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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Sat Oct 09, 2010 4:46 pm Post subject: URs and pseudo cells |
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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 |
+-----------+--------+---------------+
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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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Oct 09, 2010 5:06 pm Post subject: |
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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: 3854
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| Posted: Sat Oct 09, 2010 6:02 pm Post subject: |
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[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: 5770
Location: Rochester, NY, USA
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| Posted: Sat Oct 09, 2010 6:20 pm Post subject: |
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| 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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Oct 10, 2010 2:54 am Post subject: |
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| 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: 5770
Location: Rochester, NY, USA
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| Posted: Sun Oct 10, 2010 4:56 am Post subject: |
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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
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| Posted: Tue Oct 26, 2010 6:57 am Post subject: |
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| 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 |
+----------------------+----------------------+----------------------+ |
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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
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| Posted: Wed Feb 16, 2011 11:58 pm Post subject: |
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| 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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