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General killer Sudoku solving strategies
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Posted on: Fri May 15, 2015 3:49 am

Posts: 6
Joined: Tue Feb 10, 2015 4:06 am
General killer Sudoku solving strategies
As someone who's not new to solving normal Sudokus, what's some strategies specific to Killer Sudoku?

For instance, a useful thing to consider is square sum = row/column sum = 45. So if you have a square or row that only "leaks" one square or two, you can get (for one cell full) information out of that.

That apparently isn't the only thing to consider, though, given how long I'm staring at even the medium Killer Sudokus here. What do I miss?

Posted on: Fri May 15, 2015 1:57 pm

Posts: 226
Joined: Fri Jun 17, 2011 8:15 pm
Re: General killer Sudoku solving strategies
clm has written several great posts with solving strategies in the general strategies forum. A lot of the general strategies - parity comes readily to mind - translate directly to killer sudokus. The search option is across the top of this page just under the banner.

I know that's really general feedback, but I hope it helps.

Posted on: Sun May 17, 2015 1:41 pm

Posts: 2258
Joined: Thu May 12, 2011 11:58 pm
Re: General killer Sudoku solving strategies
fnord wrote:
As someone who's not new to solving normal Sudokus, what's some strategies specific to Killer Sudoku?

Here are some good tips:

http://killersudokuonline.com/tips.html

Posted on: Sun May 17, 2015 9:17 pm

Posts: 700
Joined: Fri May 13, 2011 6:51 pm
Re: General killer Sudoku solving strategies
fnord wrote:
As someone who's not new to solving normal Sudokus, what's some strategies specific to Killer Sudoku?

For instance, a useful thing to consider is square sum = row/column sum = 45. So if you have a square or row that only "leaks" one square or two, you can get (for one cell full) information out of that.

That apparently isn't the only thing to consider, though, given how long I'm staring at even the medium Killer Sudokus here. What do I miss?

Hi, fnord. Of course some time is required to solve the difficult KS's, so one must be patient.

As commented by jaek some general rules valid for Calcudokus can be applied to KS's since in some way a KS is a particular case of a Calcudoku where the operators are only sums and, as we know, with the additional condition/restriction that every nonet, that is, every 3x3 box, must contain the numbers {1, 2, ... , 9} only once. For instance, in the saturday's (May 16) difficult KS, it's obvious that e7 is an odd number (parity rule applied to the sum, 90, of nonets 8 and 9) but, more precisely, in this case, e7 = 3, since this is the difference between 90 and the sum of the other five cages.

The general tips provided by the page killersudokuonline.com (referenced by Patrick) are very useful and can be resumed saying that to solve the "difficult" situations we must stablish some elementary (linear) equations containing specific "innies" and "outies". Continuing with the previous example, d4 is an "outie" of nonets 4 and 7 so d4 = 5 (13 + 14 + 15 + 18 + 14 + 14 + 7 - 90), then, since d4 + f4 = 14 (45 - 7 - 14 - 10) it follows that f4 = 9 and consequently e4-e5 = [6,8] (unique) and e8-e9 = [5,9] (unique). Also d5-d6 = [3,4] (unique).

Let suppose that d4 = 5 is not so evident, anyway it would be possible to affirm that f4 = 9 following a different way. The cell h4 is an "outie" of the three top rows while f3 is an "innie" of the same rows. So we stablish this equation: 15 - h4 + f3 + 16 + 16 + 11 + 23 +16 + 10 + 14 +14 = 135 (the sum of the three upper rows or, alternately, of the three first nonets) ---> f3 - h4 + 135 = 135 ---> h4 = f3. At the same time the "innies" g4 and h4 of nonet 6 have a sum of 7 (45 - 11 - 27), that is, g4 + h4 = 7 and since h4 = f3 ---> g4 + f3 = 7 so, regardless of the particular values of g4 and f3, f4 = 16 - (g4 + f3) = 16 - 7 = 9. And consequently d4 = 5. But certainly, in this particular KS, this second way is longer and unnecessary.

And, how to continue?. For instance: c4 + c5 = 9. The pair [4,5] is not valid because it repeats a 5 in the cage "14", the pair [3,6] forces [5,9] in c6-c7 and then c8-c9, with a sum of 7, could not be accomplished. It's a little more difficult to see that the pair [1,8] is not valid thus driving to the conclusion that c4-c5 = [2,7]. If c4-c5 = [1,8] ---> c6-c7 = [5,9], c8-c9 = [3,4]. We are left with 2, 6, 7 for cells c1-c2-c3. Again we set a simple equation (considering that c1 and b3 are, respectively, an "outie" and an "innie" of the two leftmost columns): 16 - c1 + b3 + 16 + 13 + 15 + 18 + 14 = 90 ---> c1 = b3 + 2. Now we see that c1 cann't be a 2 (---> b3 = 0) or a 6 (---> b3 = 4 but c3-d3 cann't have a sum of 7, that is, cann't be [1,6], [2,5] or [3,4]) or a 7 (---> b3 = 5 and c3-d3 cann't have a sum of 6, that is, the pairs [1,5] and [2,4] are not allowed).

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