Weekly Assassin, July 14, 2006 (walkthrough)

[nd]

Despite Ruud's warning, this one requires no combination tables, either. But it's a nice puzzle that takes a little finesse after the straightforward start (the lefthand nonets crack easily at the beginning).

1. R1C12 = {(15)|(24)} -> R12C3 = {(16)|(34)} -> R23C2 = {(18)|(27)}. Hidden killer pair on {12} in the 6(2) and 9(2) cages, so R12C3 = {34}, R1C12 = {15}, R23C2 = {27}.

2. 45 rule on C1 -> R6C2 = 4, R1C12 = [51], R789C1 = {234}, R56C1 = {19}, R4C1 = 7, R23C1 = {68}, R3C3 = 9. 45 rule on N4 -> R46C3 = 10 = {28}. R5C3 = {56}, R45C2 = {356}, R9C2 = {89}, R4C4 = {28}.

3. In N7 the 1 is locked in the 20(4) cage (since it cannot go in a 19(3) cage). In turn this means R6C3 cannot be 2 (because 20 - 2 - 1 = 17 which is too high for the remaining 2 cells since one contains {567}). So R6C3 = 8, R4C34 = [28].

4. 45 rule on N3 -> R4C9 = 9, R3C7 = 2, R23C2 = [27], R1C89 = [98]. The 8 in N6 must be in the 12(3) cage so that cage = {138}. 45 rule on N6 -> R46C7 = 10 = {46}. So R6C7 = 6, R4C67 = [54]. 45 rule on N789 -> R7C5 = 8.

5. R23C8 = {(47)|(56)}. R23C9 = {(16)|(34)}, which eliminates {135} in R789C9. So R789C9 = {2(16)|(34)}, locking 2 in C9 within N9. So R78C8 = {37} ({46} is blocked by the 11(2) cage in C8), R45C8 = [18], R5C7 = 3, R12C7 = [71], R23C8 = {56}, R23C9 = {34}, R789C9 = {126}, R56C9 = {57}, R6C8 = 2, R9C8 = 4.

6. 45 rule on N369 -> R79C6 = {23}. 45 rule on N7 -> R79C4 = 9 = [45]. R9C23 = [86], R78C2 = {59}, R45C2 = [36], R5C3 = 5, R56C9 = [75] R4C5 = 6. 45 rule on N2 -> R5C5 = 4. R9C67 = [29], R7C6 = 3, R78C7 = [58], R78C8 = [73], R9C1 = 3, R9C9 = 1, R9C5 = 7.

7. In N2 the 27(5) cage in N2 must contain {46}; it must also have {27} in it (because they are blocked from the 28(5) cage). The remaining 5th cell is 27 - 2 - 4 - 6 - 7 = 8. So the cage = {24678}, R2C6 = 8, R2C4 = 7.... and the puzzle more or less finished itself from this point on.

[sudokuEd]

Not sure about your step 2 ND. How did you work out that r5c3=5?

Here's a different solving path - even less combo chart.

I'll make sure the walk-thru goes into tiny text this time.

Step 1
“45” on N1 -> 3 innies = 23 = {689} -> no 6,8 or 9 elsewhere in N1
7 now locked in 9(2) cage in N1 -> {27} only -> no 2 or 7 elsewhere in N1 or c2
->6(2) cage in N1 {15} only -> no 1 or 5 elsewhere in N1 or r1
->7(2) cage in N1 {34} only -> no 3 or 4 elsewhere in N1 or c3

“45” on N1 ->r3c3- r4c1=2
->r4c1= {467}

Step 2
“45” on c1-> 2outies =5 -> no 5 possible in r1c2 -> r1c12=[51], r6c2=4
4 in N7 now locked in 9(3) cage ->{234} only -> no 2,3 or 4 elsewhere in N7 or c1
->r56c1= {19} only -> no 1 or 9 elsewhere in c1 or N4
->r3c3=9, r4c1=7
->14(3) cage in N2 {356} only -> no 3, 5 or 6 elsewhere in N4
->r4c34 ={28} only -> no 2 or 8 elsewhere in r4

Step 3.
r46c3 is now {28} only -> no 2 or 8 elsewhere in c3
Since 1 is locked in r78c3 in 20(4) cage -> maximum possible in r78c3+r7c4={179}=17 ->min r6c3 =3
->r6c3=8, r4c34= [28]

Step 4.
17(2) = {89} -> no 8 or 9 elsewhere in r1 or N3
“45” on N3 -> 3 innies = 9 -> no 7 in r3c7 or r23c9
->16(3) cage at r2c9 = {169\259\349} = 9{16/25/34} ->r4c9 =9, r1c89=[98]
“45” on c9 ->r6c8=2
“45” on N3 ->r3c7=2, r23c2=[27]


Step 5.
In N6, r56c9 = {57} only -> no 5 or 7 elsewhere in N6 or c9
12(3) cage now {138} only ->no 1,3 or 8 elsewhere in N6
->r46c7 = [46], r4c6=5

Step 6.
In N9, 10(2) cage = {37\64} -> 9(3) cage = {126} only (no{234} since 3 or 4 needed in 10(2) cage)
-> no 1,2 or 6 elsewhere in N9 or c9
-> r23c9 = {34} only -> r12c7=[71]
->r23c8 = {56} only
->r78c8 = {37} only -> r45c8 = [18], r5c7=3, r4c2=3, r9c8=4, r4c5=6
“45” on N2 -> r5c5=4
“45” on N5 -> r7c5=8
also r5c23 = {56} only ->r56c9 = [75]

Step 7.
“45” on c789 -> r79c5=5(2) = {23} only ->r5c4=2(hidden single N5), r1c5=2
->23(4) cage in N8= {1679} only -> r79c4 = [45]
etc[/size]

[nd]

Ugh, that was a stupid mistake of mine--must have omitted a step. Try the revised version above....

[sudokuEd]

Yeah, that works now.

BTW - In step 2, how did you work out that r9c2 = {89}? Because "45" on c12 means r9c2 - r5c3 =3?

It's not important to the solving path - but you may be using a shortcut I can't see.

[nd]

Oh, I should have made that clear -- you can do it either by using the 45 rule on C12, or via the "hidden pair" on {56} in C2.