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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Mon Jul 06, 2009 2:02 am Post subject: July 6 VH |
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A multiple choice today. The old reliable or alternate routes.
Solutions: (158) xy-wing, (49) UR, or an xy-chain can solve it.
Early Earl |
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crunched
Joined: 05 Feb 2008
Posts: 168
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| Posted: Mon Jul 06, 2009 4:38 am Post subject: Re: July 6 VH |
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Wow, all kinds of ways to git er dun? Impressive! I did find the 158-xy wing, but never saw a UR, or other means to solve. Below is the grid, after basics opened up the xy wing that solved it for me. I won't claim I got all the basics; but enough to "wing it".
Now I see the UR, since Earl-y mentioning it.
| Earl wrote: | A multiple choice today. The old reliable or alternate routes.
Solutions: (158) xy-wing, (49) UR, or an xy-chain can solve it.
Early Earl |
| Code: |
+-----------+--------------+------------+
| 3 7 2 | 5 69 69 | 8 4 1 |
| 1 45 45 | 8 7 3 | 6 9 2 |
| 8 6 9 | 1 4 2 | 7 5 3 |
+-----------+--------------+------------+
| 7 15 358 | 69 18 4 | 1359 2 68 |
| 2 14 348 | 7 5 1689 | 139 13 68 |
| 6 9 58 | 3 2 18 | 15 7 4 |
+-----------+--------------+------------+
| 49 2 16 | 469 369 5 | 13 8 7 |
| 5 8 16 | 2 36 7 | 4 13 9 |
| 49 3 7 | 49 18 18 | 2 6 5 |
+-----------+--------------+------------+
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Play this puzzle online at the Daily Sudoku site |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jul 06, 2009 4:58 am Post subject: |
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| In that grid, the 18 pair in column 6 makes some eliminations. With that out of the way, a W-Wing on 15 becomes another possibility. |
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crunched
Joined: 05 Feb 2008
Posts: 168
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| Posted: Mon Jul 06, 2009 12:46 pm Post subject: |
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| Marty R. wrote: | | In that grid, the 18 pair in column 6 makes some eliminations. With that out of the way, a W-Wing on 15 becomes another possibility. |
You are right about the additional eliminations in column six. But I don't see how that helps with a w-wing on 15. I confess that w-wings are not easy for me to spot---or to even comprehend. Below is my thought on a w-wing on 15, using an xw to mark the 15s, using an x to mark the "chain" between the 15s, and using z to mark where a 5 gets eliminated, see below:
| Code: |
+-----------+--------------+------------+
| 3 7 2 | 5 69 69 | 8 4 1 |
| 1 45 45 | 8 7 3 | 6 9 2 |
| 8 6 9 | 1 4 2 | 7 5 3 |
+-----------+--------------+------------+
| 7 15xw 358 |69 18 4 | 1359 2 68|
| 2 14x 348 |7 5 1689 | 139 13x 68|
| 6 9 58z |3 2 18 | 15xw 7 4 |
+-----------+--------------+------------+
| 49 2 16 | 469 369 5 | 13 8 7 |
| 5 8 16 | 2 36 7 | 4 13 9 |
| 49 3 7 | 49 18 18 | 2 6 5 |
+-----------+--------------+------------+
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Play this puzzle online at the Daily Sudoku site |
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Roger S
Joined: 20 Jun 2009
Posts: 6
Location: UK
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| Posted: Mon Jul 06, 2009 5:11 pm Post subject: |
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I got as far as the above and then did a UR 49 in r7c1,r9c1 r7c4 r9c4. This means r7c4 must be a 6 the rest then dropped out is this valid
Thanks
Roger |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jul 06, 2009 7:55 pm Post subject: |
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| Quote: | | You are right about the additional eliminations in column six. But I don't see how that helps with a w-wing on 15. I confess that w-wings are not easy for me to spot---or to even comprehend |
Eliminating the 1 from r5c6 allows the strong link on the 1s for the W-Wing. With the 1 still there, there's no wing.
| Quote: | | I got as far as the above and then did a UR 49 in r7c1,r9c1 r7c4 r9c4. This means r7c4 must be a 6 the rest then dropped out is this valid |
Hi Roger,
This is valid. It's an example of what's known as a Type 1 UR. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Jul 09, 2009 12:12 am Post subject: |
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| Code: | .------------------.------------------.------------------.
| 3 7 2 | 5 69 69 | 8 4 1 |
| 1 45 45 | 8 7 3 | 6 9 2 |
| 8 6 9 | 1 4 2 | 7 5 3 |
:------------------+------------------+------------------:
| 7 M15 358 | 69 -18 4 | 1359 2 68 |
| 2 14 348 | 7 5 69 | 139 13 68 |
| 6 9 M58 | 3 2 M18 |M15 7 4 |
:------------------+------------------+------------------:
| 49 2 16 | 469 369 5 | 13 8 7 |
| 5 8 16 | 2 36 7 | 4 13 9 |
| 49 3 7 | 49 18 18 | 2 6 5 |
'------------------'------------------'------------------' |
since no one has pointed it out yet
the marked M-wing {1,5} solves it also.
(1=5)r4c2 - (5)r6c3 = (5-1)r6c7 = (1)r6c6; r4c5 <> 1 |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Thu Jul 09, 2009 12:52 am Post subject: |
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| storm_norm wrote: | | Code: | .------------------.------------------.------------------.
| 3 7 2 | 5 69 69 | 8 4 1 |
| 1 45 45 | 8 7 3 | 6 9 2 |
| 8 6 9 | 1 4 2 | 7 5 3 |
:------------------+------------------+------------------:
| 7 M15 358 | 69 -18 4 | 1359 2 68 |
| 2 14 348 | 7 5 69 | 139 13 68 |
| 6 9 M58 | 3 2 M18 |M15 7 4 |
:------------------+------------------+------------------:
| 49 2 16 | 469 369 5 | 13 8 7 |
| 5 8 16 | 2 36 7 | 4 13 9 |
| 49 3 7 | 49 18 18 | 2 6 5 |
'------------------'------------------'------------------' |
since no one has pointed it out yet
the marked M-wing {1,5} solves it also.
(1=5)r4c2 - (5)r6c3 = (5-1)r6c7 = (1)r6c6; r4c5 <> 1 |
Norm,
I don't see it. I see an XY-wing -158 in R4C2 R6C36. I don't see that R6C7 is needed, or that it makes an M-wing.
In my words,
a) R4C2 is <1>, or:
b) R4C2 is <5>; R6C3 is <8>; R6C6 is <1>;
=> R4C5<>1.
Sure, in b) you could stretch it to R6C7 is <5>, but that does not help, since it is already established R6C6 is <1>.
Keith |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Jul 09, 2009 4:27 am Post subject: |
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Keith,
you are right that the M-wing isn't needed and that most would see the otherwise obvious 158 xy-wing.
but the M-wing is there nonetheless.
(x=y) - y = (y-x) = x
(1=5) - 5 = (5-1) = 1
which is what we have in the marked cells.
we have the {1,5}r4c2 connected to the {1,5}r6c7 via the strong links on 5 in row 6, and the pincer 1 strongly linked to the 1 in r6c7, completing the pincer with the 1 in r4c2, thus eliminating the 1 in r4c5. |
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