Basic strategies (perhaps trivial but may be useful to someone).
Second rule: The parity (a powerful tool for puzzles like the difficult 7x7 only subtractions on wednedays, etc.).

The addition of all numbers in any row or column can be even (10, 28, 36, 78 for the 4x4, 7x7, 8x8 and 12x12) or odd (15, 21, 45, 55, 153 for the 5x5, 6x6, 9x9, 10x10 and 17x17).
If we have a cage with n- the addition of all numbers inside must be even if n is even (we will call it “cage even”) and odd if n is odd (we will call it “cage odd”). For instance, in a cage of two cells with result 2-, both numbers inside must be even or both odd, and then the addition is even (7 and 5, 2 and 4, etc.). In a cage of two cells with result 3-, the two numbers inside must be of different parity, one odd and the other even, and the addition of both will be odd. Let’s now think in a three cells cage odd-, that is, a - b - c is odd; since a - b - c = a - (b+c), then “a” and “(b+c)” must be of different parity and consequently “a” + “(b+c)” = a + b + c is odd (for instance, 9-2-4 produces 3-, odd, 9+2+4=15, odd). Same demonstration applies to a three cells cage even-. Larger cages behave similarly since a - b – c – d… = a – (b+c+d…), etc.
Next step: The additon of all numbers inside two “cages odd” is even (and obviously if the two cages are even). Iterating this procedure and by comparing the result with the parity of the line or the “rectangle”, etc., we can determine the parity of individual cells (also knowing the parity of some subareas, like those with the x sign, we can eliminate combinations, etc.).
For instance, in the Jun 08 7x7 easy, the parity of the rectangle of the top 4 rows (excluding the cell d5) (observe that the 30x is an odd cage) is odd, so to complete the even parity (4x28) of the four rows, the cell d5 must be odd and “1” is the only possible value; e5 will then be 2. And now, checking the parity of the three bottom rows, the 12x in f1-g1 must be 4x3 (cage odd) and not 6x2 (cage even).

Thanks for this tip. I have put it use and has been a definite help.

Welcome. I arrived to that conclusion when looking for some "error detection" applicable to the lines (rows or columns) (as in the computer's world) but I quickly observed that it was an all-purpose tool that could eliminate combinations in cages of the type "x" or the ":". For me many times is the key in the 9x9's. The parity rule can be combined with the "maximums and minimums in the sum of cages"; for instance, if you have s1 + s2 + s3 = 50 (where s1, s2 and s3 can be of any type of cage including "x" or ":") and you are analyzing a combination of s3 = 19, then s1 + s2 is odd and equals 31, and s1 and s2 have of course opposite parity between them, and also the only possibilities are those "pairs" of combinations for s1 and s2 that give that total of 31, and we can proceed analyzing the effect over the rest of the puzzle.

Hi, clm.

In your explanation you refer to the puzzle (Jun 08 7x7 easy).

How can I obtain this historic puzzle to better understand your explanation?

Thank you. Cheers.

Here it is:

Thank you, pnm.

Cheers.

Excuse me, pnm, but I think you made a mistake. If I'm right that puzzle refers to 7x7medium20110608.

Cheers.

Hi, jomapil. The graphic is the correct one, the one I was referring to. I cann't remember now if it was the 7x7 easy or medium for that date, but sure Patrick found the correct one. It is not possible for the moment to recall an old puzzle using the Puzzle id (identification introduced "recently" by Patrick), only the last week's puzzles can be directly seen. When I sent the post we had not defined yet the labelling for the cells (actually agreed as in Excel, after the discussion in the Forum, that is, letters for the columns and numbers for the rows, although this has not been implemented yet in the page, ... perhaps in the future upon Patrick's time availability), so I used letters for the rows and numbers for the columns in that post. And at that time I was not yet uploading graphics (sorry for that) because for me the procedure was not clear, and only after a few e-mails with Patrick, I was able to do it.

This was a very easy example for "the parity" rule. The parity of that cell must be odd, and since a cage "2:" has only (in a 7x7) the combinations 1-2, 2-4 and 3-6, that cell must contain a "1" (there is a 3 already in the column) and the roommate is a "2". Now that you know the even parity of this second cell you may determine the parity of the cage "12x" = odd (so 3-4 and not 2-6), considering the parity of the three bottom rows (even, 3 x 28 = 84; here we must be careful because, i.e., three rows will have odd parity in the case of a 5x5, 6x6, 9x9, 10x10, ..., respectively 3 x 15, 3 x 21, 3 x 45, 3 x 55, ...).

We can apply "the parity rule" to practically all puzzles, in many situations, eliminating possibilities and focusing the candidates for the cells and/or cages. The utility will be more evident with the use.

Thank you, clm, for your explanation and for the parity rule. At first sight it seems useful lowering the number of possibilities. I go to explore this concept in the next days and in the next puzzles.

Cheers and till the next.