Assassin 19

Our weekly <a href="http://www.sudocue.net/weeklykiller.php">Killer Sudokus</a> should not be taken too lightly. Don't turn your back on them.
Post Reply
Andrew
Grandmaster
Grandmaster
Posts: 300
Joined: Fri Aug 11, 2006 4:48 am
Location: Lethbridge, Alberta

Assassin 19

Post by Andrew »

It was good to have a slightly easier puzzle this time but definitely still an Assassin. I felt I had been 'hitting my head against a brick wall' with #18, which Annette and sudokuEd clearly found a lot easier than I did, and with #17v2.

Since nobody else has posted one, here is my walkthrough. As with my previous ones, it's effectively the way I solved it and some sub-steps are included for completeness of an area before I moved on, even though they may not be used immediately.

Step 1
6(3) cage in N9 = {123}, 45 rule on R9 2 outies = 12 -> R8C19 = [93], R9C89 = {12}, R9C12 = {37/46}, 32(5) cage in R9 = 589{37/46}

Step 2
45 rule on N9 3 innies R7C79 + R9C7 = 22 = 9{58/67} with the 9 in C7, no other 9 in C7, 4 in N9 locked in 17(3) cage

Step 3
R67C1 = {68}, R67C3 = {13}, 45 rule on N7 3 innies R7C13 + R9C3 = 14, R7C13 = 7 or 9 (cannot be 11 because R9C3 = 3 would then clash with R67C3) -> R9C3 = {57}, 12(3) cage in N7 = 2{37/46} -> R9C3 = 5, R67C1 = [68], R67C3 = [31]

Step 4
R34C1 = {25}/[34], R34C3 = {28/46}, 45 rule on N4 2 remaining innies R4C13 = 7, R4C1 must be an odd number since R4C3 is an even number -> R4C1 = 5, R4C3 = 2, R3C1 = 2, R3C3 = 8, 29(5) cage in N4 = {14789}, 2 in N7 locked in R78C2

Step 5
45 rule on C12 3 outies R258C3 = 19, R5C3 = {479}, R8C3 = {467}, only valid combination is {469} -> R1C3 = 7, R2C3 = {469}, R5C3 = {49}, R8C3 = {46}, R78C2 = 2{46}, R9C12 = {37}, 32(5) cage in R9 = {45689} with the 4 locked in N8 -> R78C5 = {57}

Step 6
18(3) cage in N1 = 9{36/45}, 10(3) cage in N1 = 1{36/45}, cannot have R12C1 = {36} or {45} so R12C1 = 1{34}, R1C2 = {56}, 1 in N4 locked in C2, R5C1 = {47}, 45 rule on C1 2outies – 1 innie = 5 -> R19C2 = 9 or 12 -> R19C2 = [57]/[63]

Step 7
7a. 45 rule on N2 3 remaining outies = 6 -> R1C7 + R4C46 = {114}/{123} (the first combination is possible because the outies are in 2 different rows and 3 different columns) -> R4C46 = {134}, R1C7 = {12}
7b. 45 rule on N3 3 innies = 9 = {126}/{135}/{234} -> R3C9 = {456}, R3C7 = {13} (cannot have 4 because {144} not allowed and {234} would clash with the 3 in C9, cannot have 5 because there is a 2 in R4), R34C7 = [16]/[34]
7c. R1C7 cannot be 1 because that would require [34] in R34C7 and {14} in R4C46 (two 4s in R4) -> R1C7 = 2, R4C46 = {13}
7d. R1C456 = {489}, R23C6 = 11 min and cannot be {29} -> R2C4 = 2 (hidden single in N2), R3C4 = {57}, R23C6 = 6{57}, 2 in N8 locked in C6, 2 in N5 locked in C5
[Step 7 edited. Thanks to sudokuEd for pointing out the flaw in my original logic, see step 7a above, and for providing steps 7b and 7c which saved me having to think that out for myself]

Step 8
R34C7 = [16]/[34], R23C5 = {13} so R3C57 = {13}, no other 1/3 in R3, 45 rule on N5 2 remaining innies = 13 -> R6C46 = {49/58}, 28(5) cage in N5 = 267{49/58}, R4C46 = [31] (R4C6 hidden single 1 in C6 because cannot have both 1 and 2 in R78C6) -> R3C4 = 5, R23C6 = {67}, R78C6 = [32] (hidden single 3 in N8), R6C46 = [49], R78C4 = [91], R9C7 = 9, 28(5) cage in N5 = {25678} -> R5C6 = 5 (hidden single in C6), R456C5 = {268}, R5C4 = 7, R5C1 = 4, R5C3 = 9, R456C2 = {178}, R9C12 = [73], R12C1 = {13}, R1C2 = 6, R78C2 = {24}, R8C3 = 6, R2C3 = 4, R23C2 = [59]

Step 9
5 in N3 locked in R1, 45 rule on R1 2 outies = 9 -> R2C9 = {68}, 9 in N3 locked in 22 (3) cage -> {679} -> R2C8 = 9, R2C9 = 8, R12C1 = [31], R3C9 = 4 (hidden single in N3), R4C9 = 9, R1C89 = {15}, R3C7 = 3, R4C7 = 4, R23C5 = [31]

Step 10
R67C9 = {18/27}, 45 rule on N6 2 remaining innies = 7 -> R6C79 = [52], R7C79 = [67], 25(5) cage in N6 = {13678}, R5C8 = 3 (hidden single in N6), R9C89 = [21], R1C89 = [15], R5C9 = 6, 17(3) cage in N9 = {458} -> R8C7 = 8 and carry on … with simple elimination
Ruud wrote:You need almost every trick in the book to break this Assassin. When it finally surrenders, your victory will be total.
At first I doubted that was true but, after the changes to step 7, I'm now convinced! I'll stick with my "slightly easier" comment; most steps managed to fix at least one cell.

Andrew
Oscar
Regular
Regular
Posts: 28
Joined: Wed Jan 16, 2008 10:31 am
Location: Montesson

Post by Oscar »

After the warning by Ruud, you have to proceed step by step, do not overlook any detail or opportunity, and ... then it's ok :) .
I rate it 1.0
Nothing can both be and not be
Post Reply