factorizing a tree
Jan Skibinski
jans@numeric-quest.com
Sun, 29 Apr 2001 06:27:06 -0400 (EDT)
I need some help with some optimization of certain trees:
data Tree
= Leaf Item
| Product Tree Tree
| Sum (Double, Tree) (Double, Tree)
with this additional "collating" property:
Sum (Double, Leaf Item) (Double, Leaf Item) ==> Leaf Item
I have certain control over the shape of such trees.
I love products, but sums are expensive and hence my goal is
to avoid them at all cost. If this is unavoidable then I try
to push the sums down the tree as deeply as possible. By
doing that I am sometimes able to collate a sum of two leafs
into one leaf, and that's great. But even without the collation
effect a deeply set sum is still better than the top sum.
One of the functions, f say, is given two indices i, j
from the range (1, depth of the tree). It should modify
the layer "j" based on the information contained in the layer "i"
- producing new tree of the same depth. A naive solution
looks like this:
f :: Int -> Int -> Tree -> Tree
f i j tree = Sum (1, g i j tree) (1, h i j tree)
where
g = ...
h = ...
but this is the most costly solution.
One way of pushing the sums down the tree is to do something
of this sort:
f i j (Product t1 t2)
| (i <= d) && (j <= d) = Product (f i j t1) t2
| (i > d) && (j > d) = Product t1 (f (i-d) (j-d) t2)
| ....
where
d = depth t1
depth t = ...
I wonder if there are some known practises for handling such
problems. Or am I overlooking some very simple solution?
If I did not clearly specify the problem it is probably because
I did not want to post here too much of irrelevant information.
I can clarify it if required.
Jan