수학/수치해석
[수치해석] Golden Section Method 파이썬
ntne
2022. 12. 9. 23:24
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import math
def golden(xlow, xhigh, maxit, es):
iter = 1
r = (5**0.5 -1)/2
xl = xlow
xu = xhigh
d = r * (xu - xl)
x1 = xl + d
x2 = xu - d
f1 = f(x1)
f2 = f(x2)
print(" i xl fxl x2 f(x2) x1 f(x1) xu f(xu) d ")
print("%2d %.4f %.4f %.4f %.4f %.4f %.4f %.4f %.4f %.4f" % (iter, xl, f(xl), x2, f2, x1, f1, xu, f(xu), d))
# Suppose We're finding maximum
if f1 > f2:
xopt = x1
fx = f1
else:
xopt = x2
fx = f2
while True:
d = r*d
xint = xu - xl
# min 구하면 부등호 < 로 바꾸기
if f1 > f2:
xl = x2
x2 = x1
x1 = xl + d
f2 = f1
f1 = f(x1)
else:
xu = x1
x1 = x2
x2 = xu - d
f1 = f2
f2 = f(x2)
iter += 1
# min 구하면 부등호 < 로 바꾸기
if f1 > f2:
xopt = x1
ea = (1-r)*abs((xu-xl)/xopt)*0.01
print("%2d %.4f %.4f %.4f %.4f %.4f %.4f %.4f %.4f %.4f" % (iter, xl, f(xl), x2, f2, x1, f1, xu, f(xu), d))
if iter >= maxit or ea <= es:
print("%.4f is optimal" % xopt)
break
def f(x):
# Min = 3:
# return (x**2) - (6*x) + 15
# Max = 1.4274
return 2*math.sin(x) - (x**2)/10
golden(0,4,1000,0.00001)
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