fuxino/project-euler-solutions
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Fix bugs

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This commit is contained in:
daniele 2023-06-06 22:37:04 +02:00
parent 26b940a2ac
commit b0b9705303
Signed by: fuxino
GPG Key ID: 981A2B2A3BBF5514
3 changed files with 47 additions and 25 deletions
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2
Python/p001.py
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@ -21,7 +21,7 @@ def main():
end = default_timer()
print('Project Euler, Problem 1')
print(f'Answer: {sum}')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')

2
Python/p002.py
View File

@ -33,7 +33,7 @@ def main():
end = default_timer()
print('Project Euler, Problem 2')
print(f'Answer: {sum}')
print(f'Answer: {sum_}')
print(f'Elapsed time: {end - start:.9f} seconds')

68
Python/projecteuler.py
View File

@ -4,6 +4,7 @@ from math import sqrt, floor, ceil, gcd
from numpy import zeros
def is_prime(num):
if num < 4:
# If num is 2 or 3 then it's prime.
@ -28,6 +29,7 @@ def is_prime(num):
# If no factor is found up to the square root of num, num is prime.
return True
def is_palindrome(num, base):
reverse = 0
@ -48,10 +50,13 @@ def is_palindrome(num, base):
return True
return False
# Least common multiple algorithm using the greatest common divisor.
def lcm(a, b):
return a * b // gcd(a, b)
# Recursive function to calculate the least common multiple of more than 2 numbers.
def lcmm(values, n):
# If there are only two numbers, use the lcm function to calculate the lcm.
@ -63,6 +68,7 @@ def lcmm(values, n):
# Recursively calculate lcm(a, b, c, ..., n) = lcm(a, lcm(b, c, ..., n)).
return lcm(value, lcmm(values[1:], n-1))
# Function implementing the Sieve or Eratosthenes to generate
# primes up to a certain number.
def sieve(n):
@ -75,7 +81,7 @@ def sieve(n):
# primes[3] = 1
# Cross out (set to 0) all even numbers and set the odd numbers to 1 (possible prime).
for i in range(4, n -1, 2):
for i in range(4, n - 1, 2):
primes[i] = 0
# primes[i+1] = 1
@ -96,6 +102,7 @@ def sieve(n):
return primes
def count_divisors(n):
count = 0
# For every divisor below the square root of n, there is a corresponding one
@ -113,6 +120,7 @@ def count_divisors(n):
return count
def find_max_path(triang, n):
# Start from the second to last row and go up.
for i in range(n-2, -1, -1):
@ -127,6 +135,7 @@ def find_max_path(triang, n):
return triang[0][0]
def sum_of_divisors(n):
# For each divisor of n smaller than the square root of n,
# there is another one larger than the square root. If i is
@ -147,6 +156,7 @@ def sum_of_divisors(n):
return sum_
def is_pandigital(value, n):
i = 0
digits = [0] * (n + 1)
@ -172,6 +182,7 @@ def is_pandigital(value, n):
return True
def is_pentagonal(n):
# A number n is pentagonal if p=(sqrt(24n+1)+1)/6 is an integer.
# In this case, n is the pth pentagonal number.
@ -179,6 +190,7 @@ def is_pentagonal(n):
return i.is_integer()
# Function implementing the iterative algorithm taken from Wikipedia
# to find the continued fraction for sqrt(S). The algorithm is as
# follows:
@ -205,7 +217,7 @@ def build_sqrt_cont_fraction(i, l):
while True:
mn1 = dn * an - mn
dn1 = (i - mn1 * mn1)/ dn
dn1 = (i - mn1 * mn1) / dn
an1 = floor((a0+mn1)/dn1)
mn = mn1
dn = dn1
@ -221,12 +233,12 @@ def build_sqrt_cont_fraction(i, l):
return fraction, count
# Function to solve the Diophantine equation in the form x^2-Dy^2=1
# (Pell equation) using continued fractions.
def pell_eq(d):
# Find the continued fraction for sqrt(d).
fraction, period = build_sqrt_cont_fraction(d, 100)
fraction, _ = build_sqrt_cont_fraction(d, 100)
# Calculate the first convergent of the continued fraction.
n1 = 0
@ -265,6 +277,7 @@ def pell_eq(d):
if fraction[j] == -1:
j = 1
# Function to check if a number is semiprime. Parameters include
# pointers to p and q to return the factors values and a list of
# primes.
@ -278,25 +291,28 @@ def is_semiprime(n, primes):
if primes[n//2] == 1:
p = 2
q = n // 2
return True, p, q
else:
return False, -1, -1
return False, -1, -1
# Check if n is semiprime and one of the factors is 3.
elif n % 3 == 0:
if primes[n//3] == 1:
p = 3
q = n // 3
return True, p, q
else:
return False, -1, -1
return False, -1, -1
# Any number can have only one prime factor greater than its
# square root, so we can stop checking at this point.
limit = floor(sqrt(n)) + 1
# Every prime other than 2 and 3 is in the form 6k+1 or 6k-1.
# If I check all those value no prime factors of the number
# will be missed. For each of these possible primes, check if
# If I check all those value no prime factors of the number
# will be missed. For each of these possible primes, check if
# they are prime, then if the number is semiprime with using
# that factor.
for i in range(5, limit, 6):
@ -304,26 +320,31 @@ def is_semiprime(n, primes):
if primes[n//i] == 1:
p = i
q = n // i
return True, p, q
else:
return False, -1, -1
elif primes[i+2] == 1 and n % (i + 2) == 0:
return False, -1, -1
if primes[i+2] == 1 and n % (i + 2) == 0:
if primes[n//(i+2)] == 1:
p = i + 2
q = n // (i + 2)
return True, p, q
else:
return False, -1, -1
return False, -1, -1
return False, -1, -1
# If n=pq is semiprime, phi(n)=(p-1)(q-1)=pq-p-q+1=n-(p+4)+1
# if p!=q. If p=q (n is a square), phi(n)=n-p.
def phi_semiprime(n, p, q):
if p == q:
return n - p
else:
return n - (p + q) + 1
return n - (p + q) + 1
def phi(n, primes):
# If n is primes, phi(n)=n-1.
@ -366,8 +387,8 @@ def phi(n, primes):
limit = floor(sqrt(n)) + 1
# Every prime other than 2 and 3 is in the form 6k+1 or 6k-1.
# If I check all those value no prime factors of the number
# will be missed. For each of these possible primes, check if
# If I check all those value no prime factors of the number
# will be missed. For each of these possible primes, check if
# they are prime, then check if the number divides n, in which
# case update the current ph.
for i in range(5, limit, 6):
@ -398,6 +419,7 @@ def phi(n, primes):
return ph
# Function implementing the partition function.
def partition_fn(n, partitions, mod=-1):
# The partition function for negative numbers is 0 by definition.
@ -427,12 +449,12 @@ def partition_fn(n, partitions, mod=-1):
partitions[n] = res % mod
return res % mod
else:
partitions[n] = int(res)
return int(res)
partitions[n] = int(res)
return int(res)
#
def dijkstra(matrix, distances, m, n, up=False, back=False, start=0):
visited = zeros((m, n), int)