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Nightmare, May 11

 
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Thu May 11, 2006 12:12 pm    Post subject: Nightmare, May 11 Reply with quote

Todays nightmare was a fine uniqueness exercise for me. First i came here:
Code:
 8     7     3    | 2      6     1      | 9      5     4         
 16    4     16   | 59     8     59     | 7      3     2         
 9     2     5    | 7      34    34     | 8      1     6   
-----------------------------------------------------------       
 346   1     7    | 348    9     23468  | 5      24    38         
 2     8     46   | 1345   34    3456   | 134    7     9         
 34    5     9    | 1348   7     2348   | 1234   6     138       
-----------------------------------------------------------       
 5     6     148  | 38     2     7      | 134    9     13         
 7     3     2    | 49     1     49     | 6      8     5         
 14    9     148  | 6      5     38     | 1234   24    7   

There is a UR 34 in r35c56, which allows the elimination of 34 in r5c6.

The next suspicious rectangle is in r46c14. If there would be no 1 in r6c4 and no 6 in r4c1, we could eliminate the 8 from r7c4.
We get this by setting r5c4=1, then (because of the 34-pair in r5c57) follows r5c3=6.
Now r7c4 would become 3 and r7c9=1 => r6c9<>1 and we would have this situation in band 2:
Code:
-----------------------------------------------------------       
 34    1     7    | 48     9     23468  | 5      24    38         
 2     8     6    | 1      34    56     | 34     7     9         
 34    5     9    | 48     7     2348   | 1234   6     38       
-----------------------------------------------------------   
This is a deadly UR in the 6 cells r46c149 for numbers 348.

So r5c4 cannot be 1, r6c4=1 and we come here:
Code:
 8     7     3    | 2      6     1      | 9      5     4         
 16    4     16   | 59     8     59     | 7      3     2         
 9     2     5    | 7      34    34     | 8      1     6   
-----------------------------------------------------------       
 346   1     7    | 48     9     2468   | 5      24    38         
 2     8     46   | 345    34    56     | 1      7     9         
 34    5     9    | 1      7     248    | 24     6     38       
-----------------------------------------------------------       
 5     6     48   | 38     2     7      | 34     9     1         
 7     3     2    | 49     1     49     | 6      8     5         
 14    9     148  | 6      5     38     | 234    24    7   
Here the suspicious cells are r2c46=59, r8c46=49 and r5c46. We know, that not both r5c4 and r4c6 can be 4 and 5, otherwise we would have a deadly pattern. Since there is no other 5 in row 5, the 4 must be outside the 2 cells and can be eliminated from r5c4.

Then the puzzle can be solved with an xy-wing from r4c4 (r7c4|r5c5 = 3) => r5c4<>3

BTW: Since my friend doesnt want to install this .NET stuff on her windows PC, i dont have the SudoCue program. Can someone tell me, how it solves it? I assume with disjoint subsets, but cant find effective ones.


Last edited by ravel on Fri May 12, 2006 12:09 pm; edited 2 times in total
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Thu May 11, 2006 1:02 pm    Post subject: Interesting approach Reply with quote

That's a very interesting approach to the puzzle, ravel.
Code:
 8     7     3    | 2      6     1      | 9      5     4         
 16    4     16   | 59     8     59     | 7      3     2         
 9     2     5    | 7      34    34     | 8      1     6   
-----------------------------------------------------------       
 346   1     7    | 348    9     23468  | 5      24    38         
 2     8     46   | 1345   34    56     | 134    7     9         
 34    5     9    | 1348   7     2348   | 1234   6     138       
-----------------------------------------------------------       
 5     6     148  | 38     2     7      | 134    9     13         
 7     3     2    | 49     1     49     | 6      8     5         
 14    9     148  | 6      5     38     | 1234   24    7

At this point (after eliminating {3, 4} at r5c6) we have a very nice double-implication chain.

A. r3c6 = 4 ==> r8c6 = 9 ==> r2c6 = 5 ==> r5c6 = 6
B. r3c6 = 4 ==> r3c5 = 3 ==> r5c5 = 4 ==> r5c3 = 6

So r3c5 = 4, and the puzzle solves itself. dcb Smile


Last edited by David Bryant on Thu May 11, 2006 1:40 pm; edited 1 time in total
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Thu May 11, 2006 1:11 pm    Post subject: Re: Interesting approach Reply with quote

David Bryant wrote:
That's a very interesting approach to the puzzle, ravel.

Thanks, for getting some practice i am still looking for any uniqueness solutions, before i start chaining. Of course your solution is easier Smile
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David Bryant



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Posts: 559
Location: Denver, Colorado

PostPosted: Thu May 11, 2006 4:11 pm    Post subject: Hexagons vs rectangles; avoiding uniquity Reply with quote

ravel wrote:

Code:
-----------------------------------------------------------       
 34    1     7    | 48     9     23468  | 5      24    38         
 2     8     6    | 1      34    56     | 34     7     9         
 34    5     9    | 48     7     2348   | 1234   6     38       
-----------------------------------------------------------

This is a deadly UR in the 6 cells r46c149 for numbers 348.

That's a very clever deduction, Ravel. I like the way you think!

I do have just one little nit: since the pattern 348/483 || 483/348 has six vertices, isn't it a "UH" ([non]-Unique Hexagon) and not a "UR" ([non]-Unique Rectangle)? I see that it's degenerate, since it lies within one "band", but it still has six corners.

ravel wrote:
... for getting some practice i am still looking for any uniqueness solutions, before i start chaining. Of course your solution is easier


I also like to get practice, but I usually look for ways to avoid making the uniqueness assumption. Here's another way to work this puzzle.
Code:
 8     7     3    | 2      6     1      | 9      5     4
 16    4     16   | 59     8     59     | 7      3     2
 9     2     5    | 7      34    34     | 8      1     6
-----------------------------------------------------------
 346a  1     7    | 348    9     23468* | 5      24    38a
 2     8     46   | 1345   34    3456   | 134    7     9
 34a   5     9    | 1348   7     2348*  | 1234   6     138a
-----------------------------------------------------------
 5     6     148  | 38     2     7      | 134    9     13
 7     3     2    | 49     1     49     | 6      8     5     
 14    9     148  | 6      5     38B    | 1234A  24    7


Starting from this first position, we can do quite a lot with the digit "3".

A. r9c7 = 3 ==> X-Wing in r4&6c1&9 ==> r4c6 <> 3 & r6c6 <> 3
B. r9c6 = 3 ==> r4c6 <> 3 & r6c6 <> 3

Since there are only two ways to fit a "3" in row 9 we conclude that there cannot be a "3" in either r4c6 or r6c6. Now the grid looks like this.
Code:
 8     7     3    | 2      6     1      | 9      5     4
 16    4     16   | 59     8     59     | 7      3     2
 9     2     5    | 7      34    34     | 8      1     6
-----------------------------------------------------------
 346   1     7    | 348    9     2468   | 5      24    38
 2     8     46   | 1345   34    3456   | 134    7     9
 34    5     9    | 1348   7     248    | 1234   6     138
-----------------------------------------------------------
 5     6     148  | 38     2     7      | 134    9     13
 7     3     2    | 49*    1     49     | 6      8     5     
 14    9     148  | 6      5     38     | 1234   24    7


We can eliminate more "3"s from the middle center 3x3 box by working from "alpha star" r8c4.

A. r8c4 = 9 ==> r8c6 = 4 ==> r3c6 = 3 ==> r3c5 = 4 ==> r5c5 = 3

So if r8c4 = 9 we have r5c5 = 3, and there cannot be a "3" anywhere else in the middle center 3x3 box.

B. r8c4 = 4 ==> {3, 8} pair in r4c4 & r7c4 ==> r5c4 <> 3 & r6c4 <> 3

Now we have ruled out "3" at r5c4 and at r6c4, and the grid looks like this.
Code:
 8     7     3    | 2      6     1      | 9      5     4
 16    4     16   | 5(9)   8     59     | 7      3     2
 9     2     5    | 7      34    34     | 8      1     6
-----------------------------------------------------------
 346   1     7    | 348    9     2468   | 5      24    38
 2     8     46   | 14(5)  34    3456   | 134    7     9
 34    5     9    | (1)48  7     248    | 1234   6     138
-----------------------------------------------------------
 5     6     148  | 38     2     7      | 134    9     13
 7     3     2    | 49*    1     49     | 6      8     5     
 14    9     148  | 6      5     38     | 1234   24    7


Working with the same "alpha star", and with the same chain A (r8c4 = 9 ==> r5c5 = 3) we can also rule out the "3" at r4c4:
r8c4 = 4 ==> r2c4 = 9 ==> r5c4 = 5 ==> r6c4 = 1 ==> {3, 8} pair in c9r4&6

Now we have a beautiful loop leading all the way around the middle center band -- this is your second "UR" ("UH"?), looked at another way.

r8c4 = 4 ==> {3, 8} pair in row 4 ==> r6c1 = 3 ==> r6c9 = 8 ==> r4c9 = 3 ==> r4c4 <> 3

After ruling out the "3" at r4c4 we see that the "3" at r7c4 is unique in column 4, and the rest is singles all the way. dcb
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Marty R.



Joined: 12 Feb 2006
Posts: 5176
Location: Rochester, NY, USA

PostPosted: Thu May 11, 2006 5:34 pm    Post subject: Reply with quote

Quote:
At this point (after eliminating {3, 4} at r5c6) we have a very nice double-implication chain.

A. r3c6 = 4 ==> r8c6 = 9 ==> r2c6 = 5 ==> r5c6 = 6
B. r3c6 = 4 ==> r3c5 = 3 ==> r5c5 = 4 ==> r5c3 = 6

So r3c5 = 4, and the puzzle solves itself.


David, my post serves no purpose except to provide yet another illustration of how different people take different approaches. Had I been perceptive enough to test the "4" in r3c6, I would have done one chain:

r3c6=4-->r3c5=3-->r6c5=4-->r6c3=6-->r6c6=5-->r2c6=9-->r8c6=4

This chain produces duplicate "4" in c6, so the same conclusion is reached that r3c5 must be "3."

Your method is called a DIC and is viewed positively, while mine is a chain that results in a contradiction, which is viewed negatively and is considered trial-and-error by some folks. Very interesting. Very Happy
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ravel



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Posts: 536

PostPosted: Thu May 11, 2006 10:52 pm    Post subject: Re: Hexagons vs rectangles; avoiding uniquity Reply with quote

David Bryant wrote:

I do have just one little nit: since the pattern 348/483 || 483/348 has six vertices, isn't it a "UH" ([non]-Unique Hexagon) and not a "UR" ([non]-Unique Rectangle)?

Yep, think i was too lazy to write (non)unique pattern, "UH" is fine.
Quote:

Starting from this first position, we can do quite a lot with the digit "3".

When i followed your sophisticated deductions, my grinning turned into laughing Smile
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ravel



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PostPosted: Fri May 12, 2006 11:57 am    Post subject: Reply with quote

In another forum Neunmalneun said:
Quote:

... a-b-c technique. I choose a triple and look what happens if one of the candidates was not in the cell. If the cell contains "abc" and "ab" leads to "a" then "b" can be eliminated...

Some of such chains could be found here, e.g.
Code:
 8     7     3    | 2      6     1      | 9      5     4         
 16    4     16   | 59     8     59     | 7      3     2         
 9     2     5    | 7      34    34     | 8      1     6   
-----------------------------------------------------------       
 346   1     7    | 348    9     23468  | 5      24    38         
 2     8     46   | 1345   34    56     | 134    7     9         
 34    5     9    | 1348   7     2348   | 1234   6     138       
-----------------------------------------------------------       
 5     6     148  | 38     2     7      | 134    9     13         
 7     3     2    | 49     1     49     | 6      8     5         
 14    9     148  | 6      5     38     | 1234   24    7   
r4c6<>2 => r6c6=2 => r4c8=2 => r9c8=4 => r7c79=13 => r7c4=8 => r9c6=3 => r4c6=8
=> r4c6<>346 => r4c1=6

BTW: Since my friend doesnt want to install this .NET stuff on her windows PC, i dont have the SudoCue program. Can someone tell me, how it solves it? I assume with disjoint subsets, but cant find effective ones.
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David Bryant



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Location: Denver, Colorado

PostPosted: Fri May 12, 2006 5:16 pm    Post subject: SudoCue's solution Reply with quote

ravel wrote:
BTW: Since my friend doesnt want to install this .NET stuff on her windows PC, i dont have the SudoCue program. Can someone tell me, how it solves it? I assume with disjoint subsets, but cant find effective ones.

Hi, Ravel!

I've appended the "log" file from the SudoCue program. Briefly, it uses a variant of the forcing chain I presented (showing that r3c6 <> 4) to conclude that r3c6 = 3 -- from there the solution is no problem. dcb

Opening game file 'scue051106.ss'
...
R1C9 has Digit 4 as the only remaining candidate
Box 2 found a Naked Pair with Digits {5,9}
Row 1 Digit 9 must be placed in R1C7
Row 3 Digit 9 must be placed in R3C1
Column 7 Digit 7 must be placed in R2C7
Column 8 Digit 9 must be placed in R7C8
R3C8 has Digit 1 as the only remaining candidate
Column 5 Digit 9 must be placed in R4C5
Row 4 Digit 7 must be placed in R4C3
Column 5 Digit 7 must be placed in R6C5
Row 1 Digit 7 must be placed in R1C2
R6C2 has Digit 5 as the only remaining candidate
Row 3 Digit 5 must be placed in R3C3
Box 7 Digit 5 must be placed in R7C1
Box 1 Digit 3 must be placed in R1C3
Row 8 Digit 5 must be placed in R8C9
Row 1 Digit 2 must be placed in R1C4
Box 1 Digit 2 must be placed in R3C2
Column 5 Digit 2 must be placed in R7C5
Column 2 Digit 6 must be placed in R7C2
Box 9 Digit 6 must be placed in R8C7
Column 9 Digit 6 must be placed in R3C9
Row 8 Digit 2 must be placed in R8C3
R3C7 has Digit 8 as the only remaining candidate
Row 8 only has candidates for Digit 4 in Box 8
2 candidates for Digit 3 eliminated by a template check
*** This is apparently the "X-Wing if r9c7 = 3" step I explained in an earlier post.

Placing Digit 4 in R3C6 forces 2 different cells to Digit 3 in Column 6
*** Here is the sub-explanation of this difficult step:
The conflicting forcing chains are

r3c6=4 ==> r3c5<>4 ==> r5c5=4 ==> r5c3<>4 ==> r5c3=6 ==> r4c1<>6 ==> r4c6=6 ==> r5c6<>6 ==> r5c6=3

r5c6=3 ==> r4c4<>3 ==> r4c4=8 ==> r6c6<>8 ==> r6c6=2 ==> r4c6<>2 ==> r4c8=2 ==> r6c7<>2 ==> r9c7=2
r9c7=2 ==> r9c8<>2 ==> r9c8=4 ==> r7c7<>4 ==> r7c7=3 ==> r9c7<>3 ==> r9c6=3

Row 3 Digit 4 must be placed in R3C5
R3C6 has Digit 3 as the only remaining candidate
Column 5 Digit 3 must be placed in R5C5
Row 9 Digit 3 must be placed in R9C7
Box 8 Digit 3 must be placed in R7C4
R9C6 has Digit 8 as the only remaining candidate
Row 9 Digit 2 must be placed in R9C8
Column 7 Digit 2 must be placed in R6C7
Box 9 Digit 4 must be placed in R7C7
R7C9 has Digit 1 as the only remaining candidate
Row 7 Digit 8 must be placed in R7C3
R6C6 has Digit 4 as the only remaining candidate
Column 8 Digit 4 must be placed in R4C8
Row 4 Digit 2 must be placed in R4C6
Row 6 Digit 1 must be placed in R6C4
Column 7 Digit 1 must be placed in R5C7
Row 5 Digit 4 must be placed in R5C3
Column 1 Digit 4 must be placed in R9C1
R6C1 has Digit 3 as the only remaining candidate
Column 4 Digit 4 must be placed in R8C4
R4C4 has Digit 8 as the only remaining candidate
R5C4 has Digit 5 as the only remaining candidate
R8C6 has Digit 9 as the only remaining candidate
R4C1 has Digit 6 as the only remaining candidate
Column 6 Digit 6 must be placed in R5C6
Row 6 Digit 8 must be placed in R6C9
Column 3 Digit 6 must be placed in R2C3
R9C3 has Digit 1 as the only remaining candidate
Column 1 Digit 1 must be placed in R2C1
Row 4 Digit 3 must be placed in R4C9
Column 4 Digit 9 must be placed in R2C4
Row 2 Digit 5 must be placed in R2C6
All Cells Solved
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Fri May 12, 2006 7:42 pm    Post subject: Reply with quote

Many thanks, David,

so i misunderstood, what Ruud said in the other forum:

Ruud wrote:
A nightmare is a sudoku that can be solved without guessing, but with a complicated set of solving techniques required. Some of you must have seen them. They are often used to show certain techniques.

The program will not guess when the sudoku can be solved with:
- singles
- linebox reductions
- disjoint subsets
- any single digit technique, including (finned) fish, coloring, nishio.
- remote pairs
- XY-wing
- XYZ-wing
- BUG+1
- UR types 1-4


I thought, disjoint subsets are the same as old style ALS's, but if i am not missing something, either he is using ALS chains (i am not familiar with) or "normal" chains like yours to show:
Quote:

Placing Digit 4 in R3C6 forces 2 different cells to Digit 3 in Column 6
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David Bryant



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Posts: 559
Location: Denver, Colorado

PostPosted: Fri May 12, 2006 9:44 pm    Post subject: Maybe I made a mistake Reply with quote

ravel wrote:
so i misunderstood, what Ruud said in the other forum:

Maybe not. The program, SudoCue, allows the user to turn certain algorithms on or off. The log file I presented previously had been created with the "uniqueness" tests turned off -- I was testing a puzzle the other day, and I had to turn that feature off to run the "optimize" function, and I didn't review the "options" settings before I ran this puzzle. See further remarks, below.
ravel wrote:
... either he is using ALS chains (i am not familiar with) or "normal" chains like yours to show: Placing Digit 4 in R3C6 forces 2 different cells to Digit 3 in Column 6

He's using the same kind of forcing chains you and I are familiar with. I tried to explain that in my previous post, but maybe it wasn't clear enough.

One of the nice features of Ruud's program is that, if one clicks on a line in the log file, one obtains an expanded explanation of that particular move. Here's what "SudoCue" said about the move r3c6 <> 4:

r3c6=4 ==> r3c5<>4 ==> r5c5=4 ==> r5c3<>4 ==> r5c3=6 ==> r4c1<>6 ==> r4c6=6 ==> r5c6<>6 ==> r5c6=3

r5c6=3 ==> r4c4<>3 ==> r4c4=8 ==> r6c6<>8 ==> r6c6=2 ==> r4c6<>2 ==> r4c8=2 ==> r6c7<>2 ==> r9c7=2
r9c7=2 ==> r9c8<>2 ==> r9c8=4 ==> r7c7<>4 ==> r7c7=3 ==> r9c7<>3 ==> r9c6=3

So that looks like the type of "forcing chain" I look for, but Ruud's way of presenting it is a bit verbose.

------------------------------------------------------------

OK, I made a mistake (sort of) by posting the log file that resulted when I had the "uniqueness" test turned off. Here's what the log file looks like when I have "uniqueness" turned on. Oh -- all the moves up to "2 candidates for digit 3 eliminated by a template check" are exactly the same in this version, so I'm starting this log from that point. dcb

Opening game file 'scue051106.ss'

...

2 candidates for Digit 3 eliminated by a template check
R5C6 values {3,4} eliminated by uniqueness test 1
Placing Digit 3 in R4C1 forces 2 different Digits in R4C8
Column 1 Digit 3 must be placed in R6C1
X-Wing found in Columns {1,8}, Rows {4,9} for Digit 4
Column 4 found a Naked Pair with Digits {3,8}
Row 4 found a Naked Pair with Digits {3,8}
Swordfish found in Rows {3,5,9}, Columns {5,6,7} for Digit 3
Placing Digit 1 in R5C4 eliminates all candidates for R5C5
Row 5 Digit 1 must be placed in R5C7
Column 4 Digit 1 must be placed in R6C4
Row 5 Digit 3 must be placed in R5C5
Column 7 Digit 3 must be placed in R9C7
Box 6 Digit 3 must be placed in R4C9
Column 9 Digit 1 must be placed in R7C9
R6C9 has Digit 8 as the only remaining candidate
R7C7 has Digit 4 as the only remaining candidate
Row 6 Digit 4 must be placed in R6C6
Column 4 Digit 4 must be placed in R8C4
Column 5 Digit 4 must be placed in R3C5
Row 3 Digit 3 must be placed in R3C6
Column 4 Digit 3 must be placed in R7C4
R4C4 has Digit 8 as the only remaining candidate
Row 9 Digit 2 must be placed in R9C8
Column 7 Digit 2 must be placed in R6C7
R9C6 has Digit 8 as the only remaining candidate
R7C3 has Digit 8 as the only remaining candidate
Column 3 Digit 4 must be placed in R5C3
Box 7 Digit 4 must be placed in R9C1
Box 6 Digit 4 must be placed in R4C8
Column 6 Digit 2 must be placed in R4C6
R5C4 has Digit 5 as the only remaining candidate
R8C6 has Digit 9 as the only remaining candidate
Column 4 Digit 9 must be placed in R2C4
R9C3 has Digit 1 as the only remaining candidate
Row 5 Digit 6 must be placed in R5C6
Column 3 Digit 6 must be placed in R2C3
Box 4 Digit 6 must be placed in R4C1
Column 1 Digit 1 must be placed in R2C1
Row 2 Digit 5 must be placed in R2C6
All Cells Solved
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Sat May 13, 2006 4:30 pm    Post subject: Reply with quote

Thanks again, David,

i just saw 2 more solutions for this puzzle here.
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Myth Jellies
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PostPosted: Sat May 13, 2006 10:34 pm    Post subject: Reply with quote

The link has a lot of extraneous stuff, though it does explain the molecular method. Here is my non-uniqueness chain found via the molecular method. Funny, but you don't realize how long the chain is until you show it on the grid Smile
Code:

 *--------------------------------------------------------------------*
 | 8      7      3      | 2      6      1      | 9      5      4      |
 | 16     4      16     | 59     8     K59     | 7      3      2      |
 | 9      2      5      | 7      34    I34     | 8      1      6      |
 |----------------------+----------------------+----------------------|
 | 346    1      7      | 348    9     B23468  | 5     C24     38     |
 | 2      8      46     | 1345   34    A3456   | 134    7      9      |
 | 34     5      9      | 1348   7      2348   | 1234   6      138    |
 |----------------------+----------------------+----------------------|
 | 5      6     F148    |G38     2      7      |E134    9      13     |
 | 7      3      2      | 49     1     J49     | 6      8      5      |
 | 14     9      148    | 6      5     H38     | 1234  D24     7      |
 *--------------------------------------------------------------------*

A6=B6-B2=C2-D2=D4-E4=F4-F8=G8-G3=H3-I3=I4-J4=J9-K9=K5 :
Chain implies A=6 and/or K=5, therefore A <> 5, which is enough to crack the puzzle.
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sun May 14, 2006 12:19 am    Post subject: What is "molecular method"? Reply with quote

That's fascinating, MJ.

Where can one learn more about this "molecular method"? dcb
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keith



Joined: 19 Sep 2005
Posts: 3182
Location: near Detroit, Michigan, USA

PostPosted: Sun May 14, 2006 1:25 am    Post subject: Molecular Method Reply with quote

David,

Go here:

http://www.sudoku.com/forums/viewtopic.php?t=3873

Keith
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