How to find the number of times function f is executed?

Question

The function f is defined as such:

def f(a, b):
    if a <= 0 or b <= 0:
        return a + b
    else:
        s = 0
        if b * b % (a + b) != 0:
            s += f(a, b - 3)
        if a * a % (a + b) != 0:
            s += f(a - 2, b)
        if a == b - 1:
            s += f(a - 3, b - 2) 
    return s

The question is: "How many times will a function "f" be executed, given f(4, 9)?" For example, for f(0, 0) function "f" will be executed once, since the first time is taken into account as well. Can someone explain to me, how I can find the number of executions? (Examples will be ideal.) Than you in advance!

You can always pass and return a counter variable to the function. Alternatively, have a look at this question about [Python "static" variables](http://stackoverflow.com/questions/279561/what-is-the-python-equivalent-of-static-variables-inside-a-function). — , Apr 24 '16 at 22:25

score 1 · Answer 1 · answered Apr 24 '16 at 22:43

1

You can attach an attribute to the function:

def f(a, b):
    f.num += 1
    if a <= 0 or b <= 0:
        ...

Result:

>>> f.num = 0
>>> f(0, 0)
0
>>> print(f.num)
1
>>> f.num = 0
>>> f(3, 5)
4
>>> print(f.num)
13

answered Apr 24 '16 at 22:43

TigerhawkT3

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This is correct, but is not very efficient, otherwise - not as one block of code. – L_Pav Apr 24 '16 at 22:56

Padraic Cunningham · Accepted Answer · 2016-04-24T22:31:42.657

0

You can increment a global counter, an int wrapped in a list, incrementing it in the function and printing the result after you call it:

i = [0]
def f(a, b):
    i[0] += 1
    if a <= 0 or b <= 0:
        return a + b
    else:
        s = 0
        if b * b % (a + b) != 0:
            s += f(a, b - 3)
        if a * a % (a + b) != 0:
            s += f(a - 2, b)
        if a == b - 1:
            s += f(a - 3, b - 2)

    return s

f(2, 34)
print(i[0])

edited Apr 24 '16 at 22:31

answered Apr 24 '16 at 22:28

Padraic Cunningham

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If you will admit that this is the number of executions of f, I will vote you up! – L_Pav Apr 24 '16 at 22:31
1

@L_Pav, this will increment the counter each time f is called. – Padraic Cunningham Apr 24 '16 at 22:32
1

He increments `i[0]` every time the function f is called, so it is indeed the number of executions of f. – Matt Messersmith Apr 24 '16 at 22:32
Why a list of one int; can't you use a global int variable instead? – Apr 24 '16 at 22:32
1

@Evert, yes you can but I don't like using global so I used a list – Padraic Cunningham Apr 24 '16 at 22:41
Aha; I would think that if you're using a global variable though, it's better to be explicit about it. – Apr 24 '16 at 22:44
1

@Evert, yep but I don't want to encourage or introduce someone to the global keyword, using a list also gives a good lesson in how changes to a mutable structure inside a function are reflected outside which is something one will come across a lot more often than the global keyword. – Padraic Cunningham Apr 24 '16 at 22:49

silent_dev · Answer 3 · 2016-04-24T23:26:32.873

-1

I don't trust Global variable because of the highly available scope i.e. global variables are accessible to everyone. Instead, you can keep a recursion variable which counts the number of times the function has been called.

def f(a, b, count=0):
    count+=1
    if a <= 0 or b <= 0:
        return [a + b, count]
    else:
        s = 0
        if b * b % (a + b) != 0:
            l = f(a, b - 3,count)
            s += l[0]
            count+=l[1]
        if a * a % (a + b) != 0:
            l = f(a - 2, b, count)
            s += l[0]
            count+=l[1]
        if a == b - 1:
            l = f(a - 3, b - 2, count) 
            s += l[0]
            count+=l[1]

    return [s,count]

edited Apr 24 '16 at 23:26

answered Apr 24 '16 at 22:41

silent_dev

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3

This keeps returning `0` for the count. – TigerhawkT3 Apr 24 '16 at 22:45
Yes, count was not updating. Updating the code. However, do note that you have three 'if' statements and not 'if-elif-else'. Hopefully that is intentional. – silent_dev Apr 24 '16 at 22:56
It's still returning incorrect values. With `f(0, 0)`, it'll increment `count` to 1, then immediately return the result of `[0, 2]`. – TigerhawkT3 Apr 24 '16 at 23:11
Fixed. It now returns 1 for f(0,0). – silent_dev Apr 24 '16 at 23:28

How to find the number of times function f is executed?

3 Answers3