Some cutting edge research - in July of this year, a paper was published (Louigi Addario-Berry, Serte Donderwinkel, Mickaël Maazoun, James Martin) demonstrating a new (and rather simple!) proof of Cayley's formula, by constructing a bijection between rooted labelled trees and sequences in [n] of length n-1. It's a very smart bijection, and the structure of the bijection allowed them to tackle several probabilistic problems, such as tail bounds on the height of a rooted tree.
There is a nice variant of this argument noticed by Peter Cameron (and probably others). How many functions f:[n]-->[n] are there such that f^n = fff...fff has a one-point range? Answer: This happens iff the two special vertices are the same, so the answer is n^(n-1).
I remember two of my friends figuring out a nice bijective proof of the theorem before the lecture ended. I did found and fixed a minor mistake, but that was still within an hour. Looking back, it was probably the same proof. Crazy how this proof was only found it 1981, I actually think people found it earlier and assumed the proof was known already.
The enumerator for labeled trees can be expressed as a determinant of the reduced laplacian of a complete graph. Saw this in algebraic combinatorics course material from Postnikov
Some cutting edge research - in July of this year, a paper was published (Louigi Addario-Berry, Serte Donderwinkel, Mickaël Maazoun, James Martin) demonstrating a new (and rather simple!) proof of Cayley's formula, by constructing a bijection between rooted labelled trees and sequences in [n] of length n-1. It's a very smart bijection, and the structure of the bijection allowed them to tackle several probabilistic problems, such as tail bounds on the height of a rooted tree.
I have seen this channel grow from 124 subs and now 31k! Keep going guys. We must know , we will know!
There is a nice variant of this argument noticed by Peter Cameron (and probably others). How many functions f:[n]-->[n] are there such that f^n = fff...fff has a one-point range? Answer: This happens iff the two special vertices are the same, so the answer is n^(n-1).
I remember two of my friends figuring out a nice bijective proof of the theorem before the lecture ended. I did found and fixed a minor mistake, but that was still within an hour.
Looking back, it was probably the same proof. Crazy how this proof was only found it 1981, I actually think people found it earlier and assumed the proof was known already.
The enumerator for labeled trees can be expressed as a determinant of the reduced laplacian of a complete graph. Saw this in algebraic combinatorics course material from Postnikov
Super crazy, i am writing my bachelor thesis about this formula and youtube offered me this video today.... (Algorithms at its best i guess)
Thank you for this interesting lecture
I came for Joyal, I stayed for Borcherds
Can you Imagine a world where Joyal makes videos about his ideas?
Me neither.
This is a formula that sometimes comes up in Olympiad mathematics.
I won gold, never used it, and I was never teached it at the Olympiad.
yeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee