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February 05, 2007

Comments

This is a good post.

This means, of course, that one only needs to have a defined EN or a defined BN to have a defined EBN.

Taking a union of those sets is retarded.

This is indeed a good post. But may I suggest a modification, for reasons of more-funness? Redefine x's EBP as the shortest nonrepetitive path from Erdos to x to Bacon (WLoG). Otherwise the existence of a three-edge Erdos-Bacon path (see) will mean that for the vast majority of people the w-EBN will simply be min[EN, BN]+4. For instance, my EN is 3, I have no BN, and my w-EBN is surely 7. (I had to count on my fingers to calculate it.) Even Danica McKellar has a w-EBN of min[EN, BN]+4 (probably). But the way I have defined it, EBNs are more entertaining to calculate.

That won't work, though, for the case of the X who coauthored a paper with Erdős, never starred in a movie, and never coauthored any other papers. There is no path from Erdős to X to Bacon (or the other way 'round), but there is a path containing all three. I think we still want to say that this person still has an EBN.

There is no noncyclical path, rather, and allowing the cycle would change the length.

What's the cyclical path? I guess what I meant to exclude by talking about nonrepetitive.

I was going to slang off such people as dead ends on the Erdos-Bacon tree, but I realized that I am such a person (if we change "with Erdos" to "with someone with an EN of 2") and the whole point of this was to mention my EN, so I concede.

In the example above, it would be E -> X -> E [cycle!] -> ... -> B. Whereas if you do it my way, you can just do X -> E -> ... -> B.

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