Can't force chains on this one??

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Can't force chains on this one??

Postby bradles » Mon Feb 27, 2006 7:20 am

Code: Select all
 *-----------*
 |.5.|.8.|.7.|
 |.4.|...|2..|
 |7..|.9.|..1|
 |---+---+---|
 |1..|...|.2.|
 |...|876|...|
 |.3.|...|..9|
 |---+---+---|
 |6..|.3.|..7|
 |..3|...|.8.|
 |.2.|.5.|.6.|
 *-----------*


 *-----------*
 |.5.|.8.|.7.|
 |.4.|...|2..|
 |7..|.9.|..1|
 |---+---+---|
 |1.5|.4.|.2.|
 |.9.|876|...|
 |83.|.2.|.49|
 |---+---+---|
 |6..|.3.|..7|
 |5.3|...|.82|
 |.2.|.5.|.6.|
 *-----------*

 
 *--------------------------------------------------------------------*
 | 239    5      1269   | 12346  8      1234   | 3469   7      346    |
 | 39     4      1689   | 13567  16     1357   | 2      359    568    |
 | 7      68     268    | 23456  9      2345   | 3458   35     1      |
 |----------------------+----------------------+----------------------|
 | 1      67     5      | 39     4      39     | 678    2      68     |
 | 24     9      24     | 8      7      6      | 135    135    35     |
 | 8      3      67     | 15     2      15     | 67     4      9      |
 |----------------------+----------------------+----------------------|
 | 6      18     489    | 1249   3      12489  | 1459   159    7      |
 | 5      17     3      | 14679  16     1479   | 149    8      2      |
 | 49     2      4789   | 1479   5      14789  | 1349   6      34     |
 *--------------------------------------------------------------------*


Any ideas on this tough one?

Brad
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Postby vidarino » Mon Feb 27, 2006 8:37 am

See if you can work out why R4C2 can't contain a 6.:)

(Hint: The 8s in box 3.)

Vidar
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Postby bradles » Mon Feb 27, 2006 11:50 am

vidarino wrote:See if you can work out why R4C2 can't contain a 6.:)

(Hint: The 8s in box 3.)

Vidar


Hi Vidar,

Finally, I think I have got it...after a lot of nutting out.

if r4c2=6, r6c3=7, r6c7=6, r4c9=8 therefor r3c7 would have to = 8 meaning two 8s in row 3.

Is that how you worked it out?

Brad
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Postby vidarino » Mon Feb 27, 2006 11:57 am

bradles wrote:
vidarino wrote:See if you can work out why R4C2 can't contain a 6.:)

(Hint: The 8s in box 3.)

Vidar


Hi Vidar,

Finally, I think I have got it...after a lot of nutting out.

if r4c2=6, r6c3=7, r6c7=6, r4c9=8 therefor r3c7 would have to = 8 meaning two 8s in row 3.

Is that how you worked it out?

Brad


Almost identical to my reasoning.:) I noted that the 6s would put an 8 in both R3C2 and R4C9, who in turn would eliminate all the 8s in box 3, but it's basically the same as yours.

Vidar
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Postby Carcul » Tue Feb 28, 2006 11:32 am

Hi Bradles and Vidarino.

The deduction r4c2<>6 could also be made indirectly with the following loop:

[r7c2]=8=[r3c2]-8-[r2c3]=8=[r2c9]-8-[r4c9]-6-[r4c2]-7-[r8c2]-1-[r7c2], => r7c2<>1.

However, r4c2<>6 does not solve the puzzle, with additional steps being required. So, I have tryed to spot a one step solution for this grid and here is what I have found:

[r4c7](-8-[r3c7]=8=[r2c9]-8-[r2c3])-8-[r4c9](-6-[r1c9])-6-[r4c2]-7-[r8c2]{-1-[r7c2]-8-[r3c2](-6-[r2c3])-6-[r3c7]}-1-[r8c5]=1=[r2c5]-1-[r2c3]-9-[r2c8|r3c8](-3,5-[r3c7])-3-[r1c9]-4-[r3c7].

This Single Implication Network means that, if r4c7=8 then r3c7 would be an empty cell. This is a contradiction, and so we must have r4c7<>8 which solve the puzzle.

Regards, Carcul
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Postby bradles » Tue Feb 28, 2006 10:11 pm

Carcul wrote:However, r4c2<>6 does not solve the puzzle...

Not sure what you mean by this Carcul as this was the only way I could solve the puzzle.

I get lost in your notation, due to the fact that I am not familiar with the notation. Is there somewhere that explains the notation and how to understand it?

Brad
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Postby ravel » Wed Mar 01, 2006 10:43 am

bradles wrote:I get lost in your notation, due to the fact that I am not familiar with the notation. Is there somewhere that explains the notation and how to understand it?


It is rather easy to follow the chain, if you pencil step by step on paper (mark eliminated and forced candidates)
[r4c7]-8-[r3c7] means that if r4c7=8 then r3c7<>8
[r3c7]=8=[r2c9] means that if r3c7<>8 then r2c9=8 (because it is a conjugated pair in box 3)
and so on, -6-[r3c7] is IMO not needed (there is no 6 in r3c7)
[r2c8|r3c8](-3,5-[r3c7]-3-[r1c9]-4-[r3c7] at the end means that a pair 35 is left in r2c8 and r3c8, which excludes 35 from r3c7 and 3 from r1c9, which must be 4 then (the 6 was eliminated before), so that r3c7 cannot be 4 (which is the last candidate)
Steps that are needed later are denoted bold or in brackets.

Carcul wrote:However, r4c2<>6 does not solve the puzzle...

Not sure what you mean by this Carcul as this was the only way I could solve the puzzle.

How did you proceed here?
Code: Select all
 *------------------------------------------------------------------*
 |29      5      1      | 2346   8      234    |  3469   7     346  |     
 |3       4      89     | 567    1      57     |  2      59    568  |     
 |7       6      28     | 2345   9      2345   |  3458   35    1    |     
 |----------------------+----------------------+--------------------|
 |1       7      5      | 39     4      39     |  68     2     68   |     
 |24      9      24     | 8      7      6      |  135    135   35   |     
 |8       3      6      | 15     2      15     |  7      4     9    |     
 |----------------------+----------------------+--------------------|
 |6       8      49     | 124    3      124    |  145    159   7    |     
 |5       1      3      | 479    6      479    |  49     8     2    |     
 |49      2      7      | 149    5      8      |  1349   6     34   |     
 *------------------------------------------------------------------*
ravel
 
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Postby Carcul » Wed Mar 01, 2006 12:44 pm

Ravel wrote:-6-[r3c7] is IMO not needed (there is no 6 in r3c7)


If you read carefully my post, you will conclude that I was looking for a one step solution: now, the "6" in r3c7 can be excluded with colors, but then my solution would have two steps. So, considering this puzzle after the basic steps, we do have a "6" in r3c7, and then the SIN solves the puzzle, without the need for colors to remove the "6" from r3c7.

Bradles wrote:I get lost in your notation, due to the fact that I am not familiar with the notation. Is there somewhere that explains the notation and how to understand it?


Refer here and here.

Regards, Carcul
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Postby ravel » Wed Mar 01, 2006 1:43 pm

Carcul wrote:the "6" in r3c7 can be excluded with colors, but then my solution would have two steps.

Ah, i just looked at Brad's candidate list, i did not notice that a grouped turbot fish was necessary to eliminate the 6 - sorry.
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Postby ronk » Wed Mar 01, 2006 2:06 pm

Carcul wrote:So, I have tryed to spot a one step solution for this grid and here is what I have found:

[r4c7](-8-[r3c7]=8=[r2c9]-8-[r2c3])-8-[r4c9](-6-[r1c9])-6-[r4c2]-7-[r8c2]{-1-[r7c2]-8-[r3c2](-6-[r2c3])-6-[r3c7]}-1-[r8c5]=1=[r2c5]-1-[r2c3]-9-[r2c8|r3c8](-3,5-[r3c7])-3-[r1c9]-4-[r3c7].

This Single Implication Network means that, if r4c7=8 then r3c7 would be an empty cell. This is a contradiction, and so we must have r4c7<>8 which solve the puzzle.

If I tried that on a bivalue/bilocation plot, I'd be cross-eyed from trying to follow all the lines. Is that how you worked it out?

Ron
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