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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Fri Mar 28, 2008 10:40 pm Post subject: M3853757 SH (40) |
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Is there a non-arcane, non-Medusa solution here?
| Code: |
+-------------+-------------+------------+
| 268 3678 27 | 19 4 39 | 89 568 156 |
| 68 368 4 | 5 139 2 | 7 689 16 |
| 9 1 5 | 7 8 6 | 4 3 2 |
+-------------+-------------+------------+
| 37 4 8 | 29 23 5 | 6 1 79 |
| 5 9 1 | 6 7 4 | 3 2 8 |
| 37 2 6 | 8 13 139 | 5 4 79 |
+-------------+-------------+------------+
| 1 5 9 | 3 6 8 | 2 7 4 |
| 268 68 3 | 4 29 7 | 1 589 56 |
| 4 678 27 | 129 5 19 | 89 68 3 |
+-------------+-------------+------------+
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Play this puzzle online at the Daily Sudoku site |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Sat Mar 29, 2008 12:55 am Post subject: |
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I used an UR based step in Box 1, which pins <3> in R2C2(otherwise when R1C2=3 whe have a {68}DP in R28C12) followed by a 4-cell xy-chain{29-91-19-92} with
pincer ends at R4C4 and R8C5, eliminating both <2>'s in R4C5 and R9C4, which solves the puzzle. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 29, 2008 1:20 am Post subject: |
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| Thanks Johan. I gave up on that 68 UR because neither a 6 nor an 8 were strong links in any direction. |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Sat Mar 29, 2008 1:26 am Post subject: |
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Here’s an alternative.
| Code: | +-------------+-------------+------------+
| 268 3678 27 | 19 4 39 | 89 568 156 |
| 68 368 4 | 5 139 2 | 7 689 16 |
| 9 1 5 | 7 8 6 | 4 3 2 |
+-------------+-------------+------------+
| 37 4 8 | 29 23 5 | 6 1 79 |
| 5 9 1 | 6 7 4 | 3 2 8 |
| 37 2 6 | 8 13 139 | 5 4 79 |
+-------------+-------------+------------+
| 1 5 9 | 3 6 8 | 2 7 4 |
| 268 68 3 | 4 29 7 | 1 589 56 |
| 4 678 27 | 129 5 19 | 89 68 3 |
+-------------+-------------+------------+ |
The conjugates with respect to 9 in column7 line up with the bivalents (19) in r1c4 and r9c6 to form a W-wing: 1 is eliminated from r9c4.
| Code: | +-------------+-------------+------------+
| 268 3678 27 | 1 4 39 | 89 568 56 |
| 68 368 4 | 5 39 2 | 7 689 1 |
| 9 1 5 | 7 8 6 | 4 3 2 |
+-------------+-------------+------------+
| 37 4 8 | 29 23 5 | 6 1 79 |
| 5 9 1 | 6 7 4 | 3 2 8 |
| 37 2 6 | 8 13 39 | 5 4 79 |
+-------------+-------------+------------+
| 1 5 9 | 3 6 8 | 2 7 4 |
| 268 68 3 | 4 29 7 | 1 589 56 |
| 4 678 27 | 29 5 1 | 89 68 3 |
+-------------+-------------+------------+ |
The (56) UR in r18c89 eliminates 8 from r9c8, leaving:
| Code: | +-------------+-------------+------------+
| 28 378 27 | 1 4 39 | 89 5 6 |
| 68 368 4 | 5 39 2 | 7 89 1 |
| 9 1 5 | 7 8 6 | 4 3 2 |
+-------------+-------------+------------+
| 37 4 8 | 29 23 5 | 6 1 79 |
| 5 9 1 | 6 7 4 | 3 2 8 |
| 37 2 6 | 8 13 39 | 5 4 79 |
+-------------+-------------+------------+
| 1 5 9 | 3 6 8 | 2 7 4 |
| 268 68 3 | 4 29 7 | 1 89 5 |
| 4 78 27 | 29 5 1 | 89 6 3 |
+-------------+-------------+------------+ |
Finally the (68) UR in r28c12 eliminates 6 and 8 from r2c2, bearing in mind that the vertices on the left-hand side are conjugate with respect to 6 and r1c1 contains (28).
Steve
Last edited by Steve R on Sun Mar 30, 2008 12:33 am; edited 2 times in total |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Mar 29, 2008 3:16 am Post subject: |
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| Johan wrote: | I used an UR based step in Box 1, which pins <3> in R2C2(otherwise when R1C2=3 whe have a {68}DP in R28C12) followed by a 4-cell xy-chain{29-91-19-92} with
pincer ends at R4C4 and R8C5, eliminating both <2>'s in R4C5 and R9C4, which solves the puzzle. |
Johan, like Marty I looked at the potential {68}DP in r28c12 and gave up. However unlike Marty (who seems to understand your solution), I am not able to see the how you get a DP if you do not set r2c2 to <3>. What about the <2> in r8c1?
Details appreciated ....
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 29, 2008 3:50 am Post subject: |
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Ted,
If you have that 3 in r1c2, the newly exposed 68 pair in column 2 forces r9c2 to be 7, making r9c3 = 2 and removing the 2 from r8c1. Since that 3 forces the deadly pattern, the 3 in r2c2 is the only other 3 in that box.
Steve,
Excellent, I should have caught that W-Wing, and then I would've caught the 56 UR, but I'd be lying if I said I understood the last sentence. |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Sat Mar 29, 2008 11:01 am Post subject: |
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Marty
You are right to pick me up on my last sentence. I can only plead that it was very late at night. I have now edited the post. Thanks for the correction.
Incidentally I should be interested to see the original puzzle if you still have it. In the first UR, r18c89, one of the BUG candidates (6) is missing from one of the vertices (r8c8). I have seen this before but not in a real life puzzle. In fact I was not sure it could arise naturally.
Steve |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 29, 2008 3:43 pm Post subject: |
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Steve, I don't know what was wrong with the sentence that caused you to change it, but I didn't follow the logic and I still don't.
The original is:
| Code: |
+-------+-------+-------+
| . . . | . 4 . | . . . |
| . . 4 | 5 . 2 | 7 . . |
| 9 1 . | 7 . 6 | . 3 2 |
+-------+-------+-------+
| . 4 8 | . . . | 6 1 . |
| 5 . . | . . . | . . 8 |
| . 2 6 | . . . | 5 4 . |
+-------+-------+-------+
| 1 5 . | 3 . 8 | . 7 4 |
| . . 3 | 4 . 7 | 1 . . |
| . . . | . 5 . | . . . |
+-------+-------+-------+
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sat Mar 29, 2008 7:03 pm Post subject: |
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there is several xy-chains to use after the x-wing.
there is a xyz-wing which coincides with a locked triple in box 3.
but even after the locked triple, there is still an xy-chain to use. if you want the details, I can reconstruct the path. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Sat Mar 29, 2008 8:50 pm Post subject: |
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| Code: | *-----------------------------------------------------------*
| 268 378 27 | 1 4 39 |*89 56-8 56 |
| 68 368 4 | 5 39 2 | 7 *689 1 |
| 9 1 5 | 7 8 6 | 4 3 2 |
|-------------------+-------------------+-------------------|
| 37 4 8 | 29 23 5 | 6 1 79 |
| 5 9 1 | 6 7 4 | 3 2 8 |
| 37 2 6 | 8 1 39 | 5 4 79 |
|-------------------+-------------------+-------------------|
| 1 5 29 | 3 6 8 | 29 7 4 |
| 268 68 3 | 4 29 7 | 1 *5689 56 |
| 4 678 279 | 29 5 1 | 289 68 3 |
*-----------------------------------------------------------*
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After x-wing and the w-wing 19 there is an xyz-wing 689, which takes out the 8 in r1c8 and opens the type 1 UR 56.
| Code: | *--------------------------------------------------*
| 28 378 27 | 1 4 39 | 89 5 6 |
|#68 #368 4 | 5 39 2 | 7 89 1 |
| 9 1 5 | 7 8 6 | 4 3 2 |
|----------------+----------------+----------------|
| 37 4 8 | 29 23 5 | 6 1 79 |
| 5 9 1 | 6 7 4 | 3 2 8 |
| 37 2 6 | 8 1 39 | 5 4 79 |
|----------------+----------------+----------------|
| 1 5 -29 | 3 6 8 |*29 7 4 |
|#268 #68 3 | 4 *29 7 | 1 *89 5 |
| 4 78 279 | 29 5 1 | 289 6 3 |
*--------------------------------------------------*
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Nice observations to the UR 68 by Johan and Steve R.
Alternative here: because 6 is an x-wing, it can be pinned to r2c1 and r8c2.
Then the 8 in r2c2 can be eliminated with a kite.
In r78 the UR opens an xy-wing to solve row 7.
Since in Marty's grid it already was resolved:
It could be done a half M-wing 29 in r7c7/r8c5 and the strong links in r8 (r7c7=9 => r8c1=2) |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Sat Mar 29, 2008 11:18 pm Post subject: |
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| Nice solution from Johan. Looking at other solutions makes me feel that I'm living in a parallel universe. Same W-wing, etc., but it just doesn't seem to work out as Steve in particular mentions, nor quite the same as Ravel either. Oh well, all's well that end well I guess - nice puzzle anyway. |
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