G'day Ruud + Sudo Cue'rs
In today's problem SadMan's solver claims that from this state:
http://blackcubes.dyndns.org/sudoku/sud ... 7.5.64.642.....
colouring of 3's can be used to eliminate 3 from r7c1 because a double exclusion can be found using this chain(r6c1 => r6c5 => r4c4 => r7c4).
This seems like faulty reasoning to me. I notice that Sudo Cue does not come to this conclusion.
Thoughts anyone?
jon.
(btw: the link above doesn't work in all IE6 browsers for reasons I am still not entirely clear on)
Sadman's Analysis Of SudoCue's 2006/01/03's Problem
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Hi Jon,
When I followed the link, the givens and placed digits were OK, but the candidates were not correctly shown.
This is the same situation, with all candidates for digit 3 in coloring mode. Sudo Cue suggests the same coloring elimination at this point, with the same 4 cells involved:
So, at this point Sudo Cue agrees with Sadman's solver. (as do I)
As this puzzle classifies as "Ruud's worst Nightmare", I will not yet reveal any further clues.
Ruud.
When I followed the link, the givens and placed digits were OK, but the candidates were not correctly shown.
This is the same situation, with all candidates for digit 3 in coloring mode. Sudo Cue suggests the same coloring elimination at this point, with the same 4 cells involved:
So, at this point Sudo Cue agrees with Sadman's solver. (as do I)
As this puzzle classifies as "Ruud's worst Nightmare", I will not yet reveal any further clues.
Ruud.
“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
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When you say the canidates were not correctly shown, are you referring to extra 3's in r8c7 and r9c7? I understand that these can be eliminated by the 3's in r{2,3}c7. Are there other incorrect candidates?
Are you also saying that the candidates in r1c7 are wrong? I show 5. If there were only 2, then I could understand how the colouring algorithm applies.
Or perhaps I just don't understand the colouring algorithm correctly. I thought you could only use conjugates, but as I understand the definition of conjugates, r7c1 is not conjugate with either r6c1 or r7c4.
Are you also saying that the candidates in r1c7 are wrong? I show 5. If there were only 2, then I could understand how the colouring algorithm applies.
Or perhaps I just don't understand the colouring algorithm correctly. I thought you could only use conjugates, but as I understand the definition of conjugates, r7c1 is not conjugate with either r6c1 or r7c4.
This is the candidate list at the moment coloring is suggested:
For digit 3 there are conjugate pairs in C4, B5, and R6. These are the conjugate pairs you need to build the coloring chain.
I hope this allows you to determine where the differences in candidates are.
Ruud
For digit 3 there are conjugate pairs in C4, B5, and R6. These are the conjugate pairs you need to build the coloring chain.
I hope this allows you to determine where the differences in candidates are.
Ruud
“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
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Ok
Ah, ok - I see it now!
Thanks for that.
jon.
Thanks for that.
jon.
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Coloring isn't necessary.
Hi, all!
I just posted a description of an interesting way to solve this puzzle over on the Daily Sudoku site.
Since it doesn't rely on coloring I thought I'd also post it here.
Using only the standard techniques I arrived at this position:
My attention was drawn to a peculiar form of symmetry in the bottom left 3x3 box. Suppose that the value "9" is placed at either r7c2 or r8c2.
Then the {3, 5, 7} triplet is revealed in the bottom left 3x3 box, the {1, 3, 4} triplet is created in the top left 3x3 box, the {5, 7} pair is uncovered
in row 2, and r1c3 = 9.
Armed with this information we easily see that r7c2 <> 9 and also that r8c2 <> 9:
r7c2 = 9 ==> r1c3 = 9
r7c2 = 9 ==> r8c2 = 8 ==> r8c7 = 9 ==> r8c5 = 3 ==> r6c5 = 8
But now we have the {1, 5} pair in r1c5 and r2c5, leaving no possible way to complete r9c5, which must be a "1" because of r8c5 = 3.
The argument when r8c2 = 9 is almost exactly the same:
r8c2 = 9 ==> r1c3 = 9
r8c2 = 9 ==> r8c7 = 8 ==> r8c5 = 3 ==> r6c5 = 8
But now we have the {1, 5} pair in r1c5 and r2c5, leaving no possible way to complete r9c5, which must be a "1" because of r8c5 = 3.
So we can eliminate "9" from r7c2 & r8c2, leaving the {3, 8} pair in those two cells ... the rest of the puzzle is easily solved. dcb
I just posted a description of an interesting way to solve this puzzle over on the Daily Sudoku site.
Since it doesn't rely on coloring I thought I'd also post it here.
Using only the standard techniques I arrived at this position:
Code: Select all
2 149 59 149 159 3 6 7 8
3457 134 357 6 158 248 13 24 9
6 1349 8 149 7 249 13 24 5
349 5 39 349 6 1 7 8 2
489 7 6 489 2 489 5 3 1
38 2 1 5 38 7 4 9 6
3579 389 3579 1389 4 6 2 15 37
1 389 2 7 389 5 89 6 4
357 6 4 2 13 89 89 15 37
Then the {3, 5, 7} triplet is revealed in the bottom left 3x3 box, the {1, 3, 4} triplet is created in the top left 3x3 box, the {5, 7} pair is uncovered
in row 2, and r1c3 = 9.
Armed with this information we easily see that r7c2 <> 9 and also that r8c2 <> 9:
r7c2 = 9 ==> r1c3 = 9
r7c2 = 9 ==> r8c2 = 8 ==> r8c7 = 9 ==> r8c5 = 3 ==> r6c5 = 8
But now we have the {1, 5} pair in r1c5 and r2c5, leaving no possible way to complete r9c5, which must be a "1" because of r8c5 = 3.
The argument when r8c2 = 9 is almost exactly the same:
r8c2 = 9 ==> r1c3 = 9
r8c2 = 9 ==> r8c7 = 8 ==> r8c5 = 3 ==> r6c5 = 8
But now we have the {1, 5} pair in r1c5 and r2c5, leaving no possible way to complete r9c5, which must be a "1" because of r8c5 = 3.
So we can eliminate "9" from r7c2 & r8c2, leaving the {3, 8} pair in those two cells ... the rest of the puzzle is easily solved. dcb