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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sat Jun 26, 2010 4:07 am Post subject: Puzzle 10/06/25: Advanced |
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| Code: | +-----------------------+
| . . . | . 3 . | 2 1 . |
| . 1 . | . . 6 | . 4 9 |
| . . 2 | . 9 . | 6 . 7 |
|-------+-------+-------|
| . . . | 3 . . | . . . |
| 7 . 9 | . . . | . . . |
| . 3 . | . . 9 | 4 . 2 |
|-------+-------+-------|
| 8 . 5 | . . 7 | 9 . 3 |
| 9 7 . | . . . | . . . |
| . 6 4 | . . 3 | 7 . 8 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site
| Code: | after basics
+--------------------------------------------------------------------------------+
| 46 9 678 | 478 3 48 | 2 1 5 |
| 35 1 78 | 2578 258 6 | 38 4 9 |
| 345 58 2 | 1458 9 1458 | 6 38 7 |
|--------------------------+--------------------------+--------------------------|
| 2 458 168 | 3 7 1458 | 58 9 16 |
| 7 458 9 | 124568 124568 12458 | 358 38 16 |
| 56 3 168 | 1568 1568 9 | 4 7 2 |
|--------------------------+--------------------------+--------------------------|
| 8 2 5 | 14 14 7 | 9 6 3 |
| 9 7 3 | 2568 2568 258 | 1 25 4 |
| 1 6 4 | 9 25 3 | 7 25 8 |
+--------------------------------------------------------------------------------+
# 75 eliminations remain
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Jun 26, 2010 12:23 pm Post subject: |
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Skipping the obvious Type 1 UR25, a two stepper will do the deed........
| Quote: | 1. xy-wing 35-8 with vertex r2c1; r2c3,r3c8<>8
2. Flightless xy-wing -456 with vertex r6c1: (4)r5c2 - r5c45 = (4)r4c6; r1c6<>4 |
Ted |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sat Jun 26, 2010 9:01 pm Post subject: |
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Two steps with a nice gM-wing...
| Quote: | xy-wing(35-8) r2c1; r2c3<>8, r3c8<>8
m-wing(8) (8=4)r1c6 - r4c6=(4-8)r4c2=r46c3; r1c3<>8 |
I also tried to make something of all those 25 by seeing what happened if r5c4=25 to make a loop of 25s and found this contradiction close by..
| Quote: | | (4-5)r5c2=(5-8)r3c2=r1c3-(8=4)r1c6-r4c6=r5c45 - (4)r5c2; r5c2<>4 |
which also stops the 25 loop... |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Jun 26, 2010 10:12 pm Post subject: |
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| Same as Ted, except my first XY-Wing was 58-3 rather than 35-8. |
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Luke451
Joined: 20 Apr 2008
Posts: 310
Location: Southern Northern California
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| Posted: Sat Jun 26, 2010 11:31 pm Post subject: |
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| peterj wrote: | I also tried to make something of all those 25 by seeing what happened if r5c4=25 to make a loop of 25s and found this contradiction close by..
| Quote: | | (4-5)r5c2=(5-8)r3c2=r1c3-(8=4)r1c6-r4c6=r5c45 - (4)r5c2; r5c2<>4 |
which also stops the 25 loop... |
Peter, please help me out if you would.
I don't see the (25) loop and I can't follow the disc loop (other than that I'm all over it  )
Are you starting with the "after basics" or somewhere else? |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Jun 27, 2010 7:31 am Post subject: |
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| Quote: | | Are you starting with the "after basics" or somewhere else? |
Sorry, yes I am starting from after the xy-wing! And have this grid...
| Code: | *---------------------------------------------------------------------*
| 46 9 68 | 7 3 48 | 2 1 5 |
| 3 1 7 | *25 *25 6 | 8 4 9 |
| 45 58 2 | 148 9 148 | 6 3 7 |
|----------------------+-----------------------+----------------------|
| 2 48 168 | 3 7 148 | 5 9 16 |
| 7 45 9 | *25-146 146 *25 | 3 8 16 |
| 56 3 168 | 1568 168 9 | 4 7 2 |
|----------------------+-----------------------+----------------------|
| 8 2 5 | 14 14 7 | 9 6 3 |
| 9 7 3 | 68 68 *25 | 1 *25 4 |
| 1 6 4 | 9 *25 3 | 7 *25 8 |
*---------------------------------------------------------------------* |
The potential 25 loop is marked with *. I wondered what would happen if r5c4=25 to close the loop? I saw this would make a locked pair 25 in b5 and so force 4 into r5c2.
So starting from 4 in r5c2 via SL5, SL8, SL(8=4) and a grouped SL4 I thought I convinced myself that r5c45 would be 4 contradicting my assumption.
I am now wondering whether actually this loop is a DP and 5 in r5c2 is the guardian? Or maybe I can just exclude 25 from r5c4 on that basis? I am out of my depth on DPs... |
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Luke451
Joined: 20 Apr 2008
Posts: 310
Location: Southern Northern California
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| Posted: Sun Jun 27, 2010 4:41 pm Post subject: |
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| peterj wrote: | | Quote: | | Are you starting with the "after basics" or somewhere else? |
Sorry, yes I am starting from after the xy-wing! And have this grid...
| Code: | *---------------------------------------------------------------------*
| 46 9 68 | 7 3 48 | 2 1 5 |
| 3 1 7 | *25 *25 6 | 8 4 9 |
| 45 58 2 | 148 9 148 | 6 3 7 |
|----------------------+-----------------------+----------------------|
| 2 48 168 | 3 7 148 | 5 9 16 |
| 7 45 9 | *25-146 146 *25 | 3 8 16 |
| 56 3 168 | 1568 168 9 | 4 7 2 |
|----------------------+-----------------------+----------------------|
| 8 2 5 | 14 14 7 | 9 6 3 |
| 9 7 3 | 68 68 *25 | 1 *25 4 |
| 1 6 4 | 9 *25 3 | 7 *25 8 |
*---------------------------------------------------------------------* |
The potential 25 loop is marked with *. I wondered what would happen if r5c4=25 to close the loop? I saw this would make a locked pair 25 in b5 and so force 4 into r5c2.
So starting from 4 in r5c2 via SL5, SL8, SL(8=4) and a grouped SL4 I thought I convinced myself that r5c45 would be 4 contradicting my assumption.
I am now wondering whether actually this loop is a DP and 5 in r5c2 is the guardian? Or maybe I can just exclude 25 from r5c4 on that basis? I am out of my depth on DPs... |
Thanks, I should have figured that out.
The discontinuous loop works without any DP considerations. Seems to me the DP perspective is also valid because there'd be multiple solutions with (25)r5c4.
Just an observation: from what I've seen, the term "guardian" is not so much associated with multiple solution deadly patterns as they are with zero solution illegal patterns like Broken Wings and "oddagons."
This (25) loop does not have an odd number of links. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Jun 27, 2010 6:16 pm Post subject: |
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Nice find peterj !!!
I like the DP perspective and derived r5c4=146. Combine this with r5c5=146 and r5c9=16, and we have a Naked Triple => r5c2<>4. Singles are all that's necessary to complete the puzzle.
BTW: I would have marked r5c4 as "*25+146". |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Jun 27, 2010 6:24 pm Post subject: |
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| Luke451 wrote: | | Seems to me the DP perspective is also valid because there'd be multiple solutions with (25)r5c4. |
Luke, thanks.
I have looked at this some more and as you say it is a DP. Note that with the exception of b5, r5 and c4 the pairs of 25s that make the chain are the only candidates for 2 and 5 in their common houses.
In b5 and c4 there is an additional candidate 5 in r6c4 and in r5 there is an additional candidate 5 in r5c2.
One or both of these must be a 5 as otherwise a hidden pair 25 is formed and 146 can be eliminated from r5c4 completing the DP. Placing either of these 5s to avoid the DP eliminates 25 from r5c4.
Never seen a DP loop before! |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Jun 27, 2010 6:30 pm Post subject: |
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| daj95376 wrote: | | Nice find peterj !!! |
Thanks - but I only came to the DP eventually, I first just saw it as an interesting loop to explore! |
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Luke451
Joined: 20 Apr 2008
Posts: 310
Location: Southern Northern California
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| Posted: Sun Jun 27, 2010 6:52 pm Post subject: |
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Here it is, from the Sudopedia Deadly Pattern article:
| Code: | 12 . . | 21 . . | . . .
. 21 . | . . . | 12 . .
. . . | 12 . . | 21 . .
--------- ----------- ---------
21 12 . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . . |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Sun Jun 27, 2010 7:16 pm Post subject: |
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| peterj wrote: | I have looked at this some more and as you say it is a DP. Note that with the exception of b5, r5 and c4 the pairs of 25s that make the chain are the only candidates for 2 and 5 in their common houses.
In b5 and c4 there is an additional candidate 5 in r6c4 and in r5 there is an additional candidate 5 in r5c2.
One or both of these must be a 5 as otherwise a hidden pair 25 is formed and 146 can be eliminated from r5c4 completing the DP. Placing either of these 5s to avoid the DP eliminates 25 from r5c4. |
You've idenfified a uniqueness pattern known as BUG-Lite. Specifically, it's a BUG-Lite+1 ... meaning only one cell of the pattern has extra candidates. The smallest possible BUG-Lite+1 is a Type 1 UR. Just as for a Type 1 UR, you may make the exclusions r5c4<>25. |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Jun 28, 2010 11:57 am Post subject: |
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| Danny wrote: | derived r5c4=146. Combine this with r5c5=146 and r5c9=16, and we have a Naked Triple => r5c2<>4. Singles are all that's necessary to complete the puzzle.
BTW: I would have marked r5c4 as "*25+146". |
Yes I toyed with that...v nice Peter! |
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