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Medusa 1: "Crossed Blades" - a new pattern?

 
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Bob Hanson

Joined: 05 Oct 2005
Posts: 187
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Location: St. Olaf College

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PostPosted: Tue Nov 08, 2005 1:21 pm    Post subject: Medusa 1: "Crossed Blades" - a new pattern? Reply with quote

Now that I have Sudoku Assistant http://www.stolaf.edu/people/hansonr/sudoku running smoothly, I thought it would be interesting to explore some of the implications of mixing strong and weak chains. Here is a theoretical example of a pattern I am simply calling "Medusa 1". I have no specific Sudoku to illustrate this. Please accept it simply as a theoretical idea.

I need to introduce a new notation, but I think it's pretty easy. It goes like this:

Code:
-----------1--------1--------------


This just means "there are only two 1s in some row, column, or block". Thus, this could be a row, but it wouldn't have to be. I'm calling this, for lack of anything better, a "blade." I DO SO HATE esoteric lingo, so if you have a better idea or a more standard term, let me know.

and
Code:
X....(*2)


This just means "any 2 that is a weakly connected to X" -- that is, in the same row, column or block as X. Again, not implying row here, but maybe a row, maybe a column, maybe a block.

A lot of analyses depend upon seeing cells with only two possibilities. What I'm suggesting is that it isn't so hard to spot and therefore use rows, columns, or blocks with only two occurances of the same candidate.
Here's the Medusa idea: Mix the sort of coloring attributable to cells with only two values, 12, with the sort of "X-cycle" analysis usually reserved for single-candidate cycles, like X-wings.

Medusa 1 ("crossed blades"?):

Code:


                  |
                  |
--------1--------1+----
        .         2
        .         |3
        .         |
      (*1)        |
        .         |
        .         |
        1         |
         2..(*2)..2
                  |
                  |


This says: We have two "blades" that intersect at a certain cell.
Each blade has a second value, and common to both of those OTHER
endpoints, we have a cell with only those two possibilities.

Then one can eliminate all 1s and 2s that are weak corners as indicated.

Why it works:

The blades form two strong chains with a weak link:

Code:

  blade-cell-blade
  1----1....2----2


This weak link can transmit only a T-->F, but that is fine.
In this analysis, we always start with a FALSE, because our aim
is to show that at the other end we have a T for the same candidate.
To be useful, we have to be able to put a F in at the OTHER end and
come up with a TRUE at the first end. That is, the F--->T must
correlate with a T<---F.

So here we have:

Code:

  1----1....2----2

  F--->T--->F--->T

and

  1----1....2----2

  T<---F<---T<---F


The crossed blades satisfy our two-way street requirement. But the problem is, they
don't end with the same candidate. So we add a single weakly linked 12 cell:

Code:

  blade-cell-blade--w--cell
  1----1....2----2....2----1

  F--->T--->F--->T--->F--->T

and

  T<---F<---T<---F<---T<---F


There! Now any candidate 1 weakly linked to BOTH ends may be eliminated.
That's the (*1) in the diagram above.

Same goes for 2. This time we think of the 12 cell as being connected on the left:

Code:

   cell--w--blade-cell-blade
  2----1....1----1....2----2

  F--->T--->F--->T--->F--->T

and

  T<---F<---T<---F<---T<---F


Any candidate 2 weakly linked to BOTH ends may be eliminated.
That's the (*2) in the diagram above.

Useful? Common? I don't have any idea.
So that's the theory of it.
I suppose you want a real example.... Help?
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Ruud
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Joined: 17 Sep 2005
Posts: 708
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Location: Netherlands

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PostPosted: Tue Nov 08, 2005 1:57 pm    Post subject: Reply with quote

Code:
                  |
                  |
--------1--------1+----
        .         2
        .         |3
        .         |
      (*1)        |
        .         |
        .         |
        1         |
         2..(*2)..2
                  |
                  |

When the Left-Bottom cell only has candidates 1 and 2, the Right-Top cell (in the crossing) can only support candidates 1 and 2, because a third candidate would force 1 in Left-Top and 2 in Right-Bottom, leaving no candidates for Left-Bottom. The 3 in your picture is not an option.

Doest it exist? You'd have to write some code in your program to recognize the pattern and run it against a large collection like the top870.
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Bob Hanson

Joined: 05 Oct 2005
Posts: 187
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Location: St. Olaf College

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PostPosted: Tue Nov 08, 2005 3:32 pm    Post subject: Reply with quote

OK, very good! So it is even better:

Code:

                  |
                  |
--------1--------1+----
        .         2
        .         |X
        .         |
      (*1)        |
        .         |
        .         |
        1         |
         2..(*2)..2
                  |
                  |


...and X can be eliminated as well.

Excellent! A lot like an XY-wing, I think.
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St. Olaf College
Northfield, MN
http://www.stolaf.edu/people/hansonr
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Bob Hanson

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PostPosted: Tue Nov 08, 2005 5:07 pm    Post subject: Reply with quote

Thanks, Ruud. I set Sudoku Assistant so that it doesn't connect strong edges -- only chains of length 2. Interesting, because that makes it more "human". Of course, it can't SOLVE all Sudoku puzzles in this mode, but looky here:

Code:

   |---c1--|---c2--|---c3--||---c4--|---c5--|---c6--||---c7--|---c8--|---c9--
-----------------------------------------------------------------------------
r1 |     4 |    89 |   3789 ||     2 |     1 |    58 ||    35 |     6 |    79
   |       |       |       ||       |       |       ||       |       |       
---+-------+-------+-------||-------+-------+-------||-------+-------+-------
r2 |    19 |    16 |   369 ||    39 |     7 |    45 ||    45 |     8 |     2
   |       |       |       ||       |       |       ||       |       |       
---+-------+-------+-------||-------+-------+-------||-------+-------+-------
r3 |   789 |     5 |     2 ||    39 |     6 |    48 ||   134 |    13 |    79
   |       |       |       ||       |       |       ||       |       |       
===========================||=======================||=======================
r4 |     5 |    28 |     1 ||     6 |    38 |     9 ||     7 |    23 |     4
   |       |       |       ||       |       |       ||       |       |       
---+-------+-------+-------||-------+-------+-------||-------+-------+-------
r5 |     6 |     7 |    48 ||    14 |    38 |     2 ||    13 |     9 |     5
   |       |       |       ||       |       |       ||       |       |       
---+-------+-------+-------||-------+-------+-------||-------+-------+-------
r6 |     3 |    29 |    49 ||    14 |     5 |     7 ||     8 |    12 |     6
   |       |       |       ||       |       |       ||       |       |       
===========================||=======================||=======================
r7 |   789 |    16 |   789 ||    78 |     4 |    16 ||     2 |     5 |     3
   |       |       |       ||       |       |       ||       |       |       
---+-------+-------+-------||-------+-------+-------||-------+-------+-------
r8 |    17 |     3 |   567 ||    57 |     2 |    16 ||     9 |     4 |     8
   |       |       |       ||       |       |       ||       |       |       
---+-------+-------+-------||-------+-------+-------||-------+-------+-------
r9 |     2 |     4 |    58 ||    58 |     9 |     3 ||     6 |     7 |     1
   |       |       |       ||       |       |       ||       |       |       
-----------------------------------------------------------------------------


This is a "crossed blades".

We have only two 9s in row 7, only two 8s in column 1, intersecting at r7c1. Our "12" cell is r1c2, {8,9}.

The 9s in r1c3 and r2c3 can be eliminated along with the 8 in r1c3.

I think I could spot this....
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Bob Hanson

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PostPosted: Tue Nov 08, 2005 5:16 pm    Post subject: Reply with quote

no, sorry, not the 8 in r1c3.
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Bob Hanson
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Bob Hanson

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PostPosted: Tue Nov 08, 2005 5:30 pm    Post subject: Reply with quote

I think actually, this is an example of still another pattern:

Crossed Blades II ?

Code:

                  |
                  |
--------1--------1+----
        .         2
        .         |3
        .         |
        .         |
        1         |
         2        |
          .       |
           .      |
            (*2)..2
                  |
                  |


This says: We have two "blades" that intersect at a certain cell.
Each blade has a second value, and common to ONE of those OTHER
endpoints, we have a cell with only those same two possibilities.

Then one can eliminate all 2s that are weak corners as indicated.

This is actually probably far more frequently found than Crossed Blades I,
where it is stated that the "12 cell" is weakly associated with BOTH blades.
Here it need only be associated with one, as in the actual example, above.

I do think there's more to discover here.
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http://www.stolaf.edu/people/hansonr
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