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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Mon Mar 05, 2012 6:50 am Post subject: au 3/5/12 tough |
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Code: | *-----------*
|...|3.6|.4.|
|51.|...|...|
|7..|2..|...|
|---+---+---|
|..9|8..|.21|
|...|...|...|
|24.|..9|3..|
|---+---+---|
|...|..8|..9|
|...|...|.17|
|.5.|1.3|...|
*-----------* |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Tue Mar 06, 2012 3:44 pm Post subject: |
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I had some time this morning and spent all of it chasing this "tough" puzzle. I can only assume that others will find a simpler, cleaner solution!
Code: | *-----------------------------------------------------------------------------*
| 89 289 28 | 3 17 6 | 17 4 5 |
| 5 1 346 | 479 4789 47 | 26789 6789 2368 |
| 7 36 346 | 2 14589 145 | 1689 689 368 |
|-------------------------+-------------------------+-------------------------|
| 36 367 9 | 8 34567 457 | 467 2 1 |
| 1368 3678 135678 | 467 123467 1247 | 46789 56789 468 |
| 2 4 15678 | 67 167 9 | 3 5678 68 |
|-------------------------+-------------------------+-------------------------|
| 146 267 1267 | 4567 2467 8 | 2456 3 9 |
| 34689 23689 2368 | 4569 2469 24 | 24568 1 7 |
| 4689 5 2678 | 1 24679 3 | 2468 68 2468 |
*-----------------------------------------------------------------------------*
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anp (67=1)r6c45-(1=7)*r1c5-r79c5=r7c4-(7=6)r6c4; r45c5*<>7, r6c389<>6,b5q245<>6
almost SdC(24679)r789c5, ab =(67)r6c5, cd =(24)r8c6, e=(9)r89c5, fin=(1)r6c5
(1)r6c5-(1=7)*r1c5-als(47=9)r2c64; r2c5*<>7, r2c5,r8c4<>9
anp(15=8)r3c65-(5)r3c5=r4c5-(5=47)b5q34-(67=1)r6c45-r13c5=1r3c6; r2c7<>1
Ted |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Tue Mar 06, 2012 4:53 pm Post subject: |
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I threw in the towel. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Tue Mar 06, 2012 8:01 pm Post subject: |
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While preparing a "hint" for others to use, I ran across a solution that I hadn't expected to find.
Code: | after basics
+--------------------------------------------------------------------------------+
| 89 289 28 | 3 17 6 | 17 4 5 |
| 5 1 346 | 479 4789 47 | 26789 6789 2368 |
| 7 36 346 | 2 14589 145 | 1689 689 368 |
|--------------------------+--------------------------+--------------------------|
| 36 367 9 | 8 34567 457 | 467 2 1 |
| 1368 3678 135678 | 467 123467 1247 | 46789 56789 468 |
| 2 4 15678 | 67 167 9 | 3 5678 68 |
|--------------------------+--------------------------+--------------------------|
| 146 267 1267 | 4567 2467 8 | 2456 3 9 |
| 34689 23689 2368 | 4569 2469 24 | 24568 1 7 |
| 4689 5 2678 | 1 24679 3 | 2468 68 2468 |
+--------------------------------------------------------------------------------+
# 148 eliminations remain
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This solution is something that I noticed in the diagnostics output from my first solver. It made me think of Aligned Pair/Triple Exclusion logic. However, I haven't reviewed APE logic in some time, so I could easily be mistaken.
Code: | r23c6<>4; ( r2c6=7*, r1c5=1, r3c6=5* ); r4c6<>*57=4 => r58c6<>4
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The reason I was running my first solver was to get this pattern:
Code: | Hint: strong links including the (*) cells lead to a 4-SIS chain solution
+-----------------------------------+
| . . . | . . . | . . 5 |
| 5 . . | . . . | . . . |
| . . . | . *5 *5 | . . . |
|-----------+-----------+-----------|
| . . . | . *5 *5 | . . . |
| . . 5 | . . . | . 5 . |
| . . 5 | . . . | . 5 . |
|-----------+-----------+-----------|
| . . . | 5 . . | 5 . . |
| . . . | 5 . . | 5 . . |
| . 5 . | . . . | . . . |
+-----------------------------------+
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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Tue Mar 06, 2012 8:54 pm Post subject: |
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Code: | *-----------------------------------------------------------------------------*
| 89 289 28 | 3 c17 6 | 17 4 5 |
| 5 1 346 | 479 4789 b47 | 26789 6789 2368 |
| 7 36 346 | 2 14589 d145 | 1689 689 368 |
|-------------------------+-------------------------+-------------------------|
| 36 367 9 | 8 34567 457 | 467 2 1 |
| 1368 3678 135678 | 467 123467 1247 | 46789 56789 468 |
| 2 4 15678 | 67 167 9 | 3 5678 68 |
|-------------------------+-------------------------+-------------------------|
| 146 267 1267 | 4567 2467 8 | 2456 3 9 |
| 34689 23689 2368 | 4569 2469 a24 | 24568 1 7 |
| 4689 5 2678 | 1 24679 3 | 2468 68 2468 |
*-----------------------------------------------------------------------------*
(2=4)r8c6-(4=7)r2c6-(7=1)r1c5-(1)r3c6=(1)r5c6 =>r5c6<>2 stte |
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Tue Mar 06, 2012 8:55 pm Post subject: |
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daj95376 wrote: | Code: | r23c6<>4; ( r2c6=7*, r1c5=1, r3c6=5* ); r4c6<>*57=4 => r58c6<>4
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IOW : r1c5=1 or r1c5=7 => 4 in r234c6 => -4r58c6; stte
or, in Eureka notation :
(4=157)r234c6-(1=74)r1c5.r2c6 => -4r58c6; stte
PS : sotir's 4-SIS solution is based on the XW(5C56) |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Wed Mar 07, 2012 12:51 am Post subject: |
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JC, my notation was a (feeble) attempt to reconstruct my recollection of APE logic. I do agree that your (implied) forcing network on r1c5 is simpler.
Code: | (1)r1c5 - (1=457)r234c6 => r58c6<>4
(7)r1c5 - (7=4 )r2 c6 => r58c6<>4
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ronk
Joined: 07 May 2006 Posts: 398
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Posted: Wed Mar 07, 2012 1:32 am Post subject: |
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daj95376 wrote: | This solution is something that I noticed in the diagnostics output from my first solver. It made me think of Aligned Pair/Triple Exclusion logic. However, I haven't reviewed APE logic in some time, so I could easily be mistaken.
Code: | r23c6<>4; ( r2c6=7*, r1c5=1, r3c6=5* ); r4c6<>*57=4 => r58c6<>4
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You are correct.
Sudoku Explainer wrote: | Aligned Triplet Exclusion
The cells R1C5, R4C6 and R8C6 can together accept various combinations of values. But some combinations of values can be excluded, because they would leave some cells with no possible values.
More precisely, the following combinations of values are not possible for the cells R1C5, R4C6 and R8C6:
7, 7 and 4 because the cell R2C6 must already contain 4 or 7
1, 7 and 4 because the cell R2C6 must already contain 4 or 7
7, 5 and 4 because the cell R2C6 must already contain 4 or 7
1, 5 and 4 because the cell R3C6 must already contain 1, 4 or 5
7, 4 and 4 because the same value cannot occur twice in the same row, column or block
1, 4 and 4 because the same value cannot occur twice in the same row, column or block
Because some potential values of R8C6 occur in none of the remaining combinations, they can safely be removed. |
The subsuming als-xz is certainly easier to understand. |
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