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Python: Calculate Euclid's totient function of a given integer

Python Basic - 1: Exercise-120 with Solution

In number theory, Euler's totient function counts the positive integers up to a given integer n that are relatively prime to n. It is written using the Greek letter phi as φ(n) or ϕ(n), and may also be called Euler's phi function.
Write a Python program to calculate Euclid's totient function of a given integer. Use a primitive method to calculate Euclid's totient function.

Sample Solution:

Python Code:

def gcd(p,q):
# Create the gcd of two positive integers.
    while q != 0:
        p, q = q, p%q
    return p

def is_coprime(x, y):
    return gcd(x, y) == 1

def phi_func(x):
    if x == 1:
        return 1
    else:
        n = [y for y in range(1,x) if is_coprime(x,y)]
        return len(n)
print(phi_func(10))
print(phi_func(15))
print(phi_func(33))

Sample Output:

4
8
20

Flowchart:

Flowchart: Python - Calculate Euclid's totient function of a given integer.

Python Code Editor:

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Python: Tips of the Day

How to make a flat list out of list of lists?

Given a list of lists l

flat_list = [item for sublist in l for item in sublist]

which means:

flat_list = []
for sublist in l:
    for item in sublist:
        flat_list.append(item)

is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:

flatten = lambda l: [item for sublist in l for item in sublist]

As evidence, you can use the timeit module in the standard library:

$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

Ref: https://bit.ly/3dKsNTR