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Puzzle 10/06/13: (C) XY

 
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Sun Jun 13, 2010 5:12 pm    Post subject: Puzzle 10/06/13: (C) XY Reply with quote

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 |
 +-----------------------+

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

PostPosted: Sun Jun 13, 2010 9:29 pm    Post subject: Reply with quote

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

PostPosted: Mon Jun 14, 2010 12:13 am    Post subject: Reply with quote

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

PostPosted: Mon Jun 14, 2010 11:26 am    Post subject: Reply with quote

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

PostPosted: Mon Jun 14, 2010 1:25 pm    Post subject: Reply with quote

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

PostPosted: Mon Jun 14, 2010 3:19 pm    Post subject: Reply with quote

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

PostPosted: Mon Jun 14, 2010 7:24 pm    Post subject: Reply with quote

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

PostPosted: Mon Jun 14, 2010 7:49 pm    Post subject: Reply with quote

JC Van Hay wrote:
A shorter one step proof :
Quote:
6-SIS : 2r7 5r7 5b3 15r1c6 1c4 16r7c9 : => r7c1<>6

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     |
 *-----------------------------------------------------------*

(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

PostPosted: Mon Jun 14, 2010 8:27 pm    Post subject: Reply with quote

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. Smile 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

PostPosted: Mon Jun 14, 2010 9:56 pm    Post subject: Reply with quote

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 "*".

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 Very Happy .
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
 ||
(1)r1c78-r2c9=(1)r7c9
 ||
(3)r5c78-(3=1)r5c5
 ||
(6)r8c78-(6=1)r7c9
 ||
(8)r8c8-(8=5)r8c2------(5)r8c5   
                   |    ||
                   |   (5-1)r7c5
                   |    ||
                   |   (5)r9c5- (5)r9c78
                   |             ||
                    ------------(5)r8c78
                                 ||
                                (5)r7c8-(5=16)r2c89-(1)r2c4=(1)r7c4

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

PostPosted: Tue Jun 15, 2010 1:18 am    Post subject: Reply with quote

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

PostPosted: Tue Jun 15, 2010 4:30 am    Post subject: Reply with quote

SIS, WIS, M-Rings, etc. notwithstanding Wink, 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

PostPosted: Tue Jun 15, 2010 9:55 am    Post subject: Reply with quote

Embarassed. 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 a
    3-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

PostPosted: Tue Jun 15, 2010 10:57 am    Post subject: Reply with quote

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

PostPosted: Tue Jun 15, 2010 2:38 pm    Post subject: Reply with quote

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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