I have figured out the pattern, and implemented it directly in Python. I
wonder if there is a more elegant way to implement this with sympy.
>>> var('w w0 w1 x x0 x1 y y0 y1 k')
>>> solve([Eq(w*w0+x*x0+y*y0, k), Eq(w*w1, x*x1), Eq(w*w1, y*y1)], [w, x,
y])
Result is this.
w = (k*x1*y1) / w0*x1*y1+w1*x0*y1+w1*x1*y0
x = (k*w1*y1) / w0*x1*y1+w1*x0*y1+w1*x1*y0
y = (k*w1*x1) / w0*x1*y1+w1*x0*y1+w1*x1*y0
The pattern is easier to notice when it's expanded, and then rewritten like
so.
a = w1*x1*y1
d = a*(w0/w1+x0/x1+/y0/y1)
w = (k*a/w1) / d
x = (k*a/x1) / d
y = (k*a/y1) / d
In code.
def f(l, k):
import operator
a = reduce(operator.mul, operator.itemgetter(1)(l), 1.0)
d = sum(a * i[0] / i[1] for i in l)
return [(k * a) / (i[1] * d) for i in l]
>>> f([(100, 200), (250, 100), (125, 275), (100, 125)], 500)
[0.5876068376068376, 1.1752136752136753, 0.42735042735042733,
0.94017094017094]
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