The New Sudoku Players' Forum

by **Luke** » Tue Sep 05, 2006 4:36 am

I've recently seen some documentation on ALS, which is new to me. So naturally, I'm trying to find one under every rock. Does this qualify as an ALS? Here's what I think I see:

Set A: R1C5, R2C5, R3C6
Set B: R7C6, R8C6
X = 5 (restricted common)
Z = 6 (other common)
Candidate eliminated: The 6 in R8C5, which can be seem by both other commons.

Code: Select all: *-----------------------------------------------------------------------------* | 367 678 9 | 123678 1368 23678 | 5 4 67 | | 3567 4 1 | 3678 9 35678 | 368 368 2 | | 3567 5678 2 | 4 13568 35678 | 13689 13689 679 | |-------------------------+-------------------------+-------------------------| | 567 567 3 | 689 68 1 | 2 5689 4 | | A24 A26 8 | 5 346 23469 | 7 69 1 | | 1 9 A45 | 268 7 2468 | B68 B568 3 | |-------------------------+-------------------------+-------------------------| | 9 25 7 | 136 1356 356 | 4 12356 8 | | 8 3 45 | 169 2 4569 | 169 7 569 | | 245 1 6 | 3789 3458 345789 | 39 2359 59 | *-----------------------------------------------------------------------------*

Thanks for looking.

Luke in Ca

by **Luke** » Tue Sep 05, 2006 4:46 am

Sorry, the code went awry...it's my first try at this. It looked fine in the "preview." I'll try again:

*-----------*
|..9|...|54.|
|.41|.9.|..2|
|..2|4..|...|
|---+---+---|
|..3|..1|2.4|
|..8|5..|7.1|
|19.|.7.|..3|
|---+---+---|
|9.7|...|4.8|
|83.|.2.|.7.|
|.16|...|...|
*-----------*

*-----------*
|..9|...|54.|
|.41|.9.|..2|
|..2|4..|...|
|---+---+---|
|..3|..1|2.4|
|..8|5..|7.1|
|19.|.7.|..3|
|---+---+---|
|9.7|...|4.8|
|83.|.2.|.7.|
|.16|...|...|
*-----------*

Code: Select all: *-----------------------------------------------------------------------------* | 367 678 9 | 123678 1368 23678 | 5 4 67 | | 3567 4 1 | 3678 9 35678 | 368 368 2 | | 3567 5678 2 | 4 13568 35678 | 13689 13689 679 | |-------------------------+-------------------------+-------------------------| | 567 567 3 | 689 68 1 | 2 5689 4 | | 24 26 8 | 5 346 23469 | 7 69 1 | | 1 9 45 | 268 7 2468 | 68 568 3 | |-------------------------+-------------------------+-------------------------| | 9 25 7 | 136 1356 356 | 4 12356 8 | | 8 3 45 | 169 2 4569 | 169 7 569 | | 245 1 6 | 3789 3458 345789 | 39 2359 59 | *-----------------------------------------------------------------------------*

by **Luke** » Tue Sep 05, 2006 5:00 am

Hoo-kay, I'm embarrassed! Can anyone help out a newbie?

by **udosuk** » Tue Sep 05, 2006 5:35 am

The first post you gave was clear presented enough...

Anyway it's like here (you didn't give us the original puzzle grid):

Code: Select all: *-----------* |..9|...|54.| |.41|.9.|..2| |..2|4..|...| |---+---+---| |..3|..1|2.4| |..8|5..|7.1| |19.|.7.|..3| |---+---+---| |9.7|...|4.8| |83.|.2.|.7.| |.16|...|...| *-----------* *-----------------------------------------------------------------------------* | 367 678 9 | 123678 1368 23678 | 5 4 67 | | 3567 4 1 | 3678 9 35678 | 368 368 2 | | 3567 5678 2 | 4 13568 35678 | 13689 13689 679 | |-------------------------+-------------------------+-------------------------| | 567 567 3 | 689 68 1 | 2 5689 4 | | 24 26 8 | 5 346 23469 | 7 69 1 | | 1 9 45 | 268 7 2468 | 68 568 3 | |-------------------------+-------------------------+-------------------------| | 9 25 7 | 136 1356 356 | 4 12356 8 | | 8 3 45 | 169 2 4569 | 169 7 569 | | 245 1 6 | 3789 3458 345789 | 39 2359 59 | *-----------------------------------------------------------------------------*

And your ALS demonstration could be like this:

Code: Select all: *-----------------------------------------------------------------------------* | 367 678 9 | 123678 1368 23678 | 5 4 67 | | 3567 4 1 | 3678 9 35678 | 368 368 2 | | 3567 5678 2 | 4 13568 35678 | 13689 13689 679 | |-------------------------+-------------------------+-------------------------| | 567 567 3 | 689 68 1 | 2 5689 4 | |A24 A26 8 | 5 346 23469 | 7 -69 1 | | 1 9 A45 | 268 7 2468 |B68 B568 3 | |-------------------------+-------------------------+-------------------------| | 9 25 7 | 136 1356 356 | 4 12356 8 | | 8 3 45 | 169 2 4569 | 169 7 569 | | 245 1 6 | 3789 3458 345789 | 39 2359 59 | *-----------------------------------------------------------------------------* Set A: R1C5, R2C5, R3C6 Set B: R7C6, R8C6 X = 5 (restricted common) Z = 6 (other common) Candidate eliminated: The 6 in R8C5, which can be seem by both other commons.

You could do all the editing in Notepad or other similar programs...

BTW I think it's a correct application of the ALS... And a nice puzzle to solve...

by **Luke** » Wed Sep 06, 2006 4:07 am

Thanks. I changed my screen resolution and all's OK.

On the other hand, I've looked at this for a long time
and altho I know the six has to go, I don't see the
logic of it.

Luke in Ca

by **Myth Jellies** » Wed Sep 06, 2006 7:20 am

Perhaps you can see it via an AIC...

(2&4&6)[ A ] = 5[ A ] - 5[ B ] = (6&8)[ B ]

Since a 5 anywhere in A inhibits a 5 anywhere in B, either A contains 2 & 4 & 6, or B contains 6 & 8, or both. Thus any cell which sees all the 6's in A and B cannot contain a 6.

by **daj95376** » Wed Sep 06, 2006 7:34 am

Is there any reason why eliminations from (4x) Locked Candidates, a Naked Quad, and an X-Wing were bypassed?

Code: Select all: *-----------------------------------------------------------------------------* | 367 678 9 | 123678 1368 23678 | 5 4 67 | | 3567 4 1 | 3678 9 35678 | 368 368 2 | | 3567 5678 2 | 4 3568-1 35678 | 13689 1368-9 679 | |-------------------------+-------------------------+-------------------------| | 567 567 3 | 689 68 1 | 2 5689 4 | | 24 26 8 | 5 346 369-24 | 7 69 1 | | 1 9 45 | 268 7 2468 | 68 568 3 | |-------------------------+-------------------------+-------------------------| | 9 25 7 | 136 1356 356 | 4 1236-5 8 | | 8 3 45 | 169 2 4569 | 169 7 569 | | 24-5 1 6 | 3789 3458 35789-4 | 39 23-59 59 | *-----------------------------------------------------------------------------*

Followed by:

Code: Select all: r8c9 <> 5 XY-Chain on [r8c3] r9c9 = 5 Hidden Single r8 - 169 Naked Triple *-----------------------------------------------------------------------------* | 367 678 9 | 123678 1368 23678 | 5 4 67 | | 3567 4 1 | 3678 9 35678 | 368 368 2 | | 3567 5678 2 | 4 3568 35678 | 13689 1368 679 | |-------------------------+-------------------------+-------------------------| | 567 567 3 | 689 68 1 | 2 5689 4 | | 24 26 8 | 5 346 369 | 7 69 1 | | 1 9 45 | 268 7 2468 | 68 568 3 | |-------------------------+-------------------------+-------------------------| | 9 25 7 | 136 1356 356 | 4 1236 8 | | 8 3 45 | 169 2 45 | 169 7 69 | | 24 1 6 | 3789 348 3789 | 39 23 5 | *-----------------------------------------------------------------------------*

This should be a little less cluttered to demonstrate the ALS.

by **udosuk** » Wed Sep 06, 2006 8:18 am

Luke451 wrote:On the other hand, I've looked at this for a long time
and altho I know the six has to go, I don't see the logic of it.

Code: Select all: Set A: R1C5, R2C5, R3C6 Set B: R7C6, R8C6 X = 5 (restricted common) Z = 6 (other common) Candidate eliminated: The 6 in R8C5, which can be seem by both other commons.

A={2456}, if r5c8=6, it is reduced to {245}, and the only place to hold the 5 is at r6c3.
B={568}, if r5c8=6, it is reduced to {58}, and the only place to hold the 5 is at r6c8.
So when r5c8=6, we'll have two 5s on r6! Impossible...
Therefore we can make the elimination of 6 from r5c8...
Clear?

by **Luke** » Thu Sep 07, 2006 3:19 am

The reason for the clutter was that after I solved the puzzle
I went back and coded just enuf to pose the question.

Thanks to all for yr help.

One last question: is the following statement true?

"The candidate to be eliminated in an ALS needs to be seen by EVERY 'other common' in both sets?"

Luke in Ca

by **Myth Jellies** » Thu Sep 07, 2006 5:17 am

Luke451 wrote:One last question: is the following statement true?

"The candidate to be eliminated in an ALS needs to be seen by EVERY 'other common' in both sets?"

In general, the answer is yes. However, when you link into an ALS, leaving a locked set, that locked set often contains locked subsets, such as the slight alteration to the one we have been working with shown below....

Code: Select all: *-----------------------------------------------------------------------------* | 367 678 9 | 123678 1368 23678 | 5 4 67 | | 3567 4 1 | 3678 9 35678 | 368 368 2 | | 3567 5678 2 | 4 13568 35678 | 13689 13689 679 | |-------------------------+-------------------------+-------------------------| | 567 567 3 | 689 68 1 | 2 5689 4 | |A24 A26 8 | 5 346 -23469 | 7 -69 1 | | 1 9 A45 |B2'68 7 2468 |B6'8' B56'8' 3 | |-------------------------+-------------------------+-------------------------| | 9 25 7 | 136 1356 356 | 4 12356 8 | | 8 3 45 | 169 2 4569 | 169 7 569 | | 245 1 6 | 3789 3458 345789 | 39 2359 59 | *-----------------------------------------------------------------------------* Set A: r5c12, r6c3 Set B: r6c478

(2&4&6)[ A ] = 5[ A ] - 5[ B ] = (2'&6'&8')[ B ] => r5c6 <> 2, r5c8 <> 6
Note that when B is locked into 2 & 6 & 8, you have more than just a triple--you have a hidden single (2') and a naked pair (6'8'). Since the naked pair is locked in box 6, you can make the reduction r5c8 <> 6, as well as r5c6 <> 2.

by **ronk** » Thu Sep 07, 2006 1:36 pm

Myth Jellies wrote:Note that when B is locked into 2 & 6 & 8, you have more than just a triple--you have a hidden single (2') and a naked pair (6'8'). Since the naked pair is locked in box 6, you can make the reduction r5c8 <> 6, as well as r5c6 <> 2.

But 268 is not "locked into" set B. If it were we could make the exclusion r4c8<>8 too. We only know that at least one of the sets cannot unltimately contain a 5.

And it seems simpler to use two steps ... first defining a smaller set B (r6c78=568) for the exclusion r5c8<>6 ... and then the larger set B for r5c6<>2.

by **Myth Jellies** » Thu Sep 07, 2006 4:23 pm

ronk wrote:But 268 is not "locked into" set B. If it were we could make the exclusion r4c8<>8 too. We only know that at least one of the sets cannot unltimately contain a 5.

And it seems simpler to use two steps ... first defining a smaller set B (r6c78=568) for the exclusion r5c8<>6 ... and then the larger set B for r5c6<>2.

Either 268 is locked into the larger set B, or 246 is locked into set A. Either end of the chain kills both the aforementioned 2 and 6.

I agree it is simpler to use two steps (certainly simpler to explain

), but that was not what Luke asked.

by **udosuk** » Thu Sep 07, 2006 4:38 pm

Luke451 wrote:One last question: is the following statement true?

"The candidate to be eliminated in an ALS needs to be seen by EVERY 'other common' in both sets?"

If you're referring ALS as the simple "ALS-xz" type technique like in your original example, then yes, the candidate to be eliminated (6 from r5c8 here) must be seen by all "other cells containing 6" in both sets (r5c2 from set A and r6c78 from [the old] set B)...

What MJ showed is a much more complicated application of ALS, where you're eliminating 2 different candidates from 2 different cells simultaneously... In this case, the candidates to be eliminated (2 & 6) would not need to be seen by every other cells containing them in both sets (for example, r6c4={268} from [the new] set B doesn't see r5c8)...

by **ronk** » Thu Sep 07, 2006 5:46 pm

Myth Jellies wrote:Either 268 is locked into the larger set B, or 246 is locked into set A.

Since both sets are included in this sentence it's certainly better, but I still take issue with using "locked" for the conditional state ... the "almost locked" state ... of one set. It's like saying "the prisoners are locked in jail" ... when one really means "if the guard removed the keys from the lock, the prisoners are locked in jail."

And if you use "locked" there, then what do you say for truly locked candidates in a doubly-linked ALS xz-rule? Really locked? Genuinely locked? Actually locked?

I suggest saying something like ... "either set B will be 268, or set A will be 246, or both."

The New Sudoku Players' Forum

Is this an almost locked set?

Is this an almost locked set?