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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sat Jun 05, 2010 9:22 pm Post subject: Puzzle 10/06/05: (C) Advanced |
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| Code: | +-----------------------+
| 6 . 4 | 3 . 7 | . . . |
| . 5 . | . . . | . . . |
| 3 . 2 | 4 . . | 1 . . |
|-------+-------+-------|
| 1 . 8 | . . 3 | . . 6 |
| . . . | . 1 . | . 9 . |
| 9 . . | 6 . . | 2 . . |
|-------+-------+-------|
| . . 1 | . . 4 | . 2 3 |
| . . . | . 3 . | 9 1 . |
| . . . | 2 . . | 7 . 4 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site
===== ===== ===== ===== ===== ===== ===== =====
After basics and ignoring the finned X-Wing:
| Code: | +--------------------------------------------------------------+
| 6 1 4 | 3 2 7 | *58 *58 9 |
| 78 5 79 | 1 68 689 | 34 34 2 |
| 3 89 2 | 4 58 589 | 1 6 7 |
|--------------------+--------------------+--------------------|
| 1 2 8 | 579 4579 3 | 45 47 6 |
| 4 367 3567 | 57 1 2 | *58+3 9 *58 |
| 9 37 357 | 6 4578 58 | 2 347 1 |
|--------------------+--------------------+--------------------|
| *58+7 789 1 | *58+79 579 4 | 6 2 3 |
| 2 4 67 | *58+7 3 56 | 9 1 *58 |
| *58 369 369 | 2 69 1 | 7 *58 4 |
+--------------------------------------------------------------+
# 59 eliminations remain
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Play this puzzle online at the Daily Sudoku site
One option is to use the <58> DP (*) to crack it. ___  ___ |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sat Jun 05, 2010 10:01 pm Post subject: |
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Sadly I am still at DP bootcamp .. but fwiw this does it in one step
| Quote: | | xy-chain (7=8)r2c1-(8=5)r9c1-(5=8)r9c8-(8=5)r8c9-(5=6)r8c6-(6=7)r8c3; r2c3<>7, r7c1<>7 |
Danny, look forward to seeing this play out as I need to learn this stuff! |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sat Jun 05, 2010 10:15 pm Post subject: |
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I'd love it if you could talk us through the DP diagram Danny. Like Peter, I'm on the floor giving the drill sergeant 20.
A cowardly (but different) single step xy chain is all I can offer.
| Quote: | Pincers at r2c3 and r9c5 eliminate the 9 at r9c3 thus:
(9=7)r2c3-(7=8)r2c1-(8=6)r2c5-(6=9)r9c5; r9c3 <>9 |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Jun 05, 2010 11:05 pm Post subject: |
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I was not serious about the <58> DP. That's why the "laughing" emoticon!!! I just wanted everyone to know that I found the DP.
Okay, that said, I feel that I must justify myself. It shouldn't be too difficult. _
| Code: | <58> DP (*) => r5c7=3 or r7c1|r7c4|r8c4=7 or r7c4=9
(3)r5c7 - (3)r2c7
(7)r7c1 - r7c45 = r8c4 - (7=5)r5c4 - (58=3)r5c78 - (3)r2c7
(7)r7c4 - (7=5)r5c4 - (58=3)r5c78 - (3)r2c7
(7)r8c4 - (7=5)r5c4 - (58=3)r5c78 - (3)r2c7
--------------------------------------
/ \
(9)r7c4 - (9=6)r9c5 - (6=8)r2c5 - r2c1 = (8)r3c2 - (89=7)r7c2 - (7=3)r6c2 - r6c8 = r5c7 - (3)r2c7
_________________________________________________________________________________________________
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Now, for my intended UR solution.
| Code: | r47c45 <79> UR via s-link <> 7 r4c4 (extreme UR)
r46c58 <47> UR via s-link <> 7 r4c5
r46c58 <47> UR via s-link <> 7 r6c5
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And the puzzle cracks with r5c4=7.
There's also an M-Wing that cracks the puzzle ... if you want to find it. |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sun Jun 06, 2010 7:52 am Post subject: |
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| There's also a nest of <58> classic remote pairs that threaten to make useful elimination(s) but .......... |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Jun 06, 2010 8:36 am Post subject: |
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Danny, I'd never spot that first UR in a month of Sunday! I didn't even know you could make eliminations from a AUR where one of the corners wasn't a bivalue.
Looking at it I still don't really see how you eliminate 7 from r4c4? The strong links on 9 force the opposite corners to be 9 but without a bivalue I dont see what is constraining 7 to take part at all?
Enlightenment or a link to said would be great! Thanks |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Jun 06, 2010 1:30 pm Post subject: |
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| Code: | r47c45 <79> UR via s-link <> 7 r4c4
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Hello Peter. Here are candidate grids for <7> and <9>. As you mentioned, the easy part of the explanation is for <9>.
| Code: | r4c4=7 => r4c5,r7c4=9
+-----------------------------------+
| . . . | . . . | . . 9 |
| . . 9 | . . 9 | . . . |
| . 9 . | . . 9 | . . . |
|-----------+-----------+-----------|
| . . . | *9 *9 . | . . . |
| . . . | . . . | . 9 . |
| 9 . . | . . . | . . . |
|-----------+-----------+-----------|
| . 9 . | *9 *9 . | . . . |
| . . . | . . . | 9 . . |
| . 9 9 | . 9 . | . . . |
+-----------------------------------+
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It's only slightly more tricky to explain <7>.
| Code: | r4c4=7 => ( r46c5 -or- r78c4 )<>7 => r7c5=7
+-----------------------------------+
| . . . | . . 7 | . . . |
| 7 . 7 | . . . | . . . |
| . . . | . . . | . . 7 |
|-----------+-----------+-----------|
| . . . | *7 *7 . | . 7 . |
| . 7 7 | 7 . . | . . . |
| . 7 . | . @7 . | . 7 . |
|-----------+-----------+-----------|
| 7 7 . | *7 *7 . | . . . |
| . . 7 | @7 . . | . . . |
| . . . | . . . | 7 . . |
+-----------------------------------+
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If you enter the original grid into Simple Sudoku, set it to allow invalid assignments, highlight <7>, and force r4c4=7, then the forced assignment jumps out for <7>. At least it does for me. Subsequently highlighting <9> makes the forced assignments there obvious as well.
Regards, Danny |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Jun 06, 2010 2:19 pm Post subject: |
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Danny, that is truly a wonderful ADP. I have never seen a 10-cell configuration before. And your solution was eloquent.
For fun, I tried looking at the box implication outside the cells of the ADP and found the following.
| Code: | To prevent the DP, r4c7|r7c5|r8c6=5 or r7c2=8
(5)r4c7 - (5=8)r5c9
(5)r7c5 - r7c1 = r9c1 - (5=8)r9c8 - r8c9 = (8)r5c9
(5)r8c6 - (58=7)r8c49 - (7=5+r5c4 - (5=8)r5c9
(8)r7c2 - r9c1 = r9c8 - r8c9 = (8)r5c9 |
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Jun 06, 2010 2:38 pm Post subject: |
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Nice observation Ted. It definitely makes for a shorter and simpler handling of the DP. Too bad the DP is the worst choice (overall) for solving the puzzle.
I'm still hoping that someone will track down the M-Wing ... just to show that I hadn't planned for the solution to be so complicated. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Jun 06, 2010 6:47 pm Post subject: |
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Danny, thanks for explaining the UR - I get it, but it's not yet my natural territory!
Is this your M-Wing? Nice!
m-wing(69) (9=6)r9c5-r2c5=(6-9)r2c6=r2c3; r9c3<>9 |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Jun 06, 2010 7:05 pm Post subject: |
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| peterj wrote: | Is this your M-Wing? Nice!
m-wing(69) (9=6)r9c5-r2c5=(6-9)r2c6=r2c3; r9c3<>9
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Yes, nice catch! |
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wapati
Joined: 10 Jun 2008
Posts: 472
Location: Brampton, Ontario, Canada.
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| Posted: Sun Jun 06, 2010 11:53 pm Post subject: |
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I saw a finned sword on 7s after basics.
I still needed a 4 cell xy-chain to finish.  |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jun 07, 2010 4:48 am Post subject: |
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I made a couple of eliminations from the Hidden UR (58) in boxes 25, but don't know if that impacted the next move.
| Code: | +--------------------------------------------------------------+
| 6 1 4 | 3 2 7 | *58 *58 9 |
| 78 5 79 | 1 68 689 | 34 34 2 |
| 3 89 2 | 4 58 589 | 1 6 7 |
|--------------------+--------------------+--------------------|
| 1 2 8 | 579 4579 3 | 45 47 6 |
| 4 367 3567 | 57 1 2 | *58+3 9 *58 |
| 9 37 357 | 6 4578 58 | 2 347 1 |
|--------------------+--------------------+--------------------|
| *58+7 789 1 | *58+79 579 4 | 6 2 3 |
| 2 4 67 | *58+7 3 56 | 9 1 *58 |
| *58 369 369 | 2 69 1 | 7 *58 4 |
+--------------------------------------------------------------+ |
Looking at the potential 36 UR in r5c23 and r9c23, the killers in r5 form a 57 pseudo cell which forms a naked pair with r4c4. One of r9c23 must be=9. Either way, r5c79 must be=38, finishing the puzzle. |
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