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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Sun Jun 24, 2007 3:45 am Post subject: June 23 VH |
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This VH had only one technique, an x-y wing (R5C4, R5C9, R6C6) eliminating the 9 in R6C9, and the rest of the puzzle was simple elimination.
Or did I miss something?
Earl |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Sun Jun 24, 2007 5:40 am Post subject: |
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| Quote: |
This VH had only one technique, an x-y wing (R5C4, R5C9, R6C6) eliminating the 9 in R6C9, and the rest of the puzzle was simple elimination.
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I think you mean (R5C4, R5C9, R6C5) for the x-y wing.
You need at least an X-wing to eliminate a 1 from R6C5 before the x-y wing opens up. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sun Jun 24, 2007 9:13 pm Post subject: |
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| Spent a long time staring at this one - but got totally stuck. I found the x wings on 1's and 6's easy to spot but failed to detect the xy wing given by Earl (typo corrected in the 2nd post). In fact I'm wondering if there is any other solution possible. |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Sun Jun 24, 2007 10:49 pm Post subject: |
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| Quote: |
In fact I'm wondering if there is any other solution possible.
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A solution without the XY-wing -
Starting here -
| Code: |
+-------------+------------+-----------+
| 26 | | 26 |
| 24 | 24 | |
| 124 1246 | 24 | 26 |
+-------------+------------+-----------+
| 56 | 15 | 16 |
| 23 235 | 24 259 | 49 |
| 26 | 129 124 | 46 149 |
+-------------+------------+-----------+
| | 12 | 24 14 |
| 12 12 | | |
| 34 34 | 12 | 12 |
+-------------+------------+-----------+
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Eliminate 2 from R3C1 and R3C3 with a forcing chain based on R1C2
(R1C2=6,R1C7=2,R7C7=4,R7C6=2,R2C6=4,R3C4=2 ===> R3C1 <> 2 and R3C3 <> 2)
to arrive at -
| Code: |
+-------------+------------+-----------+
| 26 | | 26 |
| 24 | 24 | |
| 14 146 | 24 | 26 |
+-------------+------------+-----------+
| 56 | 15 | 16 |
| 23 235 | 24 259 | 49 |
| 26 | 129 124 | 46 149 |
+-------------+------------+-----------+
| | 12 | 24 14 |
| 12 12 | | |
| 34 34 | 12 | 12 |
+-------------+------------+-----------+
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From there, do a little coloring starting with R1C2
| Code: |
+-------------+------------+-----------+
| +2-6 | |-2+6 |
| -2+4 | +2-4 | |
| 14 146 | 24 | 26 |
+-------------+------------+-----------+
| 56 | 15 | 16 |
| 23 235 | 24 259 | 49 |
| 26 | 129 12+4|+4-6 149 |
+-------------+------------+-----------+
| | 12 | 24 14 |
| 12 12 | | |
| 34 34 | 12 | 12 |
+-------------+------------+-----------+
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The two +4's in Row 6 means all the -'s must be true. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sun Jun 24, 2007 11:14 pm Post subject: |
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| What is a "Forcing Chain"? Hope it isn't a guess or assumption !!! |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Jun 25, 2007 2:47 am Post subject: |
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I believe R7C9 needs to be inserted between R7C7 and R7C6 in jLo's chain.
In any case, with the insertion, this is an XY-Chain. What constitutes an "acceptable" solving method is pretty much a personal judgment call in many cases. For me, the XY-Chain is a clear and simple method even though I rarely use it. It involves a sequence of bivalue cells that have a series of "strong links" with a common digit on the ends of the chain. Any cells that can "see" both of these ends cannot contain that digit. The logic is clear to me if this example is written as follows:
[26]-[62]-[24]-[41]-[12]-[24]-[42]
Notice the shared digits on each side of the "chain link" hyphens.
If the first cell is <2>, it eliminates the target candidates. If it is <6>, then there are cascading placements through the chain ending with <2> in the last cell, which also eliminates the target candidates.
Since bivalue cells are easy to spot and the links are easy to trace, applying the method is fairly mechanical. Obviously, you start by noticing elimination candidates that can "see" at least two bivalue cells that contain that digit. Then you just have to see if a chain exists.
In the case of this puzzle, I eliminated the same two <2>s due to the 12 UR in Rows 3 and 8. Because there is no other candidate for <1> in R3, I believe this allows elimination of the <2>s. I'm still trying to master URs, so maybe someone of UR authority can confirm that I'm not mistaken! |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Mon Jun 25, 2007 3:41 am Post subject: |
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| Quote: |
In the case of this puzzle, I eliminated the same two <2>s due to the 12 UR in Rows 3 and 8. Because there is no other candidate for <1> in R3, I believe this allows elimination of the <2>s. I'm still trying to master URs, so maybe someone of UR authority can confirm that I'm not mistaken!
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You can in fact eliminate both the <2>s from R3C1 and R3C3 based on the UR. If
either were a <2>, then R3C4=4 and R3C8=6 and the remaining entry would be a
<1>, leading to the deadly pattern. You could make the same argument based on no
other candidate for <1> in Box 1.
Nice find. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jun 25, 2007 3:06 pm Post subject: |
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| I couldn't find any XY-Wings. I used an X-Wing, Remote Pairs, a Type 4 rectangle and three simple colorings. |
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