The New Sudoku Players' Forum

[HATMAN]

Rainbow Killer 4
Rainbow as in "All the numbers in the rainbow"
The numbers are from 0-9
There is a zero in every nonet, row and column. Zero is chosen as it increases the killer combinations.
Each other number is missing from one and only one nonet, row and column.
This means that the nonet where a missing number row and the column with the same number missing cross, has that same number missing. I know this by checking examples but cannot see how to prove it. I used this in my solution but doubt that it was necessary.
Relatively easy - about 0.7 on the assassin scale.
Note the hidden cage (not used).

JSudoku puzzle on the assassin site.

[999_Springs]

first post here in 1.5 years

seemed like a fun puzzle, so i had a go. despite it being "relatively easy", it took me 86 minutes by hand, partly because i messed it up at one point and had to backtrack. still, it's a nice puzzle. i'm a bit out of practice on these things

pic of solution in samsung notes: Show

the nonet where a missing number row and the column with the same number missing cross, has that same number missing. I know this by checking examples but cannot see how to prove it.

proof:

suppose 1 is missing in row 1 and column 1. we will show that 1 is also missing from block 1

because 1 is missing from column 1, it has to be present in each of columns 4,5,6, which means it must appear 3 times in total in those columns. those 3 columns cover blocks 2,5,8 so it must appear 3 times in those blocks, which means 1 must appear in each of block 2, block 5 and block 8. similar reasoning in columns 7,8,9 means it has to appear in blocks 3,6,9, and by doing the same thing in the rows, it has to appear in blocks 4 and 7. that means it has to be missing from block 1

[HATMAN]

Thanks for the proof - nice and simple.

There was a nice proof long ago that Windoku X cannot be anti-Knight. It got lost, I think on DJApe's site - was that you?

Maurice

[999_Springs]

nope, that wasn't me - i didn't know djape's site used to have forums on it

but i came up with a proof anyway

proof that windoku X cannot be anti-knight:

we will prove that in an anti-knight windoku X, it is impossible for a 1-digit template to contain the cell r3c3

i don't know if there's a standard labelling for the 4 windows and 5 extra regions in it, but i'll refer to this

Code: Select all

+-------+-------+-------+
| I E E | E I F | F F I |
|  +-------+ +-------+  |
| G|A A | A|G|B | B B|G |
| G|A A | A|G|B | B B|G |
+--|----+--|-|--+----|--+
| G|A A | A|G|B | B B|G |
|  +-------+ +-------+  |
| I E E | E I F | F F I |
|  +-------+ +-------+  |
| H|C C | C|H|D | D D|H |
+--|----+--|-|--+----|--+
| H|C C | C|H|D | D D|H |
| H|C C | C|H|D | D D|H |
|  +-------+ +-------+  |
| I E E | E I F | F F I |
+-------+-------+-------+

assume that this sudoku is anti-knight windoku X with r3c3=1 and we will show that it is impossible for all 1's to be placed. cells marked with a 0 cannot contain 1

Code: Select all

+-------+-------+-------+
| 0 0 0 | 0 . . | . . . |
|  +-------+ +-------+  |
| 0|0 0 | 0|0|. | . .|. |
| 0|0 1 | 0|0|0 | 0 0|0 |
+--|----+--|-|--+----|--+
| 0|0 0 | 0|0|. | . .|. |
|  +-------+ +-------+  |
| . 0 0 | 0 0 . | . . . |
|  +-------+ +-------+  |
| .|. 0 | .|.|0 | . .|. |
+--|----+--|-|--+----|--+
| .|. 0 | .|.|. | 0 .|. |
| .|. 0 | .|.|. | . 0|. |
|  +-------+ +-------+  |
| . . 0 | . . . | . . 0 |
+-------+-------+-------+

in block 5, consider either r4c6=1 or r5c6=1. the cases r6c4=1 or r6c5=1 will be the same by diagonal symmetry

case r4c6=1:

Code: Select all

+-------+-------+-------+
| 0 0 0 | 0 . 0 | . . 0 |
|  +-------+ +-------+  |
| 0|0 0 | 0|0|0 | 0 0|. |
| 0|0 1 | 0|0|0 | 0 0|0 |
+--|----+--|-|--+----|--+
| 0|0 0 | 0|0|1 | 0 0|0 |
|  +-------+ +-------+  |
| . 0 0 | 0 0 0 | . 0 . |
|  +-------+ +-------+  |
| .|. 0 | 0|0|0 | 0 .|. |
+--|----+--|-|--+----|--+
| .|. 0 | .|.|0 | 0 .|. |
| .|0 0 | .|.|0 | . 0|. |
|  +-------+ +-------+  |
| 0 . 0 | . . 0 | . . 0 |
+-------+-------+-------+

r1c5=1 is a single in block 2
r2c9=1 is a single in block 3
r5c7=1 is a single in row 5
r9c8=1 is a single in block 9
1 in column 2 is locked in region C => r78c4=/=1
no 1 in column 4, contradiction

case r5c6=1:

Code: Select all

+-------+-------+-------+
| 0 0 0 | 0 . 0 | 0 0 . |
|  +-------+ +-------+  |
| 0|0 0 | 0|0|0 | . .|. |
| 0|0 1 | 0|0|0 | 0 0|0 |
+--|----+--|-|--+----|--+
| 0|0 0 | 0|0|0 | . 0|. |
|  +-------+ +-------+  |
| 0 0 0 | 0 0 1 | 0 0 0 |
|  +-------+ +-------+  |
| .|. 0 | 0|0|0 | . 0|. |
+--|----+--|-|--+----|--+
| .|. 0 | .|0|0 | 0 .|. |
| .|. 0 | .|.|0 | . 0|. |
|  +-------+ +-------+  |
| . . 0 | . . 0 | 0 0 0 |
+-------+-------+-------+

r1c5=1 is a single in block 2
1 in row 9 is locked in r9c24 => r8c24=/=1
r2c8=1 is a single in diagonal /
no 1 in row 4, contradiction

QED

[HATMAN]

Hi Springs
Thanks, but the first bit is actually quicker as r5c6=1 -> no 1 in C8