The New Sudoku Players' Forum

[Yogi]

.....3...1....72...7..84.6...8....93.6.349.7.93.8..62..9.73..8...59....2...4..... {edited by Moderator}
I saw this puzzle in David's article on Loops in item 5 Champagne's Virus Loop. The analysis showed the . in Box5 to be a 2, but I could not see how the exact position of this 2 was deduced, only that it had to be somewhere on the top shelf of Box5. However, continuing with this 2 in place, I was able to get to here with the usual methods known to me:
....23...1...972...7..84.6...82...93.6.349.7.93.8..62..9.732.8...59....2.2.4.....
But from there I was stuck. I was able to solve the puzzle by doing a Test or trial, which placed the 3 in Box3, but this only ran on so far. A second test to place the 2 in Box4 allowed me to finally complete the puzzle.
But I really would like to know how I could have seen the original 2 in Box5 or made further deductive moves from my second picture without having to resort to doing a test

[David P Bird]

Yogi, welcome to the forum and I hope you will enjoy exploring our pages and making your own contributions.

However you shouldn't have posted your question in the Advanced Solving Techniques section. Although it relates to an example puzzle in the Domino Loops thread, it's requesting help on the more basic solving techniqes needed to complete the solution. I am therefore asking Jason to move it over to the 'Help with puzzles and solving techniques section' {Moved as suggested. Moderator}

Now a word about posting the position you've managed to reach. Our strings of 81 characters gives only the completely solved cells. They should only contain digits or dots and should never be broken. To show the grid with cells partially solved after you have made some eliminations you must display the whole grid. In this case you have a flying start by copying the grid I posted into a plain text editor such as Word Pad and editing it. To see how to incorporate it into your post click on the 'Quote' control to see an example of where to insert the necessary [ code] and [ /code] print controls the forum software uses.

David PB
.

[Leren]

Hi Yogi. There are many ways to solve this puzzle from the second position, but I've chosen a solution path that only requires one type of move that I assume you are not familiar with.

I'll explain the first move in detail and outline the rest of the solution.

An ALS (Almost Locked Set) is a group of N cells with a total of N+1 different digits. The trick is that if any one of the digits is False in the set, then all other digits must be True in the set.

From your position ....23...1...972...7..84.6...82...93.6.349.7.93.8..62..9.732.8...59....2.2.4..... there are two basic moves which I assume that you are familiar with :

Pointing Pair of 5's r9c56 which removes the 5's from r9c789.

Box column intersection (one of r12c8 must be 5, which removes rest of the 5's from Box 3. This should get you to here :

Code: Select all

*--------------------------------------------------------------------------------*
| 4568    458     469      | 156     2       3        | 14789   145     14789    |
| 1      a458     346      | 6-5     9       7        | 2      a345    a48       |
| 235     7       239      |b15      8       4        |b139     6      b19       |
|--------------------------+--------------------------+--------------------------|
| 457     145     8        | 2       1567    156      | 145     9       3        |
| 25      6       12       | 3       4       9        | 158     7       158      |
| 9       3       147      | 8       157     15       | 6       2       145      |
|--------------------------+--------------------------+--------------------------|
| 46      9       146      | 7       3       2        | 145     8       1456     |
| 34678   148     5        | 9       16      168      | 1347    134     2        |
| 3678    2       1367     | 4       156     1568     | 1379    13      1679     |
*--------------------------------------------------------------------------------*

Now look at the 3 cells I've marked a in the PM diagram. They are an ALS because they contain a total of 4 different digits 3458. Now suppose that both of the 5's in those cells was False in the set. Then those cells must contain 348, in particular r2c8 must be 3. Now look at the 3 cells I've marked as b in the PM diagram. They are an ALS and contain 4 different digits 1359. Now suppose 3 was False in the set (because r2c8 was 3). Then it must contain 159, in particular r3c4 must be 5.

So, summarizing all this, if 5 r2c28 is false => 5 r3c4 must be True. You can reverse this argument and suppose that 5 r3c4 was False and conclude that 5 r3c28 must be True. The net result of all this is that at least one of these 3 cells must be 5.

Now look at r2c4. It can see all of these 3 cells, so it can't be 5 !

The notation of all this is ALS XZ Rule: X = 3, Z = 5: (5=3) r2c289 - (3=5) r3c479 => - 5 r2c4. Z is the pincer digit (elimination digit) and X is the digit that links the 2 ALSs (also called the restricted common digit).

The next move follows on immediately and is similar:

Code: Select all

*--------------------------------------------------------------------------------*
|b4568   b458     469      |b15      2       3        |b14789  b145    b14789    |
| 1       458     34       | 6       9       7        | 2       345    a48       |
| 235     7       239      | 15      8       4        | 139     6      a19       |
|--------------------------+--------------------------+--------------------------|
| 457     145     8        | 2       1567    156      | 145     9       3        |
| 25      6       12       | 3       4       9        | 158     7      a158      |
| 9       3       147      | 8       157     15       | 6       2      a145      |
|--------------------------+--------------------------+--------------------------|
| 4-6     9       146      | 7       3       2        | 145     8      a1456     |
| 34678   148     5        | 9       16      168      | 1347    134     2        |
| 3678    2       1367     | 4       156     1568     | 1379    13      1679     |
*--------------------------------------------------------------------------------*

ALS XZ Rule: X = 9, Z = 6: (6=9) r23567c9 - (9=6) r1c124789 => - 6 r7c1

This is followed by some basic moves that I assume you are familiar with :

Locked Triple {168} r8c256. Box/Row intersection 6r8c56. Locked Quad {2357} r3458c1. Locked Triple {168} in Cells r7c3 r8c2 r9c1.

This gets you to here, where a third ALS XZ Rule can be applied.

Code: Select all

*--------------------------------------------------------------------------------*
|b68     b458     469      | 15      2       3        | 14789   145     14789    |
| 1      b458    b34       | 6       9       7        | 2       45-3    48       |
| 235     7       239      | 15      8       4        | 139     6       19       |
|--------------------------+--------------------------+--------------------------|
| 57      145     8        | 2       1567    156      | 145     9       3        |
| 25      6       12       | 3       4       9        | 158     7       158      |
| 9       3       147      | 8       157     15       | 6       2       145      |
|--------------------------+--------------------------+--------------------------|
| 4       9       16       | 7       3       2        | 15      8       156      |
| 37      18      5        | 9       16      168      | 347     34      2        |
|a68      2       7-3      | 4      a15     a158      | 1379   a13      1679     |
*--------------------------------------------------------------------------------*

ALS XZ Rule: X = 6, Z = 3: (3=6) r9c1568 - (6=3) r1c12, r2c23 => - 3 r2c8, r9c3.

The puzzle solves in singles from there. Hope this helps.

Leren

[Yogi]

Sorry Leren I'm completely lost. I didn't know the term ALS but I think I have used this type of approach before in scanning. If there are only two locations for a number in a Region you guys call them Units?) then ignore those cells and see what's going on in the rest. I'm assuming you chose those three remaining cells in Row 2 by leaving out the possible places for the 6. What I don't understand is what drew your attention to that particular ALS. There must be lots of them in the puzzle. Why this one?
Another thing I had trouble with was the reasoning and this True / False jargon. The narrative in my head would be "If the 5 is not r2c2 or r2c8 then it must be r2c4. In this option r2c8 certainly is 3, which forces the 1,9 Pair in Row3 and leaves r3c4 to be only 5, also. So naturally the starting assumption must be false and r2c4 must be 6.
That is a result, but I still don't understand how you would spot an ALS that could help. Was the second one isolated by leaving out the candidate 2? And does it usually take two related ALSs to obtain a result?

[Leren]

Hi Yogi,

As I said before, there were many ways to solve this puzzle, most involving simpler moves but more of them. Explaining these longer solutions would have involved too much information in one post.

About the best thing I can do at this stage is to point you to some excellent teaching sites, which cover ALS chains and a lot of other techniques.

http://www.sudokuwiki.org/Strategy_Families

http://hodoku.sourceforge.net/en/techniques.php

Leren

[Yogi]

OK. Thanx.