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18 dec 06 an alternative route please

 
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csc



Joined: 03 May 2007
Posts: 14
PostPosted: Wed May 30, 2007 11:33 am    Post subject: 18 dec 06 an alternative route please Reply with quote
I proceeded from below via:
xy chain r2c8 - r3c7 - r3c3 - r3c2 - r8c2, eliminating candidate 9 in r8c8, after which an xy wing: pivot 47 in r8c8 and pincers r9c8, r8c2, eliminating candidate 9 in r9c2, and so on.

When I asked for a hint from "play online", it gave a 7 in r7c7 (ie. eliminating the 8). One of the challenges I set myself for each puzzle (I'm sure other puzzlers set such challenges for themselves too) is to solve it in the most "economic" way, ie. using the simplest and most direct techniques possible (I admit simple and direct may be partly subjective terms). In this case I'm unable to find the route to eliminate the 8 in r7c7, so if someone could show me the alternative (and possibly more economic) route, I'd appreciate it.

Thanks
scchua

Code:

+-------------+----------+--------------+
| 9    6   2  | 7   5  8 | 4   1    3   |
| 78   3   5  | 1   4  6 | 2   79   89  |
| 4    17  18 | 3   9  2 | 78  6    5   |
+-------------+----------+--------------+
| 1    2   9  | 5   7  4 | 3   8    6   |
| 6    5   4  | 8   3  9 | 1   2    7   |
| 3    8   7  | 6   2  1 | 9   5    4   |
+-------------+----------+--------------+
| 27   4   18 | 9   6  5 | 78  3    12  |
| 278  79  6  | 24  1  3 | 5   479  289 |
| 5    19  3  | 24  8  7 | 6   49   129 |
+-------------+----------+--------------+
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
PostPosted: Wed May 30, 2007 1:06 pm    Post subject: Reply with quote
The most direct route is to note the conjugates for 7 in rows 2 and 7.

Code:
+-------------+----------+--------------+
| 9    6   2  | 7   5  8 | 4   1    3   |
| 78*  3   5  | 1   4  6 | 2   79*  89  |
| 4    17  18 | 3   9  2 | 78  6    5   |
+-------------+----------+--------------+
| 1    2   9  | 5   7  4 | 3   8    6   |
| 6    5   4  | 8   3  9 | 1   2    7   |
| 3    8   7  | 6   2  1 | 9   5    4   |
+-------------+----------+--------------+
| 27*  4   18 | 9   6  5 | 78* 3    12  |
| 278  79  6  | 24  1  3 | 5   479  289 |
| 5    19  3  | 24  8  7 | 6   49   129 |
+-------------+----------+--------------+

As the first column cannot contain two 7s, one of r2c8 or r7c7 must contain a 7. So 7 can be eliminated from r8c8, placing it in r7c7.

Steve
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Mogulmeister



Joined: 03 May 2007
Posts: 1151
PostPosted: Wed May 30, 2007 1:47 pm    Post subject: Reply with quote
Just to further add that what you have here is the conjugate pattern often known as a "skyscraper" on 7's.

Since the ends or pincers are both pointing to r3c7 and r8c8 these can be eliminated.
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csc



Joined: 03 May 2007
Posts: 14
PostPosted: Thu May 31, 2007 12:25 am    Post subject: Reply with quote
Thanks, Steve and Mogulmeister.
Yes, a definitely more elegant route than my roundabout one.
regards
scchua
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