fb_1811 wrote:

... FYI, IMHO guessing is throwing in a number in and seeing what happened. Stating a systematic hypothesis and then seeking to disprove it still falls in the analytical camp. So it counts for me!

Welcome, and thank you because today I have learned this new expression, IMHO (first time for me, I had to go to Google) (I knew FYI and ASAP, etc.), in our language we use very very few phrases “compressed” and I like much those acronyms.

beaker wrote:

… I didn't think of using the total of 21 for each column (something sneaklyfox told me about months ago when I saw her video)....I am sure with this type of logic I will have fewer problems in the future.....as for the guessing discussion, I think you have taken the guessing out of the "equation" by using some established mechanisms (ie: the placement of the 5's and the total of 21 for each row)......so while thinking of the possibilities in your mind, you eliminate the possibilities that don't fit and "go" with those that, logically, do fit (the "catch 22" is the word possibilities as that word in itself suggests guessing......a real quandry!!!!!)

The new feature, “guessing” numbers in a cell (or writing the candidates), just introduced by Patrick, will help a lot to solve the puzzles on-line. It will make easy the use, in some cases, as additional tools for the calcudokus, of techniques that have its origin in the “sudokus”, like the X-ray (X-Wing), or the XY-Wing, etc..

With respect to the property of the addition of numbers in rows or columns you may find a little bit more info in this same section “Solving strategies and tips”, in the old topic “The addition of all numbers of a row or column” (the property of the product is also commented there). The “addition” is a very important property that I use continuously. The sum of numbers of a line is:

S = n * (n + 1) / 2, being n the size of the puzzle (n could be any natural number).

Then (starting from 4 which is the smaller one in the site)

4 x 4 the sum is 10

5 x 5 ----- 15

6 x 6 ----- 21

7 x 7 ----- 28

8 x 8 ----- 36

9 x 9 ----- 45

10 x 10 --- 55

11 x 11 --- 66

12 x 12 --- 78

13 x 13 --- 91

14 x 14 -- 105

15 x 15 -- 120

16 x 16 -- 136

17 x 17 -- 153

…

For puzzles with 0, we just reduce a degree the size of the puzzle as commented in the referenced topic.

Additionally, the full puzzle has a sum of n * n * (n + 1) / 2, that is:

4 x 4 ------ 40 = 4 x 10

5 x 5 ------ 75 = 5 x 15

6 x 6 ----- 126 = 6 x 21

7 x 7 ----- 196 = 7 x 28

8 x 8 ----- 288 = 8 x 36

9 x 9 ----- 405 = 9 x 45

10 x 10 --- 550 = 10 x 55

11 x 11 --- 726 = 11 x 66

12 x 12 --- 936 = 12 x 78

13 x 13 -- 1183 = 13 x 91

…

or, simply, multiplying the value of a line by n (the size of the puzzle). Very curiously, for n = 1, 5, 9, 13, 17, 21, … (when the size = multiple of 4 plus 1) the full puzzle has a sum odd, in all the other cases the puzzle sum will be even (this has an easy algebraic demo) (the sum of the full puzzle can be useful sometimes, specially in the 5x5’s or 6x6’s).

I am going to use an example to see the utility of this rule (which is also useful for the big puzzles), using the yesterday’s 9x9 (Puzzle id: 433452, solver rating 92.3):

First we write the candidates in those cages well defined including the cage “4:” (with the only combination [1,2,8]) where the 8 must go to h4.

The cell a9 is odd (parity property for column 1, with a sum of 45, which is odd) consequently it cann’t be an 8 so the 8 of column a, the leftmost column, goes to a2 (unique).

Now, let’s call “a” the value of the cell b8 as shown in the graphic, and “b” the value of the cell f9 (for simplicity); we have:

a + 45 (a plus the full row 9) = 17 + 1 + 14 + b + 9 + 4 = 45 + b (the same cells);

from this equation it follows that b = a, that is, f9 = b8. Now a 2 can not be placed in a9 or b9 (2’s already present in both columns) nor inside “14+” nor in f9 (because the value of b8 would also be 2 and that’s impossible) so, since there is already a 2 in column h, g9 = 2 (and thus h9 = 7).

Now we go back to column a: the 7 must go to a1 (a9 now forbidden) (we can define now the numbers in cage “14x” though we can do it later).

For the next step we look at the “isolated” cell g4: The sum of rows 1 thru 4 is 180 (= 4 x 45), even, so applying the “parity property” (very easy in this case since there are no uncertainties in the “addition value” of the cages) we find that g4 must be even, and its value must be 4, the only even number left (2, 6, and 8 already in row 4).

Next: We calculate the sum of cells of the cage “1-“ (knowing that four rows have a sum of 180) and we find a value of 11 so e2 = 6 and e3 = 5.

The cage “21+” would initially have these possibilities: [3,9,9], [4,8,9], [5,7,9], [5,8,8], [6,6,9] and [6,7,8] but, since the 8’s are not allowed and, in this case, the double 6 or double 9 is not permitted, the only valid combination left is the [5,7,9] so (see the orange arrows) the 5 goes to f4 and the 5 (required) inside the cage “180x” must go to g5 (and necessarily f5-f6 = [1,9]).

Again we use the property of the addition applied to the three rightmost columns (a total of 135 = 3 x 45) so the green area must sum 11 and consequently f2 = 2.

We complete the candidates for f8-f9 = [3,6]. From this point we may set, for instance, the final values in cages “14x”, “5x” and “4:” (green and gray), quickly observing as well that g2-g3 = [3,8].

Even more: In the cage “144x”, the product of d8 x e8 must be = 144 / 18 = 8 so d8-e8 = [1,8], [2,4]. Using again the parity property for the five bottom rows (a total of 225 = 5 x 45, odd) we derive that the blue area (d8-e8) must have an even sum so it must be [2,4] >>> e8 = 2, d8 = 4 and, since we have all sums of the cages in those five rows, we deduce that “3-“(+) = 9 (yellow area), then “3-“ = [3,6] and consequently a9 = 9.

... ... ...

We have seen how we have obtained a lot of information using only analytical tools, like the parity rule or the property of the addition.

The official solution: