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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Wed Sep 08, 2010 2:30 am Post subject: Puzzle 10/09/08: C |
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| Code: | +-----------------------+
| 8 . 4 | 3 2 . | . 7 9 |
| . . 2 | 5 . . | . 3 8 |
| 5 7 . | . . . | . . . |
|-------+-------+-------|
| 1 3 . | 2 . . | . 6 . |
| 2 . . | . . . | . . 4 |
| . . . | . . . | . 9 . |
|-------+-------+-------|
| . . . | . . . | 9 . . |
| 3 5 . | 9 . 2 | . 8 6 |
| 7 4 . | . 8 . | . 2 . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site
| Rating wrote: | Extreme/XY
Extreme as single-stepper; XY if other steps applied first.
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Wed Sep 08, 2010 9:10 am Post subject: |
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| Quote: | X Wing (4)C48/r37 : r37C567<>4
M Wing (47) : (47)R8C5 7R2 4C6 : (4)r8c5=(4)r4c6 : => r4c5<>4 |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Wed Sep 08, 2010 6:21 pm Post subject: |
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I played the same wing as JC initially..
I went back to find this one-step ALS move...
| Quote: | AIC with ALS (4=9)r4c5 - ANQ(1467):(9)r3c5=(1467)r1c6|r2c56|r3c5 - (4)r3c4=r7c4 ; r8c5<>4
You could also see it as an (89)AHP but to me less clear... |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Wed Sep 08, 2010 7:03 pm Post subject: |
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This one stepper perhaps:
(4=9)r4c5-r4c6=(9-8)r3c6=(8-4)r3c4=r7c4; r8c5<>4
Worked on this before seeing Peter's - similar but without the ANQ |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Wed Sep 08, 2010 7:07 pm Post subject: |
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Peter, congratulations 
Just a comment on your AIC.
From the "hub" cell R3C6 (2 spokes from 8 & 9), one can write :SIS[(8-4)r3c4=r7c4,(9-4)r4c6=r4c5] or 4-SIS AIC : 4C4 8R3 9C6 4R4 : (4)r7c4=(4)r4c5 : => r78c5<>4. Thus a 4-SIS AIC (4 strong links and 3 weak links) with 4 strengths in location instead of a 6-SIS AIC (6 strong links and 5 weak links) with 5 cells and 1 strength in location , even though the grouping of cells leads to an apparent AIC with 3 SL and 2 WL.
Regards, JC |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Wed Sep 08, 2010 8:16 pm Post subject: |
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Prior to finding the AIC above, I was working on this puzzle looking at ALS and encountered an interesting ALS xz situation. It didn't lead to a one step move but was interesting (to me) nonetheless as personally, I had not seen this situation before: (Danny and Ronk probably have a stash of them)
We have a situation where the x and z or restricted common and common candidates can do each others job - that is to say, become interchangeable:
A={1,4,6,7,9}
B={1,6}
x = 1 or 6 (r23c5)
z = 1 or 6 (r23c5)
We can therefore eliminate both 1 and 6 from r2c6 and r3c46
| Code: | +----------------------+----------------------+----------------------+
| 8 16 4 | 3 2 B16 | 5 7 9 |
| 9 16 2 | 5 A1467 47-16 | 146 3 8 |
| 5 7 3 |48-16 A1469 489-16 | 1246 14 12 |
+----------------------+----------------------+----------------------+
| 1 3 57 | 2 A49 49 | 8 6 57 |
| 2 9 567 | 1678 13567 135678 | 137 15 4 |
| 4 8 567 | 167 13567 13567 | 1237 9 12357 |
+----------------------+----------------------+----------------------+
| 6 2 8 | 147 13457 13457 | 9 145 1357 |
| 3 5 1 | 9 A47 2 | 47 8 6 |
| 7 4 9 | 16 8 1356 | 13 2 135 |
+----------------------+----------------------+----------------------+
No cigar but fun anyway!
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Wed Sep 08, 2010 8:33 pm Post subject: |
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| Mogulmeister wrote: | We have a situation where the x and z or restricted common and common candidates can do each others job - that is to say, become interchangeable:
A={1,4,6,7,9}
B={1,6}
x = 1 or 6 (r23c5)
z = 1 or 6 (r23c5)
We can therefore eliminate both 1 and 6 from r2c6 and r3c46 |
Congratulations, you've found a doubly-linked ALS-xz, this one aka a Sue De Coq. There are four more eliminations, but I won't spoil the fun by telling you what they are.  |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Sep 08, 2010 9:24 pm Post subject: |
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It's already obvious that there are better solutions available, but I still find this interesting. Besides the X-Wing (#) used by JC, there's a conjugate Swordfish (*) network that can be embedded in a discontinuous loop:
| Code: | Swordfish
*********************************
(4)r4c5 - r8c5 = r8c7 - r2c7 = (4-7)r2c6 = r2c5 - (7=4)r8c5 - (4)r4c5
+--------------------------------------------------------------------------------+
| 8 16 4 | 3 2 16 | 5 7 9 |
| 9 16 2 | 5 *1467 *1467 | *146 3 8 |
| 5 7 3 | #1468 1469 14689 | 1246 #14 12 |
|--------------------------+--------------------------+--------------------------|
| 1 3 57 | 2 *49 *49 | 8 6 57 |
| 2 9 567 | 1678 13567 135678 | 137 15 4 |
| 4 8 567 | 167 13567 13567 | 1237 9 12357 |
|--------------------------+--------------------------+--------------------------|
| 6 2 8 | #147 13457 13457 | 9 #145 1357 |
| 3 5 1 | 9 *47 2 | *47 8 6 |
| 7 4 9 | 16 8 1356 | 13 2 135 |
+--------------------------------------------------------------------------------+
# 90 eliminations remain
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Sep 08, 2010 10:11 pm Post subject: |
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I finally took some time to work on a puzzle. I did not find anything similar to the great solutions already posted, but I did find a circumstance which I do not fully appreciate/understand.
| Code: |
*-----------------------------------------------------------------------------*
| 8 16 4 | 3 2 16 | 5 7 9 |
| 9 16 2 | 5 1467 1467 | 146 3 8 |
| 5 7 3 | 1468 1469 14689 | 1246 14 12 |
|-------------------------+-------------------------+-------------------------|
| 1 3 57 | 2 49 49 | 8 6 57 |
| 2 9 567 | 1678 13567 135678 |*137 15 4 |
| 4 8 567 | 167 13567 13567 |*1237 9 12357 |
|-------------------------+-------------------------+-------------------------|
| 6 2 8 | 147 13457 13457 | 9 145 1357 |
| 3 5 1 | 9 47 2 | 47 8 6 |
| 7 4 9 | 16 8 1356 |*13 2 135 |
*-----------------------------------------------------------------------------* |
Consider the ANT(1237)r569c7, marked *.
ANT(2=137)r569c7-(7)r8c7=(7)r7c9-(7=5)r4c9-(5=1)r5c8
What now? I don't believe that we have a contradiction on 1 since r5c8=1 does not see all three occurrences of 1 in the ls:(137)r569c7. But does this circumstance cause r9c7=1? If so, then
ANT(2=137)r569c7-(7)r8c7=(7)r7c9-(7=5)r4c9-(5=1)r5c8*-LS(137)r569c7[(1)r56c7=(1)r9c7]-(1)r23c7|r3c8*=(1-2)r3c9=(2)r6c9; r6c7<>2 to complete.
Ted |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Wed Sep 08, 2010 11:05 pm Post subject: |
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| daj95376 wrote: | It's already obvious that there are better solutions available, but I still find this interesting. Besides the X-Wing (#) used by JC, there's a conjugate Swordfish (*) network that can be embedded in a discontinuous loop:
| Code: | Swordfish
*********************************
(4)r4c5 - r8c5 = r8c7 - r2c7 = (4-7)r2c6 = r2c5 - (7=4)r8c5 - (4)r4c5
+--------------------------------------------------------------------------------+
| 8 16 4 | 3 2 16 | 5 7 9 |
| 9 16 2 | 5 *1467 *1467 | *146 3 8 |
| 5 7 3 | #1468 1469 14689 | 1246 #14 12 |
|--------------------------+--------------------------+--------------------------|
| 1 3 57 | 2 *49 *49 | 8 6 57 |
| 2 9 567 | 1678 13567 135678 | 137 15 4 |
| 4 8 567 | 167 13567 13567 | 1237 9 12357 |
|--------------------------+--------------------------+--------------------------|
| 6 2 8 | #147 13457 13457 | 9 #145 1357 |
| 3 5 1 | 9 *47 2 | *47 8 6 |
| 7 4 9 | 16 8 1356 | 13 2 135 |
+--------------------------------------------------------------------------------+
# 90 eliminations remain
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Danny, nice move, congratulations Here is an equivalent interpretation based on the Finned X-Wing on 4 in the rows 2 & 8 (same set of SIS):4-SIS t chain : [4R8 4R2] 7R2 (74)R8C5 : [X Wing (4)R82/c57]=(4-7)r2c6=r2c5-(7=8)r8c5 : (4)r2c5[z term in the t chain]=(4)r8c5 : => r347c5<>4 Regards, JC |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Thu Sep 09, 2010 1:12 am Post subject: |
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Ted, nice find too. Congratulations
The first chain you wrote eliminates (1)r6c7 (Note : the SIS 7B9 is superfluous). The second chain is a continuous network.
Let me rewrite it so as to make as clear as possible the strong and the weak inferences :
| Code: | *------------------------------------------------------------------------------*
| 8 16 4 | 3 2 16 | 5 7 9 |
| 9 16 2 | 5 1467 1467 |[1*]46 3 8 |
| 5 7 3 | 1468 1469 14689 |[1*]246 [1**]4 [1][2] |
|-------------------------+-------------------------+--------------------------|
| 1 3 57 | 2 49 49 | 8 6 [57] |
| 2 9 567 | 1678 13567 135678 |[1*37] [1**5] 4 |
| 4 8 567 | 167 13567 13567 |[1*237] 9 1[2]3-5-7|
|-------------------------+-------------------------+--------------------------|
| 6 2 8 | 147 13457 13457 | 9 -145 1357 |
| 3 5 1 | 9 47 2 | 47 8 6 |
| 7 4 9 | 16 8 1356 |[1*3] 2 135 |
*------------------------------------------------------------------------------* |
7-SIS continuous t loop or Symmetric Pigeonhole Matrix :
[(2371*)R6C7 (31*)R9C7 (31*7)R5C7] (75)R4C9 (51**)R5C8 (1*1**1)B3 2C9 loop
(2)r6c7=NT(1*37)r569c7-(7=5)r4c9-(5=1**)r5c8 (pause)-{(1*)r23c7,(1**)r3c8}=(1)r3c9-(2)r3c9=(2)r6c9 loop : => (pause r6c7<>1), r6c9<>7, r6c9<>5, r7c8<>1. There are 7 WIS : [1*c7 3c7 7c7] 7b6 5b6 1**c8 r3c9. Therefore, as in a Finless Fish (equal number of SIS and WIS), the weak links become strong.
Eliminations :(2)r6c7=(1)r5c8 : => r6c7<>1 (pause)
(7)r569c7=(7)r4c9 : => r6c9<>7
(5)r4c9=(5)r5c8 : => r6c9<>5
(1**)r5c8=(1**)r3c8 : => r7c8<>1
the last WL is already strong : => no elimination Regards, JC |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Sep 09, 2010 2:19 am Post subject: |
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| tlanglet wrote: | | Code: |
*-----------------------------------------------------------------------------*
| 8 16 4 | 3 2 16 | 5 7 9 |
| 9 16 2 | 5 1467 1467 | 146 3 8 |
| 5 7 3 | 1468 1469 14689 | 1246 14 12 |
|-------------------------+-------------------------+-------------------------|
| 1 3 57 | 2 49 49 | 8 6 57 |
| 2 9 567 | 1678 13567 135678 |*137 15 4 |
| 4 8 567 | 167 13567 13567 |*1237 9 12357 |
|-------------------------+-------------------------+-------------------------|
| 6 2 8 | 147 13457 13457 | 9 145 1357 |
| 3 5 1 | 9 47 2 | 47 8 6 |
| 7 4 9 | 16 8 1356 |*13 2 135 |
*-----------------------------------------------------------------------------* |
ANT(2=137)r569c7-(7)r8c7=(7)r7c9-(7=5)r4c9-(5=1)r5c8
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As JC mentioned, I see a conclusion of r6c7<>1.
| Quote: | ANT(2=137)r569c7-(7)r8c7=(7)r7c9-(7=5)r4c9-(5=1)r5c8*-LS(137)r569c7[(1)r56c7=(1)r9c7]-(1)r23c7|r3c8*=(1-2)r3c9=(2)r6c9; r6c7<>2 to complete.
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I don't agree with your conclusion here because your initial premise is that r6c7<>2 and so you already have the strong link (2)r6c7=(2)r6c9 without running around most of [stack 3].
It does appear that you (almost) have a continuous networked loop:
...=(2)r6c9-(2)r6c7
I'm guessing this is what JC derived. I'll let you check to see if the eliminations match those from JC's handywork. Personally, I wouldn't go near a continuous networked loop with a 20-foot pole!
Regards, Danny |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Thu Sep 09, 2010 11:28 am Post subject: |
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| JC Van Hay wrote: | | Code: | *------------------------------------------------------------------------------*
| 8 16 4 | 3 2 16 | 5 7 9 |
| 9 16 2 | 5 1467 1467 |[1*]46 3 8 |
| 5 7 3 | 1468 1469 14689 |[1*]246 [1**]4 [1][2] |
|-------------------------+-------------------------+--------------------------|
| 1 3 57 | 2 49 49 | 8 6 [57] |
| 2 9 567 | 1678 13567 135678 |[1*37] [1**5] 4 |
| 4 8 567 | 167 13567 13567 |[1*237] 9 1[2]3-5-7|
|-------------------------+-------------------------+--------------------------|
| 6 2 8 | 147 13457 13457 | 9 -145 1357 |
| 3 5 1 | 9 47 2 | 47 8 6 |
| 7 4 9 | 16 8 1356 |[1*3] 2 135 |
*------------------------------------------------------------------------------* |
7-SIS continuous t loop or Symmetric Pigeonhole Matrix :
[(2371*)R6C7 (31*)R9C7 (31*7)R5C7] (75)R4C9 (51**)R5C8 (1*1**1)B3 2C9 loop
(2)r6c7=NT(1*37)r569c7-(7=5)r4c9-(5=1**)r5c8 (pause)-{(1*)r23c7,(1**)r3c8}=(1)r3c9-(2)r3c9=(2)r6c9 loop : => (pause r6c7<>1), r6c9<>7, r6c9<>5, r7c8<>1. |
Hmm, in this case two separate moves can be more elegant than one. If not more elegant, then at least easier to understand.
ALS-xz: (1=3)[r4c9,r5c78] - (3=1)r9c7 ==> r6c7<>1
AIC: (5=7)r4c9 - (7)r7c9 = (7-4)r8c7 = (4-5)r7c8 = (5)r5c8 - loop ==> r6c9<>7, r7c8=45, r6c9<>5 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Sep 10, 2010 12:47 am Post subject: |
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Type 1 UR (16) sets up
W-Wing (47)
Multi-coloring (4) + one extension |
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