<a-z>i within original selection should produce result

[vbauerster]

I wish <a-z>i would allow me to intersect two selection lists with not equal number of elements (selections), if other selections are within original one. It's hard to explain with words so here is illustration:

So selection is saved first with Z, then we have two more selections with help of <a-S> (both are within original), next we want to intersect in order to get result as depictured above on third line.

At the moment <a-z>i complains like the two selection lists don't have the same number of elements (1 vs 2).

Also imo following (same number of selections, but the second one is within the first one) should also work like this:

[mreppen]

This is the same feature I requested here: #3504
I am currently running with my own patch and love this feature. My patch is only a proof of concept. Note that the order, i.e., which selection is removed from which, is reversed.

PS.
I don't think this should be called intersection. Seeing selections as sets, it is a difference/subtraction.

[vbauerster]

You're right this is not intersection.

If we assume that selected (by anchor or cursor) cell is 1 and not selected is 0, then current intersection behaviour is AND table, survives 1 and 1:

Difference/subtraction is like negated AND table, we negate the result of AND operation:

The tricky part is that current implementation doesn't allow to combine single selection with multiple or vise versa. But I reckon for difference/subtraction operation it should.

PS: all above illustrations represent single line combined with Z and <a-z>, not three liens.

[mreppen]

@vbauerster I might be misunderstanding you, but no, it is not "negated AND". What you describe in your second example is the symmetric difference https://en.wikipedia.org/wiki/Symmetric_difference
That's actually why I called it subtraction: to avoid confusion with symmetric difference.

From your example, I'm quite convinced it's subtraction. Please have a look at the linked issue and I think you will agree.

In that issue, we also discuss paired vs non-paired operations, which likely relates to what you call single/multiple. Please check it out and join the discussion.

Also, if you want to try it out, this is my non-paired ("multiple") patched version https://github.com/mreppen/kakoune/tree/subtract
But as I said, keep in mind that I coded the order opposite of what you described (A\B vs B\A).