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Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

 Posted: Fri Sep 08, 2006 11:30 pm    Post subject: Weekly Assassin 15 8th September 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 eliminationLast edited by Andrew on Fri Sep 15, 2006 11:41 pm; edited 4 times in total
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Sat Sep 09, 2006 12:15 am    Post subject:

mmm, something doesn't look right in Step 1 Andrew.

 Quote: 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
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

Posted: Sun Sep 10, 2006 9:39 pm    Post subject:

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.

 Quote: 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
Regular

Joined: 16 Jan 2008
Posts: 28
Location: Montesson

 Posted: Wed Feb 27, 2008 9:49 pm    Post subject: 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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