Cooperative Zendo

[Unbitwise]

Ah. I'm fine with that, but I had understood it otherwise.

[Lambda]

Neither begins with d or h, nor ends with j or p.

[DanielH]

That is a superset of my already-a-superset rule.

I did in fact guess aristodemocracy because if its length.

Let p be the smallest positive prime factor of the length of the word l; if p ≥ l the word fails*. Then a word is accepted iff only its first l/p letters have ascenders and only its last l/p have descenders. This does not mean ascenders or descenders must be present, only that if they are they are in the appropriate positions.

* The only time p > l is the empty string, because 2 is a factor of zero. Since the empty string is not well-formed, it should not matter if your rule gives a different answer in that case.

[Lambda]

Edit: withdrawn, ignore me

xray
xylophone
xylophones
xerography
xylophonists
xenophobia
xenophobic

[DanielH]

All of those except xray are forbidden by my known-superset rule. Did you miss it?

[Lambda]

Uh, yes. I withdraw those.

stucco
embryo

[Lambda]

interhyperlinguomonograms
unintercameroexcuseograms

(please excuse the highly dubious wordiness of these koans)
("monograms combining multiple hyperlanguages", "written documents bearing excuses, not addressed from one house of Congress to the other")

[Unbitwise]

Let p be the smallest positive prime factor of the length of the word l; if p ≥ l the word fails*. Then a word is accepted iff only its first l/p letters have ascenders and only its last l/p have descenders. This does not mean ascenders or descenders must be present, only that if they are they are in the appropriate positions.

*squints* I think you have it.

The actual rule is:

The string may be written in a rectangle of characters (left to right and top to bottom) such that only the top row has any ascenders and only the bottom row has any descenders.

I note that I failed to write down the requirement that it be a non-degenerate rectangle (having at least two rows and at least two columns) which I was judging koans by. Without that, it trivially accepts all words as 1-high rectangles.

[DanielH]

Yeah, that seems equivalent; I specify the widest possible rectangle if you look at it that way, which always works if any rectangle does.

[Unbitwise]

Well, you've won it.

The form of your guess suggests to me that my rule was too obscure (even if the way I designed it was less mathematical than the guess). Your opinion? (Lambda?)