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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Fri Sep 24, 2010 3:53 am Post subject: Puzzle 10/09/24: A |
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| Code: | +-----------------------+
| 5 . 2 | . 1 3 | . . 8 |
| . 4 . | 9 . . | 3 . . |
| 3 . . | . 6 . | 1 . 4 |
|-------+-------+-------|
| . 3 . | 6 . . | . . . |
| 2 . 8 | . 3 . | 6 . . |
| 9 . . | . . 8 | . . 1 |
|-------+-------+-------|
| . 8 5 | . 7 . | 4 . . |
| . . . | . . . | . . . |
| 1 . 3 | . . 4 | . . 6 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Fri Sep 24, 2010 9:46 am Post subject: |
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| Quote: | Finned X Wing (5)R24 : => r5c9<>5
XY Wing (-259), pivot at R5C6 : r2c5<>2
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Fri Sep 24, 2010 2:49 pm Post subject: |
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Code after basics
| Code: |
*-----------------------------------------------------------------------------*
| 5 6 2 | 4 1 3 | 79 79 8 |
| 8 4 1 | 9 25 7 | 3 6 25 |
| 3 *79 *79 | 8 6 25 | 1 25 4 |
|-------------------------+-------------------------+-------------------------|
|#47 3 #47 | 6 259 1 | 2589 2589 259 |
| 2 1 8 | 57 3 59 | 6 4 579 |
| 9 5 6 | 27 4 8 | 27 3 1 |
|-------------------------+-------------------------+-------------------------|
| 6 8 5 | 13 7 29 | 4 129 239 |
|#47 *279 *#479 | 13 2589 6 | 25789 125789 23579 |
| 1 279 3 | 25 2589 4 | 25789 25789 6 |
*-----------------------------------------------------------------------------* |
We have two partically overlapping AURs:
A Type 3 UR(79)r38c23, marked *, and a Type 1 UR(47)r48c13, marked #, which share the common cell r8c3.
UR(79)r38c23[(2)r8c2=(4)r8c3]-(4=7)r8c1-r8c9=r5c9-(7=2)r6c7-r6c4=r9c4-r9c2=(2)r8c2; r8c2=2
Type 1 UR(47)r48c13; r8c3<>47=9
xy-wing 2-59 vertex (29)r7c6; r5c4<>5
xy-wing 25-9 vertex (25)r2c9 and pseudocell (29)r24c5; r4c9,r5c6<>9
Ted |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Fri Sep 24, 2010 3:40 pm Post subject: |
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| tlanglet wrote: | Code after basics
| Code: |
*-----------------------------------------------------------------------------*
| 5 6 2 | 4 1 3 | 79 79 8 |
| 8 4 1 | 9 25 7 | 3 6 25 |
| 3 *79 *79 | 8 6 25 | 1 25 4 |
|-------------------------+-------------------------+-------------------------|
|#47 3 #47 | 6 259 1 | 2589 2589 259 |
| 2 1 8 | 57 3 59 | 6 4 579 |
| 9 5 6 | 27 4 8 | 27 3 1 |
|-------------------------+-------------------------+-------------------------|
| 6 8 5 | 13 7 29 | 4 129 239 |
|#47 *279 *#479 | 13 2589 6 | 25789 125789 23579 |
| 1 279 3 | 25 2589 4 | 25789 25789 6 |
*-----------------------------------------------------------------------------* |
We have two partically overlapping AURs:
A Type 3 UR(79)r38c23, marked *, and a Type 1 UR(47)r48c13, marked #, which share the common cell r8c3.
UR(79)r38c23[(2)r8c2=(4)r8c3]-(4=7)r8c1-r8c9=r5c9-(7=2)r6c7-r6c4=r9c4-r9c2=(2)r8c2; r8c2=2
Type 1 UR(47)r48c13; r8c3<>47=9 |
Ted, as one of my university professors once wrote on my exam ... "you got there alrighty" ... but there's a shorter way. 
First play the Type 1 for r8c3<>47. However, instead of actually doing that, record only the r8c3<>4 and remember the r8c3<>7. You now have a Type 1 UR(79)r38c23 for r8c2<>79 ... and then apply the recalled r8c3<>7.
There's a more formal way using a UR Type 1 for r8c3<>47 and then a UR Type 1.1 for r8c2<>79. For info on the latter, search for "UR1.1" on the Players' Forums. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Sep 24, 2010 5:23 pm Post subject: |
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Congratulations Ted on spotting the overlapping URs.
As Ron pointed out ... and they would tell you at the dice table if you'd rolled double 4s ... you made your point the hard way.
When I see a scenario like this, I take a slight variation on the approach Ron described.
Since the UR Type 1 is a given for its eliminations, I examine the internal constraints on the companion UR. In this case, it results in r8c2=2 and/or r8c3=4. Now, I know the UR Type 1 is going to eliminate <4> (along with <7>) from r8c3, so that forces r8c2=2 to be true for the companion UR. I perform that assignment first, and then apply the UR Type 1 eliminations.
Note #1: My approach only works in a scenario where two alternate assignments exist in the companion UR. Fortunately, that happens often enough for it to be a technique. (see below for explanation)
Note #2: My solver uses the UR Type 1.1 logic mentioned by Ron.
Regards, Danny
As my university professor once said in a Differential Equations course: "If you do it once, it's a trick. If you do it twice, it's a technique". Boy did I hate trying to remember all of the techniques he presented those two quarters. It didn't help that I caught the flu half-way through the first course and it was too late to drop it and retake it later. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Fri Sep 24, 2010 8:17 pm Post subject: |
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Ron & Danny, thanks for the insight. I have yet to study the reference to UR1.1, but I "see" the efficient step you outlined.
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Sep 25, 2010 4:15 am Post subject: |
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The URs don't need to be overlapping and can be played in sequence.
I played the 79 UR as a Hidden, making r8c2<>7
Type 1 UR (47)
Coloring (5), r5c9<>5
M-Wing (29), r2c5<>2 |
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