The New Sudoku Players' Forum

[Yogi]

. . . 1 . . . 2 4 . 2 3 . 4 5 . . 6 . 4 . . 7 . . . 8 2 3 . . . . . 8 . . 8 7 . . . 4 5 3 . 9 . . . . 2 . 1 . . . . 5 . . 3 . 9 . . 7 3 . 8 . . 3 1 . . . 8 . . .

Can you please tell me how to spot a pair of linked ALSs which I can use to solve this puzzle. There may be other techniques but I am trying to get a handle on ALSs and so-far no-one has been able to tell me how to spot the useful ones!
Yogi

[Leren]

...1...24.23.45..6.4..7...823.....8..87...453.9....2.1....5..3.9..73.8..31...8...

Removed the blanks from the original puzzle for a more standard line format.

Unfortunately the puzzle is too easy to illustrate an ALS move, it solves with basics: a few pointing pairs, a locked triple and a locked quad.

Instead I've used Dan's puzzle for today - here it is in standard line format .....1.4..86.72.15.7..............3..95.6......2.....15....4..7.67.83...........9

After basics you get to here :

Code: Select all

*--------------------------------------------------------------*
| 2     5     39     | 6     39    1      | 7     4     8      |
| 34    8     6      | 34    7     2      | 9     1     5      |
| 14    7    a19     | 8    b49    5      | 2     6     3      |
|--------------------+--------------------+--------------------|
| 7     14    8      | 124  b24    9      | 5     3     6      |
| 13    9     5      | 13    6     8      | 4     7     2      |
| 6     34    2      | 345   345   7      | 8     9     1      |
|--------------------+--------------------+--------------------|
| 5     123   3-1    | 9    b12    4      | 6     8     7      |
| 9     6     7      | 25    8     3      | 1     25    4      |
| 8     12    4      | 7     125   6      | 3     25    9      |
*--------------------------------------------------------------*

XY Chain / ALS XZ Rule: X = 9, Z = 1: (1=9) r3c3 - (9=1) r347c5 => - 1 r7c7; stte

The cell marked a is a single cell ALS - a bi-value cell containing 19. The 3 cells marked b form a 3 cell ALS containing 1249.

The elimination digit is 1 and the linking digit (restricted common digit) is 9. If r3c3 is not 1 it must be 9, so the second ALS can't contain 9, it must contain 124. Since the only 1 in the ALS is in r7c4 it must be 1.

So r7c3 can't be 1. The puzzle solves with singles from there.

Here is a link to the part of the Hodoku site that discusses ALSs : http://hodoku.sourceforge.net/en/tech_als.php

Leren

[Yogi]

Thanx again, Leren! Sorry I still don't talk Mathematix. However - if I've got it right, this exercise started with looking at the possibility of excluding a candidate from the then bi-value cell r7c3 in Dan's puzzle, which had candidates 1,3. The method was to see if two linked ALSs could be found which could both see r7c3 in a way that the two possible options for the placement of the 1 in them would both exclude 1 from r7c3. In this case it was able to show that whether r3c3 was 1 or 9, both these options exclude 1 from r7c3, making it definitely a 3, which as you say quickly solves the puzzle.
Generalising a bit more, this suggests to me that a simple way to use this ALS strategy is to look at a known bi-value cell which has candidates a,b, and then look to see if there is another bi-value cell which can see it with candidates a,x. You then ask, 'Could the cell with a,x be used as a Single-cell ALS linked to another Group ALS (if I can find one) to eliminate candidate 'a' from the first cell I was looking at?' This can work both ways of course. If you can't assemble a Group ALS to work with the second cell you looked at, maybe the first cell (a,b) can be used as a Single-cell ALS to eliminate 'a' from the second (a,x) cell?
I actually tried to use this approach with another puzzle I was working on, but I did not get that far into it. In preparation I had identified a number of bi-value cells, but I then found that by chaining through them I could eliminate a candidate in a particular cell, which solved the puzzle anyway. Spoze I'll keep looking and trying.

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 1    b579   29     | 28    259   6      | 4     379   3578   |
|a45   b459   8      | 3     159   7      | 6     2    a15     |
| 27    6     3      | 248   1259  14     | 59    179   1578   |
|--------------------+--------------------+--------------------|
| 45    145   15     | 7     3     2      | 8     6     9      |
| 8     3     7      | 6     4     9      | 1     5     2      |
| 9     2     6      | 5     18    18     | 7     34    34     |
|--------------------+--------------------+--------------------|
| 267  b1579  29     | 24    67    45     | 3     8     457-1  |
| 367   8     15     | 9     67    345    | 2     147   1457   |
| 23   b57    4      | 1     28    38     | 59    79    6      |
*--------------------------------------------------------------*

Here is a slightly more instructive example from today's puzzle.

ALS XZ Rule: X = 4, Z = 1: (1=4) r2c19 - (4=1) r1279c2 => - 1 r7c9; stte

The way this works is as follows : The first ALS has two cells marked a in the diagram. They happen to be both bi-value cells but they don't have to be.

Between them they contain 3 digits 145. Now suppose 1 in r2c9 was false. The two a cells would become a naked pair and would have to contain just 4 and 5 - in particular r2c1 would be 4.

Now look at the second ALS - this has 4 cells marked b and between them they contain 5 digits 14579.

Now if r2c1 is 4, as we have supposed, the 4 in r2c2 would be false. Since this is the only 4 in the b cells they would be reduced to a naked quad with 4 digits 1579. Since r7c2 has the only 1 in the second ALS, it would have to be True there.

So you can eliminate 1 from r7c1 and the puzzle solves with singles from there.

The important thing to understand when forming chains of ALSs is the way the restricted common digit works - in this example it is 4. In general, all of the restricted common digits in the first ALS must see all of the restricted common digits in the next ALS in the chain. In this example there were only 2 ALSs in the chain and one instance of the restricted common digit in each ALS, but more complex chains can be formed, with, say, 3 ALSs and more than one instance of a restricted common digit in some of the ALSs.

Thus endeth the lesson

Here is a link to another good teaching site's ALS section http://www.sudokuwiki.org/Almost_Locked_Sets

Both sites I've mentioned have good graphical worked examples and are definitely worth studying if you want to learn more.

Leren