Killer #9: solution
Here’s the solution, plus a walkthrough. I’d be interested to see other solvers’ walkthroughs too.
3 2 7 1 6 8 5 9 4
6 9 5 3 4 7 8 2 1
1 8 4 5 2 9 6 3 7
2 7 1 4 9 6 3 5 8
8 5 6 2 1 3 4 7 9
9 4 3 7 8 5 2 1 6
4 3 8 9 7 2 1 6 5
5 1 9 6 3 4 7 8 2
7 6 2 8 5 1 9 4 3
Complete walkthrough for Killer #9:
If you need help with the shorthand, go to J-C Godart’s explanatory page here.
1. 45 rule on C1234 shows that R78C34 = 32 = {9986} or {9887} ({9977} is impossible because three cells are within the same cage). So R7C3 and R8C4 = {6789}, and R7C4 = R8C3 = {89}.
2. The step here is the key to the puzzle: it’s an extension of Jean-Christophe Godart’s “overlap” technique. You can get a refresher here. Basically: when there is an intersecting row and column, and all the cages fit exactly within the row and column, then you can determine the value of the cell at the point of intersection: it is 90 minus the sum of the cages. (Uh, is that clear?) In this puzzle, there is an example of “imperfect overlap”: in C1 and R9 all the cages lie within the row and column, with the exception of an innie at R6C1 and three outies at R18C2 and R8C9. So our calculation is more complex:
90 - sum of all the cages = 90 - (11 + 11+ 33 + 24) = 90 - 79 = 11
Therefore R6C1 + R9C1 = R18C2 + R8C9 + 11. This leaves three possibilities: R6C1 + R9C1 = 15, 16 or 17 (i.e. {69}, {78}, {79} or {89}), and R18C2 + R8C9 = 4, 5 or 6 (i.e. {112}, {113} or {122}, and {123} or {114}: the fact that the cells are not all in a row allows us to have duplicate numbers in the combinations). So R6C1 and R9C1 = {6789}; R1C2 and R8C9 = {123}; R8C2 = {1234}.
3. Next step: eliminate {123} from R9C4. This is easy: 45 rule on C4 reveals that R789C4 - R6C3 = 20. So R789C4 = 21..24 = {5..9}, and R6C3 = {1234}.
4. Since there are only two cages in R9, whatever number (out of {123}) is in R8C9 must also be in R9C23; and in turn it must be in R7C56. But a 26(4) cage can’t have a 1 in it. So R7C56 = {236789}, R6C5 = {6789}, R8C9 = {23}, R8C2 = {123} (and 1 is locked in R18C2), R69C1 = {(7|8)9}.
5. Now we eliminate {23} from R9C2: the 21(4) cage in N4 cannot have 1 in it and therefore must have 2 or 3 (because otherwise the maximum value of R6C1 would be 21 - {456} = 6, which contradicts what we know of R6C1). So we have a hidden {123} triplet in C2! So {123} is locked within R15678C2, R9C2 = {4..9}, R9C3 = {23}.
6. 45 rule on R9 shows R7C1 + R8C129 = 12, so R7C1 + R8C12 = 9 or 10. All possible combinations for both the 33(7) and 24(6) cages in R9 include a 1; since we know R8C9 = {23} then R9C56789 must contain a 1, and the 1 is locked within R7C1 + R8C12 within N7. Only possibilities (keeping in mind that whatever cell is in R8C9 is also in R9C3 and thus blocked from the combinations): {126}, {136} or {145}.
7. 45 rule on N14 shows R7C2 = R6C3. They must be either 3 or 4 (can’t be 1 or 2 because they are locked in the 33(7) in N7, since all combos for that cage include a 2). So R7C2 = R6C3 = {34}. We have already used the 45 rule on C4 to show that R789C4 - R6C3 = 20. So R789C4 = 23 or 24, i.e. {(6|7)89}. So {89} is locked in R789C4 within N8, and R6C5 = {89}.
8. So the only place that the 1 in C3 can go is the 11(3) cage in R345C3. This means that the 5 in C3 can’t be in the same cage, so it’s in R12C3. So R12C3 = {579}, R2C2 = {79}, R3C2 = {68}, R4C2 = {79}, R9C2 = {4568}. 45 rule on N1 shows R3C123 = 13, so R3C13 = {1234}.
9. Now we can use what we’ve learnt in previous steps in order to deal with the crux of the entire puzzle (the {23} problem in R8C9/R9C3). We can do this via a short logical branching. (If the branching bothers you, go to udosuk’s solution below for a longer but hypothetical-free step 9.) The 5 in C1 must be in either R45C1 or R78C1. So:
- If the 5 is in R78C1, then R78C1 + R8C2 = {145} = 10 (see step 6). Then (cf. step 6) R8C9 = R9C3 = 2, and (cf. step 2), R69C1 = {79}.
- If the 5 is in R45C1, then the only place in C1 for the 7 is R69C1. So we get the identical situation (cf. step 2): R69C1 = {79} and R8C9 = R9C3 = 2.
So in either case, we end up with the same result: R8C9 = R9C3 = 2, R1C2 = 2, R8C2 = 1, R69C1 = {79}. This breaks the deadlock in the puzzle!
10. So R7C56 = {27}, which means the hidden 32(4) in R78C34 is {6899}. So R7C4 = R8C3 = 9, R6C5 = 8, R69C1 = [97], R7C3 = 8, R89C4 = [68], R345C3 = {146}, R6C3 = R7C2 = 3, R78C1 = {45}, R9C2 = 6, R12C3 = {57}, R234C2 = [987], R3C13 = [14], R12C1 = {36}, R45C1 = {28}; R56C2 = {45}, R8C56 = {34}, R78C1 = [45], R9C56 = {15}, R9C789={349}; R8C78= {78}; R7C789 = {156}.
11. The cells R6C67 + R7C7 add to 23 - 15 = 8. But 3 is blocked from all these cells, so R6C67 = {(1|5)2}; R7C7 = {15}, and the 6 is locked in R7C89 within N9. R456C4 = {(15|24)7}, So {369} is locked within the section of the 25(4) cage in N5, and that cage cannot have 8; the only possible combo for that cage is now {123469} (with {369} excluded from the cells in N2). So R3C5 = 2, R7C56 = [72], R6C7 = 2, R123C4 = {135}, R456C4 = {247}, R2C5 = 4, R6C6 = 5, R7C7 = 1, R56C2 = [54], R6C4 = 7, R45C4 = {24}, R8C56 = [34], R9C56 = [51], R6C89 = [16], R7C89 = [65], R2C8 = 2.
12. 45 rule on N6 shows that R4C7 = 3, so R5C6 = 3. 45 rule on N3 shows that R3C6 - R1C7 = 4, and the only possibility is R3C6 = 9, R1C7 = 5. So R4C6 = R1C5 = 6, R23C7 = [86], R12C6 = [87], R12C3 = [75], etc. From this point on the puzzle is more or less fill-in-the-blank.
April 19th, 2006 at 1:26 am
Thanks for the walkthrough.
At step 5 you say “So we now know that {123} is locked in C2 within R14567″ Isn’t it R15678 ? Also, it could help reminding us that R18C2 is {1(2|3)}, which forms a killer (or complex) naked pair on {23} with R567C2.
April 19th, 2006 at 1:29 am
Right — fixed!
April 19th, 2006 at 9:34 am
Tried everywhere to find my way…
Just found some interactions…
We know Cage 21/3 in N1 = {579}
R78C3 = (15|17)/2 = {69|78|89} = {(7|9)…}
Killer naked pair on {79} with R12C3 -> nowhere else in C3
-> Cage 11/3 in C3 = {128|146} (can’t have 3)
-> 3 of C3 in R69C3
We know that R6C3 = R7C2 = {34}
-> 3 of N7 in R7C2 or R9C3 -> nowhere else in N7
Then in C4, N8 (not useful at this stage)
R789C4 = 23|24 = {89(6|7)}
R456C4 (cage 16/4) = {157|156|247|237|256} = {(6|7)…)
-> cage 9/3 in R123C4 = {135|234} = {3…}
In N8
Must have 2 or 3 in R7C56
-> Cage 26/4 = {2789|3689} = {89(27|36)} -> R7C56 = {27|36}
Also killer naked on {67} with R789C4
-> R89C56 can’t have {6789}
We know R8C9 = R9C3. So the outies of R9 could well be used for R7C1+R8C12+R9C3 (therefore, numbers can’t repeat)
In N7, R7C1+R8C12 = 9|10 = {126|145}
{127} can’t work, because -> R8C9 = 2 -> R9C3 = 2 (two 2 in N7)
-> 7 of cage 33/7 in R9C14 -> Cage 24/6 can’t have 7
Could also have deduced that using R9, since R8C9 = R9C3 = {23} -> R9C124 = 45-24 = 21 = {579|678} = {7…}
-> Cage 24/6 must have 4 in R9
-> 4 of C2 is in Cage 21/4
Since Cage 21/4 must have 4, either the 4 is in R56C2 or in R7C2
If R7C2 = 4 -> R6C3 = 4 -> 4 of N4 is locked in these.
Also in N8
R78C3+R8C4 = 23|24 -> R8C45 = 6|7 = {15|24|25|34} = {(1|2|3)…}
Killer naked triplet on {123} in R8C2569 -> each in R8C14578 >= 4
No way ! Needed some T&E
R8C9 = {23}
If R8C9 = 3 -> R18C2 = {12}, R9C3=3, R69C1 = {89}
-> R7C2 = R6C3 = 4 -> cage 11/3 in C3 = {128}
-> 7 of C1 in cage 11/3 -> R4C2 = 9
-> R69C1 = [89] -> R8C3 = 8 wich conflicts with cage 11/3 in C3
So R8C9 = 2 -> R8C2 = 1, R1C2 = {23}, R9C3=2
Cage 11/3 in C3 = {146}, R3C3 = 4, R45C3 = {16}
R6C3 = R7C2 = 3, R1C2 = 2
R7C56 = {27} -> R7C3 & R7C3 = {689} -> R7C4 = 9, R8C3 = 9, R8C56 = {34}, R8C56 = {15}
…
April 21st, 2006 at 10:55 am
It doesn’t affect the logic of the walkthrough because you give the conclusions inline but in step 2 the following sentence “Therefore R6C1 + R9C1 = R18C2 + R8C9 - 11.” have an error. Shouldn’t the 11 be positive?
Great puzzle, how much wait until the 10th?
Cheers,
Nuno
April 21st, 2006 at 3:17 pm
Agh, those things are hard to proof — it’s fixed now!
Haven’t started on #10 yet, though I have a few ideas…
April 24th, 2006 at 9:55 pm
Just changed step 9 to make it a little more elegant — see what you think, guys.
May 4th, 2006 at 8:40 am
I think I’ve (finally!) worked an alternative approach to breaking the deadlock at ND’s step 8.
This follows on from ND’s step 8.
9)Lets look at C3. I chose this one because it is linked into the hidden 32(4) cage (step 1) and because it is links into the ‘overlap’ of step 2 through r9c3 and because r6c3=r7c3.
a. first we can see that 3 in c3 must go in r69. Here’s how. If r12 is {57} then r6789 = 45-12-11 = 22. This could be [3982] [3892] [4693]. If r12 is {59} then r6789=45-14-11=20. This could be only be [3782] ([4682] is not an option since r78c3 cannot be {68} in the hidden 32(4) cage in r78c34 since that would require {99} in r78c4). So, all options in c36789 have 3 in r69 ->no 3 elsewhere in c3.
b. In addition, we can now see that the 3 in N7 can only go in r9c3 or r7c2. If the 3 in c3 is not in r9 it must be in r6. And since we know that r6c3=r7c2, then r7c2 must also be in r7c3. This means that 3 cannot be anywhere else in N7 -> r8c2 is {12}. This also means the only combinations possible in r78c1and r8c2 are {126} or {145} (cf. Step 6).
10. “45” on c12 means that r12c3+r9c34=22. If r12c3 is {57} then r9c34=10, can only be {28} or {37}. If r12c3 is {59} then r9c34=8, can only be {26}. Therefore, there is no 9 in r9c4.
11. We know from step 7 that r789c4 has {(6\7)89}. We know from my step 10 that the 9 can only be in r78. This means we can now eliminate {9887} from the hidden 32(4) cage in r78c34. Here’s how.
a. If the 9 of {9887} is in r8c4 then 8 must be in r6c4 & r8c3, so the 7 must be in r7c3. This means that r9c4 can only be 7 (if it was 6 then no 6 or 7 would be available for the r7c56 which is {27\36}). When r9c4 is 7 then r9c3 can only be 3 (my step 10 above). When r9c3 is {3} then r12c3 is {57}. But this cannot be since now 2 7’s are now in c3.
b. If the 9 of {9887} is in r7c4 then the 2 8’s must be in the one cage which cannot be.
12. We are now only left with {9986} for r78c34 (from step 1). Since there must be two 9’s, they are placed in r8c3 and r7c4 since 2 9’s cannot go in the same cage. This unlocks the puzzle!
May 4th, 2006 at 9:49 am
Nate,
Since posting the walkthru V2 above I realize that my step 9 is not really needed. Nothing in step 10-12 depends on step 9 (what you’ve been telling me the whole time!!!). So, if step 10-12 is valid then these could become steps 9-11. If that gets changed around then some of my references back to my steps will have to get edited.
This time I really am going to bed!
May 4th, 2006 at 2:11 pm
Ed — yes that works, though I think that the chain of hypotheticals in 10a is a pretty long one — so far I haven’t seen anything that’s quicker than the two-step branch in my revised step 9. — Yeah, neither deduction in your step 9 really affects the solving-path. FWIW I think it’d be much easier to eliminate {137} in the 11(3) cage in C3 simply by eliminating the 7 rather than the 3: if R12C3={59} then R78C3={78}, & so whether R12C3={57} or {59} the 7 is accounted for. But it doens’t really advance the solution, I think.
June 25th, 2006 at 5:22 am
I think I’ve nailed the most logical steps to resolve the puzzle after step 8…
Here is how the grid looks like after step 8:
http://i4.tinypic.com/15mntqh.png
We have just eliminated the possibility of the 11/3 cage of r345c3={137}. (If you can’t see that, just notice that r78c3 must contain either a 7 or a 9, so forming a killer naked pair with r12c3.) So r345c3={1(28|46)}.
Next, do a complex innie-outie on c12, which gives r9c34=r2c2+1. So r9c34=8 or 10. Therefore r9c4 cannot be 9, and the 9 on c4 is locked within r78c4, so r78c4={(6|8)9}, and 7 is eliminated from r8c4.
Next, we already know that r7c1+r8c129=12 (outies from r9), and r8c9=r9c3. So r7c1+r8c12+r9c3 forms a 12/4 cage within n7, which must be {12(36|45)}. In other words, 3 & 6 must be both inside or outside this 12/4 cage.
Now look at r7c3, we know it could not be a 6 because it would force r9c3 to be 2, violating r345c3={1(28|46)}. Therefore r78c3={8(7|9)}, and 8 is locked within these 2 cells, forcing r345c3 to be {146}, and in turn r6c3=r7c2=3 and r8c9=r9c3=2 and the rest of the puzzle should be solved relatively easily…
I don’t know if you guys think this is good enough to be considered non-hypothetical… And I know it’s a bit awkward to post all these after the puzzle’s released for 3 months, but I was taking a semi-retirement from killer solving activities… Well it seems I still got something left inside…
June 25th, 2006 at 1:01 pm
udosuk — many thanks for the alternate path. That’s longer than my step 9 but neatly removes the need for any use of extended hypotheticals. — When I actually made this puzzle I got to the key point (the one charted in your tinyurl image) & initially thought that I still needed to add some more “helpful” features to the puzzle to prevent it from being indeterminate or impossible to solve without T&E. But then after staring a while at it & thinking about it I realized that there was indeed only one solution & managed to get over the hurdle, so left it as it was without further “help” — with that little crisis-point still in there. Hm: a question occurs to me — is this puzzle at all solvable without the use of the “overlap” method at the start? Most times when one invokes the overlap technique it’s just a more efficient form of the vanilla 45 rule; but I think in this case there’s simply no way around using it.
Still have that draft of #10 kicking around here though I haven’t turned attention to it lately — will dig it out…
June 25th, 2006 at 6:27 pm
I think it’s pretty cool for the puzzle to required the “overlap”… And I wouldn’t stretch my brain too much to find the solution without it…
Anyway great puzzle and look forward to your #10!
June 26th, 2006 at 3:36 pm
Nice to hear from you again Udosuk
I also had a look at this puzzle a few days ago !
Starting after step 8 of Nate :
As I (and Udosuk) already mentioned earlier :
R12C3 = {5(7|9)}, R78C3 = 15|17 = {69|78|89} = {(7|9)…} forms a killer naked pair -> no {79} elsewhere in C3
-> Cage 11/3 in C3 = {128|146}
45 on N1 -> R3C123 = 13
Min R3C2 = 6 -> Max R3C13 = 7 -> each in R3C13 <= 4 (cannott repeat 6, 5 locked in Cage 21/3)
Cage 11/3 in C3 = {128|146} = {(6|8)…} -> R45C3 = {(6|8)…}
-> Cage 21/4 in N4 cannot have both {68}
-> Cage 21/4 in N4 = {2469|2478|3459} = {(7|9)…}
Since R24C2 = {79} -> R6C1 = {79} -> naked pair within N4
-> {79} of C1 locked in R69C1 (hidden pair)
This fixes the complex cage from the overlap of C1+R9
-> R1C2 = 2, R8C29 = [12]
…
June 29th, 2006 at 8:17 am
Great spot JC!
As we can see from my last posted pic, obviously r45c3 must have either 6 or 8. Then it’s a matter of seeing the 21/4 cage must contain 7 or 9 (otherwise the largest it could get is {3456}=18 or {3458}=20), so a killer naked pair of {79} is formed and r45c1 cannot have 7… Hence the hidden pair of {79} in r69c1…