20.181/Lecture3: Difference between revisions
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(in pseudocode) | (in pseudocode) | ||
<blockquote><tt> | |||
for each possible tree: | for each possible tree: | ||
:calculate the score of (tree,data)<br> | :calculate the score of (tree,data)<br> | ||
return tree with BEST score | return tree with BEST score | ||
</tt></blockquote> | |||
===Possible trees=== | ===Possible trees=== | ||
*how many trees are there? | *how many trees are there? | ||
**how does the number of possible trees increase with the number of leaves | **how does the number of possible trees increase with the number of leaves... linearly? ...exponentially? | ||
***start with the simplest unrooted tree, it has three leaves | ***start with the simplest unrooted tree, it has three leaves | ||
***how many ways are there to add another leaf? there are 3 ways- by adding the new leaf attached to each of the 3 existing branches (ignore the center leaf for now because we want to stick to binary trees) | ***how many ways are there to add another leaf? there are 3 ways- by adding the new leaf attached to each of the 3 existing branches (ignore the center leaf for now because we want to stick to binary trees) | ||
***now there are 5 places to add a leaf to a 4-leaf tree | ***now there are 5 places to add a leaf to a 4-leaf tree | ||
***every time you put a new branch down, you gain 2 more places to put a new branch: one from splitting an existing branch into two parts, and one from the new branch itself | ***every time you put a new branch down, you gain 2 more places to put a new branch: one from splitting an existing branch into two parts, and one from the new branch itself | ||
** f_trees(n) = f_trees(n-1) * (2n-5) | ** <tt>f_trees(n) = f_trees(n-1) * (2n-5) | ||
** for n leaves, f_trees(n) = (2n-3)! | ** for n leaves, f_trees(n) = (2n-3)!! | ||
** | ** f_trees(n=10)= 34*10^6 f_trees(n=50) = 2.7*10^76 </tt> | ||
**Enumerating trees is not possible, so we are going to look only at a small number of possible trees. We need a search strategy. And the optimal search strategy will depend on the topology of the space you're looking at. | **Enumerating trees is not possible, so we are going to look only at a small number of possible trees. We need a search strategy. And the optimal search strategy will depend on the topology of the space you're looking at. | ||
Revision as of 08:38, 13 September 2006
Phylogenetic trees
Input: (a multiple sequence alignment)
- AATGC
- TATGC
- GGTGG
- ACTCG
Output: tree, an abstract representation of the same data ((1,4),(2,3))
Overview of Approach
(in pseudocode)
for each possible tree:
- calculate the score of (tree,data)
return tree with BEST score
Possible trees
- how many trees are there?
- how does the number of possible trees increase with the number of leaves... linearly? ...exponentially?
- start with the simplest unrooted tree, it has three leaves
- how many ways are there to add another leaf? there are 3 ways- by adding the new leaf attached to each of the 3 existing branches (ignore the center leaf for now because we want to stick to binary trees)
- now there are 5 places to add a leaf to a 4-leaf tree
- every time you put a new branch down, you gain 2 more places to put a new branch: one from splitting an existing branch into two parts, and one from the new branch itself
- f_trees(n) = f_trees(n-1) * (2n-5)
- for n leaves, f_trees(n) = (2n-3)!!
- f_trees(n=10)= 34*10^6 f_trees(n=50) = 2.7*10^76
- Enumerating trees is not possible, so we are going to look only at a small number of possible trees. We need a search strategy. And the optimal search strategy will depend on the topology of the space you're looking at.
- how does the number of possible trees increase with the number of leaves... linearly? ...exponentially?
Tree Data Structure
- For each node, we need to store:
- names (and sequences?)
- pointers to its left and right subchildren
- we're going to use a built-in dictionary as our data structure
