这里有一种方法可以枚举三元组,而不是穷举测试它们,使用数论,如此描述:
https://mathoverflow.net/questions/29644/enumerating-ways-to-decompose-an-integer-into-the-sum-of-two-squares
由于这个数学方法花了我一些时间去理解和实现(收集了一些在上面提到的代码),而且我对这个主题并没有太多的权威感,所以我将其留给读者自己研究。这基于将数字表示为高斯整数共轭。
(a + bi)*(a-bi)= a ^ 2 + b ^ 2
。我们首先将数字
z ^ 2-1
分解成质数,将质数分解为高斯共轭,并找到不同的表达式,我们扩展和简化以获得
a + bi
,然后可以升级为
a ^ 2 + b ^ 2
。
阅读
平方和函数的好处是发现我们可以通过使用一个检查来排除任何包含形式为
4k + 3
的质数的奇次幂的候选者
z^2 - 1
。仅仅使用这个检查,我就能够使用下面的Rosetta质因数分解代码将Prune在10^5上的循环从214秒减少到19秒(在repl.it上)。
这里的实现只是一个演示。它没有处理或优化限制
x
和
y
。相反,它只是在枚举过程中进行。在
这里尝试一下。
Python代码:
def mods(a, n):
if n <= 0:
return "negative modulus"
a = a % n
if (2 * a > n):
a -= n
return a
def powmods(a, r, n):
out = 1
while r > 0:
if (r % 2) == 1:
r -= 1
out = mods(out * a, n)
r /= 2
a = mods(a * a, n)
return out
def quos(a, n):
if n <= 0:
return "negative modulus"
return (a - mods(a, n))/n
def grem(w, z):
(w0, w1) = w
(z0, z1) = z
n = z0 * z0 + z1 * z1
if n == 0:
return "division by zero"
u0 = quos(w0 * z0 + w1 * z1, n)
u1 = quos(w1 * z0 - w0 * z1, n)
return(w0 - z0 * u0 + z1 * u1,
w1 - z0 * u1 - z1 * u0)
def ggcd(w, z):
while z != (0,0):
w, z = z, grem(w, z)
return w
def root4(p):
if p <= 1:
return "too small"
if (p % 4) != 1:
return "not congruent to 1"
k = p/4
j = 2
while True:
a = powmods(j, k, p)
b = mods(a * a, p)
if b == -1:
return a
if b != 1:
return "not prime"
j += 1
def sq2(p):
if p % 4 != 1:
return "not congruent to 1 modulo 4"
a = root4(p)
return ggcd((p,0),(a,1))
from math import floor, sqrt
def fac(n):
step = lambda x: 1 + (x<<2) - ((x>>1)<<1)
maxq = long(floor(sqrt(n)))
d = 1
q = n % 2 == 0 and 2 or 3
while q <= maxq and n % q != 0:
q = step(d)
d += 1
return q <= maxq and [q] + fac(n//q) or [n]
from collections import Counter
from itertools import product
from sympy import I, expand, Add
def valid(ps):
for (p, e) in ps.items():
if (p % 4 == 3) and (e & 1):
return False
return True
def get_sq2(p, e):
if p == 2:
if e & 1:
return [2**(e / 2), 2**(e / 2)]
else:
return [2**(e / 2), 0]
elif p % 4 == 3:
return [p, 0]
else:
a,b = sq2(p)
return [abs(a), abs(b)]
def get_terms(cs, e):
if e == 1:
return [Add(cs[0], cs[1] * I)]
res = [Add(cs[0], cs[1] * I)**e]
for t in xrange(1, e / 2 + 1):
res.append(
Add(cs[0] + cs[1]*I)**(e-t) * Add(cs[0] - cs[1]*I)**t)
return res
def get_lists(ps):
items = ps.items()
lists = []
for (p, e) in items:
if p == 2:
a,b = get_sq2(2, e)
lists.append([Add(a, b*I)])
elif p % 4 == 3:
a,b = get_sq2(p, e)
lists.append([Add(a, b*I)**(e / 2)])
else:
lists.append(get_terms(get_sq2(p, e), e))
return lists
def f(n):
for z in xrange(2, n / 2):
zz = (z + 1) * (z - 1)
ps = Counter(fac(zz))
is_valid = valid(ps)
if is_valid:
print "valid (does not contain a prime of form\n4k + 3 with an odd power)"
print "z: %s, primes: %s" % (z, dict(ps))
lists = get_lists(ps)
cartesian = product(*lists)
for element in cartesian:
print "prime square decomposition: %s" % list(element)
p = 1
for item in element:
p *= item
print "complex conjugates: %s" % p
vals = p.expand(complex=True, evaluate=True).as_coefficients_dict().values()
x, y = vals[0], vals[1] if len(vals) > 1 else 0
print "x, y, z: %s, %s, %s" % (x, y, z)
print "x^2 + y^2, z^2-1: %s, %s" % (x**2 + y**2, z**2 - 1)
print ''
if __name__ == "__main__":
print f(100)
输出:
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 3, primes: {2: 3}
prime square decomposition: [2 + 2*I]
complex conjugates: 2 + 2*I
x, y, z: 2, 2, 3
x^2 + y^2, z^2-1: 8, 8
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 9, primes: {2: 4, 5: 1}
prime square decomposition: [4, 2 + I]
complex conjugates: 8 + 4*I
x, y, z: 8, 4, 9
x^2 + y^2, z^2-1: 80, 80
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 17, primes: {2: 5, 3: 2}
prime square decomposition: [4 + 4*I, 3]
complex conjugates: 12 + 12*I
x, y, z: 12, 12, 17
x^2 + y^2, z^2-1: 288, 288
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 19, primes: {2: 3, 3: 2, 5: 1}
prime square decomposition: [2 + 2*I, 3, 2 + I]
complex conjugates: (2 + I)*(6 + 6*I)
x, y, z: 6, 18, 19
x^2 + y^2, z^2-1: 360, 360
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 33, primes: {17: 1, 2: 6}
prime square decomposition: [4 + I, 8]
complex conjugates: 32 + 8*I
x, y, z: 32, 8, 33
x^2 + y^2, z^2-1: 1088, 1088
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 35, primes: {17: 1, 2: 3, 3: 2}
prime square decomposition: [4 + I, 2 + 2*I, 3]
complex conjugates: 3*(2 + 2*I)*(4 + I)
x, y, z: 18, 30, 35
x^2 + y^2, z^2-1: 1224, 1224
j*j
和i*i
,而不是在最内层的k循环中执行。 - Peter Bk
重新计算j*j
和i*i
(这些与k
无关)。 - Peter B