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| DATE | 2017-01-26 |
| FROM | ruben safir
|
| SUBJECT | Re: [Learn] Felsenstein Phylogenies
|
From learn-bounces-at-nylxs.com Thu Jan 26 06:46:43 2017
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From: ruben safir
To: "learn-at-nylxs.com"
Message-ID: <429a9d55-1da8-b068-0049-4029944f897c-at-mrbrklyn.com>
Date: Thu, 26 Jan 2017 06:46:40 -0500
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Subject: Re: [Learn] Felsenstein Phylogenies
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On 01/25/2017 09:35 PM, John Harshman wrote:
> On 1/25/17 3:53 PM, Ruben Safir wrote:
>> Does anyone have the above text handy? I think what he wrote with
>> regard to Subtree Programming and Grafting is incorrect.
>>
>> If you have 2 subtrees n1 and n2,the number of neighbors should be (2n1
>> -4) * (2n2 - 4) --- not addition
>>
>> each spot has 2n-3 - 1 permutations.
>>
>> He doesn't explain what external branches are either.
>>
>>
> The book you're referring to is called Inferring Phylogenies and the
> procedure you're talking about is called subtree *pruning* and
> *regrafting*. The number of rearrangements given a particular subtree
> should be equal to the number of branches on the second subtree, which
> is twice the number of taxa minus 3.
>
> I don't currently have a copy handy. Please explain more clearly what
> Felsenstein says about it and what you think it should say.
Correct, what it does say is that once you divide the tree there would
be 2n1 - 3 - 1 reassertion points for the tree. Then after that he is
not clear to me. He says
"In fact considering both subtrees (no having n1 species and the one
having n2 species, there are
(2n1-3-1) + (2n2-3-1 ) = (2n-3-1) = 2n-8
neighbors generated at each interior branch."
This assumes n1 + n2 = n.
I guess that is all the possible combinations assuming the same
attachment locations for the trees, examining one tree at a time.
Then he states that external nodes (which is not defined) is 2n-6.
Without proof I'll accept that for a moment (and I think it corresponds
to binary tree theory), but I'm not sure that an exterior node is. That
is a node that connects to leafs?
Finally, the last unclear sentence, at least to me, states:
"Thus, as there are n exterior branches on an unrooted bifurcating tree
and n-3 interior branches, the total number of neighbors examined by SPR
will be
n(2n-6)+(n+3)(2n-8)"
That is where he lost me. Then he follows up
and he says tha there are 288 neighbors for n=11
and
"Of course, 2(n-3)=16 of them are the same as NNI"
For TBR he says that there is no general formula for the number of
neighbors that will be examine. That made be stand on the edge of my
seat? Say what? Then what are we doing?
:)
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