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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sun Jun 13, 2010 5:12 pm Post subject: Puzzle 10/06/13: (C) XY |
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Code: | +-----------------------+
| 4 . . | . 8 . | . . . |
| . 7 9 | . 4 . | 2 . . |
| . 2 1 | . . 7 | 8 . 3 |
|-------+-------+-------|
| . . . | 9 6 . | . . . |
| 9 6 . | 5 . 2 | . . . |
| . . 2 | . 7 4 | . . 5 |
|-------+-------+-------|
| . 9 7 | . . . | 4 . . |
| . . . | . . . | . . . |
| . . 4 | . . 6 | . . 2 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site
Code: | after basics
+--------------------------------------------------------------+
| 4 3 6 | 2 8 15 | 157 157 9 |
| 8 7 9 | 13 4 135 | 2 156 16 |
| 5 2 1 | 6 9 7 | 8 4 3 |
|--------------------+--------------------+--------------------|
| 7 4 5 | 9 6 13 | 13 2 8 |
| 9 6 8 | 5 13 2 | 137 137 4 |
| 3 1 2 | 8 7 4 | 69 69 5 |
|--------------------+--------------------+--------------------|
| 26 9 7 | 13 1235 8 | 4 1356 16 |
| 26 58 3 | 4 125 9 | 156 1568 7 |
| 1 58 4 | 7 35 6 | 359 3589 2 |
+--------------------------------------------------------------+
# 45 eliminations remain
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sun Jun 13, 2010 9:29 pm Post subject: |
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This was a very interesting puzzle with lots of xy action!
This was my natural solve path for 3 steps...
Quote: | x-wing(1) r27c49; r2c68<>1, r7c58<>1
kite(1) in b5; r8c7<>1
w-wing(56) (5=6)r2c8 - r2c9=r7c9 - (6=5)r8c7; r1c7<>5, r789c8<>5 |
After step 2 there was also this lovely xy-cycle which makes 5 eliminations over 3 digits (but does not crack it!)..
Quote: | (5=8)r8c2 - (8=5)r9c2 - (5=3)r9c5 - (3=1)r7c4 - (1=6)r7c9 - (6=5)r8c7 - (5)r8c2; r9c78<>5, r7c5<>3, r78c8<>6 |
Here's a one-stepper..
Quote: | AIC (3=1)r2c4 - (1=3)r7c4 -(3=5)r9c5 - (5)r7c5=r7c8 - (5=16)r2c89 - (1=57)r1c78 -(5=1)r1c6 - (1=3)r4c6; r2c6<>3 |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Mon Jun 14, 2010 12:13 am Post subject: |
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I am not a xy-chain type of person. I only find them when looking to extend a potential xy-wing such as occurred in my last step for this puzzle.
My first step is something of a mess. I hope the post is understandable since I am not aware of proper notation. Finally the deletion provide by this step is not effective, and would have been accomplished by a following step in any case, but it was mountain that had to be climbed.
Code: | AMUG157 r158c78. To prevent the deadly pattern:
(1)r1c78 - r2c9 = r7c9
(3)r5c78 - (3=1)r5c5
(6)r8c78 - (6=1)r7c9
(8)r8c8 - (8=5)r8c2 - r8c578* = (5)r79c5
If r7c5=5: (5)r7c5 - (5=3)r9c5 - (3=1)r5c5
If r9c5=5: (5)r9c5 - r9c78 = r7c8* - (5=16)r2c89 - (1)r1c78 = r1c6 - r4c6 = r5c5
Thus these pincers force r7c5<>1
Note: the first "*" is intended to indicate a deletion that is utilized at the second "*". |
Quote: | AUR58 r89c28 with x-wing 8 overlay; r89c8<>5
x-wing 1 r27c49; r2c68,r7c8<>1
(5=3)r2c6 - r2c4 = r7c4 - (3=5)r9c5 - r7c5 = (5)r7c8; r2c8<>5 | .
Actually, only the last two steps are need to complete the puzzle.
Ted |
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Mon Jun 14, 2010 11:26 am Post subject: |
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A shorter one step proof :
Quote: | 6-SIS : 2r7 5r7 5b3 15r1c6 1c4 16r7c9 : => r7c1<>6 |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Mon Jun 14, 2010 1:25 pm Post subject: |
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JC - I have not been initiated into the world of SIS and appropriate notation! Can you point me somewhere where I can learn?
Peter |
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Mon Jun 14, 2010 3:19 pm Post subject: |
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Peter,
SIS = Strong (Inference) Set = set in which at least one element is true.The candidates in a cell, the candidates for a single digit in a unit are said to be native SIS while the guardians in a Broken Wings pattern, the endpoints of an AIC chain ... are said to be derived SIS WIS = Weak (Inference) set = set in which at most one element is true.The candidates in a cell, the candidates for a single digit in a unit, or a subset of these are native WIS; derived WIS are used in full tagging, for example. Some references :If further information is wanted, don't hesitate to ask.
JC |
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Mon Jun 14, 2010 7:24 pm Post subject: |
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Can you put it in pictures while we grapple with the notation JC ? |
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Luke451
Joined: 20 Apr 2008 Posts: 310 Location: Southern Northern California
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Posted: Mon Jun 14, 2010 7:49 pm Post subject: |
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JC Van Hay wrote: | A shorter one step proof :
Quote: | 6-SIS : 2r7 5r7 5b3 15r1c6 1c4 16r7c9 : => r7c1<>6 |
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After a bit of grappling, I'd say it's no coincidence that the SIS components form an AIC:
Code: | *-----------------------------------------------------------*
| 4 3 6 | 2 8 15 | 157 157 9 |
| 8 7 9 | 13 4 135 | 2 156 16 |
| 5 2 1 | 6 9 7 | 8 4 3 |
|-------------------+-------------------+-------------------|
| 7 4 5 | 9 6 13 | 13 2 8 |
| 9 6 8 | 5 13 2 | 137 137 4 |
| 3 1 2 | 8 7 4 | 69 69 5 |
|-------------------+-------------------+-------------------|
| 26 9 7 | 13 1235 8 | 4 1356 16 |
| 26 58 3 | 4 125 9 | 156 1568 7 |
| 1 58 4 | 7 35 6 | 359 3589 2 |
*-----------------------------------------------------------*
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(2)r7c1=(2-5)r7c5=r7c8-r12c8=r1c7-(5=1)r1c6-r2c4=r7c4-(1=6)r7c9
=>r7c1<>6 |
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ronk
Joined: 07 May 2006 Posts: 398
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Posted: Mon Jun 14, 2010 8:27 pm Post subject: |
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Luke451 wrote: | After a bit of grappling, I'd say it's no coincidence that the SIS components form an AIC:
...
(2)r7c1=(2-5)r7c5=r7c8-r2c8=r1c7-(5=1)r1c6-r2c4=r7c4-(1=6)r7c9
=>r7c1< >6 |
Definitely not coincidence. When a SIS component has three or more elements of its own though, writing an AIC becomes a bit tougher.
Last edited by ronk on Tue Jun 15, 2010 11:56 am; edited 1 time in total |
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Luke451
Joined: 20 Apr 2008 Posts: 310 Location: Southern Northern California
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Posted: Mon Jun 14, 2010 9:56 pm Post subject: |
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tlanglet wrote: |
Code: | AMUG157 r158c78. To prevent the deadly pattern:
(1)r1c78 - r2c9 = r7c9
(3)r5c78 - (3=1)r5c5
(6)r8c78 - (6=1)r7c9
(8)r8c8 - (8=5)r8c2 - r8c578* = (5)r79c5
If r7c5=5: (5)r7c5 - (5=3)r9c5 - (3=1)r5c5
If r9c5=5: (5)r9c5 - r9c78 = r7c8* - (5=16)r2c89 - (1)r1c78 = r1c6 - r4c6 = r5c5
Thus these pincers force r7c5<>1
Note: the first "*" is intended to indicate a deletion that is utilized at the second "*". |
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WOW! He also wrote: | ...I am not aware of proper notation... but it was a mountain that had to be climbed. |
You must be exhausted, Sir Tedmund Hillary .
BUT....
Since you went through all that trouble for one unnecessary elim, and since I'm on vacation...
...I'll be your Sherpa.
Here's a stab at the notation for your net (no correctness guarantee.)
Code: | MUG157r158c78 =>r7c5<>1
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(1)r1c78-r2c9=(1)r7c9
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(3)r5c78-(3=1)r5c5
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(6)r8c78-(6=1)r7c9
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(8)r8c8-(8=5)r8c2------(5)r8c5
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| (5-1)r7c5
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| (5)r9c5- (5)r9c78
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------------(5)r8c78
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(5)r7c8-(5=16)r2c89-(1)r2c4=(1)r7c4
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Code: | +--------------------------------------------------------------+
| 4 3 6 | 2 8 15 | 157 157 9 |
| 8 7 9 | 13 4 135 | 2 156 16 |
| 5 2 1 | 6 9 7 | 8 4 3 |
|--------------------+--------------------+--------------------|
| 7 4 5 | 9 6 13 | 13 2 8 |
| 9 6 8 | 5 13 2 | 137 137 4 |
| 3 1 2 | 8 7 4 | 69 69 5 |
|--------------------+--------------------+--------------------|
| 26 9 7 | 13 1235 8 | 4 1356 16 |
| 26 58 3 | 4 125 9 | 156 1568 7 |
| 1 58 4 | 7 35 6 | 359 3589 2 |
+--------------------------------------------------------------+ |
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ronk
Joined: 07 May 2006 Posts: 398
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Posted: Tue Jun 15, 2010 1:18 am Post subject: |
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An m-ring between 2 wings:
Quote: | x-wing: (1)c49\r27; r2c68,r7c58<>1
m-ring: (5=3)r9c5 - (3)r7c45 = (3-5)r7c8 = (5)r7c5 - loop; r8c5<>5, r7c8<>6
w-wing: (5=3)r7c8 - (3)r7c4 = (3)r2c4 - (3=5)r2c6; r2c8<>5 |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Tue Jun 15, 2010 4:30 am Post subject: |
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SIS, WIS, M-Rings, etc. notwithstanding , either way of breaking up the 17 DP in boxes 36 results in r5c5=1, which solves the puzzle. |
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Tue Jun 15, 2010 9:55 am Post subject: |
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. Forgive me for the grappling. I thought I were using well-known notations.
Of course, here, as noted by Luke451 and Ronk, the ordered list of strong sets is a concise way to represent an AIC chain. It is exactly similar to a path in a B/B plot. By application of the Sudoku rules, the AIC chain is easily reconstructed. In case of an AIC net, the reconstruction is much more involved and an ad-hoc diagram is needed as in the 9-SIS MUG above. For further examples of notations and diagrams, see Allan Barker's website at http://sudokuone.com/.
As an application, Marty's move may be written as a3-SIS : 1b5 5r1 (5)r1c78&(3)r5c78 : => -(3)r5c5
[i.e. all the 1 in box b5, all the 5 in row 1, the digits 5 and 3 in the cells r15c78 preventing the DP] or
(1)r5c5=r4c6-r1c6=(5)r1c6-r1c78=(3)r5c78 : => r5c5<>3, r5c5=1. Here the reconstruction needed the applications of the Sudoku rules 4 times and the uniqueness rule once.
JC |
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Tue Jun 15, 2010 10:57 am Post subject: |
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Thanks JC and thanks Luke - for me certainly a new way of looking at sudoku. Will go to that Alan Barker page you recommended JC. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Tue Jun 15, 2010 2:38 pm Post subject: |
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I migrated from NL notation to AIC notation because I found it easier to understand what was happening in a chain. Now, with the use of SIS notation, I see that a chain can be reduced to its fundamental strong links and still be understandable. It is certainly a more compact notation than either of the two former notations.
What my solver found:
Code: | X-Wing <1>, Kite <1>, <58> UR Type 4, and W-Wing
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