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Unformatted text preview: (* Straightforward implementation of fib has exponential number of * recursive calls. Requires n non-negative. *) let rec fib(n) = if n<2 then 1 else fib(n-1) + fib(n-2) (* However most of these recursive calls share common subproblems - * there are only n distinct subproblems to solve, one for each value * of n. Idea of memoization is to keep track of these results as so * they can be looked up rather than re-computed. *) let fibm(n) = let memo: int option array = Array.create (n+1) None in let rec f_mem(n) = match memo.(n) with Some result -> result (* computed already! *) | None -> let result = if n<2 then 1 else f_mem(n-1) + f_mem(n-2) in memo.(n) <- (Some result); (* record in table *) result in f_mem(n) (* Maximum weight independent set in a tree, office party optimization * problem. *) module Unmemoized = struct type tree = Empty | Node of int * tree * tree (* Returns optimum fun for t. *) let rec party(t) = max (party_in t) (party_out t) (* Returns optimum fun for t assuming the root node of t * is included. *) and party_in(t) = match t with Empty -> 0 | Node(v,left,right) -> v + party_out(left) + party_out(right) (* Returns optimum fun for t assuming the root node of t * is excluded. *) and party_out(t) = match t with Empty -> 0 | Node(v,left,right) -> party(left) + party(right) end e module Memoized = struct (* This version memoizes the optimal fun value for each tree node....
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This note was uploaded on 10/25/2009 for the course PHYS 2214 at Cornell.