" as to what can be eliminated when a pseudo cell leads to an invalidity.
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sun Oct 10, 2010 2:54 am    Post subject: Reply with quote

Quote:
Is it possible to dismiss the logic in the first two paragraphs above? In other words, if the 1289 quad forces a 6 and that leads to an invalidity, can we eliminate that 6 right then and there?


Marty, no. You can only conclude that my pseudo-triple, or your pseudo-quad, is not true.

If you conclude that the consequences of the pseudo-group in R9 are false, the UR says the condition in R1 must be true: One of R1C89 is 1, thus R3C8 is 8, and the puzzle is solved immediately.

Of course, if a 6 in R9C7 forces an invalidity, then R9C7 is not 6. That has nothing to do with how you decided to test that value, whether by guessing or some higher justification.

I hope you are enjoying our perfect weather this weekend.

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5179
Location: Rochester, NY, USA

PostPosted: Sun Oct 10, 2010 4:56 am    Post subject: Reply with quote

Thanks Keith, that's a big help.

Yes, I'm enjoying the weather this weekend and enjoyed a very interesting football Saturday.
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Tue Oct 26, 2010 6:57 am    Post subject: Reply with quote

Quote:
After basics:
After basics and an X-wing on 1:
Code:
+----------------------+----------------------+----------------------+
| 59     19     56     | 8      7      4      | 2      136    136    |
| 238    238    268    | 9      5      1      | 68     7      4      |
| 48     148    7      | 26     3      26     | 5      18     9      |
+----------------------+----------------------+----------------------+
| 345    6      1      | 345    8      39     | 7      49     2      |
| 234578 2348   258    | 2345   12     2379   | 1468   4689   68     |
| 2478   248    9      | 124    6      27     | 3      5      18     |
+----------------------+----------------------+----------------------+
| 6      7      3      | 12     4      8      | 9      12     5      |
| 1      5      28     | 36     9      36     | 48     248    7      |
| 289    289    4      | 7      12     5      | 168    2368   368    |
+----------------------+----------------------+----------------------+


Marty,

If you are ever tempted to go after UR naked pseudo triples and quads, I suggest you look for the UR hidden singles instead. Take a look at box 3 and box 9 and notice the only spots where 3's and 6's can escape the UR pattern.

(6)r2c7 = (6&3)r1c89 -UR- (6&3)r9c89 = (6)r9c7

Right away you can see an immediate elimination, r5c7 <> 6.

Add a little winglike extension to each side and you have

(8)r3c8 = (8 - 6)r2c7 = (6&3)r1c89 -UR- (6&3)r9c89 = (6 - 1)r9c7 = (1)r7c8 => r3c8 <> 1 + lots of deductions.

BTW, go Sparty!
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Wed Feb 16, 2011 11:58 pm    Post subject: Reply with quote

While Myth's approach is nicely elegant, Keith's more complex chain makes the same elimination. Because the two "pincer" <8>s are peers in Box 3, it is a continuous AIC loop and all the links become conjugate. (That's why that "and" is not valid.) So, the <6> weak link exploited in r29c7 becomes conjugate, eliminating that <6> in r5c7.
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