Killer Sudoku Solutions: November
Here are last month’s puzzle solutions…
#1:
7 6 9 2 3 4 5 8 1
3 1 4 8 6 5 7 9 2
5 2 8 1 7 9 3 6 4
1 9 7 6 2 8 4 5 3
8 3 5 9 4 1 2 7 6
2 4 6 3 5 7 8 1 9
4 7 3 5 1 6 9 2 8
9 5 1 4 8 2 6 3 7
6 8 2 7 9 3 1 4 5
#2:
9 8 2 5 1 4 6 3 7
7 1 4 3 6 9 8 2 5
5 3 6 8 2 7 4 9 1
8 4 7 1 9 3 2 5 6
2 6 9 4 8 5 7 1 3
1 5 3 2 7 6 9 8 4
3 7 8 9 4 1 5 6 2
6 2 5 7 3 8 1 4 9
4 9 1 6 5 2 3 7 8
#3:
2 9 3 6 1 4 7 5 8
7 4 6 2 5 8 9 3 1
5 8 1 7 9 3 6 4 2
1 7 5 3 4 2 8 9 6
3 2 4 9 8 6 1 7 5
9 6 8 1 7 5 3 2 4
8 1 7 5 2 9 4 6 3
6 5 9 4 3 1 2 8 7
4 3 2 8 6 7 5 1 9
#4:
9 3 5 2 8 7 4 1 6
1 4 7 9 6 5 8 3 2
8 2 6 4 3 1 5 9 7
6 8 9 7 2 3 1 4 5
5 7 2 1 4 8 9 6 3
4 1 3 5 9 6 7 2 8
2 6 1 8 7 4 3 5 9
7 9 4 3 5 2 6 8 1
3 5 8 6 1 9 2 7 4
Since #4 is a very hard one I’ll post the initial steps to a solution (just enough to get you over the hump if you’re stuck).
Step 1. 45 rule applied externally to nonet 1 reveals that r4c1/2/3 adds to 23, thus combo 6-8-9. Internal 45-sum rule on nonet 1 reveals that r4c3 – r3c2 = 7. So r4c3 = 8/9, r3c2 = 1/2.
Step 2. Internal 45-sum rule on lefthand nonets 1, 4, 7 reveals r5/6/7/8c3 adds to 10, thus 1-2-3-4. Internal 45-sum rule on nonet 7 reveals r7c1 + r7c3 + r8c3 = 7, thus 1-2-4. The presence of the 11(4) cage in nonet 4 (i.e. 1-2-3-5) means the 3 must go in r5/6c3, the 4 in r7/8c3. r7c1 = 1 or 2. A 5 must go in r5/6c4. The four cells of the 19(5) cage in nonet 4 must thus contain either 1-4-5-7 or 2-4-5-7.
Step 3. Nonet 5: use 45-sum rule on rows 1-4 to split 18(5) cage into 6(3) on top and 12(2) on bottom. 6(3) is of course 1-2-3; the 12(2) subcage must be 4-8 (there are no other possibilities, since both 3 and 5 are blocked). Since r5/6c4 must be 1-5 or 2-5, r3c6 must be 1 or 2 to avoid contradiction. The remaining three cells in nonet 5 must be 6-7-9, so obviously r4c4 must be 7 (because the 6-8-9 combo to its left rules out 6-9), and r6c5/6 must be 6-9.
Step 4. Note that r7/8c6 must be a 6(2) cage, i.e. 1-5 or 2-4. Thus it contains either a 1 or 2; so in conjunction with r3c6 (which must also be 1 or 2) it blocks out 1-2 from r4c6 leaving only 3 in that cell.
Step 5. This step is courtesy udosuk (I did this one differently originally but this is the most efficient method). Use the “coloring” technique on the 1-2 pair in the 18(5) cage in the central nonet (the chain involved is: r3c6; r4c5; the pair r5/6c4; the pair r7/8c6) & you can eliminate 1 and 2 from r7/8/9c4, leaving only 3-6-8-9 as possibilities. The only possible combos in this cage is then: 1-4 on the left / 3-6-8 on the right.
(If you need more description of coloring, then this is a good spot.)
The rest of the puzzle should be self-explanatory, though still not exactly easy going.
January 17th, 2006 at 6:31 pm
Hi! I’ve been enjoying your puzzles quite a bit, but I found myself stuck on this one. Reading through the solution bit by bit, I can’t figure out a part of Step 3.
You state that “Since r5/6c4 must be 1-5 or 2-5, r3c6 must be 1 or 2 to avoid contradiction.” What’s the reasoning behind a 3 being contradictory in that spot, given only the information up to that point?
Thanks for the help - your tutorials have been extremely useful!
Aubrey
January 18th, 2006 at 1:47 am
Aubrey — glad you’re enjoying them! & that the walk-throughs help: I should really write out “full” walkthroughs I suppose but usually just give the opening steps that are enough to “crack”, not solve, the puzzle. But udosuk’s walkthrough on #7 is quite thorough & illuminating.
Re: that particular step: We’ve already shown that the combination {1,2,3} is in the three cells r3c6 and r4c5/6. If we put the 3 in r3c6 & put the 1 & 2 in r4c5/6, then there would be no valid combination left over for r5/6c4 — since we have already shown it must have either 1-5 or 2-5. So r3c6 must be 1 or 2.
January 18th, 2006 at 6:20 pm
Oh, for the love of… I mis-read it as r6c3. Makes a bunch more sense now!
I’ll brush up on my coloring, since that’s the spot where I got stuck.
Thank you!
January 19th, 2006 at 12:45 am
Aubrey — just realized I make a small typo in step 2 (put the 4 as a possibility in 4 cells instead of the correct 2). Have fixed it up. If you’re still stuck there’s a diagram of the grid at step 4 on the new tips page part two which I just created.
May 23rd, 2006 at 8:57 am
Gah. I breezed through puzzles 1 to 3 in the last few days so I thought I’d try 4 and I was utterly humbled. I gave up after several hours having only managed to place one digit (the 7 in r4c4). I don’t know how anyone manages to do these!
May 23rd, 2006 at 11:28 am
Michael — yeah, it’s a hard one! But do, even if you find yourself stumped, try working through the walkthroughs on the solutions pages. & maybe you’ll have better luck with #5!
May 25th, 2006 at 6:28 pm
Thanks!
Well, I solved #5, but there were a couple of places where I could prove it was impossible to progress so I had to compare my grid with the solution to know which numbers to rub out in backtracking. Is that cheating, do you think? — because it means I didn’t solve it completely on my own.
PS. I just noticed another Michael has been posting comments on your later puzzles. Just wanted to clarify that I’m not the same Michael!
May 26th, 2006 at 12:11 am
Well, I think everyone gets stuck (I know I do, with some puzzles) — you can learn a lot from trying to work things out knowing the solution in advance, or just doublechecking every so often to make sure you’ve not gone wrong. — #5 is a fun one — I did that on a long busride (to/from Chicago).
August 17th, 2006 at 4:53 pm
Just finished #4. Actually found it easier than #3, but there seem to be two unique solutions for it. I was approaching the end and was left with the following choices:
r1c2=1/3
r1c8=1/6
r1c9=3/6
r2c1=1/3
r2c8=1/3
r5c1=1/5
r5c4=1/5
r5c8=3/6
r5c9=3/6
r6c2=1/5
r6c4=1/5
r9c1=3/5
r9c2=3/5
You can solve this by putting either a 3 or a 6 into r5c9 and following the domino effect through to the end and both seem to come out right. Or have I missed something????
I liked the 45+out-in technique; that has helped me on a number of puzzles recently.
August 18th, 2006 at 1:07 pm
“Two unique solutions”? It’s either unique or it’s not!
I haven’t worked out your alternate solution in detail but it seems clear you’re getting the alternate solution by following “normal” sudoku rules rather than killer sudoku rules: i.e. ignoring cage values, which place an additional constraint on cell candidates. For instance, if I follow your list of candidates correctly, you’re saying that you have already placed the 9 and 8 in the 18(3) cage in N1, but that the third cell is {13}. But it must be 1 (18-9-8).
August 19th, 2006 at 9:08 am
D’oh! Stupid of me. You’re quite right. Funny how on the home straight I forgot the cage values!