Memoization and Decorators
Definition of Memoization
The term "memoization" was introduced by Donald Michie in the year 1968. It's based on the Latin word memorandum, meaning "to be remembered". It's not a misspelling of the word memorization, though in a way it has something in common. Memoisation is a technique used in computing to speed up programs. This is accomplished by memorizing the calculation results of processed input such as the results of function calls. If the same input or a function call with the same parameters is used, the previously stored results can be used again and unnecessary calculation are avoided. In many cases a simple array is used for storing the results, but lots of other structures can be used as well, such as associative arrays, called hashes in Perl or dictionaries in Python.Memoization can be explicitly programmed by the programmer, but some programming languages like Python provide mechanisms to automatically memoize functions.
Memoization with Function Decorators
In our previous chapter about recursive functions, we worked out an iterative and a recursive version to calculate the Fibonacci numbers. We have shown that a direct implementation of the mathematical definition into a recursive function like the following has an exponential runtime behaviour:def fib(n): if n == 0: return 0 elif n == 1: return 1 else: return fib(n1) + fib(n2)
We also presented a way to improve the runtime behaviour of the recursive version by adding a dictionary to memorize previously calculated values of the function. This is an example of explicitly using the technique of memoization, but we didn't call it like this. The disadvantage of this method is that the clarity and the beauty of the original recursive implementation is lost.
The "problem" is that we changed the code of the recursive fib function. The following code doesn't change our fib function, so that its clarity and legibility isn't touched. To this purpose, we use function which we call memoize. memoize() takes a function as an argument. The function memoize uses a dictionary "memo" to store the function results. Though the variable "memo" as well as the function "f" are local to memoize, they captured by a closure through the helper function which is returned as a reference by memoize(). So, the call memoize(fib) returns a reference to the helper() which is doing what fib() would have done plus a wrapper which is saving the results, which are still not stored, in the memo dictionary and is preventing recalculation of results, which are already in memo.
def memoize(f): memo = {} def helper(x): if x not in memo: memo[x] = f(x) return memo[x] return helper def fib(n): if n == 0: return 0 elif n == 1: return 1 else: return fib(n1) + fib(n2) fib = memoize(fib) print(fib(40))
Let's look at the line in our code were we assign memoize to fib:
fib = memoize(fib)Doing this, we turn memoize into a decorator. One says that the fib function is decorated by the the memoize() function.
We will illustrate with the following diagrams how the decoration is accomplished. The first diagram illustrates the state before the decoration, i.e. before we call
fib = memoize(fib)
. We can see the function
names referencing their bodies:
After having executed
fib = memoize(fib)
fib points to the body
of the helper function, which had been returned by memoize.
We can also perceive that the code of the original fib function can only be reached via the "f" function of the helper function. There is no way anymore to call the original fib directly, i.e. there is no other reference to it.
The decorated Fibonacci function is called in the return statement return fib(n1) + fib(n2)
, this means the code of the helper function which had been returned by memoize:
Another point in the context of decorators deserves special mention: We don't usually write a decorator for just one use case or function. We rather use it multiple times for different functions. So we could imagine having further functions func1, func2, func3 and so on, which consume also a lot of time. Therefore, it makes sense to decorate each one with our decorator function "memoize":
fib = memoize(fib) func1 = memoize(func1) func2 = memoize(func2) func3 = memoize(func3) # and so on
Memoize as a Class
The subchapter can be skipped without problems by those who don't know about object orientation so far.We can encapsulate the caching of the results in a class as well, as you can see in the following example:
class Memoize: def __init__(self, fn): self.fn = fn self.memo = {} def __call__(self, *args): if args not in self.memo: self.memo[args] = self.fn(*args) return self.memo[args]
As we are using a dictionary, we can't use mutable arguments, i.e. the arguments have to be immutable.
The Usual Syntax for Decorators in Python
Summarizing we can say that a decorator in Python is a callable Python object that is used to modify a function, method or class definition. The original object, the one which is going to be modified, is passed to a decorator as an argument. The decorator returns a modified object, e.g. a modified function, which is bound to the name used in the definition.The decoration in Python is usually not performed in the way we did it in our previous example, even though the notation
fib = memoize(fib)
is catchy and easy to grasp. The decoration is in the line before the function header. The "@" is followed by the decorator function name.
We will rewrite now our initial example. Instead of writing the statement
fib = memoize(fib)we can write
@memoizeBut this line has to be directly in front of the decorated function, in our example fib(). The complete example looks like this now:
def memoize(f): memo = {} def helper(x): if x not in memo: memo[x] = f(x) return memo[x] return helper @memoize def fib(n): if n == 0: return 0 elif n == 1: return 1 else: return fib(n1) + fib(n2) #fib = memoize(fib) print(fib(40))
Further Examples for Decorators
Checking Arguments with a Decorator
In our chapter about recursive functions we introduced the factorial function. We wanted to keep the function as simple as possible and we didn't want to obscure the underlying idea, so we hadn't incorporated any argument checks. So, if somebody had called our function with a negative argument or with a float argument, our function would have got into an endless loop.The following program uses a decorator function to ensure that the argument passed to the function factorial is a positive integer:
def argument_test_natural_number(f): def helper(x): if type(x) == int and x > 0: return f(x) else: raise Exception("Argument is not an integer") return helper @argument_test_natural_number def factorial(n): if n == 1: return 1 else: return n * factorial(n1) for i in range(1,10): print(i, factorial(i)) print(factorial(1))
Counting Function Calls with Decorators
The following example uses a decorator to count the number of times a function has been called.def call_counter(func): def helper(*args, **kwargs): helper.calls += 1 return func(*args, **kwargs) helper.calls = 0 helper.__name__= func.__name__ return helper @call_counter def succ(x): return x + 1 print(succ.calls) for i in range(10): print(succ(i)) print(succ.calls)The output looks like this:
0 10
Exercise

Our exercise is an old riddle, going back to 1612. The French Jesuit ClaudeGaspar
Bachet phrased it. We have to weigh quantities (e.g. sugar or flour) from 1 to 40 pounds.
What is the least number of weights that can be used on a balance scale to way any of
these quantities.
The first idea might be to use weights of 1, 2, 4, 8, 16 and 32 pounds. This is a minimal number, if we restrict ourself to put weights on one side and the stuff, e.g. the sugar, on the other side. But it is possible to put weights on both pans of the scale. Now, we need only for weights, i.e. 1,3,9,27
Write a Python function weigh(), which calculates the weights needed and their distribution on the pans to weigh any amount from 1 to 40.
Solution

We need the function linear_combination() from our chapter
"Linear Combinations".
def factors_set(): factors_set = ( (i,j,k,l) for i in [1,0,1] for j in [1,0,1] for k in [1,0,1] for l in [1,0,1]) for factor in factors_set: yield factor def memoize(f): results = {} def helper(n): if n not in results: results[n] = f(n) return results[n] return helper @memoize def linear_combination(n): """ returns the tuple (i,j,k,l) satisfying n = i*1 + j*3 + k*9 + l*27 """ weighs = (1,3,9,27) for factors in factors_set(): sum = 0 for i in range(len(factors)): sum += factors[i] * weighs[i] if sum == n: return factors
With this, it is easy to write our function weigh().def weigh(pounds): weights = (1,3,9,27) scalars = linear_combination(pounds) left = "" right = "" for i in range(len(scalars)): if scalars[i] == 1: left += str(weights[i]) + " " elif scalars[i] == 1: right += str(weights[i]) + " " return (left,right) for i in [2,3,4,7,8,9,20,40]: pans = weigh(i) print("Left pan: " + str(i) + " plus " + pans[0]) print("Right pan: " + pans[1] + "\n")