Calm down, everyone. Brackets form a tree structure, and can be represented by a free magma, while strings with concatenation are equivalent to a free monoid. You’re essentially asking for the two respective common involutory operations to be connected by this map, just because they’re involutory, which put that way is a wild guess at best. In fact, reversing this string produces something outside the range of the map entirely, which is injective and so can’t be surjective for combinatorical reasons.

… Yeah I might be the only person that finds that useful.

the part about the “wild” guess is, but this is a counterexample, and something like the reciprocal vs the negative of reals or rationals when moved across the log map would be an example. So, either you’re a galaxybrain that just instantly knows if the transformation is structure-preserving in that way, or you’re guessing to some degree as well.

The symbols and abstractions have touched me in no-no ways. I miss okaybuddyphd on r*ddit, they knew the pain.

I suppose I could also just say that characters which aren’t just drawn asymmetrical, but actually point in a direction as part of their function, look wrong when reversed like this. So, (e) -> )e( is no good, but bed -> deb is fine.

I was saying unipotent at first instead of involutory, which was actually the wrong jargon because of the context, but I’ve fixed that now. Yes, they’re all real.

Map, although in this context I could probably have just said function. I go with map by default when thinking bidirectionally.

I think most people here will know combinatorics, the study of the different possible configurations of something. The number of n-length strings with two possible characters is 2^{n}, as coders should all know, and the number of trees turns out to be Catalan numbers, many of which have prime factors other than 2. This is an injective map from n node trees to 2n character strings, so it’s possible, but you’ll (almost?) never get a perfect match, so by the pigeonhole principle it can’t be surjective.

I’m wondering now if Catalan numbers are O(n!). The equation has a lot of n! but it also has a certain smell like it might depend on big or little o.

Edit: D’oh, they must grow no faster than 2^{2n}; I just wrote that. So, exponential.

Calm down, everyone. Brackets form a tree structure, and can be represented by a free magma, while strings with concatenation are equivalent to a free monoid. You’re essentially asking for the two respective common involutory operations to be connected by this map, just because they’re involutory, which put that way is a wild guess at best. In fact, reversing this string produces something outside the range of the map entirely, which is injective and so can’t be surjective for combinatorical reasons.

… Yeah I might be the only person that finds that useful.

yeah but that’s just like your opinion man

## I mean,

the part about the “wild” guess is, but this is a counterexample, and something like the reciprocal vs the negative of reals or rationals when moved across the log map would be an example. So, either you’re a galaxybrain that just instantly knows if the transformation is structure-preserving in that way, or you’re guessing

to some degreeas well.The symbols and abstractions have touched me in no-no ways. I miss okaybuddyphd on r*ddit, they knew the pain.

I suppose I could also just say that characters which aren’t

justdrawn asymmetrical, but actually point in a direction as part of their function, look wrong when reversed like this. So, (e) -> )e( is no good, but bed -> deb is fine.I’m just going to assume those 4 dollar words are real and you aren’t just misspelling normal words to fuck with us.

Non surjective free magma? What about the doblastic amortized basalt?

I was saying unipotent at first instead of involutory, which was actually the wrong jargon because of the context, but I’ve fixed that now. Yes, they’re all real.

A glossary:

Involution

Surjective

Injective

Free magma

Free monoid

Map, although in this context I could probably have just said function. I go with map by default when thinking bidirectionally.

I

thinkmost people here will know combinatorics, the study of the different possible configurations of something. The number of n-length strings with two possible characters is 2^{n}, as coders should all know, and the number of trees turns out to be Catalan numbers, many of which have prime factors other than 2. This is an injective map from n node trees to2ncharacter strings, so it’s possible, but you’ll (almost?) never get a perfect match, so by the pigeonhole principle it can’t be surjective.I’m wondering now if Catalan numbers are O(n!). The equation has a lot of n! but it also has a certain smell like it might depend on big or little o.

Edit: D’oh, they must grow no faster than 2

^{2n}; I just wrote that. So, exponential.You are awesome.

Lemmy is better for you being here. Thanks for the reading material!

You’re very welcome! How kind.