Killer Sudoku Online Weekly No. 381  Flexing those flabby algebra muscles
Walkthrough by Steve Metzler (2013)
Last updated: 11th May, 2013
Update: reader Patrick Kelly informed me that working too far ahead on paper to find anamolies is frowned upon by the
purists. Always aspiring to be the purist myself, I reworked the puzzle to remove the wildassed guessing, and that
eliminated four diagrammes as well :)
Introduction
In my previous killer sudoku article:
Killer
Sodoku Online Weekly No. 371  Mind Bending
We discussed the ground rules for this type of sudoku puzzle, tackled a really tough puzzle, and won out in
the end. The subject of this article is an equally tough puzzle, but for different reasons. Here, the solution
lies in figuring out a relationship between cells that is not exactly obvious... unless you know what to look
for, that is :)
Solution
Here is the puzzle in question, with the initial 'gimme' cages filled in:

Figure 1: Killer Sudoku Online Weekly No. 381  Mind Bending. Initial stage 
The link for playing this puzzle online is at:
Killer Sudoku Online Weekly No. 381  Mind Bending
Since we covered all the background last time, we're going to just dive right in this time. The key to this puzzle lies
straightaway with nonet F and its 'overflow' cells. Note that there are two 'outs' here: r3c9 and r6c6. And a single 'in':
r4c7. Since the sum of the cells in any nonet must be 45, there is an algebraic relationship between these 3 cells:
28 + 30 + r4c7  (r3c9 + r6c6) = 45
Which reduces to:
r4c7 + 13 = r3c9 + r6c6
In plain English, this means that r3c9 + r6c6 must be 13 greater than r4c7. The fact that the {89} cage on r3 limits what
can go in r3c9 helps a lot to cut down the possibilities in this respect. Think about it. The least we can have in r4c7 is a 1,
so r3c9 + r6c6 must be at least 14. And since r3c9 can't be an 8 or a 9, that leaves us with:

Figure 2: Algebra. It works, bitc...! Ahem. 
Next we note that r3c9 is part of the 'ins' in nonet B + C. So r3c4 + r3c5 (part of the 12 x 3 cage) + r3c9 = 90  (13 + 16
+ 10 + 15 + 5 + 17) = 14. According to the Sum Calculator, the possibilities for 14 x 3 excluding 8 and 9 are:
14 x 3 (x8,9)

167
257
347
356
We can immediately rule out {347} as a possibility here, because that makes the 5 x 2 cage in nonet B unviable. So now let's
look at all the possible permutations for filling in that 'hidden' 14 x 3 cage:
r3c4 + r3c5  r3c9
 
16 7
17 6
25 7 (ruled out because we can't have two 5s in the 12 x 3 cage)
27 5
35 6
36 5 (ruled out because we can't have two 3s in the 12 x 3 cage)
Note that the overflow from nonet B is: (13 + 16 + 5 + 12 + 17)  45 = 18. From the above, we know that r4c5 must be {345}.
But we can immediately rule out 3 as a possibility. If r4c5 is a 3, the remainder of the 12 x 3 cage can only be {27}. That
means the 5 x 2 cage must be {14}. Note that if r4c5 is a 3, then the 3 in the 4 x 2 cage below must be in c6. And that leaves
the only possible place for a 3 in c4 at r1c4, which is part of a 13 x 2 cage, and so is not possible. Now we can fill in all
the permutations of 18 in those three cells leaving us with {456} in r2c7. But the only way the 6 can get there is if r4c5 is
a 4. If r5c4 is a 4, then the remainder of the 12 x 3 cage is 8. And because of the hidden 14 x 3 cage on r3, r3c9 would have
to be a 6. Therefore, r2c7 cannot be a 6. So now we have:

Figure 3: The overflow from nonet B 
The 5 is now gone from r3c9 since r4c5 cannot be a 3, and a bit more deduction allows us to get rid of the 7 in r6c6. Do you
see it? Right. If r3c9 is a 6, then r6c6 cannot be a 7, as then those two cells would not be 13 greater than r4c7. If r3c9 is a
7, then r6c6 cannot be a 7 as well or there would be no possible way to put a 7 in nonet F :) I thought I'd show you the state
of the board at this point in Figure 3 before we go further in narrowing down nonet B. Ready? OK then... if r2c7 is a 4, then the only
possible way to make 12 to finish off the 16 x 3 cage is with {57} (note how the two {89} cells in c6 block the only other
possibility of {39}). Likewise, if r2c7 is a 5, the only way to make the remaining 11 is with {47}. So the 16 x 3 cage must contain
{457}, and that limits the 13 x 2 cage in nonet B to {4589}. Since either the 13 x 2 cage or the nonet B portion of the 16 x 3
cage must contain a 4, then the 5 x 2 cage has to be {23}. That leaves {16} over to complete the 12 x 3 cage, and we have now
fleshed out the top 3 nonets considerably:

Figure 4: Top 3 nonets taking shape 
Two items of note there, the first being that since the 19 x 4 cage in nonet A cannot contain {234}, then it must contain
the 5 on r3. Secondly, we have eliminated the 1 from r4c7 bcause of the '13 greater' rule. Now let's turn our attention to
the bottom of the board. The overflow from nonet H is: (11 + 9 + 24 + 4 + 11)  45 = 14. That, coupled with the fact that
there is a hidden 20 x 3 cage (45  (18 + 7)) in nonet G gives us the following possibilities there (because of the 20 x 3
cage, the minimum value r7c3 can have is 3):

Figure 5: On the fringes of nonet H 
Took a few liberties there while I was at it. r7c3 cannot be a 7, because the overflow out of nonet H would then exceed 14.
That also tells us that r7c1 must be at least a 7, because r7c3 + r8c3 can be no greater than 13. Next, the geometry of that
32 x 7 cage in the middle of the puzzle helps narrow things down quite a bit more. The Sum Calculator says:
32 x 7

1234589
1234679
1235678
Very handy indeed. Since every permutation contains {123}, that means that whatever is in r6c5 must also be in r3c7 as
there is no place else it can go. That eliminates the 1 from r6c5, and we are left with:

Figure 6: c5 nearly cracked 
Take a close look at the 11 x 3 cage hanging out of nonet H. Note that there's no way to make 8 with what's already in nonet H,
so we can safely discard the 3 in r6c5. Then the only way to make 9 is with {45}. And there's another hidden cage in r7 to help us
close this one down: 90  (18 + 9 + 24 + 4 + 7 + 11) = 17 x 3. So r7c1 must be an 8. All of this leads to a vastly improved view of
the board as below in Figure 7:

Figure 7: Broken! 
The 21 x 4 cage in nonet I leads to a lot of that, and is nailed simply by running 21 x 4 (x3,5) through the Sum Calculator
after you determine that the r7 values in nonet I must be {1279}. Standard techniques will get you the rest. My work is done here.
Copyright © Steve Metzler 2013. All rights reserved. 