mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-24 05:21:09 +00:00
8d94f7745f
* Replacing the generator with numpy vector operations from lu_decomposition.
* Revert "Replacing the generator with numpy vector operations from lu_decomposition."
This reverts commit ad217c6616.
* Application of vector operations to reduce calculation time and refactoring numpy.
* [pre-commit.ci] auto fixes from pre-commit.com hooks
for more information, see https://pre-commit.ci
---------
Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
51 lines
1.4 KiB
Python
51 lines
1.4 KiB
Python
"""
|
|
Problem 72 Counting fractions: https://projecteuler.net/problem=72
|
|
|
|
Description:
|
|
|
|
Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1,
|
|
it is called a reduced proper fraction.
|
|
If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we
|
|
get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7,
|
|
3/4, 4/5, 5/6, 6/7, 7/8
|
|
It can be seen that there are 21 elements in this set.
|
|
How many elements would be contained in the set of reduced proper fractions for
|
|
d ≤ 1,000,000?
|
|
|
|
Solution:
|
|
|
|
Number of numbers between 1 and n that are coprime to n is given by the Euler's Totient
|
|
function, phi(n). So, the answer is simply the sum of phi(n) for 2 <= n <= 1,000,000
|
|
Sum of phi(d), for all d|n = n. This result can be used to find phi(n) using a sieve.
|
|
|
|
Time: 1 sec
|
|
"""
|
|
|
|
import numpy as np
|
|
|
|
|
|
def solution(limit: int = 1_000_000) -> int:
|
|
"""
|
|
Returns an integer, the solution to the problem
|
|
>>> solution(10)
|
|
31
|
|
>>> solution(100)
|
|
3043
|
|
>>> solution(1_000)
|
|
304191
|
|
"""
|
|
|
|
# generating an array from -1 to limit
|
|
phi = np.arange(-1, limit)
|
|
|
|
for i in range(2, limit + 1):
|
|
if phi[i] == i - 1:
|
|
ind = np.arange(2 * i, limit + 1, i) # indexes for selection
|
|
phi[ind] -= phi[ind] // i
|
|
|
|
return np.sum(phi[2 : limit + 1])
|
|
|
|
|
|
if __name__ == "__main__":
|
|
print(solution())
|