The New Sudoku Players' Forum

[sudoku]

Hi
I`m new to this...
can someone please help us with the attached puzzle? We have spent weeks sitting in front of the fire trying it. Whats the next step and WHY?
hope the attachment works.
Thanks a mil
Valli

sudoku.odt

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[JasonLion]

An .odt file isn't the best way to share a puzzle. A simple text representation of the puzzle would be much better, or failing that an image.

board1.png (8.08 KiB) Viewed 772 times

[daj95376]

Code: Select all

 +--------------------------------------------------------------------------------+
 |  235     359     4       |  1       29      6       |  235     7       8       |
 |  25      6       8       |  247     247     3       |  1245    9       125     |
 |  237     379     1       |  8       5       249     |  234     23      6       |
 |--------------------------+--------------------------+--------------------------|
 |  135     135     359     |  6       12389   2589    |  7      e1238    4       |
 |  4       8       379     |  237     12379   29      |  6       5      c1239    |
 |  6       1357    2       |  347     134789  4589    | d13     e138    c139     |
 |--------------------------+--------------------------+--------------------------|
 |  8       135     35      |  234     6       7       |  9      f123-4  b1235    |
 |  1357    4       357     |  9       238     28      |  1235    6      b1235    |
 |  9       2       6       |  5       34      1       |  8      a34      7       |
 +--------------------------------------------------------------------------------+
 # 99 eliminations remain


This puzzle seems to have a complex solution whose next step needs a chain. Are you aware of Eureka notation and how to read it?

Code: Select all

 (4=3)r9c8 - r78c9 = r56c9 - (3=1)r6c7 - r46c8 = (1)r7c8  =>  -4 r7c8


This chain doesn't crack the puzzle, but it is useful and among the least difficult to understand. There are more difficult chains that crack the puzzle.

_

[Leren]

This was a tough puzzle to solve compared to the inquiries we normally get (but not really tough by the standards of this forum). There are many solution paths but I've chosen one that solves in 2 non-basic moves (to simplify the explanation).

The first one is an ALS chain :

Code: Select all

*--------------------------------------------------------------------------------*
| 235     359     4        | 1       29      6        | 235     7       8        |
| 25      6       8        | 247     247     3        | 1245    9      c125      |
| 237     379     1        | 8       5       249      | 234    a23      6        |
|--------------------------+--------------------------+--------------------------|
| 135     135     359      | 6       12389   2589     | 7      a1238    4        |
| 4       8       379      | 237     12379   29       | 6       5      b1239     |
| 6       1357    2        | 347     134789  4589     |b13     a138    b139      |
|--------------------------+--------------------------+--------------------------|
| 8       135     35       | 234     6       7        | 9       124-3  c1235     |
| 1357    4       357      | 9       238     28       | 1235    6      c1235     |
| 9       2       6        | 5       34      1        | 8       4-3     7        |
*--------------------------------------------------------------------------------*

ALS XY Wing: (3=1) r346c8 - (1=2) r5c9, r6c79 - (2=3) r278c9 => - 3 r79c8;

An ALS (Almost Locked set) is a set of cells that holds one more digit than the number of cells. If it is assumed that all instances one of the digits are False in the set, it becomes a locked set and one of all of the other digits must be True somewhere in the set. For example in the ALS marked a in the diagram there are 3 cells holding 4 different digits 1238. If all the 3's are assumed False then the set becomes locked and must hold 128 in some order.

So in the above chain we assume that 3 is False in the cells marked a then all of the other digits must be True in that set - in particular 1 must be True somewhere in that set.

The ALS cells marked b can see all of the 1's in the first set, so 1 must be False in that set so all of the other digits must be True in that set - in particular 2 must be True somewhere in that set.

The ALS cells marked c can see all of the 2's in the second set, so 2 must be False in that set so all of the other digits must be True in that set - in particular 3 must be True somewhere in that set.

You can repeat this argument in the reverse order by assuming that all of the 3's are False in ALS c and you will conclude that 3 must be True in ALS a.

What all this proves is that at least one of r346c8 or r78c9 is 3. Since cells r79c8 can see all of these cells, 3 can be removed from them. In particular this solves r9c8 and is followed by some simple moves involving pointing pairs and box/row column intersections (I'm assuming you are familiar with this technique), which gets you to here.

Code: Select all

*--------------------------------------------------------------------------------*
| 235     359     4        | 1      d29      6        | 25      7       8        |
| 25      6       8        | 27      247     3        | 1245    9       15       |
| 27      79      1        | 8       5       24-9     | 24      3       6        |
|--------------------------+--------------------------+--------------------------|
| 135     135     359      | 6       128-9   2589     | 7       128     4        |
| 4       8       79       | 237     127-9  a29       | 6       5       1239     |
| 6       157     2        | 37      1478-9  4589     | 13      18      139      |
|--------------------------+--------------------------+--------------------------|
| 8       135     35       | 4       6       7        | 9       12      1235     |
| 1357    4       357      | 9      c28     b28       | 135     6       135      |
| 9       2       6        | 5       3       1        | 8       4       7        |
*--------------------------------------------------------------------------------*

(9=2) r5c6 - r8c6 = r8c5 - (2-9) r1c5 => - 9 r3c6, r456c5; stte

This logic chain is called a W Wing and is based on 2 cells that hold the same 2 candidates (in this case r5c6 (cell a) and r1c5 (cell d) both hold 29).

In addition there is a Strong link on 2 in Row 8 (in cells b and c - one of them must be 2).

So the W Wing starts by assuming 9 is False in the cell marked a. So 2 must be True in a, False in b, True in c (because of the Strong link in Row 8) and False in d, so 9 must be True in cell d.

You can reverse this logic by assuming 9 is False in cell d and following the chain in the direction d-c-b-a you deduce that 9 must be True in cell a.

What all this proves is that at least one of cells a or d must be 9. They might both be 9 but they can't both not be 9.

Now 6 cells, r123c6 and r456c5 can see both of cells a and d, so none of them can be 9. That eliminates four 9's as indicated in the diagram. In particular it removes all but one of the 9's in Column 5, so r1c5 = 9.

You will find that the puzzle then solves trivially !!

Here is a link to a site that discusses a whole variety of solving techniques, including the ones I have shown. It's one of the best teaching sites around (in my opinion).

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

Phew, that was a long post - hopefully I've not made too many typos.

Leren