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Assassin 16

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

PostPosted: Sat Sep 16, 2006 5:19 am    Post subject: Assassin 16 Reply with quote

Quote:
a tough nut to crack.


Indeed it was. Took ages to find the right nut-cracker to use - lots of nooks and crannies to search through. Very enjoyable. A walk-through follows - but if you want just a hint on the nook to start with - in tiny text here 4 innies of N89

Assassin 16 –

Step 1
“45” on N89 -> 4 innies = 29
Max r7c45 + r9c4 = 7+8+9 = 24 -> min r7c9 = 5
->r67c9 = [15] and r7c45 & r9c4 = {789} only -> no 7,8 or 9 elsewhere in N8
“45” on N89 -> 2 outies = 10 -> r9c3 = {678} r6c4 = {234}

Step 2
“45” on c89 -> 2 innies = 3 -> r47c8 = [21]
“45” on r12 -> 2 innies = 3 -> r2c47 = [21]
r3c9 = 2 (hidden single r3)
9(2) cage in N3 {36} only -> no 3 or 6 elsewhere in N3 or c9
->14(2) cage in N23 = {59} or [68]

Step 3
“45” on r6789 -> 4 innies = 29 -> r6c5678 = {5789} only -> no 5,7,8 or 9 in r6c1234
->17(3) cage in N56 = {359/458} (not 368/467 since 3,4 & 6 only in r5c7)
-> 17(3) cage = 5{39/48} with r5c7 = {34} only, r6c67 = {589} with 5 locked in r6-> no 5 elsewhere in r6
“45” on N36 -> 2 outies = 13 -> r16c6 = [58] ([85] not possible since 8 has been eliminated from r1c6 in step 2), r16c7 = [95], r5c7 = 4, r34c7 = 13 = [76], r5c8 = 3 (hidden single N6), rest of 19(3) cage {79}
12(2) cage in N3 = {48} only -> no 4 or 8 elsewhere in N3 or c8 -> r3c8 = 5, r4c9 = 8

Step 4
“45” on r89 -> 2 innies = 5 -> r8c36 each max 4
-> 5 in N7 locked in 13(2) cage = {58} -> no 5 or 8 elsewhere in N7 or c1
-> r9c34 = [69/78] -> 7 in N8 only in r 7 (no 7 elsewhere in N8 - see step 1)-> no 7 elsewhere in r7
9(3) cage in N47 now {234} only
-> 6 in r6 now locked in 18(4) cage in N47 ->18(4) cage {1269} only
-> r78c3 = [91], r7c6 = 6 (hidden single r7), -> rest of 14(4) cage in N89 = 7 -> r7c7 = 3, r8c6 = 4, r89c7 = {28} = 10 -> r9c6 = 1, r9c9 = 4 (hidden single N9)

Step 5
r7c12 = {24} = 6 -> r6c1 = 3, r6c4 = 4
10(2) cage in N7 = {37} only -> no 3 or 7 elsewhere in N7 or c2 -> r9c34 = [69]
2 in N5 only in 29(5) cage -> no 1 in 29(5) cage (since the last 3 cells in that cage could not total 26) -> r4c4 = 1 (hidden single N5),
“45” on N5 -> r4c6 = 3
the rest is pretty straightforward



Edited again to make some points clearer (thanks again Andrew) and fix more typos


Last edited by sudokuEd on Wed Sep 20, 2006 9:07 pm; edited 3 times in total
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Andrew
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Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

PostPosted: Mon Sep 18, 2006 3:22 am    Post subject: Reply with quote

Ruud wrote:
A slightly altered version of my recent Ruudiculous Killer. It is easier, but still a tough nut to crack.
I agree completely with sudokuEd, a tough nut to crack which is hardly surprising since it's stated to have come from a Ruudiculous Killer. I found it the hardest Weekly Assassin so far.

I'm also posting my walkthrough for this puzzle. While it does, of course, follow roughly the same method as used by sudokuEd there are differences which I think come from our different solving approaches. sudokuEd puts in all possible candidates very early while I take work from the opposite approach and avoid putting in candidates until necessary and never more than four candidates in a cell unless I'm stuck and forced to put in more. That explains why my walkthroughs sometimes state the obvious in the final cage of a nonet before I move on to another part of the puzzle.

Step 1
45 rule on R12 2 innies = 3 -> R2C47 = {12}, 45 rule on C89 2 innies = 3 -> R47C8 = {12}, 45 rule on R6789 4 innies = 29 -> R6C5678 = {5789}, 9 in R7 locked in R7C345 because 14(4) cage in N89 cannot contain 9 -> 9 in N9 locked in 26(4) cage

Step 2
6(2) cage in N69 = {15/24}, 45 rule on N89 4 innies = 29 = {5789}* -> R67C9 = [15] -> R7C45 + R9C4 = {789}, R4C8 = 2, R2C47 = [21], R7C8 = 1, R6C4 = {34}

* This was “loose thinking” by me. Thanks to sudokuEd for pointing this out to me in a private message. Fortunately it happens to be correct in this case because the 5 is in N9 and the {789} are in N8. However there are cases when applying the 45 rule to multiple nonets where this would not be true. For example, with two nonets, alternative candidates for 4 innies totalling 29 could be {4799}, {4889} or {3899} where the repeated digits are not in the same row/column/nonet while it is possible with three/four nonets for the combination to be {2999}!

Step 3
6 in R6 locked in N4, it cannot be in R6C1 because {12} in R7C12 would clash with the 1 in R7C8 so the 6 is in R6C23, 9(3) cage in N47 = {234}, 6 in R7 must be in R7C67 because there is already a 6 in the 18(4) cage in N47, 45 rule on R89 2 innies = 5 = {14/23} -> R8C6 = {234}, R8C3 = {123}, 14(4) cage in N89 must be {1346} (it cannot be {1256} because 5 cannot go in any cell of this cage), 2 in R7 is locked in N7 -> R8C36 = [14], R7C67 = {36}, R7C12 = {24}, R6C1 = 3, R6C4 = 4, R6C23 = {26}, R7C3 = 9, R7C45 = {78}, R9C4 = 9, R9C3 = 6, R6C23 = [62], R89C1 = {58}, R89C2 = {37}

Step 4
45 rule on C12 2 innies = 7 -> R3C2 = 1, R5C1 = 1, 45 rule on N8 2 innies = 7 -> R79C6 = [61], R7C7 = 3, 10(3) cage in N8 = {235} with the 2 in C5, R5C6 = 2, R5C7 = {1234} from the valid combinations in R6C67 = 4 because {123} are all blocked, R6C67 = {58}, 45 rule on N5 4 innies = 16, because R6C6 is {58} the remaining 3 innies must total 8 or 11 and cannot contain 9 -> the 9 in N5 must be in the 29(5) cage in C5, R34C7 = {67} because {58} would clash with the {58} in R6C7, R89C7 = {28}, R6C67 = [85], R1C67 = [59] (hidden single in C7), 26(4) cage in N9 = {4679} with R8C89 = {69} and R9C89 = {47}, R89C2 = [73]

Step 5
45 rule on N3 2 outies = 14 -> R4C79 = [68], R3C7 = 7, neither the 7(2) sub-cage nor the 12(2) cage in N3 can contain 6 -> R12C9 = {36}, R3C89 = [52], R12C8 = {48}, R9C89 = [74], R6C58 = [79], R8C89 = [69], R5C89 = [37], the remaining 3 cells in the 29(5) cage in N5 are {569} with the 9 in C5 and 6 in R5, R4C46 = [13], 45 rule on N4 -> R4C1 = 7, R3C1 = 6, R3C56 = [49], R45C2 = [49] because 9 in N4 blocked from C3, R45C3 = [58], R3C34 = [38] and carry on, the rest is filling in the remaining candidates and simple elimination

This was written more as how I solved this puzzle, with minor rearrangement of sub-steps, rather than deliberately as a walkthrough.

As an example of the different approaches, sudokuEd was able to fix a hidden single early in his step 2 but I didn't fill that cell until my step 5.

My later message in this thread has now been deleted. It is no longer relevant now that my walkthrough has been changed to normal text.


Last edited by Andrew on Fri Sep 22, 2006 12:05 am; edited 2 times in total
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sudokuEd
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Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

PostPosted: Mon Sep 18, 2006 4:18 am    Post subject: Reply with quote

Andrew wrote:
sudokuEd puts in all possible candidates very early


SumoCue does that for me right at the beginning when I tell it to 'mark-up'. I never do that when paper solving - but always pencil in each eliminated candidate (at the bottom of the cell). If I'm looking for hidden singles/locked candidates etc, I look at the eliminated marks. I only put in the remaining possible candidates much further on.

I'd call myself an elimination solver. So, SumoCue feels quite natural to me because I tell each cell to eliminate candidates as I go.

Would love to know what methods others use with marks.

Thanks a lot Andrew for posting your walkthrough as well. The harder puzzles always seem to have a number of different variations in solving strategy. I enjoy finding out how others approached it.
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Oscar
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Joined: 16 Jan 2008
Posts: 28
Location: Montesson

PostPosted: Sun Mar 02, 2008 11:42 am    Post subject: Reply with quote

After 4 days, I just couldn't make it without the "hint on the nook" from sudokuEd Embarassed .
It is narrow gorge that allows to pass to the other side of the mountain.
I would rate this one 1.25
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