The New Sudoku Players' Forum

[Condor]

If you have checked the rows, you know that the every digit is present exactly 9 times. So, when you check the columns afterwards, you only
have to check the first 8 columns, as the last column must obviously contain the remaining digits. Same for the blocks.

That method takes 25 checks. That is a saving of only 2 checks.

It can be done in 18 checks as I showed here

[algernon]

It can be done in 18 checks as I showed here

I admit that your method uses fewer "checks", but I think each of your
checks is a bit more expensive, as you not only verify that a certai digit
exists in a box, but has to check in what line therein it lies.

Not that I think this is illegitimate (I often prefer to count in a similar way myself), it might just be a matter of taste.

[udosuk]

You can also do this in 9 expensive "checks"...

Look at block 1, try to locate the digit "1". Then look towards the right for the "1"s in blocks 2 and 3, and make sure they don't appear on the same row as the "1" in block 1 or each other...

Repeat this for blocks 4-6 and 7-9, but add this operation: after you've locate the "1"s, make sure there isn't another "1" appearing above it on the same column.

Now that's one "check". Repeat this for the other 8 digits...