Re: Binary Tree and Pairs of Nodes

On Oct 9, 3:20 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
Therefore I do not express a criterion, but show that there can no
path start at the root and lead out of the conquered domain without
getting its node. That is a *forcing* logical element.

You HAVE NOT DONE this!
All you have said is "Take node 1. Give it a path.
Take node 2. Give it a path, it doesn't matter which.
Take node 3, give it a path. It doesn't matter which".
THAT IS ALL you have said! That obviously DOES NOT NECESSARILY
use all paths! I could do THAT using ONLY paths of all 1s followed by
all 0's!
I take node 0. I give it the all 0's path.
I take node 1. I give it the one 1 followed by all 0's path.
I take node 2. I give it the two 1's followed by all 0's path.
I can do this for ALL natnums. When I get to the end I will have
used up all the natnums, BUT NOWHERE NEAR all the paths!
In fact, NO MATTER WHAT way you assign the paths, YOU WILL NEVER
COME CLOSE to using all the paths!

The easy way to prove this is just (even though you can't finish)
to ASSUME THAT YOU HAVE finished and given EVERY node a path!
THEN, WE CAN CONSTRUCT a path that YOU DID NOT include!
It is just the path whose nth step is THE OPPOSITE WAY from the
nth step of the path that YOU mapped to node n!
THAT path IS NOT IN YOUR map!

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