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1. When I solve Kakuro, I start by placing a set of up to 9 dots in each square, representing the numbers 1-9 as possibilities for that square. The dots are arranged in a grid, like so: * * * * * * * * * Representing the numbers as follows 1 2 3 4 5 6 7 8 9 When I eliminate a number as a possibility, I erase the dot (unlike crosswords, I use a pencil for Kakuro). If I solve the square, I erase the dot from other cells within the word. After some experience with the following rules, you can start most squares with fewer than 9 dots. 2. For any pair for which you've eliminated a dot in one square, you can eliminate the corresponding dot in the other square. For example, if I have a pair which sums to 8, and one box has the following dots: . . * . . * . . . The the other box can only have the following dots: . * . . * . . . . 3. For pairs which sum to an even number, you can eliminate the digit which corresponds to the number/2. For all pairs, you can eliminate dots which are equal to, or higher than the sum. You can also eliminate dots which correspond to X if sum-X is > 9. 4. For each clue length, there are certain sums that have a restricted number of possibilities (these sums are always on the outer end of ranges of minimum and maximum sums). For example, a 3-digit word which sums to 6 can only contain the numbers 1, 2, and 3. And a 2-digit word which sums to 16 must contain the digits 9 and 7. The complete table of such restricted clues is as follows. Also note that that there are also clue-length/sum combinations that produce a restricted range of numbers.
If the length of the word is 8, the sum is going to be 45 - X in which X
is the missing digit. You can eliminate this missing digit from all the
cells.
If the length of the word is 9, the sum is always 1+2+3+4+5+6+7+8+9 = 45.
More generally, for any word, the sum of the digits which don't appear in
the word is going to be 45 minus the sum of the word. Any digits excluded
from the missing digits must be in the word. Any digits which must be in
the missing digits cannot be in the word.
For example: If the sum of a 7 digit word is 41, the two digits which
don't appear in the word = 45 - 41 = 4. Since the two missing digits
sum to 4, they must be 1 and 3. Therefore, the word contains the digits
2,4,5,6,7,8, and 9.
If the sum of a 7 digit word is 37, the missing two digits must sum to 8,
and we know that 4,8 and 9 cannot be one of the missing two digits,
therefore they must be present somewhere in the word.
5. When there are two pairs (A+B) crossed by a pair (C), as in the
following:
A B
C _ _
D _ _ _ _
Then you can treat the first two digits of D as a pair using the sum
A + B - C
The remaining digits of D can be treated as a smaller clue - minus
the sum of the pair.
6. You can sometimes use a similar technique to determine the value of
corner squares (by subtracting horizontal sums from vertical sums, or
vice versa). For example, in a puzzle fragment with the following pattern:
A B C
D _ _ _
E _ _ _
F _ _ _
You can compute the value of the first square of F using (A+B+C) - (D+E).
7. The "Naked Subset" rule used to solve Sudoku applies to Kakuro for all
words. For example, if you have a 6 digit clue, and two of the words have
the same 2-dot pattern of possibilities, then those two words must
contain those two numbers, and those numbers can be eliminated from all
other cells in the word. You also know the sum of those two cells,
and can infer the sum of the remaining cells, which can be treated like
a smaller word. The same rule applies to 3 cells which have the same
3 dot pattern etc.
8. You can use the "Hidden Subset" rule from Sudoku,but only if you are
dealing with a sum that has a known set or subset of digits, either due
to tip #4, or to elimination from other rules. So, if you know that a
7 letter clue must must contain 4 8 and 9 (because it sums to 41) and
only 3 cells have 4 8 or 9 as possibilities, then only those 3 cells
can contain 4 8 and 9. You can eliminate all other digits as
possibilities for those cells (and you know the sum of the those 3 cells
(21) and the sum of the remaining cells (41-21 = 20).
Other fun stuff...
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