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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Mon Apr 26, 2010 11:38 am Post subject: Puzzle 10/04/26 ___ BBDB |
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| Code: | +-----------------------+
| 1 3 . | . . 8 | . 4 2 |
| 8 . 6 | . . . | . . . |
| . 7 4 | . . 5 | . . . |
|-------+-------+-------|
| . . . | 6 . . | 2 . . |
| . . . | . . . | . . 6 |
| 6 . 7 | . . . | . 1 9 |
|-------+-------+-------|
| . . . | 4 . . | 6 . 1 |
| 7 . . | . . 6 | . 5 . |
| 5 . . | . 8 7 | 3 . . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Mon Apr 26, 2010 3:29 pm Post subject: |
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Woohoo! I've been looking for one of these for weeks... makes it a two-stepper
| Quote: | Finned x-wing(5) r24c29 fin r4c4; r56c2<>5
ALS-XZ ALS(2589)r267c2, ALS(59)r1c3, X=5, Z=9; r79c3<>9 |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Apr 26, 2010 3:54 pm Post subject: |
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One stepper (I think) [Edit - no it isn't!!]
| Quote: | If you start r6c7 as 5 it goes around the strong links and shows that 5 can't exist there. r6c7 <>5 solves the puzzle.
Diagram to follow
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Last edited by Mogulmeister on Mon Apr 26, 2010 4:00 pm; edited 1 time in total |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Apr 26, 2010 4:00 pm Post subject: |
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r6c7(4=5)-r1c7=r2c9-(5=2)r2c2-r6c2-(2=58)r6c2c4-(5)r6c7 so r6c7<>5
Nope doesn't solve it...my mistake ! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Apr 27, 2010 3:53 pm Post subject: |
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| I used a half-dozen of the usual steps. |
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oaxen
Joined: 10 Jul 2006
Posts: 96
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| Posted: Tue Apr 27, 2010 4:17 pm Post subject: |
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[quote="peterj"]Woohoo! I've been looking for one of these for weeks... makes it a two-stepper]
Why a two stepper? There are many bivalues from where successful onestepper chains can start. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Tue Apr 27, 2010 4:31 pm Post subject: |
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| Oaxen, Iwas mostly happy to find an ALS-XZ - that it reduced the number of steps significantly was icing on the cake! |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Apr 27, 2010 10:23 pm Post subject: |
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A classic example of an ALS-XZ  Great find.
Ted |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Apr 27, 2010 11:37 pm Post subject: |
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I found a (terminology again #@-!...) flightless finned xy-wing 2-58 with vertex 28 in r6c2, pincer 25 in r2c2, pincer 58 in r6c4 and fin 5 in r6c2.
If xy-wing is true; no deletions but (5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
If fin is true, (5)r6c2; r6c7<>5. (Fun but not very productive.)
Next an ALS-XZ pattern which has no direct deletions, but with a little transport makes a score
ALS_A 159 in r19c3
ALS_B 259 in b1
Exclusive=5
Common=9
No direct deletions are possible with this configuration, but when all the (9s) in ALS_B are transported:
(9)r3c1 - r3c45 = r1c4 - r9c4 = (9)r9c3; r457c3<>9.
Note: I realize that a x-wing 9 or an AIC will make the identical deletions, but I want to insure my ALS operations are valid.)
ALS239[(2)r67c6 = (9)r7r6]r67c6 - r7c12 = r9c3 - (9=5)r1c3 - (5=2)r2c2; r2c6<>2
At this point, I sort of gave in and used the ALS-XZ posted by Peterj to delete (9) in r9c3 to complete the puzzle
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Apr 28, 2010 12:05 am Post subject: |
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| tlanglet wrote: | I found a (terminology again #@-!...) flightless finned xy-wing 2-58 with vertex 28 in r6c2, pincer 25 in r2c2, pincer 58 in r6c4 and fin 5 in r6c2.
If xy-wing is true; no deletions but (5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
If fin is true, (5)r6c2; r6c7<>5. (Fun but not very productive.)
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Alternate perspectives:
(58=2)als:r6c24 - (2=5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
np(58)r6c24 = (2)r6c2 - (2=5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
Luke: The [r6] unit containing the np(58 ) cells is also the unit of the elimination. This is acceptable use! |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Wed Apr 28, 2010 2:46 am Post subject: |
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| Do you mean Luke or Ted Danny ? |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Apr 28, 2010 4:19 am Post subject: |
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| Mogulmeister wrote: | | Do you mean Luke or Ted Danny ? |
Luke !
I was carrying over a point from another thread. Namely, an example where I felt the use of naked subset was applicable in a chain. |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Wed Apr 28, 2010 5:39 am Post subject: |
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Good - I knew I was missing context somewhere..........  |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Apr 28, 2010 12:23 pm Post subject: |
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| daj95376 wrote: | | tlanglet wrote: | I found a (terminology again #@-!...) flightless finned xy-wing 2-58 with vertex 28 in r6c2, pincer 25 in r2c2, pincer 58 in r6c4 and fin 5 in r6c2.
If xy-wing is true; no deletions but (5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
If fin is true, (5)r6c2; r6c7<>5. (Fun but not very productive.)
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Alternate perspectives:
(58=2)als:r6c24 - (2=5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
np(58)r6c24 = (2)r6c2 - (2=5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
Luke: The [r6] unit containing the np(58 ) cells is also the unit of the elimination. This is acceptable use! |
Danny, thanks for the feedback. I appreciate the different views.
Ted |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Apr 28, 2010 1:02 pm Post subject: |
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Danny & Luke, at this point I have still another question on a related issue; naked pair (np) vs hidden pair (hp).
Danny, for this puzzle you suggested an alternate step using a np: np(58)r6c24 = (2)r6c2 - (2=5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
Does the pair become hidden if recast to: hp(58)r6c25 = (5)r6c7 - r4c9 = r2c9 - (5=2)r2c2; r6c7<>5
I have seen references to both terms and, being a crude mechanic, do not understand the finer points such as terminology. In the two forms noted here, the only difference I see is whether the extra digit causing the "almost" condition is, or is not, one of the digits of the pair.
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Apr 28, 2010 2:09 pm Post subject: |
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For anyone who might be interested in the grid under discussion.
| Code: | *--------------------------------------------------------------------*
| 1 3 59 | 79 6 8 | 57 4 2 |
| 8 25 6 | 237 234 234 | 1 9 57 |
| 29 7 4 | 129 129 5 | 8 6 3 |
|----------------------+----------------------+----------------------|
| 349 1589 3589 | 6 479 149 | 2 38 57 |
| 2349 12589 23589 | 58 2479 1249 | 457 38 6 |
| 6 258 7 | 58 234 234 | 45 1 9 |
|----------------------+----------------------+----------------------|
| 239 289 2389 | 4 5 239 | 6 7 1 |
| 7 4 123 | 123 123 6 | 9 5 8 |
| 5 6 19 | 19 8 7 | 3 2 4 |
*--------------------------------------------------------------------*
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| tlanglet wrote: | Danny, for this puzzle you suggested an alternate step using a np: np(58)r6c24 = (2)r6c2 - (2=5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
Does the pair become hidden if recast to: hp(58)r6c25 = (5)r6c7 - r4c9 = r2c9 - (5=2)r2c2; r6c7<>5
I have seen references to both terms and, being a crude mechanic, do not understand the finer points such as terminology. In the two forms noted here, the only difference I see is whether the extra digit causing the "almost" condition is, or is not, one of the digits of the pair.
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Although your chain is incorrect, it starts off with an acceptable strong link IMO.
I need for Luke/Norm/Ron to check my reply because I'm always getting the np() and hp() usage backwards. In fact, I originally wrote hp() in my alternate perspectives, and had to go back and correct it. Here's how I believe it works. I'm going to reverse the terms in the strong link to match the definition:
| Code: | If r6c2<>2 then naked pair r6c24=58 remains: (2)r6c2 = np(58)r6c24
If r6c7<>5 then hidden pair r6c24=58 remains: (5)r6c7 = hp(58)r6c24
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This can be made more confusing by adding a weak link:
| Code: | (2)r6c2 = np(58)r6c24 - (5)r6c7
(5)r6c7 = hp(58)r6c24 - (2)r6c2
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However, it's now clear that the np() usage would perform an elimination elsewhere in [r6], and the hp() usage would perform an elimination within the cells of the hp().
Regards, Danny |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Wed Apr 28, 2010 5:09 pm Post subject: |
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| tlanglet wrote: | Danny, for this puzzle you suggested an alternate step using a np: np(58)r6c24 = (2)r6c2 - (2=5)r2c2 - r2c9 = (5)r4c9; r6c7<>5
Does the pair become hidden if recast to: hp(58)r6c24 = (5)r6c7 - r4c9 = r2c9 - (5=2)r2c2; r6c7<>5 |
I doubt that's true in general. but in this case it is. Your chain is correct, but the exclusion is actually r6c2<>2.
Last edited by ronk on Thu Apr 29, 2010 1:29 am; edited 1 time in total |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Apr 28, 2010 9:23 pm Post subject: |
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Thanks Danny & Ronk for the feedback.
Danny, I think I now understand the key to what is naked and what is hidden.
Ronk, I was not sure how to interpret the chain. The chain as posted ends with r2c2=2 given the initial strong inference hp(58)r6c25 = (5)r6c7, but I did not even include (2) in the initial inference even though I assumed that the conflict was between the two (2s) in r26c2.
How could I write the chain to make the exclusion obvious?
Ted |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Thu Apr 29, 2010 1:28 am Post subject: |
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| tlanglet wrote: | I was not sure how to interpret the chain. The chain as posted ends with r2c2=2 given the initial strong inference hp(58)r6c24 = (5)r6c7, but I did not even include (2) in the initial inference even though I assumed that the conflict was between the two (2s) in r26c2.
How could I write the chain to make the exclusion obvious? |
Well, you could do what [edit: Asellus] IIRC, daj95376 sometimes, and I sometimes do, i.e., write the inference stream for the discontinuous loop rather than the chain. When written for the loop, the discontinuity cell(s) appear(s) at both ends. In this instance, that would be ...
hp(58)r6c24 = (5)r6c7 - r4c9 = r2c9 - (5=2)r2c2 - (2)r6c2; r6c2<>2
Unfortunately, other than those using nice-loop (NL) notation and the above mentioned for AIC notation, there are precious few doing so.
Last edited by ronk on Tue May 11, 2010 9:12 pm; edited 1 time in total |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Apr 29, 2010 7:01 pm Post subject: |
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I appreciate the timely feedback.
Ted |
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