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functions.py
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functions.py
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import random
def prime_sieve(n):
"""
Returns a prime sieve with the numbers up to n
"""
m = n+1
res = [True]*m
res[0] = False
res[1] = False
for i in range(2, int(n**0.5) + 1):
if res[i]:
for j in range(i*i, m, i):
res[j] = False
return res
# TODO: For 1000 this returns 961 as prime?
def optimised_sieve(n):
"""
Returns a prime sieve with the numbers up to n, only containing the odd
numbers
res[k] corresponds to the number 2*k + 1
e.g: res[3] = 2*3 + 1 = 7 = True (prime)
To check if an ODD integer k is prime, we have to look at index k//2
Be careful because number TWO is not included in the sieve
"""
sieve_bound = n//2
res = [True]*sieve_bound
res[0] = False
crosslimit = int(((n**0.5) - 1)//2)
for i in range(1, crosslimit):
if res[i]:
for j in range(2*i*(i+1), sieve_bound, 2*i + 1):
res[j] = False
return res
def primes_up_to(n):
"""
Returns a list containing all primes up to n
"""
sieve = prime_sieve(n)
res = [2]
for i in range(len(sieve)):
if sieve[i]:
res.append(i)
return res
def pythTriplet(m, n, d = 1):
"""
Returns a Pythagorean triplet obtained from the values m and n.
It will be primitive if and only if exactly one of m, n is even and
gcd(m, n) = 1.
"""
if m < n:
m, n = n, m
return [(m*m - n*n)*d, (2*m*n)*d, (m*m + n*n)*d]
def gcd(a, b):
"""
Returns the greatest common divisor of two numbers.
a is assumed to be greater than b
"""
while b != 0:
a, b = b, a%b
return a
def prime_factorization(n):
"""
Returns a list containing the prime factorization of a given number n
[[prime1, exp1], [prime2, exp2], ... [primeN, expN]]
"""
res = []
if n%2 == 0:
lastFactor = 2
n = n//2
cnt = 1
while n%2 == 0:
n = n//2
cnt += 1
res.append([lastFactor, cnt])
else:
lastFactor = 1
factor = 3
maxFactor = int(n**0.5)
while n > 1 and factor <= maxFactor:
if n%factor == 0:
n = n//factor
lastFactor = factor
cnt = 1
while n%factor == 0:
n = n//factor
cnt += 1
res.append([lastFactor, cnt])
maxFactor = int(n**0.5)
factor += 2
if n != 1:
res.append([n, 1])
return res
def number_of_divisors(n):
"""
Returns the number of divisors of n
"""
prime_fac = prime_factorization(n)
res = 1
for fac in prime_fac:
res *= (fac[1] + 1)
return res
def totient(n):
"""
Returns the Euler's totient function of n, that is, the count of the
positive integers up to n that are relatively prime to it
"""
prime_fac = prime_factorization(n)
res = n
for fac in prime_fac:
res = int(res*(fac[0] - 1)/fac[0])
return res
def ap_sqrt(n, steps):
"""
Approximates the sqrt values of some number following some magic algorithm
Only displays digits
"""
a, b = 5*n, 5
for i in range(steps):
if a >= b:
a, b = a-b, b+10
else:
a, b = 100*a, 10*b - 45
return b
def continued_fraction(x):
"""
Returns the continued fraction of x as a list of tuples (a,b,c),
where:
a + (sqrt(x)+b / c)
represents each step.
If we are interested on the notation of the form [4;(1,3,1,8)],
we must take the first value from each tuple.
Note: Be careful if the value passed is a perfect square.
"""
res = []
a = floor(x**0.5)
b = -floor(x**0.5)
c = 1
t = (a,b,c)
res.append(t)
while True:
a, b, c = res[-1][0], res[-1][1], res[-1][2]
a1 = floor(c*(x**0.5 - b)/(x - b*b))
b1 = -b * c
c1 = x - b*b
b1 -= a1*c1
_gcd = gcd(b1, c1)
_gcd = gcd(_gcd, c)
b1 //= _gcd
c1 //= _gcd
t = (a1, b1, c1)
if t in res: # All continued fractions are periodic
res.append(t)
break
res.append(t)
return res
def solve_pell(n):
"""
This function solves the Pell equation of the form:
a*a - n * b*b = 1
Input
-----
n - Integer
Value of n. >= 2
Output
------
None
If n is a perfect square (it has no solution)
(a, b) - Tuple of integers
Solutions for which the Pell equation is checked
This function has been taken from the p066 solution forum
https://projecteuler.net/thread=66
"""
n1, d1 = 0, 1
n2, d2 = 1, 0
# These are the two bounding fractions
while True:
a = n1 + n2
b = d1 + d2
# a/b is the new candidate somewhere in the middle
t = a*a - n*b*b # See how close a^2/b^2 is to n
if t == 1: # You have your pell solution (a,b)
return (a, b)
elif t == 0: # N was a square = (a/b)^2
return None;
else: # Not there yet - adjust low or hi bound
if t > 0:
n2 =a
d2 =b
else:
n1 =a
d1 =b
def is_probable_prime(n):
"""
Miller-Rabin primality test.
A return value of False means n is certainly not prime. A return value of
True means n is very likely a prime.
"""
_mrpt_num_trials = 5 # number of bases to test
if n < 2:
return False
# special case 2
if n == 2:
return True
# ensure n is odd
if n % 2 == 0:
return False
# write n-1 as 2**s * d
# repeatedly try to divide n-1 by 2
s = 0
d = n-1
while True:
quotient, remainder = divmod(d, 2)
if remainder == 1:
break
s += 1
d = quotient
assert(2**s * d == n-1)
# test the base a to see whether it is a witness for the compositeness of n
def try_composite(a):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2**i * d, n) == n-1:
return False
return True # n is definitely composite
for i in range(_mrpt_num_trials):
a = random.randrange(2, n)
if try_composite(a):
return False
return True # no base tested showed n as composite