beaker wrote:
Columns "e" and "j" had many small #'s in some of the cages.....when I get too many small # cages in the same column, I then work them out with pencil and paper.......and then the rest is not so bad .......and if I can do the puzzle, then it cannot be too bad.......I actually enjoy the challenge that this one plus the +3 -3 puzzle of the day before presents me
Here are a couple of images to show what could be done. I have put in more pencil marks than is really needed, but it helps to demonstrate what could be done with column I.
The first image just shows the first steps to get started: 14x cage = 2&7, then 5+ can only be 1,4, then 28x cage and then 32x cage.
This leaves only 3 unused values in row 10 (2,3,8), and without needing any pencil marks we can see that 8 can only be in i8.
Now if we consider column I we can see that 4 is only possible in the cell i7, I have put in all the pencil marks in the image, but even without the pencil marks it is clear that it can not be in any other cell in that column, the first cage is 7- so that will not work with a 4, the next 2 cages in the column are 5/ and 3/ and 4 is not going to fit in either of them, and we already have 4 in the lower 3 rows, so that leave i7 as the only possible 4.
Now with 4 in i7, i8 must be 9, i9 must be 7 and that leaves 3 in i1 & 10in i2 as the only possible values in the 7- cage. It is not needed but just to tidy up the remaining pencil marks: we are left with 1,2,5 & 6, and that gives 5/ cage = 1&5 and the 3/ cage = 2&6.
This demonstrates the importance of looking for the basic things like a value being in only 1 cell in a row or column, the 8 in row 10 and the 4 in column i make it all come together easily, but it is easy to miss in a 10x10 puzzle.
For column e, I don't think I used any pencil marks, after a few more solved cells I started from the top and everything just fell into place with only 1 option for each cage.