fuxino/project-euler-solutions
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Use timing decorator for problems 71-80

Browse Source
This commit is contained in:
daniele 2023-06-07 20:39:55 +02:00
parent f54083389b
commit e8098ed5e5
Signed by: fuxino
GPG Key ID: 981A2B2A3BBF5514
10 changed files with 103 additions and 161 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

28
Python/p071.py
View File

@ -12,22 +12,22 @@
# of the fraction immediately to the left of 3/7.
from math import gcd
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p071():
N = 1000000
max_ = 0
# For each denominator q, we need to find the biggest numerator p for which
# p/q<a/b, where a/b is 3/7 for this problem. So:
# pb<aq
# pb<=aq-1
# p<=(aq-1)/b
# So we can set p=(3*q-1)/7 (using integer division).
# For each denominator q, we need to find the biggest numerator p for which
# p/q<a/b, where a/b is 3/7 for this problem. So:
# pb<aq
# pb<=aq-1
# p<=(aq-1)/b
# So we can set p=(3*q-1)/7 (using integer division).
for i in range(2, N+1):
j = (3 * i - 1) // 7
@ -36,20 +36,16 @@ def main():
d = i
max_ = n / d
# Reduce the fractions if it's not already reduced.
# Reduce the fractions if it's not already reduced.
if gcd(i, j) > 1:
n /= gcd(i, j)
d /= gcd(i, j)
max_n = n
end = default_timer()
print('Project Euler, Problem 71')
print(f'Answer: {max_n}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p071()

25
Python/p072.py
View File

@ -10,33 +10,26 @@
#
# How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?
from timeit import default_timer
from projecteuler import sieve, phi
from projecteuler import sieve, phi, timing
def main():
start = default_timer()
@timing
def p072():
N = 1000001
count = 0
primes = sieve(N)
# For any denominator d, the number of reduced proper fractions is
# the number of fractions n/d where gcd(n, d)=1, which is the definition
# of Euler's Totient Function phi. It's sufficient to calculate phi for each
# denominator and sum the value.
# For any denominator d, the number of reduced proper fractions is
# the number of fractions n/d where gcd(n, d)=1, which is the definition
# of Euler's Totient Function phi. It's sufficient to calculate phi for each
# denominator and sum the value.
for i in range(2, N):
count = count + phi(i, primes)
end = default_timer()
print('Project Euler, Problem 72')
print(f'Answer: {count}')
print(f'Elapsed time: {end - start:.9f} seconds')
print(f'Answer: {int(count)}')
if __name__ == '__main__':
main()
p072()

22
Python/p073.py
View File

@ -11,32 +11,28 @@
# How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?
from math import gcd
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p073():
count = 0
# For each denominator q, we need to find the fractions p/q for which
# 1/3<p/q<1/2. For the lower limit, if p=q/3, then p/q=1/3, so we take
# p=q/3+1. For the upper limit, if p=q/2 p/q=1/2, so we take p=(q-1)/2.
# For each denominator q, we need to find the fractions p/q for which
# 1/3<p/q<1/2. For the lower limit, if p=q/3, then p/q=1/3, so we take
# p=q/3+1. For the upper limit, if p=q/2 p/q=1/2, so we take p=(q-1)/2.
for i in range(2, 12001):
limit = (i - 1) // 2 + 1
for j in range(i//3+1, limit):
# Increment the counter if the current fraction is reduced.*/
# Increment the counter if the current fraction is reduced.*/
if gcd(j, i) == 1:
count = count + 1
end = default_timer()
print('Project Euler, Problem 73')
print(f'Answer: {count}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p073()

34
Python/p074.py
View File

@ -23,7 +23,8 @@
# How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
from math import factorial
from timeit import default_timer
from projecteuler import timing
N = 1000000
@ -42,23 +43,23 @@ def len_chain(n):
tmp = 0
count = count + 1
# Generate the next number of the chain by taking
# the digits of the current value, calculating the
# factorials and adding them.*/
# Generate the next number of the chain by taking
# the digits of the current value, calculating the
# factorials and adding them.
while value != 0:
tmp = tmp + factorial(value % 10)
value = value // 10
# If the chain length for the new value has already been
# calculated before, use the saved value (only chains for
# values smaller than N are saved).*/
# If the chain length for the new value has already been
# calculated before, use the saved value (only chains for
# values smaller than N are saved).
if tmp < N and chains[tmp] != 0:
return count + chains[tmp]
value = tmp
# If the current value is already present in the chain,
# the chain is finished.*/
# If the current value is already present in the chain,
# the chain is finished.
for i in range(count):
if chain[i] == value:
finished = 1
@ -70,24 +71,19 @@ def len_chain(n):
return count
def main():
start = default_timer()
@timing
def p074():
count = 0
# Simple brute force approach, for every number calculate
# the length of the chain.
# Simple brute force approach, for every number calculate
# the length of the chain.
for i in range(3, N):
if len_chain(i) == 60:
count = count + 1
end = default_timer()
print('Project Euler, Problem 74')
print(f'Answer: {count}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p074()

36
Python/p075.py
View File

@ -18,27 +18,27 @@
# Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?
from math import gcd
from timeit import default_timer
from projecteuler import timing
def main():
start = default_timer()
@timing
def p075():
N = 1500000
l = [0] * (N+1)
# Generate all Pythagorean triplets using Euclid's algorithm:
# For m>=2 and n<m:
# a=m*m-n*n
# b=2*m*n
# c=m*m+n*n
# This gives a primitive triple if gcd(m, n)=1 and exactly one
# of m and n is odd. To generate all the triples, generate all
# the primitive one and multiply them by i=2,3, ..., n until the
# perimeter is larger than the limit. The limit for m is 865, because
# when m=866 even with the smaller n (i.e. 1) the perimeter is greater
# than the given limit.
# Generate all Pythagorean triplets using Euclid's algorithm:
# For m>=2 and n<m:
# a=m*m-n*n
# b=2*m*n
# c=m*m+n*n
# This gives a primitive triple if gcd(m, n)=1 and exactly one
# of m and n is odd. To generate all the triples, generate all
# the primitive one and multiply them by i=2,3, ..., n until the
# perimeter is larger than the limit. The limit for m is 865, because
# when m=866 even with the smaller n (i.e. 1) the perimeter is greater
# than the given limit.
for m in range(2, 866):
for n in range(1, m):
if gcd(m, n) == 1 and ((m % 2 == 0 and n % 2 != 0) or (m % 2 != 0 and n % 2 == 0)):
@ -68,13 +68,9 @@ def main():
if l[i] == 1:
count = count + 1
end = default_timer()
print('Project Euler, Problem 75')
print(f'Answer: {count}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p075()

21
Python/p076.py
View File

@ -11,28 +11,21 @@
#
# How many different ways can one hundred be written as a sum of at least two positive integers?*/
from timeit import default_timer
from projecteuler import partition_fn
from projecteuler import partition_fn, timing
def main():
start = default_timer()
@timing
def p076():
partitions = [0] * 101
# The number of ways a number can be written as a sum is given by the partition function
# (-1 because the partition function includes also the number itself).
# The function is implemented in projecteuler.py.
# The number of ways a number can be written as a sum is given by the partition function
# (-1 because the partition function includes also the number itself).
# The function is implemented in projecteuler.py.
n = partition_fn(100, partitions) - 1
end = default_timer()
print('Project Euler, Problem 76')
print(f'Answer: {n}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p076()

26
Python/p077.py
View File

@ -10,16 +10,13 @@
#
# What is the first value which can be written as the sum of primes in over five thousand different ways?
from timeit import default_timer
from projecteuler import is_prime
from projecteuler import is_prime, timing
primes = [0] * 100
# Function using a simple recursive brute force approach
# to find all the partitions.
# Function using a simple recursive brute force approach to find all the partitions.
def count(value, n, i, target):
for j in range(i, 100):
value = value + primes[j]
@ -36,10 +33,9 @@ def count(value, n, i, target):
return n
def main():
start = default_timer()
# Generate a list of the first 100 primes.
@timing
def p077():
# Generate a list of the first 100 primes.
i = 0
j = 0
@ -51,9 +47,9 @@ def main():
i = 2
# Use a function to count the number of prime partitions for
# each number >= 2 until the one that can be written in over
# 5000 ways is found.
# Use a function to count the number of prime partitions for
# each number >= 2 until the one that can be written in over
# 5000 ways is found.
while True:
n = count(0, 0, 0, i)
@ -62,13 +58,9 @@ def main():
i = i + 1
end = default_timer()
print('Project Euler, Problem 77')
print(f'Answer: {i}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p077()

18
Python/p078.py
View File

@ -13,32 +13,24 @@
#
# Find the least value of n for which p(n) is divisible by one million.
from timeit import default_timer
from projecteuler import partition_fn
from projecteuler import partition_fn, timing
def main():
start = default_timer()
@timing
def p078():
N = 1000000
partitions = [0] * N
i = 0
# Using the partition function to calculate the number of partitions,
# giving the result modulo N.*/
# Using the partition function to calculate the number of partitions, giving the result modulo N.
while partition_fn(i, partitions, N) != 0:
i = i + 1
end = default_timer()
print('Project Euler, Problem 78')
print(f'Answer: {i}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p078()

35
Python/p079.py
View File

@ -10,13 +10,13 @@
import sys
from itertools import permutations
from timeit import default_timer
from projecteuler import timing
def check_passcode(passcode, len_, logins, n):
# For every login attempt, check if all the digits appear in the
# passcode in the correct order. Return 0 if a login attempt
# incompatible with the current passcode is found.
# For every login attempt, check if all the digits appear in the passcode in the correct order. Return 0 if a login attempt
# incompatible with the current passcode is found.
for i in range(n):
k = 0
for j in range(len_):
@ -32,9 +32,8 @@ def check_passcode(passcode, len_, logins, n):
return 1
def main():
start = default_timer()
@timing
def p079():
try:
with open('p079_keylog.txt', 'r', encoding='utf-8') as fp:
logins = fp.readlines()
@ -48,7 +47,7 @@ def main():
for i in logins:
keylog = int(i)
# Check which digits are present in the login attempts.
# Check which digits are present in the login attempts.
while True:
digits[keylog % 10] = digits[keylog % 10] + 1
keylog = keylog // 10
@ -58,8 +57,7 @@ def main():
j = 0
for i in range(10):
# To generate the passcode, only use the digits present in the
# login attempts.
# To generate the passcode, only use the digits present in the login attempts.
if digits[i] > 0:
passcode_digits[j] = i
j = j + 1
@ -70,18 +68,17 @@ def main():
while not found:
passcode = [0] * len_
# For the current length, generate the first passcode with the
# digits in order.
# For the current length, generate the first passcode with the digits in order.
for i in range(len_):
passcode[i] = passcode_digits[i]
# Check if the passcode is compatible with the login attempts.
# Check if the passcode is compatible with the login attempts.
if check_passcode(passcode, len_, logins, 50):
found = 1
break
# For the given length, check every permutation until the correct
# passcode has been found, or all the permutations have been tried.
# For the given length, check every permutation until the correct
# passcode has been found, or all the permutations have been tried.
passcodes = permutations(passcode, len_)
for i in passcodes:
@ -90,16 +87,12 @@ def main():
res = ''.join(map(str, i))
break
# If the passcode has not yet been found, try a longer passcode.
# If the passcode has not yet been found, try a longer passcode.
len_ = len_ + 1
end = default_timer()
print('Project Euler, Problem 79')
print(f'Answer: {res}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p079()

19
Python/p080.py
View File

@ -8,9 +8,10 @@
# For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits
# for all the irrational square roots
from timeit import default_timer
from mpmath import sqrt, mp
from projecteuler import timing
def is_square(n):
p = sqrt(n)
@ -19,18 +20,16 @@ def is_square(n):
return bool(p == m)
def main():
start = default_timer()
# Set the precision to 100 digits
@timing
def p080():
# Set the precision to 100 digits
mp.dps = 102
sum_ = 0
for i in range(2, 100):
if not is_square(i):
# Calculate the square root of the current number with the given precision
# and sum the digits to the total.
# Calculate the square root of the current number with the given precision and sum the digits to the total.
root = str(sqrt(i))
sum_ = sum_ + int(root[0])
@ -38,13 +37,9 @@ def main():
for j in range(2, 101):
sum_ = sum_ + int(root[j])
end = default_timer()
print('Project Euler, Problem 80')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')
if __name__ == '__main__':
main()
p080()