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Fix code style

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This commit is contained in:
daniele 2023-06-07 22:06:22 +02:00
parent f827fd5dd6
commit cc90c8aa0c
Signed by: fuxino
GPG Key ID: 981A2B2A3BBF5514
1 changed files with 80 additions and 83 deletions
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Python/projecteuler.py
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@ -9,26 +9,26 @@ from numpy import zeros
def is_prime(num):
if num < 4:
# If num is 2 or 3 then it's prime.
# If num is 2 or 3 then it's prime.
return num == 2 or num == 3
# If num is divisible by 2 or 3 then it's not prime.
# If num is divisible by 2 or 3 then it's not prime.
if num % 2 == 0 or num % 3 == 0:
return False
# Any number can have only one prime factor greater than its
# square root. If we reach the square root and we haven't found
# any smaller prime factors, then the number is prime.
# Any number can have only one prime factor greater than its
# square root. If we reach the square root and we haven't found
# any smaller prime factors, then the number is prime.
limit = floor(sqrt(num)) + 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. If a factor is found, the number is not prime
# and the function returns 0.
# 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. If a factor is found, the number is not prime
# and the function returns 0.
for i in range(5, limit, 6):
if num % i == 0 or num % (i + 2) == 0:
return False
# If no factor is found up to the square root of num, num is prime.
# If no factor is found up to the square root of num, num is prime.
return True
@ -37,17 +37,17 @@ def is_palindrome(num, base):
tmp = num
# Start with reverse=0, get the rightmost digit of the number using
# modulo operation (num modulo base), add it to reverse. Remove the
# rightmost digit from num dividing num by the base, shift the reverse left
# multiplying by the base, repeat until all digits have been inserted
# in reverse order.
# Start with reverse=0, get the rightmost digit of the number using
# modulo operation (num modulo base), add it to reverse. Remove the
# rightmost digit from num dividing num by the base, shift the reverse left
# multiplying by the base, repeat until all digits have been inserted
# in reverse order.
while tmp > 0:
reverse = reverse * base
reverse = reverse + tmp % base
tmp = tmp // base
# If the reversed number is equal to the original one, then it's palindrome.
# If the reversed number is equal to the original one, then it's palindrome.
if num == reverse:
return True
@ -61,41 +61,40 @@ def lcm(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.
# If there are only two numbers, use the lcm function to calculate the lcm.
if n == 2:
return lcm(values[0], values[1])
value = values[0]
# Recursively calculate lcm(a, b, c, ..., n) = lcm(a, lcm(b, c, ..., 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.
# Function implementing the Sieve or Eratosthenes to generate primes up to a certain number.
def sieve(n):
primes = [1] * n
# 0 and 1 are not prime, 2 and 3 are prime.
# 0 and 1 are not prime, 2 and 3 are prime.
primes[0] = 0
primes[1] = 0
# Cross out (set to 0) all even numbers and set the odd numbers to 1 (possible prime).
# Cross out (set to 0) all even numbers and set the odd numbers to 1 (possible prime).
for i in range(4, n, 2):
primes[i] = 0
# If i is prime, all multiples of i smaller than i*i have already been crossed out.
# if i=sqrt(n), all multiples of i up to n (the target) have been crossed out. So
# there is no need check i>sqrt(n).
# If i is prime, all multiples of i smaller than i*i have already been crossed out.
# if i=sqrt(n), all multiples of i up to n (the target) have been crossed out. So
# there is no need check i>sqrt(n).
limit = floor(sqrt(n))
for i in range(3, limit, 2):
# Find the next number not crossed out, which is prime.
# Find the next number not crossed out, which is prime.
if primes[i] == 1:
# Cross out all multiples of i, starting with i*i because any smaller multiple
# of i has a smaller prime factor and has already been crossed out. Also, since
# i is odd, i*i+i is even and has already been crossed out, so multiples are
# crossed out with steps of 2*i.
# Cross out all multiples of i, starting with i*i because any smaller multiple
# of i has a smaller prime factor and has already been crossed out. Also, since
# i is odd, i*i+i is even and has already been crossed out, so multiples are
# crossed out with steps of 2*i.
for j in range(i * i, n, 2 * i):
primes[j] = 0
@ -104,10 +103,10 @@ def sieve(n):
def count_divisors(n):
count = 0
# For every divisor below the square root of n, there is a corresponding one
# above the square root, so it's sufficient to check up to the square root of n
# and count every divisor twice. If n is a perfect square, the last divisor is
# wrongly counted twice and must be corrected.
# For every divisor below the square root of n, there is a corresponding one
# above the square root, so it's sufficient to check up to the square root of n
# and count every divisor twice. If n is a perfect square, the last divisor is
# wrongly counted twice and must be corrected.
limit = floor(sqrt(n))
for i in range(1, limit):
@ -121,11 +120,11 @@ def count_divisors(n):
def find_max_path(triang, n):
# Start from the second to last row and go up.
# Start from the second to last row and go up.
for i in range(n-2, -1, -1):
# For each element in the row, check the two adjacent elements
# in the row below and sum the larger one to it. At the end,
# the element at the top will contain the value of the maximum path.
# For each element in the row, check the two adjacent elements
# in the row below and sum the larger one to it. At the end,
# the element at the top will contain the value of the maximum path.
for j in range(0, i+1):
if triang[i+1][j] > triang[i+1][j+1]:
triang[i][j] = triang[i][j] + triang[i+1][j]
@ -136,11 +135,11 @@ def find_max_path(triang, n):
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
# a divisor of n, so is n/i. Checking divisors i up to square
# root of n and adding both i and n/i is sufficient to sum
# all divisors.
# For each divisor of n smaller than the square root of n,
# there is another one larger than the square root. If i is
# a divisor of n, so is n/i. Checking divisors i up to square
# root of n and adding both i and n/i is sufficient to sum
# all divisors.
limit = floor(sqrt(n)) + 1
sum_ = 1
@ -148,8 +147,7 @@ def sum_of_divisors(n):
for i in range(2, limit):
if n % i == 0:
sum_ = sum_ + i
# If n is a perfect square, i=limit is a divisor and
# has to be counted only once.
# If n is a perfect square, i=limit is a divisor and has to be counted only once.
if n != i * i:
sum_ = sum_ + n // i
@ -183,8 +181,8 @@ def is_pandigital(value, n):
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.
# A number n is pentagonal if p=(sqrt(24n+1)+1)/6 is an integer.
# In this case, n is the pth pentagonal number.
i = (sqrt(24*n+1) + 1) / 6
return i.is_integer()
@ -236,10 +234,10 @@ def build_sqrt_cont_fraction(i, l):
# 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).
# Find the continued fraction for sqrt(d).
fraction, _ = build_sqrt_cont_fraction(d, 100)
# Calculate the first convergent of the continued fraction.
# Calculate the first convergent of the continued fraction.
n1 = 0
n2 = 1
d1 = 1
@ -250,14 +248,14 @@ def pell_eq(d):
d3 = fraction[j] * d2 + d1
j = j + 1
# Check if x=n, y=d solve the equation x^2-Dy^2=1.
# Check if x=n, y=d solve the equation x^2-Dy^2=1.
sol = n3 * n3 - d * d3 * d3
if sol == 1:
return n3
# Until a solution is found, calculate the next convergent
# and check if x=n and y=d solve the equation.
# Until a solution is found, calculate the next convergent
# and check if x=n and y=d solve the equation.
while True:
n1 = n2
n2 = n3
@ -281,11 +279,11 @@ def pell_eq(d):
# pointers to p and q to return the factors values and a list of
# primes.
def is_semiprime(n, primes):
# If n is prime, it's not semiprime.
# If n is prime, it's not semiprime.
if primes[n] == 1:
return False, -1, -1
# Check if n is semiprime and one of the factors is 2.
# Check if n is semiprime and one of the factors is 2.
if n % 2 == 0:
if primes[n//2] == 1:
p = 2
@ -295,7 +293,7 @@ def is_semiprime(n, primes):
return False, -1, -1
# Check if n is semiprime and one of the factors is 3.
# Check if n is semiprime and one of the factors is 3.
elif n % 3 == 0:
if primes[n//3] == 1:
p = 3
@ -305,15 +303,15 @@ def is_semiprime(n, primes):
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.
# 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
# they are prime, then if the number is semiprime with using
# that factor.
# 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
# they are prime, then if the number is semiprime with using
# that factor.
for i in range(5, limit, 6):
if primes[i] == 1 and n % i == 0:
if primes[n//i] == 1:
@ -346,11 +344,11 @@ def phi_semiprime(n, p, q):
def phi(n, primes):
# If n is primes, phi(n)=n-1.
# If n is primes, phi(n)=n-1.
if primes[n] == 1:
return n - 1
# If n is semiprime, use above function.
# If n is semiprime, use above function.
semi_p, p, q = is_semiprime(n, primes)
if semi_p:
@ -358,8 +356,8 @@ def phi(n, primes):
ph = n
# If 2 is a factor of n, multiply the current ph (which now is n)
# by 1-1/2, then divide all factors 2.
# If 2 is a factor of n, multiply the current ph (which now is n)
# by 1-1/2, then divide all factors 2.
if n % 2 == 0:
ph = ph * (1 - 1 / 2)
@ -369,8 +367,7 @@ def phi(n, primes):
if n % 2 != 0:
break
# If 3 is a factor of n, multiply the current ph by 1-1/3,
# then divide all factors 3.
# If 3 is a factor of n, multiply the current ph by 1-1/3, then divide all factors 3.
if n % 3 == 0:
ph = ph * (1 - 1 / 3)
@ -380,16 +377,16 @@ def phi(n, primes):
if n % 3 != 0:
break
# Any number can have only one prime factor greater than its
# square root, so we can stop checking at this point and deal
# with the only factor larger than sqrt(n), if present, at the end
# Any number can have only one prime factor greater than its
# square root, so we can stop checking at this point and deal
# with the only factor larger than sqrt(n), if present, at the end
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
# they are prime, then check if the number divides n, in which
# case update the current ph.
# 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
# they are prime, then check if the number divides n, in which
# case update the current ph.
for i in range(5, limit, 6):
if primes[i]:
if n % i == 0:
@ -410,9 +407,9 @@ def phi(n, primes):
if n % (i + 2) != 0:
break
# After dividing all prime factors smaller than sqrt(n), n is either 1
# or is equal to the only prime factor greater than sqrt(n). In this
# second case, we need to update ph with the last prime factor.
# After dividing all prime factors smaller than sqrt(n), n is either 1
# or is equal to the only prime factor greater than sqrt(n). In this
# second case, we need to update ph with the last prime factor.
if n > 1:
ph = ph * (1 - 1 / n)
@ -421,16 +418,16 @@ def phi(n, primes):
# Function implementing the partition function.
def partition_fn(n, partitions, mod=-1):
# The partition function for negative numbers is 0 by definition.
# The partition function for negative numbers is 0 by definition.
if n < 0:
return 0
# The partition function for zero is 1 by definition.
# The partition function for zero is 1 by definition.
if n == 0:
partitions[n] = 1
return 1
# If the partition for the current n has already been calculated, return the value.
# If the partition for the current n has already been calculated, return the value.
if partitions[n] != 0:
return partitions[n]
@ -443,7 +440,7 @@ def partition_fn(n, partitions, mod=-1):
res = res + pow(-1, k+1) * partition_fn(n-k*(3*k-1)//2, partitions)
k = k + 1
# Give the result modulo mod, if mod!=-1, otherwise give the full result.
# Give the result modulo mod, if mod!=-1, otherwise give the full result.
if mod != -1:
partitions[n] = res % mod