The New Sudoku Players' Forum

[pjaj]

I use SadMan Sudoku to set me problems.
Recently I thought I'd cracked colouring and can usualy solve most problems that don't need forcing chains, this includes the following problem rated as "fiendish".

Code: Select all

+----------------+----------------+----------------+
|           3    | 5    1         |      8         |
| 8              | 6              | 4              |
|      1    9    |                |                |
+----------------+----------------+----------------+
|      8         |           5    |      2         |
|           5    |      6         | 1              |
|      3         | 4              |      6         |
+----------------+----------------+----------------+
|                |                | 3    1         |
|           4    |           2    |           6    |
|      6         |      5    9    | 8              |
+----------------+----------------+----------------+

I got stuck at this position:

Code: Select all

+----------------+----------------+----------------+
| 6    47   3    | 5    1    47   | 2    8    9    |
| 8    257  27   | 6    9    3    | 4    57   1    |
| 457  1    9    | 28   27   478  | 6    3    57   |
+----------------+----------------+----------------+
| 1    8    6    | 9    3    5    | 7    2    4    |
| 47   247  5    | 28   6    78   | 1    9    3    |
| 9    3    27   | 4    27   1    | 5    6    8    |
+----------------+----------------+----------------+
| 25   9    8    | 7    4    6    | 3    1    25   |
| 3    57   4    | 1    8    2    | 9    57   6    |
| 27   6    1    | 3    5    9    | 8    4    27   |
+----------------+----------------+----------------+


and went for a hint and got

Colouring 7: double exclusion found, eliminating 7 from r1c2,
(r1c6 => r3c5 => r3c9 => r9c9 => r9c1 => r8c2)

But this does not seem correct, how can r1c6 => r3c5 when r3c6 also
contains a 7 in the same block?

Similarly how can r3c5 => r3c9 when r3 contains 4 possible 7s?

Am I missing something, or has SadMan thrown a cog?

[MCC]

Hi pjaj

I'm not to sure about the colouring but I'm sure someone will explain.

In the meantime:

Code: Select all

+----------------+----------------+----------------+
| 6    47   3    | 5    1    47   | 2    8    9    |
| 8    257  27   | 6    9    3    | 4    57   1    |
| 457  1    9    | 28   27   478  | 6    3    57   |
+----------------+----------------+----------------+
| 1    8    6    | 9    3    5    | 7    2    4    |
| 47   247  5    | 28   6    78   | 1    9    3    |
| 9    3    27   | 4    27   1    | 5    6    8    |
+----------------+----------------+----------------+
| 25   9    8    | 7    4    6    | 3    1    25   |
| 3    57   4    | 1    8    2    | 9    57   6    |
| 27   6    1    | 3    5    9    | 8    4    27   |
+----------------+----------------+----------------+

If r8c2=7=>r1c2=4
If r8c2=5=>r2c2<>5=>r2c23(naked pair)=>r1c2<>7=>r1c2=4
Therefore r1c2=4

MCC

[Jeff]

I use SadMan Sudoku to set me problems.
Recently I thought I'd cracked colouring and can usualy solve most problems that don't need forcing chains.........

Welcome to the forum, Pjaj.

Firstly, all colouring deductions can be expressed as forcing chains. It is easy to understand colouring through forcing chain; but not easy the other way around.

There are 4 types of colourings, namely:

Type 1: Colouring with pure conjugate links on a filtered digit.
Type 2: x-cycle type colouring on a filtered digit.
Type 3: Advanced colouring involving all 9 digits.
Type 4: Grouped colouring on a filtered digit (grouped x-cycle).

SadMan Sudoku uses the term "simple colouring" to collectively describe both type 1 and 2. The colouring deduction hinted by SadMan Sudoku is type 2. With type 2 colouring, only alternative links are required to be conjugate links. I think your understanding with colouring is up to type 1.

[pjaj]

Thanks Jeff and MCC. I guess I didn't understand colouring as well as I thought I did
I must now go away and find more information on Types 2,3 and 4.
There are some interesting articles on SadMan's links page

http://www.simes.clara.co.uk/programs/sudokulinks.htm

including a link to this group of forums.

As for forcing chains, well they are still a black art to me, but I've noticed a thread in this forum that looks very promising, so I'm off to read it.

[JeffInCA]

pjaj,

I haven't really gotten my mind around coloring based on cycles yet either; however, there are many coloring-based eliminations that can be made based on what I consider easier coloring techniques, and with these you can quickly arrive at a solution

One semi-advanced coloring technique is called Multi-Colors by Simple Sudoko, which I believe simply refers to the fact that more than one conjugate chain is colored at the same time and the deductions are made accordingly.

Here's a great post by angusj on this subject

http://www.setbb.com/phpbb/viewtopic.php?p=2575&mforum=sudoku#2575

Back to your puzzle, coloring on ths 7s in this sudoku yields the following:

Code: Select all

 .  7C .  | .  .  7c | .  .  . 
 .  7  7B | .  .  .  | .  7a . 
 7  .  .  | .  7b 7  | .  .  7A
----------|----------|-----------
 .  .  .  | .  .  .  | .  .  . 
 7  7  .  | .  .  7b | .  .  .
 .  .  7b | .  7B .  | .  .  . 
----------|----------|-----------
 .  .  .  | .  .  .  | .  .  .
 .  7a .  | .  .  .  | .  7A .
 7A .  .  | .  .  .  | .  .  7a


1) Take a look at the 7 in r3c1.

We can see that it is in the same row as the 7 at r3c5 (7b), and in the same box as the 7 at r2c3 (7B). Since r3c5 and r2c3 are opposite colors, one of them must be 7 so we can eliminate 7 as a candidate for r3c1.

2) Now take a look at aboxes 1 and 2

This is an example of what angusj calls Type 2b multi-coloring in the post I linked to above.

r1c2 (7C) and r2c3 (7B) cannot both be 7, therefore either B or C is false. This means that either 'b or 'c' must be true. Hence, one of r1c6 (7c) or r3c2 (7b) must be true. As a result, r3c6 must not be 7 and 7 can be eliminated here.

Note that the exact reverse logic can be used to eliminate 7 in r2c2. (And in fact it also eliminates 7 from r3c1 as well, even though we already did that in 1) )

3) Now that we've made some eliminations, we can recolor the 7s using only one conjugate chain as follows:

Code: Select all

 .  7- .  | .  .  7+ | .  .  . 
 .  .  7+ | .  .  .  | .  7- . 
 .  .  .  | .  7- .  | .  .  7+
----------|----------|-----------
 .  .  .  | .  .  .  | .  .  . 
 7- 7  .  | .  .  7- | .  .  .
 .  .  7- | .  7+ .  | .  .  . 
----------|----------|-----------
 .  .  .  | .  .  .  | .  .  .
 .  7- .  | .  .  .  | .  7+ .
 7+ .  .  | .  .  .  | .  .  7-


From this we can see that in box 4, 7- appears twice (in r5c1 and r6c3). Since both of these cells cannot be 7, then '-' must represent the false condition and thus 7 can be eliminated from all cells marked '7-'. and conversely, all '7+' cells can be set to 7.

From here the puzzle solution is trivial.

In my opinion coloring can be an extremely powerful technique and you really only need to understand simple coloring on one chain and the multiple chain techniques outlined by angus, both of these just filtering on a single digit. With this coloring along with understansing x-wing, swordfish, and xy-wing, I've been able to solve all but the most difficult of puzzles.

If you haven't checked it out yet, Simple Suduko has a really nice interface for coloring and filtering on a single digit.

Cheers,

Jeff

[pjaj]

JeffInCA that is a brilliantly clear explanation, thank you. Your link to the other post clarifies things nicely as well.
Although not strictly colouring I have found that, even with my limited ability , colouring can often help you spot an x-wing as the 4 cells involved are frequently picked up as two conjugate pairs.

[Havard]

...or you could find the funky swordfish:



and the nice skyscraper:



and that would do exactly the same!

havard

[TKiel]

It appears to me that all the colouring exclusions can be made using only two conjugate chains, not three.

Code: Select all

 
 *--------------------------------------------------*
 | 6    47A   3   | 5    1    47a  | 2    8    9    |
 | 8    257  27B  | 6    9    3    | 4    57   1    |
 | 457  1    9    | 28   27   478  | 6    3    57   |
 |----------------+----------------+----------------|
 | 1    8    6    | 9    3    5    | 7    2    4    |
 | 47   247  5    | 28   6    78b  | 1    9    3    |
 | 9    3    27b  | 4    27B   1   | 5    6    8    |
 |----------------+----------------+----------------|
 | 25   9    8    | 7    4    6    | 3    1    25   |
 | 3    57   4    | 1    8    2    | 9    57   6    |
 | 27   6    1    | 3    5    9    | 8    4    27   |
 *--------------------------------------------------*

Since A & B share a group (Box 1) then both can't be true, which means either a or b is true and any cell sharing a group with them can be eliminated. This excludes the 7 in r3c6.

Recolouring with a single conjugate chain shows this

Code: Select all

*--------------------------------------------------*
 | 6    47A   3   | 5    1    47a  | 2    8    9    |
 | 8    257  27a  | 6    9    3    | 4    57   1    |
 | 457  1    9    | 28   27   48   | 6    3    57   |
 |----------------+----------------+----------------|
 | 1    8    6    | 9    3    5    | 7    2    4    |
 | 47   247  5    | 28   6    78A  | 1    9    3    |
 | 9    3    27A  | 4    27a   1   | 5    6    8    |
 |----------------+----------------+----------------|
 | 25   9    8    | 7    4    6    | 3    1    25   |
 | 3    57   4    | 1    8    2    | 9    57   6    |
 | 27   6    1    | 3    5    9    | 8    4    27   |
 *--------------------------------------------------*


which excludes the 7 in r2c2 and r3c1, since they share a group with both A & a.

The chain can then be lengthened

Code: Select all

 
 *--------------------------------------------------*
 | 6    47A   3   | 5    1    47a  | 2    8    9    |
 | 8    25   27a  | 6    9    3    | 4    57A   1   |
 | 45   1    9    | 28   27   48   | 6    3    57a  |
 |----------------+----------------+----------------|
 | 1    8    6    | 9    3    5    | 7    2    4    |
 | 47   247  5    | 28   6    78A  | 1    9    3    |
 | 9    3    27A  | 4    27a   1   | 5    6    8    |
 |----------------+----------------+----------------|
 | 25   9    8    | 7    4    6    | 3    1    25   |
 | 3    57   4    | 1    8    2    | 9    57   6    |
 | 27a   6   1    | 3    5    9    | 8    4    27A  |
 *--------------------------------------------------*


to make the exclusion of 7 in r5c1, because it shares a group with both A (r5c6) and a (r9c1), which solves the puzzle.

Tracy