trjhtr

Here's a snippet of code I use to examine the functionality of Python decorator(@memorize). An example of Fibonacci computation:

def memorize(f):
 memo = 
 def helper(*args):
 if args not in memo:
 memo[args] = f(*args)
 return memo[args]
 return helper

def fib(n):
 if n==0:
 return 0
 elif n==1:
 return 1
 else:
 return fib(n-1) + fib(n-2)

Here's the problem:

Different naming is causing a huge speed difference, why is it?

• 
  

  Execution 1:

  
  
  

  s = time.time()
  fib = memorize(fib)
  a = fib(40)
  e = time.time()
  print(a)
  print(e-s)
  

  
  
  

  returns

  
  
  

  
  

  102334155

  
  
  

  7.319450378417969e-05

  
  

  
  


• 
  

  Execution 2:

  
  
  

  s = time.time()
  memo_fib = memorize(fib)
  a = memo_fib(40)
  e = time.time()
  print(a)
  print(e-s)
  

  
  
  

  returns

  
  
  

  
  

  102334155

  
  
  

  46.79982662200928

  
  

  
  

Edited:

A screen copy from running the code

Edited:

Unless running the two executions seperately, to obtain aforementioned results, "Execution 2" must be run before "Execution 1".

fib works by recursively calling the function named fib:

def fib(n):
 if n==0:
 return 0
 elif n==1:
 return 1
 else:
 return fib(n-1) + fib(n-2)

In your "execution 1" the function named fib is the memoized version of the function, because you have assigned it like this:

fib = memorize(fib)

But (assuming that you haven't run "execution 1" yet), in your "execution 2" the function named fib is the original function (not the memoized version of the function, which you have assigned to memo_fib), so when you call memo_fib it calls the original fib and when that recurses it calls the original fib, bypassing the memoization.