euler/python/e075.py

178 lines
4.7 KiB
Python

from math import sqrt
from lib_misc import gcd
def proper_divisors(n):
"""
Returns the list of divisors for n excluding n.
"""
if n < 2:
return []
divisors = [1, ]
d = 2
while d * d <= n:
if n % d == 0:
divisors.append(d)
d += 1
return divisors
def triples_brute_force(b_max):
squares = {n * n: n for n in range(1, b_max)}
triples = []
for b in range(3, b_max):
for a in range(2, b):
c_squared = a * a + b * b
if c_squared in squares:
triples.append((a, b, squares[c_squared]))
return triples
def dicksons_method(r):
triples = []
assert(r % 2 == 0)
st = r * r // 2
for d in proper_divisors(st):
s = d
t = st // s
a = r + s
b = r + t
c = r + s + t
L = a + b + c
triples.append(L)
return sorted(triples)
def dickson_method_min_max_perimeter(r):
""" I used this method to check that the min and max
perimeter for increasing r are also increasing strictly
monotonic. """
assert(r % 2 == 0)
st = r * r // 2
L_min = 10**8
L_max = 0
for d in proper_divisors(st):
s = d
t = st // s
L = r + s + r + t + r + s + t
if L > L_max:
L_max = L
if L < L_min:
L_min = L
return (L_min, L_max)
def dicksons_method_efficient(r):
L_max = 1500000
perimeters = []
assert(r % 2 == 0)
st = r * r // 2
s_max = int(sqrt(st))
for s in range(s_max, 0, -1):
if st % s == 0:
t = st // s
L = r + s + r + t + r + s + t
if L > L_max:
break
perimeters.append(L)
else:
pass
return perimeters
def euler_075_dicksons():
"""
We are using Dickson's method to get all Pythagorean triples. The
challenge is that this method requires finding the divisors for a
given r. To make it more efficient we stop iterating after the
perimeter exceeds the threshold L_max for any given r.
In addition we found a r_max of 258000 empirically.
This runs for 15 minutes with cpython and in under 5 minutes in pypy.
It would have been better to use a generator function that finds native
solutions and then extrapolate with factors k in range(2, 1500000)
while L < L_max.
https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples#Dickson's_method
XXX: Can be strongly optimized.
"""
r_max = 258000
perimeter_counts = {}
for r in range(2, r_max, 2):
if r % 10000 == 0:
print(r)
for perimeter in dicksons_method_efficient(r):
try:
perimeter_counts[perimeter] += 1
except KeyError:
perimeter_counts[perimeter] = 1
one_integer_sided_right_triangle_count = 0
for perimeter, count in perimeter_counts.items():
if count == 1:
one_integer_sided_right_triangle_count += 1
return one_integer_sided_right_triangle_count
def triples_euclids_formula(m, n):
"""
Generates only primitive triples.
https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
"""
assert(m > n)
assert(n > 0)
assert(gcd(m, n) == 1)
assert(m % 2 == 0 or n % 2 == 0)
a = m * m - n * n
b = 2 * m * n
c = m * m + n * n
if a < b:
return (a, b, c)
else:
return (b, a, c)
def euler_075():
"""
The range for m was chosen empirically (aka by trial and error).
I tried terminating when L > L_max the first time, but there might
be combinations for m = m + 1 that yield lower L. I guess the right
criteria to return would be if L > L_max and n == 1.
This is good enough for now. I still like my other solution, but it
is so much slower because finding the divisors is so expensive.
"""
L = 0
L_max = 1500000
m = 1
perimeter_counts = {}
for m in range(1, 1000):
for n in range(1, m):
try:
L = sum(triples_euclids_formula(m, n))
L_k = L
while L_k <= L_max:
try:
perimeter_counts[L_k] += 1
except KeyError:
perimeter_counts[L_k] = 1
L_k += L
except AssertionError:
pass
one_integer_sided_right_triangle_count = 0
for perimeter, count in perimeter_counts.items():
if count == 1:
one_integer_sided_right_triangle_count += 1
return one_integer_sided_right_triangle_count
if __name__ == "__main__":
print("e075.py: " + str(euler_075()))
assert(euler_075() == 161667)