# The Enormous TREE(3): A Mathematical Game

Submitted by: fuad119 11 months ago in
2

I still don't know how to play this.
2
Male 12,258
you could easily figure this out this iterating through a collection of trees
Male 4,319
monkwarrior If it's so easy; we're waiting then...

Also, recursively doing it would be far, far, far more efficient.
Male 12,258
DuckBoy87 Yeah i suppose you're right, i wrote a pgrogram in c++ when i was in school to hunt the way through a maze that could be used here.  You basically create a linked node with each node holding a tree collection, which is basically recursion.  Then it would be just a matter of creating a search algorithm to build a tree to just before it breaks and fill it in the node, then keep doing that until no more possibilities, then just iterate over the linked list to find which node has the biggest collection of trees.

At first glance it seems it would just require about 2 node classes, a tree filling class with some methods, a search class with a few methods, and the driver program.  I don't know how to solve this with just math though, and i think this is more of a programming solution because there's so many possibilities that it doesn't seem sensible to find the max on paper.  I dont know what the max is because i didn't write it, and even though it's easy to figure out, it would still require a few hours to write code to find the max, but i know the min would be 1 (before 1 green dot | 1 green dot)

Male 654
Nice post.  The first 5+ min post I've actually sat through entirely.  Proofs are amazing things imo.
Male 6,713
Well ok, so i understood the video, googled for more info and this kinda shit comes up:

First, let us assume that for every limit ordinal (a limit ordinal is an ordinal with no predecessor) in a sufficiently large initial segment of ordinals, we have defined a fundamental sequence; that is, an increasing sequence of ordinals whose limit is the original ordinal. So for
ω
ω we can take the sequence {0, 1, 2, ...}; for
ω∗2
ω∗2 we can take the sequence
ω,ω+1,ω+2,…,
ω,ω+1,ω+2,…,; for
ω2
ω2, we can take the sequence
0,ω,ω∗2,…,
0,ω,ω∗2,…,; and so on. For an ordinal
α
α, define
α[n]
α[n] to be the
n
nth element of the fundamental sequence of
α
α when
α
α is a limit ordinal, and the predecessor of
α
α when
α
α is a successor ordinal.

SO, ok im clearly way out of depth, so i will just check back in a few years and see if some super computer has solved it for us.

Female 5,268
after about 30 seconds I zoned out. too much thinky.