Sadman's Analysis Of SudoCue's 2006/01/03's Problem

[jon.seymour]

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)

[Ruud]

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.

[jon.seymour]

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.

[Ruud]

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

[jon.seymour]

Ah, ok - I see it now!

Thanks for that.

jon.

[David Bryant]

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:

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

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