The New Sudoku Players' Forum

[mikebot]

can someone type this game into simple sudoku:

Code: Select all

382 6*1 9*5
94* 8*3 *2*
6** 2** 8*3

19* 5** 234
7*4 *2* 158
25* **4 *9*

5** **8 4*2
42* 7*5 389
8*9 4*2 56*


and press 'f7', and explain to me how you can remove that '1' based on multi. colors? thanks.

[Sped]

Code: Select all

 
 *-----------------------------------------------------------*
 | 3     8     2     | 6     47    1     | 9     47    5     |
 | 9     4     157A  | 8     57    3     | 67    2     167a  |
 | 6     17    157   | 2     4579  79    | 8     147A  3     |
 |-------------------+-------------------+-------------------|
 | 1     9     68    | 5     78    67    | 2     3     4     |
 | 7     36    4     | 39    2     69    | 1     5     8     |
 | 2     5     38    | 13    138   4     | 67    9     67    |
 |-------------------+-------------------+-------------------|
 | 5     1367 (1)367 | 139   1369  8     | 4     17a   2     |
 | 4     2     16    | 7     16    5     | 3     8     9     |
 | 8     137   9     | 4     13    2     | 5     6     17    |
 *-----------------------------------------------------------*


It's simple colors, not multi colors.

The conjugate 1s in row 2 are marked "A" and "a". Conjugate in this case means that there are just two 1s in the row such that one must be true for 1 and the other false.

In box 3 the conjugate 1s are marked "a" (r2c9) and "A" (r3c8). Either all the "A" cells are 1 or all the "a" cells are 1.

The conjugate 1s in column 8 are r3c8 "A" and r7c8 "a".

It so happens that r7c3 sees both the "A" in r2c3 and the "a" in r7c8.

Since either all the "A" cells are 1 or all the "a" cells are 1, we know that a cell that sees both "a" and "A" cannot be 1.

Therefore the 1 can be excluded from r7c3.

[tso]

It's the 1 from r7c4 that Mikebot is asking about.

Code: Select all

 *-----------------------------------------------------------*
 | 3     8     2     | 6     47    1     | 9     47    5     |
 | 9     4    +157   | 8     57    3     | 67    2    -167   |
 | 6     17    157   | 2     4579  79    | 8    +147   3     |
 |-------------------+-------------------+-------------------|
 | 1     9     68    | 5     78    67    | 2     3     4     |
 | 7     36    4     | 39    2     69    | 1     5     8     |
 | 2     5     38    | 13    18    4     | 67    9     67    |
 |-------------------+-------------------+-------------------|
 | 5     1367  367   |x19    1369  8     | 4    -17    2     |
 | 4     2    ^16    | 7    v16    5     | 3     8     9     |
 | 8     137   9     | 4     13    2     | 5     6     17    |
 *-----------------------------------------------------------*

If r2c3<>1, follow the chain clockwise:

r2c3<>1 -> r2c9=1 -> r3c8<>1 -> r7c8=1 -> r7c4<>1

If r2c3=1, follow it counter-clockwise.

r2c3=1 -> r8c3<>1 -> r8c5=1 -> r7c4<>1

[emm]

Here’s how the chain is explained by multiple colouring.

You have two different chains - labelled Aa and Bb - where the first cell of each chain (A & B) share a group

Code: Select all

 . . . | . . . | . . .
 . . 1A| . . . | . . 1a
 . 1 1 | . . . | . 1A.
-------+-------+------
 . . . | . . . | . . .
 . . . | . . . | . . .
 . . . | 1 1 . | . . .
-------+-------+------
 . 1 . | 1*1 . | . 1a.
 . . 1B| . 1b. | . . .
 . 1 . | . 1 . | . . 1A

A and B share a group so they can’t both be true so either a or b are true and any cell that shares a group with both a and b is false ie r7c4

This can also be visualised with strong links - see this topic

You can remove 1s from both r7c4 and r7c5 as they see both ends of the links at r7c8 and r8c5

Code: Select all

 . . . | . . . | . . .
 . . 1-|-------|-----1
 . 1 1 | . . . | . 1 .
-------+-------+---|---
 . . . | . . . | . | .
 . . . | . . . | . | .
 . . . | 1 1 . | . | . 
-------+-------+---|--
 . 1 . | 1*1*. | . 1 .
 . . 1-|---1 . | . . .
 . 1 . | . 1 . | . . 1

[keith]

Why not just use the UR's to eliminate <7> from R3C5,8, and also <67> from R2C9?

Then, an XY-wing solves the puzzle.

Is it that Simple Sudoku does not do Unique Rectangles?

Keith

Code: Select all

+----------------+----------------+----------------+
| 3    8    2    | 6    47   1    | 9    47   5    |
| 9    4    157  | 8    57   3    | 67   2    167  |
| 6    17   157  | 2    4579 79   | 8    147  3    |
+----------------+----------------+----------------+
| 1    9    68   | 5    78   67   | 2    3    4    |
| 7    36   4    | 39   2    69   | 1    5    8    |
| 2    5    38   | 13   138  4    | 67   9    67   |
+----------------+----------------+----------------+
| 5    1367 1367 | 139  1369 8    | 4    17   2    |
| 4    2    16   | 7    16   5    | 3    8    9    |
| 8    137  9    | 4    13   2    | 5    6    17   |
+----------------+----------------+----------------+

[udosuk]

Is it that Simple Sudoku does not do Unique Rectangles?

Simple Sudoku doesn't do Unique Rectangles... If it does than all of us who use it to solve killer sudoku, sudoku X or other variants will be in big trouble...

[daj95376]

I probably have an incorrect understanding on how Multi-Coloring works, but I'm going to present it anyway.

Below should be the grid where you encounter Multi-Coloring. The hint is to set [r7c4]<>1. With my interpretation, [r7c5]<>1 should also be indicated, but it isn't ... and it isn't critical.

Use Blue/Green to indicate the conjugate pairs chain in [r2], [b3], and [b9] for <1>. This represents an either/or mapping on where <1> can be a solution in these cells. Use Pink/Amber to indicate the conjugate pair in [r8] for <1>. Again, an either/or relationship.

Now, the key to understanding Multi-Coloring in this grid is to realize that [c3] has two ends of different chains in common. Since Blue and Pink can not both be <1>, this translates into Blue and Amber having an equivalent relationship to Pink. I now treat Blue and Amber as being equivalent. This being the case, then any <1> in the overlap of Amber and Green can be removed. Thus, [r7c45]<>1. From here, the puzzle is solved with Singles.

Code: Select all

 *-----------------------------------------------------------*
 | 3     8     2     | 6     47    1     | 9     47    5     |
 | 9     4     157B  | 8     57    3     | 67    2     167G  |
 | 6     17   ?157   | 2     4579  79    | 8     147B  3     |
 |-------------------+-------------------+-------------------|
 | 1     9     68    | 5     78    67    | 2     3     4     |
 | 7     36    4     | 39    2     69    | 1     5     8     |
 | 2     5     38    | 13    18    4     | 67    9     67    |
 |-------------------+-------------------+-------------------|
 | 5     1367  367   |*19   *1369  8     | 4     17G   2     |
 | 4     2     16P   | 7     16A   5     | 3     8     9     |
 | 8     137   9     | 4     13    2     | 5     6     17B   |
 *-----------------------------------------------------------*


Note: It's tempting to think of Pink and Green as being equivalent. If they are equivalent, then [r3c3]<>1 can also be deduced.

[ravel]

Note: It's tempting to think of Pink and Green as being equivalent. If they are equivalent, then [r3c3]<>1 can also be deduced.

No, there is still a valid template with r3c3=1:

Code: Select all

 . . . | . . 1 | . . .
 . . . | . . . | . . 1
 . . 1 | . . . | . . .
-----------------------
 1 . . | . . . | . . .
 . . . | . . . | 1 . .
 . . . | 1 . . | . . .
-----------------------
 . . . | . . . | . 1 .
 . . . | . 1 . | . . .
 . 1 . | . . . | . . .


Since blue and pink share a unit, they cannot both be true. That means, that either green or amber (or both) must be true.
But green and amber dont share a unit.

[udosuk]

Since blue and pink share a unit, they cannot both be true...

Blue and pink cannot both be true, but they can be both false, which enables r3c3 to be 1.

Green and amber cannot be both false, so either one of them must be true, and also they can be both true (since there were no interaction between their cells).

Blue and amber are not equivalent... You can only conclude that if blue is true, amber must be true... If blue is false, amber can be true or false... So the true relationship is "blue => amber" instead... And the relationship between pink and green is if pink is true then blue must be false and green must be true, i.e. "pink => green"...

And if you look at it the other way it was the possibility of r3c3 being 1 that forced you to label r8c35 with another set of colours instead of using simple colouring in the first place...

[daj95376]

Thanks for the clarification ravel & udosuk !!!

So, any cell in the overlap of the non-linked colors -- Amber and Green in this case -- can have the designated candidate eliminated. At least, this still means that <1> can be eliminated from [r7c45].

[mikebot]

thanks a lot for the help everyone, i get it now. it is way harder to find multi colors than it is to get them once you see them.