Solving Kakuro with arithmetics (cont’d)
2008-02-29As discussed in previous post, kakuro can be tackled using simple addition and subtraction during initial stage, particularly if the kakuro puzzle contains some certain arrangement pattern. As an example, see the figure below:
With arithmetics, one can deduce A=1 and B=3. This method, though less likely mentioned in general kakuro strategy guide, has been presented in Wikipedia page. Generally, this method is only usable when an area is enclosed, and there is only one single outlet among all those squares. In the above puzzle the outlet is square A. Thus such method is not useful in the next puzzle below:
Does it mean addition and subtraction will be completely useless then? Not entirely, depends on situation. Under rare circumstances, addition and subtraction might be useful even during mid-games or end-games. The diagram below shows the stage when large part of above puzzle is solved:
Here comes the tricky bit: concentrate on upper right area. We are not getting the value of a single square, not even sum of two squares, but three.
Take summation of all red squares horizontally, then deduct all deeper red squares vertically, one gets A+B+C=21. So what? 21-in-3 has too many combinations, and it’s still hard to try one by one. The key here is, value of A has great limitation. See the final figure below, and observe the square marked “x”:
Within the column with sum=32, there is a ‘2′, thus this column has only one single combination: (2,6,7,8,9), which means x=6 or 8 or 9. Is it true? No. If x=6, then A=6, and violates kakuro rule.
That means x=8 or 9, implying A=3 or 4. But recall that A+B+C=21. If A=3, then B=C=9, game over. So we get the answer: (A,B,C) = (4,9,8).
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