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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Thu Jun 10, 2010 11:36 am Post subject: Puzzle 10/06/10: (A) |
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Code: | +-----------------------+
| . . . | . 9 . | . . 2 |
| . 6 9 | 2 . 3 | 4 . 1 |
| . 1 . | 4 . . | . 5 . |
|-------+-------+-------|
| . 8 1 | 7 2 . | . . . |
| 2 . . | 1 . 8 | . . . |
| . 5 . | . 3 9 | . . . |
|-------+-------+-------|
| . 4 . | . . . | 2 . . |
| . . 7 | . . . | . . 4 |
| 9 2 . | . . . | . 8 . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site
Code: | after basics
+--------------------------------------------------------------+
| 4 7 358 | 58 9 1 | 368 36 2 |
| 58 6 9 | 2 58 3 | 4 7 1 |
| 38 1 2 | 4 67 67 | 389 5 39 |
|--------------------+--------------------+--------------------|
| 36 8 1 | 7 2 5 | 369 4 369 |
| 2 9 36 | 1 4 8 | 57 36 57 |
| 7 5 4 | 6 3 9 | 1 2 8 |
|--------------------+--------------------+--------------------|
| 1568 4 568 | 3589 5678 67 | 2 19 356 |
| 1568 3 7 | 589 568 2 | 56 19 4 |
| 9 2 56 | 35 1 4 | 3567 8 3567 |
+--------------------------------------------------------------+
# 55 eliminations remain
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Thu Jun 10, 2010 4:04 pm Post subject: |
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Home from work two-step...
Quote: | type 1 UR(67); r7c5<>67
kite(8) r2c5=r2c1-r1c3=r7c3; r7c5<>8 |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Thu Jun 10, 2010 5:05 pm Post subject: |
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Another variation on the two stepper............
Quote: | Type 1 UR67 r37c56 r7c5<>67
ARP58 r2c1,r2c5,r7c5 & r7c3 plus fin (6)r7c3 - r7c9 = r4c9 - (6=3)r4c1 - (3=8)r3c1; r78c1<>8 completes the puzzle.
I initially found this step as an "Almost Remote Pair" (APR). However in transporting the fin I realized the fin could be removed using a conflicting chain: (6)r7c3 - r7c9 = r4c9 - r4c1 = (6)r5c3; r7c3<>6 which also completes the puzzle. |
Ted |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Jun 11, 2010 1:10 am Post subject: |
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I didn't see the 67 UR at the start. The implications of the 39 UR solved a few cells. Then I played the 67 UR, coloring on 5 and XY-Wing on 685. |
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Luke451
Joined: 20 Apr 2008 Posts: 310 Location: Southern Northern California
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Posted: Sat Jun 12, 2010 6:19 am Post subject: |
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tlanglet wrote: | Another variation on the two stepper............
Quote: | Type 1 UR67 r37c56 r7c5<>67
ARP58 r2c1,r2c5,r7c5 & r7c3 plus fin (6)r7c3 - r7c9 = r4c9 - (6=3)r4c1 - (3=8)r3c1; r78c1<>8 completes the puzzle.
I initially found this step as an "Almost Remote Pair" (APR). However in transporting the fin I realized the fin could be removed using a conflicting chain: (6)r7c3 - r7c9 = r4c9 - r4c1 = (6)r5c3; r7c3<>6 which also completes the puzzle. |
Ted |
What's up, my friend?
Man, you are downright prolific these days. . . it’s got to be hurting your golf handicap.
A twist on the ARP idea:
I’m into remote pairs, like you and others. As I’ve mentioned, with remote pair possibilities also come guardian possibilities. After the (67) UR:
Code: | *-----------------------------------------------------------*
| 4 7 *58+3 |*58 9 1 | 368 36 2 |
| 58 6 9 | 2 *58 3 | 4 7 1 |
| 38 1 2 | 4 7 6 | 389 5 39 |
|-------------------+-------------------+-------------------|
| 36 8 1 | 7 2 5 | 369 4 369 |
| 2 9 36 | 1 4 8 | 7 36 5 |
| 7 5 4 | 6 3 9 | 1 2 8 |
|-------------------+-------------------+-------------------|
| 1568 4 *58+6 | 3589 *58 7 | 2 19 36 |
| 18 3 7 | 89 6 2 | 5 19 4 |
| 9 2 56 | 35 1 4 | 36 8 7 |
*-----------------------------------------------------------* |
The illegal chain of five (58)s is prevented only by the guardians, +3 and +6. Together with (36)r5c3 it’s safe to say r9c3<>6. The above pattern is illegal because it would result in zero solutions.
Funny how some patterns are “deadly” while others are called “illegal.” I’m guessing that maybe “deadly” is to multiple solution patterns as “illegal” is to zero solution patterns. |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Sat Jun 12, 2010 1:46 pm Post subject: |
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Luke451 wrote: | A twist on the ARP idea:
I’m into remote pairs, like you and others. As I’ve mentioned, with remote pair possibilities also come guardian possibilities. After the (67) UR:
Code: | *-----------------------------------------------------------*
| 4 7 *58+3 |*58 9 1 | 368 36 2 |
| 58 6 9 | 2 *58 3 | 4 7 1 |
| 38 1 2 | 4 7 6 | 389 5 39 |
|-------------------+-------------------+-------------------|
| 36 8 1 | 7 2 5 | 369 4 369 |
| 2 9 36 | 1 4 8 | 7 36 5 |
| 7 5 4 | 6 3 9 | 1 2 8 |
|-------------------+-------------------+-------------------|
| 1568 4 *58+6 | 3589 *58 7 | 2 19 36 |
| 18 3 7 | 89 6 2 | 5 19 4 |
| 9 2 56 | 35 1 4 | 36 8 7 |
*-----------------------------------------------------------* |
The illegal chain of five (5s is prevented only by the guardians, +3 and +6. Together with (36)r5c3 it’s safe to say r9c3<>6. The above pattern is illegal because it would result in zero solutions.
Funny how some patterns are “deadly” while others are called “illegal.” I’m guessing that maybe “deadly” is to multiple solution patterns as “illegal” is to zero solution patterns. |
Howdy Luke,
Great to see that you have not totally abandon the DailySudoku site. Your posts are always interesting and of merit.
Regarding this "Guardian" stuff, it is a completely new concept for me. I have noticed references to it on Eureka but have not pursued the concept.
My first reaction to your post was "How do I know that the 5-cell pattern is illegal?" I have never given that serious consideration other than in a contradiction chain which does not require any forethought anyway. I assume it is foolish to ask if guardian issues are presented somewhere but I will ask it just because. In the meantime, I will ponder the idea of illegal patterns and guardians while I "putter" around in my garden, which is as close as I will ever get to a golf course.
Ted |
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Luke451
Joined: 20 Apr 2008 Posts: 310 Location: Southern Northern California
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Posted: Sat Jun 12, 2010 3:54 pm Post subject: |
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Ted wrote: | My first reaction to your post was "How do I know that the 5-cell pattern is illegal?" |
I’m reluctant to try to explain anything, because I usually have some nuance wrong. The following is just my understanding, and hopefully others with more experience can throw in.
Short answer: demonstrate it for yourself.
Here’s the pm without the guardians:
Code: | *-----------------------------------------------------------*
| 4 7 *58 |*58 9 1 | 368 36 2 |
| 58 6 9 | 2 *58 3 | 4 7 1 |
| 38 1 2 | 4 7 6 | 389 5 39 |
|-------------------+-------------------+-------------------|
| 36 8 1 | 7 2 5 | 369 4 369 |
| 2 9 36 | 1 4 8 | 7 36 5 |
| 7 5 4 | 6 3 9 | 1 2 8 |
|-------------------+-------------------+-------------------|
| 1568 4 *58 | 3589 *58 7 | 2 19 36 |
| 18 3 7 | 89 6 2 | 5 19 4 |
| 9 2 56 | 35 1 4 | 36 8 7 |
*-----------------------------------------------------------* |
Place a 5 or 8 in any cell of the pattern and note what happens to the other cells.
I’ll go out on a limb and say that these things could be called Broken Wings. In the Sudopedia article Broken Wings are only presented as single digit patterns, but the same concept holds true for identical bivalues that form odd-sized, conjugate-pairs-only loops.
Of course, just as with DPs, if there are only two guardians, then one can utilize the strong link they create: one must be true. If there are three guardians, a “tri-SIS” can be used, and so on (SIS=strong inference set.)
Anything you can do with a DP can be done to a Broken Wing.
BTW, it’s good that you putter around in the garden and not on the links. At the end of the day you have something to show for your efforts! |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sat Jun 12, 2010 7:41 pm Post subject: |
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If five cells are connected by five edges that are strong links on a digit, then you have an invalid/illegal pattern. Look here for details on why it's an invalid/illegal pattern.
Luke's use of pairs allows him to exert the constraint that another candidate must exist and be true in one of the five cells.
If r1c3<>3 and r7c3<>6 occurs, then the <58> pairs reduce the grid to a state where both <5> and <8> contain this invalid/illegal pattern in the (*) cells.
Code: | *-----------------------------------------------------------*
| 4 7 *58 |*58 9 1 | 36 36 2 |
| 58 6 9 | 2 *58 3 | 4 7 1 |
| 38 1 2 | 4 7 6 | 389 5 39 |
|-------------------+-------------------+-------------------|
| 36 8 1 | 7 2 5 | 369 4 369 |
| 2 9 3 | 1 4 8 | 7 36 5 |
| 7 5 4 | 6 3 9 | 1 2 8 |
|-------------------+-------------------+-------------------|
| 16 4 *58 | 39 *58 7 | 2 19 36 |
| 18 3 7 | 89 6 2 | 5 19 4 |
| 9 2 6 | 35 1 4 | 36 8 7 |
*-----------------------------------------------------------*
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Thus, Luke can assert that r1c3=3 and/or r7c3=6. |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Mon Jun 14, 2010 4:05 am Post subject: |
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Luke and Danny,
Thanks to both of you for you feedback. I have read both references but am still trying to absorb it all.
Regarding the term "guardians", at first I thought it made reference to arbitrary patterns of which remote pairs was one, but I now understand it applies to a very limited set of patterns which are readily determine to be legal/illegal. I do not emphasis single digit patterns currently, but will try to incorporate the guardian concept into them.
Thanks again....
Ted |
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Luke451
Joined: 20 Apr 2008 Posts: 310 Location: Southern Northern California
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Posted: Wed Jun 23, 2010 1:50 pm Post subject: |
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Just for interest on this topic:
Today on Eureka Steve Kurzhals posted a guardian based solution. He calls them "impossible oddagons."
Here is the link for anyone interested (post dated June 22, 2010.) |
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