Killer Solutions: #7
#7 was so hard that I’m posting the answer now (with a walkthrough of the opening) rather than waiting till month’s end.
#7
7 9 5 1 6 8 2 3 4
3 8 4 9 5 2 7 6 1
1 2 6 7 4 3 5 9 8
4 3 9 2 8 7 6 1 5
5 7 8 3 1 6 9 4 2
2 6 1 4 9 5 3 8 7
9 1 2 6 7 4 8 5 3
8 4 7 5 3 9 1 2 6
6 5 3 8 2 1 4 7 9
Initial steps:
Step 1: 45-sum rule on leftmost 4 columns gets you r1/5/6c4 = 8; 45-sum on bottom 4 rows gets you r6c4/5/8/9 = 28. This leaves only two candidates for r6c4, {4,5}. r6c5/8/9 must be {6,7,8,9}, with {8,9} “necessary” inclusions.
Step 2: split the 23(4) right of centre into two cages: cage A (r4/5c7) and cage B (r6c6/7).
cage A: max value is 16 = {7,9} (can’t be 17 as {8,9} conflicts with the deductions at the end of step 1)
cage B: min value is therefore 23-16=7
cage C (r6c1/2/3): max value is 45-(28+7) = 10
cage D (r4/5c3): min value is 26-10=16.
In other words cage D must = either 16 or 17. Either {7,9} or {8,9}. So now we get (working our min/max in reverse)
cage D: 16 or 17
cage C: 9 or 10
cage B: 7 or 8
cage A: 15 or 16
Step 3: possible combos for cage A are thus {6,9}, {7,8}, {7,9}. We have already shown in step 1 that the r6c8/9 candidates are {6,7,8,9}; now the same goes for r4/5c7, producing a “four of a kind” which excludes those possibilities in the rest of nonet 6, & also excludes them in cage B (because that cage is doubly determined both by cage A and r6c8/9). So B can only contain numbers 1 through 5. This leaves only these possibilities for cages B and C:
C = 10, B = 7: 6-3-1 / 2-5 (C = 5-4-1 is impossible as it would contradict the {4,5} in r6c4. C = 7-2-1 is impossible as then D would be equal to 16 = {7,9}, resulting in two 7s in the same cage.)
C = 9, B = 8: 6-2-1 / 3-5
In either case the total combination includes 1-2-3-6-5. So r6c4 = 4 and r6c5/8/9 is {7,8,9}. This in turn leaves r4/5c7 as {6,9}, r6c8/9 as {7,8}, r6c5 = 9. And now in turn cage B = {3,5}, cage C = {1,2,6}
Step 4: The next step involves using a lot of X-wing techniques to deal with columns 8 & 9. In other words: if you can figure out that a given number MUST go in columns 8 & 9 in two particular nonets, then you can eliminate that candidate for the third nonet in columns 8 & 9 (& pencil it in for column 7 instead).
- Since the bottom half of the 25(5) cage in the bottom right corner = 10, it can only be 6-2-1, 5-4-1, 5-3-2. In other words, this cage must have a 1 OR 2 in it. The four cells r4/5c8/9 must have 1 AND 2 between them. So (using X-wing) the 14(4) cage in the top right corner cannot have any combination that has BOTH 1 AND 2 in it. That leaves only two combos: 1-3-4-6 or 2-3-4-5. This cage must therefore have 3 AND 4 in it.
- We’ve already shown (see step 3) that r6c7 cannot = 4. Therefore r4/5c8/9 contains a 4 somewhere in it. So we can X-wing on the number 4 in columns 8 and 9, showing that r7/8/9c8/9 cannot have a 4 in it. So the bottom half of the 25(5) in the lower right corner, which adds to 10, cannot have 5-4-1 in it. This leaves only 6-3-1 and 5-3-2.
- X-wing again! Both 6-3-1 and 5-3-2 have a 3 in there. So does the 14(4) at the top right. So now we can eliminate 3 from all cells of nonet 6 except for r6c7, so pencil a 3 in there. This means that r6c6 is 5.
This should be enough to get you over the hump, though it still takes a bit of work to figure out nonets 8 and 9. You can start by deducing that r7/8/9c4 must (via 45-sum rule) add to 19; the possibilities for combinations are constrained by the presence of 1-3-4 in the rest of the column, so there are only two choices: 9-8-2 or 8-6-5. Both of these contain an 8 so the next step is to observe that the already partially-filled-out 23(4) cage next to it must have a 3-7 combination in cells r7/8c5 (since they must add to 10 and cannot include a 4, 8 or 9).And so on… anyone who wants more walkthrough, just ask me…
January 9th, 2006 at 12:07 pm
Here is my complete walkthrough of puzzle 7 which doesn’t make use of ND’s X-wing techniques.
I think it is quite simple and clear enough and hopefully it’s understandable to everybody!
Step 1:
Since R[156]C4=8 and R6C[4589]=28, R6C4={45}, R[15]C4={12} or {13}, R6C[589]={689} or {789}.
Since R6C[12367]=45-28=17, R[45]C3+R[45]C7=26+23-17=32=15+17 or 16+16 or 17+15. So R[45]C3 & R[45]C7 are both from {6789}.
Since R6C[89] and R[45]C7 are both from {6789}, they form a “naked quad” in nonet 6, so the remaining 5 cells are {12345}.
Since R[345]C[89]=14+15=29 and R3C[89] is at most 17, R[45]C[89]=(1+2+3+4+5)-R6C7 is at least 12, so R6C7=1, 2 or 3.
Step 2:
Since R6C[89]+R[45]C7=6+7+8+9=30 and R[45]C7 ranges from 15 to 17, R6C[89] ranges from 13 to 15.
On the other hand, R6C[45] cannot exceed 5+9=14, so R6C[89] must be at least 14 to make up the total of 28.
So R6C[89]=14..15, R[45]C7=15..16, R6C[67]=7..8 to make up the 23/4 cage.
Now that R6C7={123}, R6C6 must be from {4567} (1,2,3 are eliminated to make sure the pair sums to at least 7).
Further, 6,7 are also eliminated because R[45]C7 & R6C[89] are in the same cage/row of R6C6 and they already cover {6789}.
So R6C6={45}, forming a naked pair with R6C4!
Step 3:
Last step we had R[45]C7=15..16, so R[45]C3=16..17 to make up 32, i.e. {79} or {89}.
Also, R6C[123]=9..10 to make up the 26/5 cage which now must be from {1236789}.
“Subtraction combo” shows that a pair totalling 10 must be eliminated.
These can be {19},{28} or {37}. But 6 must stay in the cage, which must be within R6C[123]. So R6C[123]={126} or {136}.
Step 4:
Now R6C[89]=14..15 coming from {789}.
This can only be {78}, forcing R[45]C7={69}, R6C5=9, R6C4=28-7-8-9=4, R6C6=5, R6C7=23-6-9-5=3.
Also, Now that R6C[123] must be {126}, R[45]C3 must be 26-1-2-6=17, i.e. {89}.
Step 5:
Now we look at nonet 3. Since R[45]C[89]={1245}=12, R3C[89]=14+15-12=17={89}.
Also, R[123]C7=45-14-17=14 and must be from {12457} now. Subtraction combo eliminates {14}, leaving {257}.
Comparing the 17/3 cage and these 3 cells, R1C6=R3C7+3, and the only remaining possibility is 8=5+3. So R1C6=8, R3C7=5.
Step 6:
Focus on C4 now. First we have R[15]C4={13} to make up the 8 with R6C4.
Next, we must have R[789]C4=19. However, R[78]C4 belongs to the 14/4 cage, and cannot exceed 11.
So R9C4=8..9, R[78]C4=10..11, R7C[23]=3..4 i.e. {12} or {13}, so the 1 in R7 must remain in these 2 cells.
Step 7:
Next we go to nonet 9. We already have the naked triple {148} in R[789]C7.
Comparing with the 13/3 cage, we have R7C6=R9C7 and the only common candidate is 4. So R[789]C7=[814] & R9C7=4.
Now we have the part of the 25/5 cage in this nonet summing to 10, which is easily determined as {235}.
The 22/3 cage is of course what’s left, {679}.
Step 8:
Since R[345]C6=21-5=16 comes from {12367}, subtraction combo quickly eliminates 1 & 2, leaving the naked triple {367}.
Going down to the 16/4 cage at the bottom, we should see that the 1 in R9 must be in R9C5 or R9C6.
This together with R8C6 being {29} now fixes the cage as {1249}. So 1,2,9 could be eliminated from the rest of nonet 8.
Therefore R9C4=8, R[78]C4=11={56}, R7C[23]=3={12}.
Now we have the naked pair {35} on R7C[89] which basically cracks down the bottom 3 rows!
Still a few twists and turns before we get to the finish line…
Step 9:
Look at the 16/4 cage in the centre. Now it must be from {123468}.
Subtraction combo eliminates a pair summing to 8, i.e. 2 & 6, leaving {1348}.
So, R[45]C5={18} and this naked pair almost cleans up the central 3 columns!
Step 10:
Go to nonet 1 now. R3C[123] is the naked pair {126}.
More importantly, since R[45]C[12]=45-26=19, comparing to the 17/4 cage gives R4C1=R3C2+2.
With R4C1={3457}, we must have R4C1={34}, R3C2={12}, giving us a naked pair in C2 and forcing R6C2=6.
This in turn gives us the naked pair R[67]C3={12} which forces R3C3=6.
To complete the 36/6 cage, we must have R2C[23]=12, and 8+4 is the only option left!
The puzzle is as good as finished now!
January 9th, 2006 at 2:41 pm
Thanks! I look forward to working through your solution to compare it with mine. I can see some interesting divergences in the order of solution…
January 9th, 2006 at 3:17 pm
Thanks for cleaning up the mess I made…
FYI I just posted a better version of the walkthrough in your thread in the sudoku.com forums… Fixed quite a few typos, so perhaps you would like to go there and retract that version instead!
January 9th, 2006 at 3:45 pm
Great Puzzle!
I have a difficulty however with step 3 of your solution. If C=10, B=7 can the combination not be 7-2-1, 4-3?
January 9th, 2006 at 5:36 pm
Glad you enjoyed it! — Re: your question: it can’t because then the other part of the 25(5) cage would have to add to 16, i.e. 7-9, thus producing two 7s in a single cage. I’ve now added this in for clarity’s sake above.
January 9th, 2006 at 5:42 pm
Just pasted in the fixed version of udosuk’s solution.
January 10th, 2006 at 10:17 pm
Just went through udosuk’s walkthrough — nice to watch a master at work! I think that your path through the puzzle once you get past the “range modification” bit is more efficient than mine; I think I’ve got a slightly simpler presentation of the initial steps up to that point (in other words, at the point where our solutions diverge, at my Step 4 — you move up into the left side of nonet 3 & into nonet 2, where I move down into nonet 9). Perhaps an optimal walkthrough would combine features from my opening & your middle- & endgame. Anyway, both of them eventually join up again by converging on the problem of the hidden 19(3) in nonet 8.
Getting time for me to update the tips page: there are now a good many more solving techniques & bits of terminology worked out since I wrote that.
January 13th, 2006 at 8:34 am
You’re probably right ND. I’ve been always thinking my strengths are in middle stage & endgame, so I usually spent more than 50% of the solving time in the beginning moves and once I crack down about 10-20 cells the rest would be flowing nicely and easily. Interestingly that’s also the case for my chess/shogi/go gameplay - the beginning is always the hardest part for me.
January 24th, 2006 at 11:04 am
Any idea on when number 8 is coming ?
May 29th, 2006 at 6:44 am
Done this one too. My solution involves, I think, an easier but more tedious way of working out the “range modification” bit:
1. As before, deduce that r6c4 = 4 or 5 and that r6c5/8/9 contains 8 and 9. These cannot both be in nonet 6, else both would be barred from the 23(4), which is impossible. So r6c5 = 8 or 9.
2. r4c3 + r5c3 is at most 17, so r6c1 + r6c2 + r6c3 is at least 9. Now list all four possible combinations for r6c4 and r6c5, with what each implies about the rest of the row; it turns out that only r6c4 = 4 and r6c5 = 9 is compatible with this.
3. r6c8 and r6c9 are {7,8}, so the 23(4) must be {3,5,6,9}. The 1 and 2 in row 6 must be in the left three cells, so these are {1,2,6}, and the 23(4) must have {6,9} in the upper half and {3,5} in the lower.
4. r6c7 is not 5, else the remaining cells of nonet 6 would total 10 and r3c8 + r3c9 would have to be 19. So r6c7 = 3 (and r6c6 = 5 and r3c8 and r6c9 are {8,9}).
5. The 27(4) and 23(4) in nonet 7 can have only one number in common. If this is 7 or less, the remaining seven numbers total 43 or more, an impossibility. So r9c4 is 8 or 9.
6. If r9c4 is 9, r7c2 and r7c3 are {1,3}, r7c4 and r8c4 must be {2,8}, and r7c5 and r8c5 must be {3,7}. By the 45-rule, r7c6 = r9c7 and therefore cannot be 8, but by X-wing there is an 8 in column 7 in nonet 9, which thus must be in the 13(3). Thus r7/8/9c9 = {1,4,8}, making the 10(3) {2,3,5}. We now have the impossibility of three 3s in rows 7 and 8.
7. So r9c4 is 8. Now go through a series of deductions parallel to step 6 to work out pretty much everything about the bottom three rows. The rest is straightforward.
September 8th, 2006 at 4:33 pm
Hi
Are you still creating Killer Sudoku puzzles…it seems the 1-9 were done in November 2005..??
Thanks
Andy
September 10th, 2006 at 1:54 am
I’ve not been doing a lot — there’s a puzzle I’ve been tinkering with for months but haven’t got a “release” version yet. I’m not home right now but will see if I can get a workable version up shortly.
February 21st, 2008 at 6:36 am
There seems to be an alternative solution:
793158264
584962731
216743598
378216945
459387612
162495387
921674853
847539126
635821479
The Sudoku conditions are fulfilled and each cage has the right value. Where is the error?
February 24th, 2008 at 10:35 am
The problem is that you’ve got two 3s in the 29(6) cage. Your solution only works if you permit doubled numbers in a single cage, which isn’t permitted in normal killer sudoku rules.