Two-Clue Deductions
qqwrefPosted: Mon Dec 23 2013 05:52 pm

I'm sure you guys have all seen situations where just having two clues intersecting, without any numbers solved, lets you fill in a number. I'll be using the notation X/Y for a clue of Y numbers that must sum to X. Perhaps the easiest examples of this type of logic are 3/2 and 4/2 (the intersection must be a 1), or 16/2 and 17/2 (the intersection must be a 9). I wrote a little program to find all possible deductions of this type. There are 138 possibilities, which may seem like a lot, but if you want to be really serious about fast kakuro solving they're probably worth trying to memorize.

Here goes...

3/2 + 4/2 => 1

3/2 + 11/2 => 2

3/2 + 19/3 => 2

3/2 + 26/4 => 2

3/2 + 32/5 => 2

3/2 + 37/6 => 2

3/2 + 41/7 => 2

3/2 + 43/8 => 1

3/2 + 44/8 => 2

4/2 + 6/2 => 1

4/2 + 11/2 => 3

4/2 + 12/2 => 3

4/2 + 7/3 => 1

4/2 + 19/3 => 3

4/2 + 20/3 => 3

4/2 + 26/4 => 3

4/2 + 27/4 => 3

4/2 + 32/5 => 3

4/2 + 33/5 => 3

4/2 + 37/6 => 3

4/2 + 38/6 => 3

4/2 + 42/7 => 3

4/2 + 42/8 => 1

4/2 + 44/8 => 3

5/2 + 13/2 => 4

5/2 + 21/3 => 4

5/2 + 28/4 => 4

5/2 + 34/5 => 4

5/2 + 38/6 => 3

5/2 + 39/6 => 4

6/2 + 14/2 => 5

6/2 + 22/3 => 5

6/2 + 29/4 => 5

6/2 + 34/5 => 4

6/2 + 35/5 => 5

6/2 + 38/6 => 5

7/2 + 15/2 => 6

7/2 + 23/3 => 6

7/2 + 29/4 => 5

7/2 + 30/4 => 6

8/2 + 16/2 => 7

8/2 + 23/3 => 6

8/2 + 24/3 => 7

9/2 + 16/2 => 7

9/2 + 17/2 => 8

12/2 + 6/3 => 3

12/2 + 7/3 => 4

13/2 + 7/3 => 4

13/2 + 10/4 => 4

13/2 + 11/4 => 5

14/2 + 16/2 => 9

14/2 + 8/3 => 5

14/2 + 11/4 => 5

14/2 + 15/5 => 5

14/2 + 16/5 => 6

14/2 + 22/6 => 5

15/2 + 9/3 => 6

15/2 + 12/4 => 6

15/2 + 16/5 => 6

15/2 + 21/6 => 6

15/2 + 22/6 => 7

16/2 + 17/2 => 9

16/2 + 10/3 => 7

16/2 + 11/3 => 7

16/2 + 23/3 => 9

16/2 + 13/4 => 7

16/2 + 14/4 => 7

16/2 + 17/5 => 7

16/2 + 18/5 => 7

16/2 + 22/6 => 7

16/2 + 23/6 => 7

16/2 + 28/7 => 7

16/2 + 36/8 => 7

16/2 + 38/8 => 9

17/2 + 11/3 => 8

17/2 + 14/4 => 8

17/2 + 18/5 => 8

17/2 + 23/6 => 8

17/2 + 29/7 => 8

17/2 + 36/8 => 8

17/2 + 37/8 => 9

6/3 + 20/3 => 3

6/3 + 27/4 => 3

6/3 + 33/5 => 3

6/3 + 38/6 => 3

6/3 + 41/7 => 2

6/3 + 42/7 => 3

7/3 + 20/3 => 4

7/3 + 21/3 => 4

7/3 + 27/4 => 4

7/3 + 28/4 => 4

7/3 + 33/5 => 4

7/3 + 34/5 => 4

7/3 + 39/6 => 4

7/3 + 42/7 => 4

8/3 + 22/3 => 5

8/3 + 29/4 => 5

8/3 + 34/5 => 4

8/3 + 35/5 => 5

9/3 + 23/3 => 6

9/3 + 29/4 => 5

9/3 + 30/4 => 6

10/3 + 23/3 => 6

10/3 + 24/3 => 7

21/3 + 10/4 => 4

21/3 + 11/4 => 5

22/3 + 11/4 => 5

22/3 + 15/5 => 5

22/3 + 16/5 => 6

23/3 + 12/4 => 6

23/3 + 13/4 => 6

23/3 + 16/5 => 6

23/3 + 17/5 => 6

23/3 + 21/6 => 6

23/3 + 28/7 => 6

24/3 + 13/4 => 7

24/3 + 17/5 => 7

24/3 + 22/6 => 7

24/3 + 28/7 => 7

24/3 + 29/7 => 8

10/4 + 28/4 => 4

10/4 + 34/5 => 4

10/4 + 38/6 => 3

10/4 + 39/6 => 4

11/4 + 28/4 => 5

11/4 + 29/4 => 5

11/4 + 35/5 => 5

11/4 + 39/6 => 5

12/4 + 29/4 => 5

12/4 + 30/4 => 6

29/4 + 15/5 => 5

29/4 + 21/6 => 5

30/4 + 16/5 => 6

30/4 + 21/6 => 6

30/4 + 22/6 => 7

15/5 + 34/5 => 4

15/5 + 35/5 => 5

16/5 + 35/5 => 6


Last Edited: Mon Dec 23 2013 05:53 pm
arbor8Posted: Tue Dec 24 2013 07:38 am

Are there any combinations where both parts would have at least 3 alternatives each?

( and yet would give only 1 possible intersection)


DarkladyPosted: Tue Dec 24 2013 08:57 am

I think an easier way to remember most of these is to learn the maximum and minimum totals for any clue length: for instance, for a clue that's 5 long, the minimum is 15 and the maximum is 35. If a clue is at the min or max, or 1 away from either value, then there's only one possibility for the set of digits that make up the answer.

Furthermore, if it's close to the min, then the largest possible digit is (length) + (difference from min) - e.g., a 4-length 12 can have no digit larger than (4) + (12-10) = 6. Close to the max, the smallest possible digit can be figured out by reversing this. This will let you easily make most of the deductions on this list even if you haven't memorized every possibility yet. :)

If you also remember that a 2-length even clue cannot have a digit that's half the total, and that a clue 1 away from the min or max will have a known "hole" in the sequence of digits (e.g. 5-length 34: 9,8,7,6,4 - no 5), that may cover every possibility on this list.

Also, one more handy thing: for "close to min", if the difference from min is smaller than the length, then the difference between those is the number of consecutive digits (starting at 1) which must be in the answer: e.g., for that 4-length 12, we get (4) - (12-10) = 2, so the answer must have a 1 and a 2 in it. As before, simply reverse for "close to max" clues.