Weekly Assassin 15 8th September

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Andrew
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Weekly Assassin 15 8th September

Post by Andrew »

Here is my walkthrough. I've edited it for clarity and added more notes of explanation based on comments received from sudokuEd. Thanks for those comments.

1. R34C5 = {12}, R67C5 = {34}, R5C12 = {89}, 7 in N4 locked in 20(3) cage which must have {49/58} in the other 2 cells with R3C2 = {89} to avoid clashing with R5C12, 45 rule on R123 3 innies R3C258 = 11 -> R3C2 = 8, R3C58 = {12}, 45 rule on N6 2 outies = 6 -> R7C8 = {45}, R5C12 = [89], R4C12 = {57}, 45 rule on R12 2 outies R3C19 = 8 -> R3C19 = [35] because the 21(3) cage in N3 cannot contain a 3, 45 rule on N4 2 outies = 10 -> R7C2 = 2

2. R3C34 = {47}, 45 rule on N1 2 innies = 16 -> R13C3 = [97], R3C4 = 4, 45 rule on N3 3 innies = 10 -> max 7 in each innie -> R3C67 = [96], 45 rule on N3 again 2 remaining innies = 4 -> R1C7 = 3, R37C8 = [15], R34C5 = [21], 2 in N4 locked in C3, 2 in N1 locked in C1 -> R12C1 = {26}, 10(3) cage in N1 = {145}, R12C9 = [79], 14(3) cage in N3 = {248}, R4C89 = {68}, R6C89 = [93], R5C89 = [41], R2C7 = 4, R12C8 = {28}, R1C2 = 4, R4C89 = [68], 14(3) cage in N6 = {257} with the 2 in R4C7, R67C5 = [43], R6C12 = [16], R456C3 = [432]

3. 45 rule on R89 2 outies = 11 -> R7C19 = [74], R7C67 = [69], R4C12 = [57], R89C1 = {49}, R7C34 = {18}, 45 rule on N7 3 innies = 11 -> R9C3 = {18}, 14(3) cage in N7 = {356} with the 6 in R8C3, R89C9 = [26], 45 rule on N9 3 innies = 15 -> R9C7 = 1, 18(3) cage in N9 = {378} with the 8 in R8C7, R79C3 = [18], R7C4 = 8

4. R1C456 = {156} so 6 in R1 locked in N2 -> R12C1 = [26], R12C8 = [82], R2C456 = {378}, 1 in N8 locked in R8 so R8C456 = {179}, R9C456 = [254] and carry on, the rest is simple elimination
Last edited by Andrew on Fri Sep 15, 2006 11:41 pm, edited 4 times in total.
sudokuEd
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Post by sudokuEd »

mmm, something doesn't look right in Step 1 Andrew.
R37C8 = {15/24}, Killer pair {12} in R3C58 so no other {12} in R3
For r3c58 to be a Killer pair in r3, 1 or 2 must be in every combination in both those cells - but this is not possible since r3c8 can still be 4 or 5. So, I don't think you can eliminate 1 and 2 from elsewhere in r 3 at this point.

The next part of Step 1 is dependant on this elimination - so haven't gone any further into the walkthrough.
Andrew
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Post by Andrew »

As sudokuEd has pointed out, my original walkthrough was flawed. It would have been OK if the hidden split cage (is that an existing description or have I just made it up?) had been in the same row as the other {12} cell but doesn't work when the hidden split cage is at right angles to the row.
This Assassin is like a safe that requires the right combination to be opened. Should you fail, a timing device opens it after a week.
I've now changed the walkthrough having hopefully found Ruud's combination to crack the safe. This doesn't depend on killer pairs in that row. Fortunately the changes only affect step 1 of the walkthrough.

Thanks sudokuEd for pointing out the flaw in my logic. I'm happy to learn from my errors, hopefully it will make me a better solver.
Oscar
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Post by Oscar »

very easy one, just follow naked pairs as they appear.
with some 40 minutes work I would rate it as 0.50
Nothing can both be and not be
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