I am pleased to be able to host a new explanation for the Setti strategy written by Petra from Germany. It is far better written than anything I've come up with so I am replacing my text with this excellent document. Setti is named for the user who first suggested the idea on this website the Weekly Extreme no 25, on 15th of December 2010.

The rule makes use of the observation that, regardless of compartments, every digit appears in exactly the same number of rows and columns. This is due to the Str8ts rule that says numbers must be unique to each row and column. For example, if 8 appears in four columns, then it must appear in exactly four rows as well.

## Comments

## ... by: hone

I have two remarks.

1. Example 6 (#246) is actually solvable with an easy oddness consideration, where only odd-length compartments need to be examined:

The six horizontal ones are all fixed and yield a surplus of two odd numbers.

Out of the 4 vertical ones A4..J4 and B9..D9 are fixed with a neutral result. So the remaining two (A5..C5 and G7..J7) need to be odd to match the horizontal surplus.

Those conditions are very rare. But the oddness can be regarded as a standard distributional tool like settis and wings. There are no street interactions ("look around the corner") involved, which is the case in the first example (#247).

2. The point of view that black cell clues count towards settis is a bit risky (at leasr my solver would mess up). IMHO it is not forbidden to have such a clue more than once in a line. In example 2 there could be the digit 1 also in E9 in addition to E5.

Thank you for your feedback.

I've only had time to consider point 2 so far: You're right: I didn't think of the (evil!) possibility of a puzzle creator adding the same white-on-black-clue twice in a row or column. In that case counting the clues towards settis would only work if they were counted two/three/... times in rows/cols where they appear two/three/... times. Counting them as missing would work correctly.

However: In a BCA analysis both ways wouldn't work correctly, though one could repair the method "counting white-on-black-digits as missing" by counting them two/three/... times in rows/cols where they appear two/three/... times. (Writing two/three/... times the digit in red behind the row/col.)

Hm, I shall think about what to do with that. Maybe a footnote warning user's of the danger of evil puzzle creators might do the trick :-)

First of all: All examples in my tutorial are of course chosen to demonstrate something. The solutions are not about solving the puzzle the quickest way but to explain a certain strategy.

Second: I wouldn't agree that the described oddness consideration is easier or quicker than a combined setti on the even numbers, actually for me the opposite is true. I would always start with a BCA in order to get an overview about missing digits, before even thinking of doing an oddness argument. And having done the BCA the combined setti leaps to the eye and is counted much quicker than the oddness argument - but that's of course a subjective choice.

Plus I cannot really see the advantage of using an oddness argument. As I see it: If the oddness argument works a combined setti on even or odd numbers will work as well, not vice versa.

But I will modify my note that a combined setti on even/odd numbers is called an oddness argument, because I didn't know that for the latter you just look at the odd compartments and do not count the missing odd or even digits but the differences between them.