The New Sudoku Players' Forum

[Carcul]

Let's suppose that we have two almost locked sets A and B with at least two common candidates "x" and "y": A (x,y,a,b,c,...); B (x,y,d,e,f,...).
The ALS A is located in unit "UA" and B in "UB". Now, if "x" can occur for both A and B only in unit "U1", and if "y" can occur for both A and B only in unit "U2" (where U1 and U2 are different units, or we would have a N-tuple), then we can make the following eliminations:

- all other "x" in U1 not belonging to A and B can be eliminated;
- all other "y" in U2 not belonging to A and B can be eliminated;
- all other a,b,c,... in UA not belonging to A can be eliminated;
- all other d,e,f... in UB not belonging to B can be eliminated.

The first two eliminations are due to the ALSs xz-rule of Bennys.
Now, I apologize for illustrating this concept with the following silly example (silly because is from a very simple puzzle, and also because the concept is not even necessary for solving the grid), but let's focus just in the idea. The initial puzzle:

Code: Select all

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


and let's concentrate only in boxes 1 and 2:

Code: Select all

 145    25    3      | 12456  12456   16  |
 459    259   24569  | 7      8       69  |
 145789 25789 1245689| 124569 1234569 1369|   
---------------------+--------------------+-


The locked candidate "6" on row 1, box 2 will provide a straightforward solution for the puzzle. But please note the following: we have two almost locked sets in the conditions described above - A (r1c1, r1c2, r2c1, r2c2), and B (r1c6, r2c6). For these, x = 1 and y = 9, and the following eliminations can be made:

r2c3<>2,4,5,9
r3c1<>4,5
r3c2<>2,5
r3c3<>2,4,5
r1c4<>1,6
r1c5<>1,6
r3c4<>6
r3c5<>6
r3c6<>6

This makes a total of 18 eliminations with a single logical argument.

If someone find a really useful example of this concept (which I guess might be very difficult), please post in this thread.

Carcul

[ronk]

Let's suppose that we have two almost locked sets A and B with at least two common candidates "x" and "y": A (x,y,a,b,c,...); B (x,y,d,e,f,...).
The ALS A is located in unit "UA" and B in "UB". Now, if "x" can occur for both A and B only in unit "U1", and if "y" can occur for both A and B only in unit "U2", then we can make the following eliminations:

- all other "x" in U1 not belonging to A and B can be eliminated;
- all other "y" in U2 not belonging to A and B can be eliminated;
- all other a,b,c,... in UA not belonging to A can be eliminated;
- all other d,e,f... in UB not belonging to B can be eliminated.

The first two eliminations are due to the ALSs xz-rule of Bennys.

Very nice description, but I suggest you add that U1 and U2 can be one and the same unit.

The first time I saw this rule was here where ...

IV. Almost-locked sets mutual exclusion rule
-------------------------------

If A and B are almost-locked sets
x,y restricted common to A,B

any z common to any other candidates of A OR B may be eliminated

and

any x or y common to BOTH A AND B may be eliminated

Bob was obviously trying to follow bennys' style of writing ALS rules.

Now, I apologize for illustrating this concept with the following silly example (silly because is from a very simple puzzle, and also because the concept is not even necessary for solving the grid), but let's focus just in the idea.
....................
If someone find a really useful example of this concept (which I guess might be very difficult), please post in this thread.

LOL! Doing so allowed you to show 18 eliminations, didn't it?

Bob's example only yielded 3 eliminations, but they were "useful", i.e., they occurred at a point where "basic techniques" could provide no further progress. For example, Angus Johnson's SS displays "No hint available".

Code: Select all

  459      1     47  |   78      3     89A |    2     58      6 
   59    569    679  |  278    289      4  |   35   1358     13 
    8      3      2  |    5      6      1  |    7      9      4 
---------------------+---------------------+--------------------
  139      7    139  |    6      4      5  |    8    123   1239 
    6     28    134  |    9     28      7  |   34     13      5 
   24    289      5  |  238      1    238* |   49      6      7 
---------------------+---------------------+--------------------
    7      4     39  |   23B   259*     6  |    1    235      8 
12359   2569   1369  |    4   2589*  2389B | 3569      7    239 
 2359   2569      8  |    1      7    239B | 3569      4    239 

sets A: {r1c6} = {89}
     B: {r7c4,r8c6,r9c6} = {2389}
     x,y = 8,9
     d,e = 2,3
     yielding eliminations(*) r6c6<>8, r7c5<>2, and r8c5<>2

Additional advanced technique(s) are required to solve the puzzle.

[Carcul]

Hi Ronk.

Very nice description, but I suggest you add that U1 and U2 can be one and the same unit.

Thanks. Very good sugestion, I didn't thought in that possibility (original post already edited).

The first time I saw this rule was here

I didn't know that someone have already described this. Even then, it is sometimes good to find things by ourselves.

Regards, Carcul

[ronk]

Even then, it is sometimes good to find things by ourselves.

I agree and am glad you started a thread on this variant of the almost-locked-set xz rule. The topic deserves its own thread, and hopefully we'll see active participation and meaningful contributions.

Regards, Ron

[Carcul]

Bob's example only yielded 3 eliminations, but they were "useful", i.e., they occurred at a point where "basic techniques" could provide no further progress. For example, Angus Johnson's SS displays "No hint available".

Code: Select all

  459      1     47  |   78      3     89A |    2     58      6 
   59    569    679  |  278    289      4  |   35   1358     13 
    8      3      2  |    5      6      1  |    7      9      4 
---------------------+---------------------+--------------------
  139      7    139  |    6      4      5  |    8    123   1239 
    6     28    134  |    9     28      7  |   34     13      5 
   24    289      5  |  238      1    238* |   49      6      7 
---------------------+---------------------+--------------------
    7      4     39  |   23B   259*     6  |    1    235      8 
12359   2569   1369  |    4   2589*  2389B | 3569      7    239 
 2359   2569      8  |    1      7    239B | 3569      4    239 

sets A: {r1c6} = {89}
     B: {r7c4,r8c6,r9c6} = {2389}
     x,y = 8,9
     d,e = 2,3
     yielding eliminations(*) r6c6<>8, r7c5<>2, and r8c5<>2

Additional advanced technique(s) are required to solve the puzzle.

This is a very good example. However, it is not needed to solve the grid:

[r1c3]=4=[r1c1]-4-[r6c1]-2-[r5c2]-8-[r5c5]-2-[r2c5|r1c6]-8-[r1c4]-7-[r1c3], => r1c3<>7

which solve the puzzle.

Regards, Carcul

[ronk]

This is a very good example. However, it is not needed to solve the grid:

[r1c3]=4=[r1c1]-4-[r6c1]-2-[r5c2]-8-[r5c5]-2-[r2c5|r1c6]-8-[r1c4]-7-[r1c3], => r1c3<>7

which solve the puzzle.

Nice find, but nothing is ever needed to solve a puzzle IMO. There are "50 ways to leave your lover", as Paul Simon wrote.

Whether someone would first spot a nice loop with 8 cells (and with an almost-locked-set) ... or an almost-locked-set xz rule with 4 cells ... is subject to experience and proclivity ... and is certainly debatable.

In any case, I'm surprised you would go off-topic on your own thread.

Regards, Ron

P.S. I notice your nice loop uses the bivalue r6c2=24, as originally posted by Bob Hanson ... but I have no idea how the elimination r6c1<>9 was made. [edit: r6c1 was typo'd r6c2]

[Carcul]

P.S. I notice your nice loop uses the bivalue r6c2=24, as originally posted by Bob Hanson ... but I have no idea how the elimination r6c2<>9 was made.

Do you mean r6c1, right? R6c1<>9 due to the following:

[r6c1]=4=[r6c7]-4-[r5c7]-3-[r2c7]-5-[r2c1]-9-[r6c1].

In fact, we could also prove that r6c2<>9:

[r6c2]=8=[r6c4|r6c6]-8-[r5c5]-2-[r2c5|r1c6]-8-[r1c4]-7-[r1c3]-4-[r5c3]
=4=[r5c7]-4-[r6c7]-9-[r6c2].

Regards, Carcul

[ronk]

Very nice description, but I suggest you add that U1 and U2 can be one and the same unit.

Thanks. Very good suggestion, I didn't [edit: think of] that possibility (original post already edited).

Hi Carcul. At the time I didn't realize that U1 and U2 being one and the same unit was a degenerative case, namely that of a naked N-tuple (naked pair, triple, quad, etc.). You may wish to reconsider your edit.

Ron

[Carcul]

Hi Ronk.

Thanks again. In your first sugestion I believed it and didn´t bother to confirm, but its very good that you have made now this conclusion, namely, that if the units were the same then it were a simple N-tuple. I guess that, for all of us (at least certainly for me), in sudoku puzzles is sometimes very easy to not see the obvious.

Regards, Carcul