Sudoku Players' Forum (clone)

RW used the following technique to solve one of the puzzles in the effortless extreme thread. Havard suggested that I post the approach on the advanced solving techniques so that others might see it. The approach takes advantage of a sparsely populated box combined with a bivalue cell (or an ALS in the more general form) so it is fairly easy to spot. The puzzle is

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.4..9...76.85....2..5....8....2..7..41..36........1.3.2..3...1.9...4.........9.2.
+---------------------+-------------------+--------------------+
|    13      4      2 |   168      9   38 |   1356   56      7 |
|     6    379      8 |     5     17  347 |   1349   49      2 |
|   137    379      5 |  1467      2  347 |  13469    8   3469 |
+---------------------+-------------------+--------------------+
|  *358  *3568    369 |     2    #58 -458 |      7  469      1 |
|     4      1     79 |   789      3    6 |      2   59    589 |
|  -578      2    679 |  4789    578    1 |   4689    3   4689 |
+---------------------+-------------------+--------------------+
|     2   5678    467 |     3   5678  578 |  45689    1  45689 |
|     9   3568    136 |    18      4    2 |   3568    7   3568 |
|  3578  35678  13467 |   178  15678    9 |  34568    2  34568 |
+---------------------+-------------------+--------------------+

Row 4, box 3 are almost locked candidates for both 5 and 8 except for the values located in r6c1. Because of the bivalue ("58" in r4c5 no other cells in the row besides the bivalue or those of the almost locked candidates may contain "5" or "8" (if so then r6c1 must contain both). Likewise r6c1 must contain "5" or "8" (if not then r4c5 is empty).

I should let Havard come up with designations for the approach ("almost locked candidates"?) as he has a much more colorful way of expressing these concepts.

More generally, consider a box with 2 candidates, "a" and "b", existing only in a box-line plus one cell and both candidates in the line restricted common to an ALS:

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+--------------------------+--------------------------
|                         a,b
| (a)(b)X (a)(b)Y (a)(b)Z=====ALS(a,b,...) (a)(b)U ...
|   abW       -       -    |
|    -        -       -    |
+--------------------------+--------------------------

where "(a)" implies "a" is optional and "-" implies "a" and "b" are not candidates in the cell. Then the cell in the box, "abW", must be either "a" or "b". In addition, "a" and "b" can be eliminated from all other cells, "(a)(b)U", common to the box-line and the ALS. In the above example r4c6<>5,8 and r6c1<>7 where ALS(a,b,...)=r4c5="58".

This is sort of the hidden ALS counterpart to the normal ALS

Normal ALS has something like n+1 digits in n cells in a group

Hidden ALS has n digits that can only lie in n+1 cells in a group

An ALS containing those n digits can force one of those n digits out of some of the hidden ALS cells that share a group with the ALS.

In this example:

Hidden ALS = 58 in r4c12 & r6c1.
Normal ALS = 58 in r4c5

Therefore r6c1 = 58 & r4c125 = (5&8)

Very cool!

Mike Barker wrote:RW used the following technique to solve one of the puzzles in the effortless extreme thread.
(...)

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.4..9...76.85....2..5....8....2..7..41..36........1.3.2..3...1.9...4.........9.2.
+---------------------+-------------------+--------------------+
|    13      4      2 |   168      9   38 |   1356   56      7 |
|     6    379      8 |     5     17  347 |   1349   49      2 |
|   137    379      5 |  1467      2  347 |  13469    8   3469 |
+---------------------+-------------------+--------------------+
|  *358  *3568    369 |     2    #58 -458 |      7  469      1 |
|     4      1     79 |   789      3    6 |      2   59    589 |
|  -578      2    679 |  4789    578    1 |   4689    3   4689 |
+---------------------+-------------------+--------------------+
|     2   5678    467 |     3   5678  578 |  45689    1  45689 |
|     9   3568    136 |    18      4    2 |   3568    7   3568 |
|  3578  35678  13467 |   178  15678    9 |  34568    2  34568 |
+---------------------+-------------------+--------------------+

Row 4, box 3 are almost locked candidates for both 5 and 8 except for the values located in r6c1. Because of the bivalue ("58" in r4c5 no other cells in the row besides the bivalue or those of the almost locked candidates may contain "5" or "8" (if so then r6c1 must contain both). Likewise r6c1 must contain "5" or "8" (if not then r4c5 is empty).

I think RW's viewpoint is appropriate when solving without pencilmarks. When using candidates (pencilmarks), however, it's easier for me to see a doubly-linked ALS xz-rule, specifically ...

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Set A = {r4c5} = {58}
    B = {r4c123,r56c3} = {356789}
    x,z = 5,8 and/or x,z = 8,5

    for exclusions r4c6<>5, r4c6<>8, and r6c1<>7

It may also be viewed as a SueDeCoq using three ALSs.

One other way to see this elimination is by a Continuous Nice Loop with grouped strong links:

[r6c1]=5=[r4c12]-5-[r4c5]-8-[r4c12]=8=[r6c1] => r6c1<>7, r4c6<>5,8

I think it is in order to pass the cells r4c12 twice in the loop.

/Viggo

There is a "hidden ALS" in cell pair r5c6 and r6c4 of puzzle 207 of the top1465.

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53..2.9...24.3..5...9..........1.827...7.........981.............64....91.2.5.43.

 5     3     178   | 168   2     1467  | 9     14678 1468
 678   2     4     | 1689  3     1679  | 67    5     168
 678   1678  9     | 1568  4678  14567 | 23    14678 23
-------------------+-------------------+------------------
 3469  4569  35    | 356   1     3456  | 8     2     7
 23468 14568 1358  | 7     46    23456 | 356   9     3456
 23467 4567  357   | 2356  9     8     | 1     46    3456
-------------------+-------------------+------------------
 49    49    3578  | 123   678   123   | 2567  1678  12568
 378   578   6     | 4     78    123   | 257   178   9
 1     78    2     | 689   5     679   | 4     3     68

There is another in cell pair r4c12. How would you deduce the corresponding exclusions?

Nice find.

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------------
53.|.2.|9..
.24|.3.|.5.
..9|...|...
------------
...|.1.|827
...|7..|.9.
...|.98|1..
------------
...|...|...
..6|4..|..9
1.2|.5.|43.
------------


2,3,5 have 4 cells in box 5. Since 2 has to go to r5c6 or r6c4, there must be a 3 or 5 in r4c46, which pairs with 35 in r4c3 - eliminate 35 from r4c12. Since not both r4c4 and r4c6 can be 3 and 5, 46 can be eliminated from r5c6 and r6c4 (one must be 2, the other 3 or 5).