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Use timing decorator for first 10 problems

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This commit is contained in:
daniele 2023-06-07 17:46:10 +02:00
parent 955dc12737
commit 885084211a
Signed by: fuxino
GPG Key ID: 981A2B2A3BBF5514
10 changed files with 65 additions and 123 deletions

17
Python/p001.py

@ -4,27 +4,22 @@
#
# Find the sum of all the multiples of 3 or 5 below 1000.
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p001():
sum_ = 0
# Simple brute-force approach: try every number between 3 and 999,
# check if it's a multiple of 3 or 5, if yes add it to the total.
# Simple brute-force approach: try every number between 3 and 999,
# check if it's a multiple of 3 or 5, if yes add it to the total.
for i in range(3, 1000):
if i % 3 == 0 or i % 5 == 0:
sum_ += i
end = default_timer()
print('Project Euler, Problem 1')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p001()

18
Python/p002.py

@ -6,13 +6,11 @@
#
# By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p002():
N = 4000000
fib1 = 1
@ -20,8 +18,8 @@ def main():
fibn = fib1 + fib2
sum_ = 2
# Simple brute-force approach: generate every value in the Fibonacci
# sequence smaller than 4 million and if it's even add it to the total.
# Simple brute-force approach: generate every value in the Fibonacci
# sequence smaller than 4 million and if it's even add it to the total.
while fibn < N:
if fibn % 2 == 0:
sum_ = sum_ + fibn
@ -30,13 +28,9 @@ def main():
fib2 = fibn
fibn = fib1 + fib2
end = default_timer()
print('Project Euler, Problem 2')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p002()

24
Python/p003.py

@ -4,27 +4,24 @@
#
# What is the largest prime factor of the number 600851475143?
from timeit import default_timer
from projecteuler import is_prime
from projecteuler import is_prime, timing
# Recursive approach: if num is prime, return num, otherwise
# recursively look for the largest prime factor of num divided
# by its prime factors until only the largest remains.
def max_prime_factor(num):
# Use function defined in projecteuler.py to check if a number is prime.
# Use function defined in projecteuler.py to check if a number is prime.
if is_prime(num):
return num
# If num is even, find the largest prime factor of num/2.
# If num is even, find the largest prime factor of num/2.
if num % 2 == 0:
return max_prime_factor(num // 2)
i = 3
# If num is divisible by i and i is prime, find largest
# prime factor of num/i.
# If num is divisible by i and i is prime, find largest prime factor of num/i.
while True:
if num % i == 0:
if is_prime(i):
@ -32,22 +29,17 @@ def max_prime_factor(num):
i = i + 2
# Should never get here
# Should never get here
return -1
def main():
start = default_timer()
@timing
def p003():
res = max_prime_factor(600851475143)
end = default_timer()
print('Project Euler, Problem 3')
print(f'Answer: {res}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p003()

22
Python/p004.py

@ -4,32 +4,26 @@
#
# Find the largest palindrome made from the product of two 3-digit numbers.
from timeit import default_timer
from projecteuler import is_palindrome
from projecteuler import is_palindrome, timing
def main():
start = default_timer()
@timing
def p004():
max_ = 0
# Using a brute-force approach: generate every product of 3-digit numbers
# and check if it's palindrome. If the product found is greater than the
# current maximum, save the current product.
# Using a brute-force approach: generate every product of 3-digit numbers
# and check if it's palindrome. If the product found is greater than the
# current maximum, save the current product.
for i in range(999, 99, -1):
for j in range(i, 99, -1):
num = i * j
# Use the function defined in projecteuler.py to check if a number is palindrome.
# Use the function defined in projecteuler.py to check if a number is palindrome.
if num > max_ and is_palindrome(num, 10):
max_ = num
end = default_timer()
print('Project Euler, Problem 4')
print(f'Answer: {max_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p004()

16
Python/p005.py

@ -4,26 +4,20 @@
#
# What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
from timeit import default_timer
from projecteuler import lcmm
from projecteuler import lcmm, timing
def main():
start = default_timer()
@timing
def p005():
values = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20)
# Function define in projecteuler.py to find the least common multiple of multiple numbers.
# Function define in projecteuler.py to find the least common multiple of multiple numbers.
res = lcmm(values, 20)
end = default_timer()
print('Project Euler, Problem 5')
print(f'Answer: {res}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p005()

15
Python/p006.py

@ -12,29 +12,24 @@
#
# Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p006():
sum_squares = 0
square_sum = 0
# Straightforward brute-force approach.
# Straightforward brute-force approach.
for i in range(1, 101):
sum_squares = sum_squares + i * i
square_sum = square_sum + i
square_sum = square_sum * square_sum
end = default_timer()
print('Project Euler, Problem 6')
print(f'Answer: {square_sum - sum_squares}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p006()

22
Python/p007.py

@ -4,32 +4,26 @@
#
# What is the 10 001st prime number?
from timeit import default_timer
from projecteuler import is_prime
from projecteuler import is_prime, timing
def main():
start = default_timer()
@timing
def p007():
count = 1
n = 1
# Brute force approach: start with count=1 and check every odd number
# (2 is the only even prime), if it's prime increment count, until the
# target prime is reached.
# Brute force approach: start with count=1 and check every odd number
# (2 is the only even prime), if it's prime increment count, until the
# target prime is reached.
while count != 10001:
n = n + 2
# Use the function in projecteuler.py to check if a number is prime.
# Use the function in projecteuler.py to check if a number is prime.
if is_prime(n):
count = count + 1
end = default_timer()
print('Project Euler, Problem 7')
print(f'Answer: {n}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p007()

17
Python/p008.py

@ -25,12 +25,11 @@
#
# Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p008():
number = '73167176531330624919225119674426574742355349194934' +\
'96983520312774506326239578318016984801869478851843' +\
'85861560789112949495459501737958331952853208805511' +\
@ -51,12 +50,12 @@ def main():
'05886116467109405077541002256983155200055935729725' +\
'71636269561882670428252483600823257530420752963450'
# Transform the string into a list of integers
# Transform the string into a list of integers
number = list(map(int, number))
max_ = 0
# Calculate all the 13-digit products, and save the maximum
# Calculate all the 13-digit products, and save the maximum
for i in range(1000-13):
curr = number[i:i+13]
prod = 1
@ -67,13 +66,9 @@ def main():
if prod > max_:
max_ = prod
end = default_timer()
print('Project Euler, Problem 8')
print(f'Answer: {max_}')
print(f'Elapsed time: {end - start:.9f} seconds'.format(end - start))
if __name__ == '__main__':
main()
p008()

17
Python/p009.py

@ -11,18 +11,17 @@
# Find the product abc.
from math import gcd
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p009():
found = 0
m = 2
# Brute force approach: generate all the Pythagorean triplets using
# Euclid's formula, until the one where a+b+c=1000 is found.
# Brute force approach: generate all the Pythagorean triplets using
# Euclid's formula, until the one where a+b+c=1000 is found.
while not found:
for n in range(1, m):
if found == 1:
@ -58,13 +57,9 @@ def main():
m = m + 1
end = default_timer()
print('Project Euler, Problem 9')
print(f'Answer: {a * b * c}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p009()

20
Python/p010.py

@ -4,32 +4,26 @@
#
# Find the sum of all the primes below two million.
from timeit import default_timer
from projecteuler import sieve
from projecteuler import sieve, timing
def main():
start = default_timer()
@timing
def p010():
N = 2000000
# Use the function in projecteuler.py implementing the
# Sieve of Eratosthenes algorithm to generate primes.
# Use the function in projecteuler.py implementing the
# Sieve of Eratosthenes algorithm to generate primes.
primes = sieve(N)
sum_ = 0
# Sum all the primes
# Sum all the primes
for i in range(N):
if primes[i] == 1:
sum_ = sum_ + i
end = default_timer()
print('Project Euler, Problem 10')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p010()