## Jigsaw Killer

For fans of Killer Sudoku, Samurai Sudoku and other variants

### Jigsaw Killer

Here is a Jigsaw Killer Sudoku I designed. A pretty hard one, but it can be solved by logic using techniques for both jigsaw and killer. I'll publish a walkthrough in a few days.

Enjoy

Jean-Christophe

Posts: 149
Joined: 22 January 2006

I've written a page on my site about combining techniques for jigsaw (Law of Leftovers) and killer sudoku (45 AKA innies-outies...) :
http://sudoku.apinc.org/?p=109

Here is my walkthrough in tiny text you may copy-paste to a text editor like note pad

If the notation isn't clear, please have a look at my killer lingo page :
http://sudoku.apinc.org/?page_id=3

Nonets are numbered from top to bottom and from left to right :
N1..4 = top 4 nonets from left to right
N5 = Central nonet
N6 = bottom left nonet
N7 = bottom right nonet (because it's higher than the other 2)
N89 are central bottom nonets
LOL means "Law Of Leftover"
{R56C6} means the set of values in R56C6, here a pair of values

Step 1
LOL on C12 -> {R8C2} = {R5C3}
We may use 45 on N6 (bottom left nonet) using R8C2 in place of R5C3
45 on N6 -> R45C1 = 10, R5C2+R6C1 = 10
LOL on C89 -> {R2C8} = {R5C7}
45 on N4 -> R56C9 = 10, R4C9+R5C8 = 6 = {15|24}

Step 2
LOL on C6789 -> {R56C6} = {R89C5}
Each part have the same pair of values. Although we don't know yet their sum, since each part have the same pair of values, they also add up to the same sum which we call Y.
Plugging this relation into the cages -> (Y+R9C4)-(Y+R5C7) = 22-15 -> R9C4-R5C7 = 7
This is similar to outies minus innies
Max R9C4 = 9 -> Max R5C7 = 9-7 = 2
Min R5C7 = 1 -> Min R9C4 = 1+7 = 8
-> R9C4 = {89}, R5C7 = {12}
LOL on C89 -> R2C8 = {12}

Step 3
45 on N79 (bottom right nonets) -> R7C5+R9C4 = 15
Since R9C4 = {89} -> R7C5 = {67}
Cage 22/3 in N89 (bottom center) = {589|679}
Since R7C5 = {67} -> Cage 22/3 cannot be {679} -> Cage 22/3 = {589} -> R89C5 = {5(8|9)}
-> no 5 elsewhere in C5, N9
LOL on C6789 -> R56C6 = {5(8|9)} -> no 5 elsewhere in N5, C6

Step 4
LOL on R1234 -> {R3C5+R4C45} = {R5C178}
Since {R3C5+R4C45} cannot have 5 -> {R5C178} cannot have 5
LOL on C1234 -> {R45C4} = {R12C5}
Plugging this relation into the cages -> R1C6 = R5C3 (which = R8C2)
R1C6 cannot have 5 -> R5C3 and R8C2 cannot have 5
-> R5C6 = 5 (HS in R5)

Step 5
45 on R123 -> R3C28 = 11, R4C238 = 23 = {689}
45 on N12 -> R1C6+R3C5 = 8 -> R3C5 = {12467}, R1C6 = {12467}
-> R5C3 = R8C2 = {12467}
45 on N1 -> R4C3-R1C2 = 6 -> R4C3 = {89}, R1C2 = {23}

Step 6
Cage 21/3 in N12 = {489|678} -> R3C2 = {47} -> R3C8 = {47} (naked pair in R3)
-> R3C5 = {126} -> R1C6 = R5C3 = R8C2 = {267}

Step 7
Cage 14/2 in N2 = {59|68} = {(8|9)...} forms a killer naked pair on {89} with R4C3 -> not elsewhere in N2, C3
Cage 14/3 in N23 = {167|347} -> R1C6 = R5C3 = R8C2 = {67}, R12C5 = R45C4 = {13467}
-> R3C5 = {12}
Since R8C2 = {67} -> Cage 21/3 in N12 cannot be {678}
-> Cage 21/3 = {489}, R3C2 = 4, R4C23 = {89}, R34C8 = [76]

Step 8
Cage 6/2 at R6C2 = {15}
-> R5C2+R6C1 cannot be {19}
Cage 16/3 in N68 (bottom left) with R8C2 = {67} -> {268|367}
-> 9 of N6 locked in Cage 13/2 = {49}
-> R45C1 = {37}, R1C2 = 2, R5C2 = 8, R6C1 = 2, R9C12 = [63], R8C2 = R5C3 = R1C6 = 7, R3C5 = 1
R45C1 = [73], R4C23 = [98], R2C2 = 6
R45C4 = [34], R23C4 = [72], R2C8 = 2, R5C78 = [21]
...

Jean-Christophe

Posts: 149
Joined: 22 January 2006