## Puzzle and ALS

Post the puzzle or solving technique that's causing you trouble and someone will help

### Puzzle and ALS

I have been working on several puzzles trying to use ALS (almost locked sets). I was wondering if someone could help me with this puzzle - ALS doesn't seem to work.

The orginal puzzle is as follows:

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`. . . | .  .  3| 6 . . . 2 . | .  5  4| . . 3 1 . . | . . .  | 8 . . ------+--------+------  . . .| 9 . .  | . . . 4 . . | . . .  | . . 7 . . . | . .  6 | . . . ------+-------+------ . . 8 | . . .  | . . 5 2 . . | 7  1 . | . 4 . . . 9 | 8  . . | . . . `

The marked puzzle is as follows:

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`9     57      457    |2      8        3     |6        15     148      2       6     |1      5        4     |79       79      31     53      345    |6      9        7     |8        25      24---------------------+----------------------+-----------------------6     3578   572     |9      247      18    |2345    238     124     9      125     |35     23       18    |1235    6         735   3578    1257    |345    247       6    |12345   1238     9------------------+-------------------+-----------------------37   14       8      |34     6      29      |1237     12379   52     6        35    |7       1     59      |39        4      8357   14       9     |8       34    25      |1237     1237    6`
]

The first ALS is as follows:
Set 1 - r7c1:r7c2:r9c2 {37, 14,14} = 1347
Set 2 - r7c4:r7c6:r8c6:r9c6 {34,29,59,25} = 23459

The 2 commons are 3 & 4 with 3 being the strong link (all 3's in set 1 can see all 3's in set 2). You should conclude that all 3's that can see 3's in each set can be eliminated - that is the 3's in r7c7 and r7c8. However to solve the puzzle a 3 must be in r7c8 (i have solved the puzzle with trial and error) .

The second ALS set is as follows:

Set 1 - r1c2,r1c3 {57,457} = 457
Set 2 - r3c2,r7c2,r9c3 {35,14,14} = 1345

The 2 commons are 4 & 5 with 5 being the stong link. You should be able to eliminate the 5 in r3c3, however a 5 is required in r3c3 to solve puzzle.

Ok, so where did I make a mistake? I believe the marked puzzle contains correct numbers in each square (that is I can find all of the required numbers to solvle puzzle in each square prior to the two above ALS elimnations).

Any thoughts? Are my ALS assumption incorrect?

Thank you
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al_bell

Posts: 4
Joined: 01 December 2006

Your pencilmark is already at a stage where the puzzle is quite easy to solve (nothing worse than a naked pair I think).

Aside from that, your first ALS xz-rule application is flawed. 3 is a restricted common candidate. You would need another common candidate (4 for instance) on which to make exclusions. You can elimate a 4 from any cell that sees all of the 4's in your two sets. Unfortunately, there aren't any. The same applies to your second example. The eliminations should be made on the 4 in your second example (though there aren't any to be made).

Check out the original rules again.
re'born

Posts: 551
Joined: 31 May 2007

al_bell, I don't know how you eliminated the 3 from r4c5, but here is how I did it...

This is the original puzzle grid:
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` *-----------* |...|..3|6..| |.2.|.54|..3| |1..|...|8..| |---+---+---| |...|9..|...| |4..|...|..7| |...|..6|...| |---+---+---| |..8|...|..5| |2..|71.|.4.| |..9|8..|...| *-----------*`

After SSTS:
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` *--------------------------------------------------------------------* | 9      7      4      | 2      8      3      | 6      5      1      | | 8      2      6      | 1      5      4      | 79     79     3      | | 1      35     35     | 6      9      7      | 8      2      4      | |----------------------+----------------------+----------------------| | 6      358    7      | 9     *34     1      |*345   -38     2      | | 4      9      1      | 35     2      8      |*35     6      7      | | 35     358    2      | 345    7      6      | 1345   138    9      | |----------------------+----------------------+----------------------| | 37     14     8      | 34     6      29     | 12379  1379   5      | | 2      6      35     | 7      1      59     | 39     4      8      | | 357    14     9      | 8      34     25     | 127    137    6      | *--------------------------------------------------------------------*`

We have an xyz-wing in r4c5+r45c7={345} => r4c8<>3

SSTS then solves the rest...
udosuk

Posts: 2698
Joined: 17 July 2005

Redoing it a 2nd time, after singles, naked pairs, locked candidates:
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` *--------------------------------------------------------------------* | 9      7      4      | 2      8      3      | 6      5      1      | | 8      2      6      | 1      5      4      | 79     79     3      | | 1      35     35     | 6      9      7      | 8      2      4      | |----------------------+----------------------+----------------------| | 6      358    7      | 9     *34     1      |*345   -38     2      | | 4      9      1      |*35     2      8      |*35     6      7      | | 35     358    2      | 345    7      6      |-1345  -138    9      | |----------------------+----------------------+----------------------| | 37     134    8      | 34     6      29     |-12379  1379   5      | | 2      6      35     | 7      1      59     |-39     4      8      | | 357    1345   9      | 8      34     25     |-1237   137    6      | *--------------------------------------------------------------------*`

It's easy to see that one of r45c7 must be 3, so we can eliminate 3 from 6 cells, and the rest are all singles...

But I don't know what exactly is this technique called... Any advice?
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:It's easy to see that one of r45c7 must be 3, so we can eliminate 3 from 6 cells, and the rest are all singles...

But I don't know what exactly is this technique called... Any advice?

Myth Jellies' first real world example in his Combined Overlapping ALS thread looks just like that. For your example,

{ALS:r45c7=3|4=r4c7}-4-r4c5-3-r5c4-5-{ALS:r5c7=5|3=r5c7}

Any candidate 3 that sees all the 3s in the combined ALS (r45c7 and r5c7) may be eliminated.

[edit: last r5c7 was typo c5c7]
Last edited by ronk on Mon Dec 04, 2006 10:43 pm, edited 1 time in total.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Thank you for your reply. I can see how you get the puzzle to this point (that is the point below), but I dont see the next step - ie I can see that r5c5 needs to be a 2- but what steps did you use?

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`9     7       4      |2      8        3     |6         5      18      2       6     |1      5        4     |79       79      3 1     35      35     |6      9        7     |8        2       4  ---------------------+----------------------+----------------------- 6     358    17      |9      347      18    |1345    138     2 4     9      12      |35     23       18    |135     6       7 35   358     127     |345    247       6    |1345    138     9 ------------------+-------------------+----------------------- 37   14       8      |34     6      29      |1237     1379   5 2     6        35    |7       1     59      |39        4      8 357   14       9     |8       34    25      |1237     137     6`
al_bell

Posts: 4
Joined: 01 December 2006

Al_bell, you have an hidden single in r5c6.

Carcul
Carcul

Posts: 724
Joined: 04 November 2005

ronk wrote:Myth Jellies' first real world example in his Combined Overlapping ALS thread looks just like that. For your example,

{ALS:r45c7=3|4=r4c7}-4-r4c5-3-r5c4-5-{ALS:r5c7=5|3=r5c7}

Any candidate 3 that sees all the 3s in the combined ALS (r45c7 and c5c7) may be eliminated.

Thanks ronk for the explanation... Didn't realise I accidentally hit on the COALS... But the notation is quite complicated...

I think it can also be represented as a forcing net:

r5c7=3 or r5c7=5 => r5c4=3 => r4c5=4 => r4c7=3
So one of r45c7 must be 3...

But this representation doesn't look too elegant...
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:Didn't realise I accidentally hit on the COALS... But the notation is quite complicated...

I think it can also be represented as a forcing net:

r5c7=3 or r5c7=5 => r5c4=3 => r4c5=4 => r4c7=3
So one of r45c7 must be 3...

But this representation doesn't look too elegant...

I used verbose notation to highlight the two ALSs of the the "CoALS." If we dispense with that, we have simply ...

-3-r45c7-4-r4c5-3-r5c4-5-r5c7-3-

Its elegance is debatable, but the latter notation allows one to see that r4c5, r5c4 and r5c7 are bivalue cells ... and easily deduce that r45c7 is a trivalued ALS ... all without looking at a grid.

I get none of that from your "foricing net notation."
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA