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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Sat Jan 09, 2010 2:51 pm Post subject: Jan 9 DB |
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The Jan 9 DB has a tempting xyz-wing that goes nowhere so I had to resort to the old chain gang.
A Solution: An xy-chain eliminates the 3 in R9C9 and solves the puzzle.
Earl
| Code: |
+-------+-------+-------+
| . . . | 7 . . | 9 . . |
| 6 5 . | . 1 . | 3 . 2 |
| 7 . . | . . . | . 6 . |
+-------+-------+-------+
| . . 5 | 6 . . | 8 . 9 |
| . . . | . . . | . . . |
| 4 . 3 | . . 9 | 1 . . |
+-------+-------+-------+
| . 6 . | . . . | . . 7 |
| 8 . 7 | . 2 . | . 9 1 |
| . . 1 | . . 4 | . . . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat Jan 09, 2010 3:28 pm Post subject: |
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| The 543 xyz-wing did it for me. Left one pair and the rest singles. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Jan 09, 2010 5:13 pm Post subject: |
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Yes. After basics:
| Code: | +----------------+----------------+----------------+
| 3 4 2 | 7 6 5 | 9 1 8 |
| 6 5 9 | 4 1 8 | 3 7 2 |
| 7 1 8 | 239 39 23 | 5 6 4 |
+----------------+----------------+----------------+
| 1 2 5 | 6 34 7 | 8 34 9 |
| 9 8 6 | 123 345 123 | 7 2345 35 |
| 4 7 3 | 28 58 9 | 1 25 6 |
+----------------+----------------+----------------+
| 5 6 4 | 1389 389 13 | 2 38 7 |
| 8 3 7 | 5 2 6 | 4 9 1 |
| 2 9 1 | 38 7 4 | 6 358 35 |
+----------------+----------------+----------------+ |
There is a -345 XYZ-wing in R5 that makes a pair 12 in B5R5.
I don't recall ever seeing an XYZ-wing that makes two eliminations. (There is a 34 UR that I didn't see. It destroys the XYZ-wing but reveals another.)
Keith |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Jan 10, 2010 1:14 am Post subject: |
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I believe, but am not confident, that the logic I employed is valid.
| Code: |
*-----------------------------------------------------------*
| 3 4 2 | 7 6 5 | 9 1 8 |
| 6 5 9 | 4 1 8 | 3 7 2 |
| 7 1 8 | 239 39 23 | 5 6 4 |
|-------------------+-------------------+-------------------|
| 1 2 5 | 6 #34 7 | 8 #34 9 |
| 9 8 6 | 123 #345 123 | 7 #2345 3-5 |
| 4 7 3 | 28 58 9 | 1 @25 6 |
|-------------------+-------------------+-------------------|
| 5 6 4 | 1389 389 13 | 2 38 7 |
| 8 3 7 | 5 2 6 | 4 9 1 |
| 2 9 1 | 38 7 4 | 6 358 35 |
*-----------------------------------------------------------* |
First, note the type 3 UR34 in r45c58, marked #, creates a pseudocell 25 to prevent the DP. Now observe the 25 in r5c8, marked @. Both of these see the 5 in r5c9, which can therefore be deleted. Question to those who understand the fundamentals: is this valid?
If so, the puzzle is solved.
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Jan 10, 2010 1:47 am Post subject: |
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Ted: I'm just throwing in my opinion ... no expertise implied.
Although there is a UR Type 4 that results in r5c58<>3, I don't think your logic qualifies for a UR Type 3. However, basic (forcing chain) UR logic does apply for your elimination.
| Code: | r5c58=5 => r5c9<>5
r5c 8=2 => r6c8=5 => r5c9<>5
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Regards, Danny
Addendum: the <25> pseudo-cell in the UR is restricted to [r5], but your second <25> is not in [r5]. (You have a typo saying it is in r5c8.) |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Jan 10, 2010 2:14 am Post subject: |
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Danny, thanks for the feedback, and sorry about the typo. The reason that I assumed the deletion is valid is because both the pseudocell in row5 and the 25 in r6c8 "see" the 5 in r5c9. Maybe that is saying the same as your forcing chain.
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Jan 10, 2010 2:29 am Post subject: |
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| tlanglet wrote: | Danny, thanks for the feedback, and sorry about the typo. The reason that I assumed the deletion is valid is because both the pseudocell in row5 and the 25 in r6c8 "see" the 5 in r5c9. Maybe that is saying the same as your forcing chain.
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Ted, I'm not sure if it's saying the same thing or not. I do know that I had fun with your observation.
If you don't like forcing chains:
| Code: | UR[(5)r5c58 = (2)r5c8] - (2=13)r5c46 - (3=5)r5c9 - (5=2)r6c8 - UR[(2)r5c8 = (5)r5c58]
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| Code: | (5)r5c58 = (5)r5c9 - (5=2)r6c8 - (2)r5c8 = DP <34> in r45c58 => r5c58=5
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Jan 10, 2010 4:22 am Post subject: |
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Ted,
Your logic is correct, but it is not a Type-3 UR.
One of R5C58 is 5, or R5C8 is 2. Either way, R5C9 (is not 5) is 3.
Keith |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Jan 10, 2010 7:45 pm Post subject: |
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Ted,
One way of looking at your UR elimination is that it is really just an ALS elimination founded upon the UR. The two UR cells r5c58 contain 4 digits. However, we know that (at least) one of the UR digits must be false in both cells. Whichever it is, a 2-cell ALS results that contains the two non-UR digits. Thus, they can be used in any available ALS elimination, such as the one you make with the 25 bivalue ALS in r6c8. The shared exclusive (weak inference) digit is <2> and the shared common digit is <5>. Or, in Eureka:
34UR[(5)r5c58=(2)r5c8] - (2=5)r6c8; r5c9<>5
If r5c58 had been an actual ALS (no UR involved) the AIC would be identical except for the notation "ALS" in place of "34UR".
In fact, seeing the inherent UR strong inference and the weakly linked <2>s in c8 makes the same elimination evident without having to think about the potential ALS. (That is just another way to "find" this inherent strong inference.) Still, both points of view are interesting and valuable. |
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