
An unsolved problem in graph theory - prideout
https://prideout.net/blog/graceful/
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noud
The definition of graceful graph is wrong. Edges E_{i,j} should have unique
values as well, i.e. {E_{i,j}} = {1, 2, ..., n}. Otherwise it's trivial to
label every graph into a graceful graph.

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Syzygies
Yes, thank you. I kept looking at the date, wondering if it was already April
1st, was this a brilliant troll? Then I checked Wikipedia.

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cschmidt
A month ago [1] there was a post about the proof of Ringel’s conjecture, that
a complete graph can be covered by certain trees. Sadly it was a non-
constructive proof.

In a little googling of the Graceful Tree Conjecture, it turns out that
Ringel's conjecture inspired the Graceful Tree Conjecture [2]

It was shown that the GTC implies Ringel's conjecture. Fun to see the
connection in the last two math problems I dipped into.

[1]
[https://news.ycombinator.com/item?id=22373701](https://news.ycombinator.com/item?id=22373701)

[2]
[https://tspace.library.utoronto.ca/bitstream/1807/13623/1/MQ...](https://tspace.library.utoronto.ca/bitstream/1807/13623/1/MQ53395.pdf)
page 13 in their numbering

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sdenton4
I, uh, wasted a bit of time once on the Tree Packing conjecture of Gyarfas:
[https://arxiv.org/abs/1104.0642](https://arxiv.org/abs/1104.0642)

In your left hand, place a complete graph, which has n-choose-2 edges.

In your right hand, a collection of trees on 2, 3, 4, ..., n-1 vertices,
which, by miraculous concidence, also has n-choose-2 edges.

Q: Can you pack the trees into a complete graph with no overlapping edges?

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gweinberg
Pretty cool. I would think there would either be an algorithm to label the
vertices or an example of one that cannot be elegantly labelled, but I guess
not. I's pretty obvious that node n must be adjacent to node 0, since that's
the only way to get edge n. And it's pretty obvious that if you have one
labeling you can get another by replacing m with n-m, so all the edges keep
the same labels. But that's about as far as I get.

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empath75
Numberphile video:
[https://www.youtube.com/watch?v=v5KWzOOhZrw](https://www.youtube.com/watch?v=v5KWzOOhZrw)

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jjtheblunt
little mistake:

    
    
       "Vi <> Vj for all i,j" 
    

fails if i == j, which isn't forbidden.

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prideout
Nice one! Fixing...

