Hi, richmize. Besides the useful ideas given by skeeter84 it is interesting to comment that there is another useful application: “A free auxiliary program” by jomapil (in this same section “Solving strategies and tips”) (it goes up to 16x16).
Probably each puzzler prefers to follow a different “strategy” or “tactic” to deal with every type of Calcudoku. In my opinion the tools you use are the correct ones (specially the analysis for every situation) but I would like to underline the interest of the “multiplication rule”: Similarly to the addition rule, the product of all numbers in every line is constant and equal to n! (factorial of n) being n the size of the puzzle; i.e., in a 6x6, this product is 6! = 720 = 6 x 5 x 4 x 3 x 2 x 1.
I assume Patrick has no objections to the comments below with respect to the Nov 21, 2013, 6x6 difficult (around 20:00, anyway more than 200 players have solved the puzzle at this time).
“54x” = [3,3,6] (unique) and “32x” = [1,1,2,4,4] (unique, since it’s not possible to have a number more than twice in this type of Lshape cage, so the 2 goes to f1 and the 1’s and 4’s go to the branches). Now if you consider the three top rows, the product of cells c3 and d3 is equal to (720 x 720 x 720) / (2160 x 32 x 10 x 54) = 10, then c3d3 = [2,5].
Now observe that from the five possible combinations for “13+” ([1,6,6], [2,5,6], [3,4,6], [3,5,5] and [4,4,5]) only the [2,5,6] is valid, as shown. The product of numbers in this cage is 60 so applying again the multiplication rule to the three rightmost columns we obtain that the product of cells d3 and d4 is equal to (720 ^ 3) / (32 x 54 x 60 x 720) = 5, then d3d4 = [1,5].
Consequently d3 = 5, d4 = 1, c3 = 2 >>> c4 = 6. Without 6’s, “48x” = [3,4,4]; the cage “10x” is immediately solved and also f6 = 6 (only place for a 6 in column f).
Once more, we can obtain the product of numbers in “12+” using, for instance, the three bottom rows: (720 ^ 3) / (48 x 6 x 60 x 720) = 30, so “12+” = [1,1,2,3,5] (being [1,1,1,5,6] invalid).
We also have b6 = 5 (1, 2, 3 and 4 already in column b and f6 = 6).
And why have I placed a 3 in a6? (this means: a4 + a5 + c6 = 4 = [1,1,2] >>> c6 = 1, etc.). Because the corners in this 6x6 puzzle are: 2, 3, 5 and 6, so the 3 is the only possibility for this corner (you can refer to various posts in this section, for instance, “Segmentation”, to clarify these ideas: The sum of the inner part, 58 in our case, is equal to the sum of the corners, which is 16, plus two complete lines, which sum is 42; and, what is more relevant, the” inner part”, that is, the 4x4 interior “box”, contains the four corners plus the numbers of “two complete lines”, that is, plus twice the set 1 thru 6, so we just count the numbers inside the inner area and find that the numbers repeated three times are the 2, 3, 5 and 6, while the 1 and 4, which cann’t be corners, are present only twice).
The multiplication rule is useful in many different occasions, for instance, sometimes in the 9x9’s (specially those very difficult on Tuesdays) many product cages are given in, i.e., the three leftmost columns with, i.e., only one “addition cage”; then, just dividing (9!) ^ 3 by all known products (including the individual cells and the products of numbers in the 2cell cages) you arrive to the product of the operands of that “addition cage” and then simply you make the factoring and find the appropriate combination for the sum.
