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RobertRattley
Joined: 24 Jun 2007
Posts: 118
Location: Australia
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 Posted: Thu Aug 20, 2015 8:35 am Post subject: Aug 20 VH |
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Interesting: after basics (including one slightly "hard" step) I could see 3 independent (non-overlapping) xy wings. One of them didn't help much. Each of the other two was needed.
I'll let someone else post details. |
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hughwill
Joined: 05 Apr 2010
Posts: 424
Location: Birmingham UK
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| Posted: Thu Aug 20, 2015 9:19 am Post subject: Aug 20 VH |
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After (sufficient) basics: | Code: |
+----------------+------------+----------------+
| 1267 9 1267 | 48 3 246 | 5 28 1248 |
| 26 5 8 | 1 9 246 | 3 7 24 |
| 4 3 12 | 58 25 7 | 18 6 9 |
+----------------+------------+----------------+
| 12567 8 12567 | 9 157 3 | 167 4 12 |
| 9 27 1257 | 6 4 8 | 17 25 3 |
| 1567 4 3 | 2 157 15 | 16789 589 18 |
+----------------+------------+----------------+
| 3 6 24 | 7 12 124 | 89 89 5 |
| 8 1 45 | 45 6 9 | 2 3 7 |
| 257 27 9 | 3 8 25 | 4 1 6 |
+----------------+------------+----------------+
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Play this puzzle online at the Daily Sudoku site
This is an XY-Wing fest. But the 12-8 on r13 and the 28-4 on r12 don't set
it (even in combination). The 17-2 on r34 looks more promising but it is the which sets it in one.
Those were all I could find, but....... |
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bat999
Joined: 09 Jul 2015
Posts: 55
Location: UK
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| Posted: Thu Aug 20, 2015 1:26 pm Post subject: |
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| Code: | .--------------------.----------------.-------------------.
| 1267 9 b1267 | 48 3 246 | 5 28 1248 |
| a26 5 8 | 1 9 246 | 3 7 24 |
| 4 3 1-2 | f58 f25 7 | e18 6 9 |
:--------------------+----------------+-------------------:
| 12567 8 c12567 | 9 157 3 | d167 4 12 |
| 9 27 1257 | 6 4 8 | d17 25 3 |
| 1567 4 3 | 2 157 15 | 16789 589 18 |
:--------------------+----------------+-------------------:
| 3 6 24 | 7 12 124 | 89 89 5 |
| 8 1 45 | 45 6 9 | 2 3 7 |
| 257 27 9 | 3 8 25 | 4 1 6 |
'--------------------'----------------'-------------------' |
(2=6)r2c1 - r1c3 = r4c3 - (6=17)r45c7 - (1=8)r3c7 - (8=2)r3c45 => -2 r3c3; stte
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Aug 20, 2015 6:12 pm Post subject: |
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A typo, Hugh, it's a 25-1. |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Fri Aug 21, 2015 5:06 pm Post subject: |
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| I have another blind spot that I need help with! I'm still working on my Eureka so I like to check out chains that are posted. I'm reading Bat's Eureka expression as follows: If r2c1 is not 2, then it's 6 which means r1c3<>6 so r4c3=6 so r4c7<>6 leaving the naked 17 pair in r45c7 so r3c7<>1 so it's 8 so r3c4=5 and r3c5=2 and therefore r3c3<>2 but I don't see how that concludes anything (again)! I don't see an obvious contradiction and it doesn't look like the chain ends contain pincers like I missed last time. What am I missing? Thanks in advance, Don. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Aug 21, 2015 6:23 pm Post subject: |
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| dongrave wrote: | | I have another blind spot that I need help with! I'm still working on my Eureka so I like to check out chains that are posted. I'm reading Bat's Eureka expression as follows: If r2c1 is not 2, then it's 6 which means r1c3<>6 so r4c3=6 so r4c7<>6 leaving the naked 17 pair in r45c7 so r3c7<>1 so it's 8 so r3c4=5 and r3c5=2 and therefore r3c3<>2 but I don't see how that concludes anything (again)! I don't see an obvious contradiction and it doesn't look like the chain ends contain pincers like I missed last time. What am I missing? Thanks in advance, Don. |
I agree with your interpretation. The last term could've been written as (8=52)r3c45. This is one of those solutions that doesn't conclude with pincers, requiring the reader to look more closely and figure out what's going on.
In summary, the chain says if r2c1 is <>2 and =6, a 2 in proven in r3c5, proving r3c3<>2. This is an either/or, so it's either (2=6) or (6=2). Each one proves r3c3<>2, with no chain needed for the 2nd term, thus the conclusion of -2r3c3. |
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bat999
Joined: 09 Jul 2015
Posts: 55
Location: UK
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| Posted: Fri Aug 21, 2015 6:36 pm Post subject: |
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| dongrave wrote: | .. the chain ends contain pincers...
...What am I missing? ... |
Hi Don
You read the chain correctly. 
If r2c1=2 then r3c3<>2 (obviously).
If r2c2<>2 then (through the chain) r3c5=2 ... and r3c3<>2 again.
Cells r2c1 and r3c5 are the pincers, one of them is 2 so r3c3 can't be 2. |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Fri Aug 21, 2015 7:54 pm Post subject: |
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| Oh! Obviously! Thanks again for your help Marty and Bat! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Aug 21, 2015 8:09 pm Post subject: |
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I blew it by failing to recognize the last term as the 2nd pincer.  |
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