Efficiency Announcements Measuring Efficiency Recursive - PowerPoint PPT Presentation

Efficiency

Announcements

Measuring Efficiency

Recursive Computation of the Fibonacci Sequence def fib (n): Our first example of tree recursion: if n == 0 : return 0 elif n == 1 : fib(5) return 1 else : return fib(n- 2 ) + fib(n- 1 ) fib(3) fib(4) fib(1) fib(2) fib(2) fib(3) fib(0) fib(1) 1 fib(0) fib(1) fib(1) fib(2) 0 1 fib(0) fib(1) 0 1 1 0 1 (Demo) 4 http://en.wikipedia.org/wiki/File:Fibonacci.jpg

Memoization

Memoization Idea: Remember the results that have been computed before def memo(f): Keys are arguments that map to return values cache = {} def memoized(n): if n not in cache: cache[n] = f(n) return cache[n] return memoized Same behavior as f, 
 if f is a pure function (Demo) 6

Memoized Tree Recursion Call to fib fib(5) Found in cache Skipped fib(3) fib(4) fib(1) fib(2) fib(2) fib(3) fib(0) fib(1) 1 fib(0) fib(1) fib(1) fib(2) 0 1 fib(0) fib(1) 0 1 1 0 1 7

Exponentiation

Exponentiation Goal: one more multiplication lets us double the problem size def exp(b, n): � if n == 0: 1 if n = 0 b n = return 1 b · b n − 1 otherwise else: return b * exp(b, n-1) def exp_fast(b, n): if n == 0: return 1 elif n % 2 == 0:  1 if n = 0 return square(exp_fast(b, n//2))  b n =  1 else: 2 n ) 2 ( b if n is even return b * exp_fast(b, n-1)  b · b n − 1 if n is odd  def square(x): return x * x (Demo) 9

Exponentiation Goal: one more multiplication lets us double the problem size def exp(b, n): Linear time: if n == 0: • Doubling the input 
 return 1 doubles the time else: • 1024x the input takes 
 return b * exp(b, n-1) 1024x as much time def exp_fast(b, n): Logarithmic time: if n == 0: • Doubling the input 
 return 1 increases the time 
 elif n % 2 == 0: by a constant C return square(exp_fast(b, n//2)) • 1024x the input 
 else: increases the time 
 return b * exp_fast(b, n-1) by only 10 times C def square(x): return x * x 10

Orders of Growth

Quadratic Time Functions that process all pairs of values in a sequence of length n take quadratic time 3 5 7 6 def overlap(a, b): count = 0 0 0 0 0 4 for item in a: for other in b: 0 1 0 0 5 if item == other: count += 1 0 0 0 1 return count 6 overlap([3, 5, 7, 6], [4, 5, 6, 5]) 0 1 0 0 5 (Demo) 12

Exponential Time def fib (n): Tree-recursive functions can take exponential time if n == 0 : return 0 elif n == 1 : fib(5) return 1 else : return fib(n- 2 ) + fib(n- 1 ) fib(3) fib(4) fib(1) fib(2) fib(2) fib(3) fib(0) fib(1) 1 fib(0) fib(1) fib(1) fib(2) 0 1 fib(0) fib(1) 0 1 1 0 1 13 13 http://en.wikipedia.org/wiki/File:Fibonacci.jpg

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Order of Growth Notation