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Aug 20 VH

 
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RobertRattley



Joined: 24 Jun 2007
Posts: 118
Location: Australia

PostPosted: Thu Aug 20, 2015 8:35 am    Post subject: Aug 20 VH Reply with quote

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

PostPosted: Thu Aug 20, 2015 9:19 am    Post subject: Aug 20 VH Reply with quote

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    |
+----------------+------------+----------------+

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
Quote:
12-5 pivot r9c6
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

PostPosted: Thu Aug 20, 2015 1:26 pm    Post subject: Reply with quote

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
Cool
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Thu Aug 20, 2015 6:12 pm    Post subject: Reply with quote

Quote:
12-5 pivot r9c6


A typo, Hugh, it's a 25-1.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Fri Aug 21, 2015 5:06 pm    Post subject: Reply with quote

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

PostPosted: Fri Aug 21, 2015 6:23 pm    Post subject: Reply with quote

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

PostPosted: Fri Aug 21, 2015 6:36 pm    Post subject: Reply with quote

dongrave wrote:
.. the chain ends contain pincers...
...What am I missing? ...

Hi Don
You read the chain correctly. Very Happy
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. Cool
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Fri Aug 21, 2015 7:54 pm    Post subject: Reply with quote

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

PostPosted: Fri Aug 21, 2015 8:09 pm    Post subject: Reply with quote

I blew it by failing to recognize the last term as the 2nd pincer. Embarassed
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