KiCo #1

For fans of Killer Sudoku, Samurai Sudoku and other variants

KiCo #1

Here is a (new ?) Killer variant. I called it KiCo (Killer by Color) since it requires techniques similar to Hatman’s KiMo (Killer Modulo).

The rules :
All rules for “vanilla” sudoku apply : fill in the grid such that each row, column and 3×3 nonet/block contains all numbers from 1 to 9.
All rules for killer sudoku apply : the numbers within a cage should add up to the sum; no number may be repeated within a cage.
Each cage has a color out of : green, blue, pink, yellow.
No cage may “touch” another cage of the same color, not even at the corners.
Some cages are given in the grid with their sums and colors.
The layout of the other cages is given in the grid, but without their sum nor color.
Below the grid, you’ll find a list of cages with number of cells, sum and color.

It can be solved by logic since I designed it by logic
As usual, I’ll publish the solution and my walktrough in a few days.
Jean-Christophe

Posts: 149
Joined: 22 January 2006

Finding the colors is straightforward, and some of the number placements (like the blues) are easy, but things seem to come to a halt there. Any technique suggestions for us first timers (especially those who haven't worked KiMo before)?

P.S. Beautiful puzzle. I love that it makes me think of the 4-color theorem.
re'born

Posts: 551
Joined: 31 May 2007

Here are some hints in tiny text you may copy-paste to a text editor like note pad.

<SPOILER MODE>

Find out all colors : consider N1&9, R5, N3,5&7 and progress from there toward N2,4,6&8
This should already give you a few sums
Use 45 (innies/outies) on N5 and consider what sum for the big cage can work with what you deduced earlier.
See what can work for cage 20/3 in R4C4+R5C34 with some other cage in R5. This should fix R4C4
Use 45 on N3 which should fix R2C67, R3C67 and R4C9
...

</SPOILER MODE>
Jean-Christophe

Posts: 149
Joined: 22 January 2006

I really enjoyed this killer with a twist. I got there without the spoiler but with one false start. I hope there’ll be a KiCo#2.
emm

Posts: 987
Joined: 02 July 2005

As promised, here is my walkthrough in tiny text you may copy-paste to a text editor like note pad

I'll use an abbreviated notation for the colors : G = green, B = blue, P = pink, Y = yellow
When the sum is determined, the color is followed by the sum and number of cells eg Y 11/2 means a yellow cage of 2 cells adding up to 11
When the sum is not yet determined, the color is followed by number of cells eg Y/2 means a yellow cage of 2 cells.
When the color is not yet determined, a slash preceede the number of cells eg /2 means a cage of 2 cells.
Two possible colors are separated by a vertical bar eg P|B/3 means a cage of 3 cells either pink or blue.

Step 1
There are 4 cages of 4 cells to place in N1 and N9. Two of them are P and can't touch P 12/3 in C2.
-> P 18/4 & P 20/4 go in N9
/4 in C34 <> G = B 20/4
/4 in C1 = G 15/4

Step 2
The two /2 in R5 must be Y
There are only three Y/2, two of them are Y 16/2 = {79} and can't go both in R5
-> one of the /2 in R5 = Y 16/2 = {79}, the other = Y 11/2 = {38|56}
/2 in R4C78 = B. The only B/2 = B 5/2 = {14|23}

Step 3
The B 28/5 can't go in N7 nor in N5
-> B 28/5 go in N3
/5 in N5 & N7 = Y/5
/2 in R3C67 = G/2
/2 in R2C67 = Y/2 = the other Y 16/2 = {79}
/3 in N2 = P 13/3
/2 in R1C56 = G/2
/3 in R1C234 = Y/3

Step 4
/2 in R6C23 = P/2
/2 in R7C34 = G/2
/2 in R8C34 = P/2
/3 in N7 = B 14/3
/2 in R9C45 = G/2
/3 in R9C678 = Y/3
We now have found the colors of all cages, but only a few sums.

Step 5
In N5, /5 is either Y 20/5 or Y 29/5
See what the outies gives if it were Y 29/5
45 on N5 -> R5C37 = 20+29+12 - 45 = 16 = {79}
This can't work with the Y 16/2 either in R5C12 or R5C89
-> Y 20/5 go in N5, Y 29/5 go in N7
45 on N5 -> R5C37 = 7/2
45 on R5 -> R5C456 = 11/3

Step 6
In R5, Since either R5C12 or R5C89 = Y 16/2 = {79} -> G 20/3 in R45C4+R5C3 <> {479}
-> R5C567 = {124} = 7 (hidden triplet in R5)
45 on R5 -> R4C4 = 9, R5C34 = 11 = {38|56}

Step 7
In N3, we already found that R2C67 = Y 16/2 and B 28/5 go in N3
45 on N3 -> R2C6+R4C9 - R3C7 = 28+16+17 - 45 = 16
Max R2C6+R4C9 = 9+8 = 17 -> Max R3C7 = 17-16 = 1 = Min R3C7
-> R3C7 = 1, R2C67 = [97], R4C9 = 8
Max sum of G/2 in R3C67 = 1+8 = 9
There is only one G/2 <= 9
-> R3C67 = G 5/2, R3C6 = 4
R3C89 = {36}

Step 8
G/2 in R1C56 = 12|13 = {57|58|67} = {(5|6)...} = {(5|7)...}
-> P 13/3 in N2 <> {256|157} = {238}
-> R2C5 = 3, R3C45 = {28}
R1C56 = {57|67} = {7(5|6)}
R12C4 = {1(5|6)}

Step 9
G 12/3 in N5 = {147|246} = {4...}
-> R5C7 = 4, R5C56 = {12}, R6C6 = {67}
Since R5C37 = 7/2 -> R5C34 = [38], R3C45 = [28]
Y 11/2 in R5 = {56}
B 5/2 in R4C78 = {23}
R6C4 = 3

Step 10
P/2 in R6C23 = 7|11
Since R5C3 = 3 -> R6C23 <> {38}
-> R6C1 = 8, R6C23 = {2...} = {25|29}
R6C5 = 4

Step 11
45 on N7 -> R78C3 = 6 = {15|24}
G/2 in R7C34 = 12|13 = [48|57|49|58]
Since R45C4 = [98] -> R8C34 = [57] = G 12/2
Since R78C3 = 6 -> R8C3 = 1
P/2 in R8C34 = 7|11 -> R8C34 = 7/2, R8C4 = 6

Step 12
P/2 in R6C23 = 11/2 = {29}
Y/2 in R5C12 = 11/2 = {56}
Y/2 in R5C89 = 16/2 = {79}
R6C789 = {156}, R56C6 = [17], R5C5 = 2
R4C123 = {147}

Step 13
R12C4 = {15}, R1C56 = [76], R4C56 = [65]
G/2 in R9C45 = G 13/2 = [49]
R78C5 = [15], R7C6 = 8
R7C12 = {46}

Step 14
R2C3 = 6 (hidden single in C3)
B 20/4 in C34 = {1469} -> R34C3 = [94], R12C4 = [51]
R6C23 = [92], R19C3 = [87]
P 12/3 in R234C2 = [471]
G 15/4 in R1324C1 = [1257]
R89C1 = [93], R89C6 = [32]
Several nacked and hidden singles

Step 15
Y/3 in R1 = [385] = Y 16/3
-> Y/3 in R9 = Y 13/3 -> R9C78 = {56}
Fix all nacked and hidden singles until R17C79 = {29}
One of the P/4 in N9 = 18/4, the other = 20/4
If you read until here, I guess you can findout this yourself !

Unique solution
138576942
246139785
579284163
714965328
653821497
892347516
465718239
921653874
387492651
Jean-Christophe

Posts: 149
Joined: 22 January 2006