本文实例讲述了python中黄金分割法实现方法。分享给大家供大家参考。具体实现方法如下：
''' a,b = bracket(f,xstart,h) finds the brackets (a,b) of a minimum point of the user-supplied scalar function f(x). the search starts downhill from xstart with a step length h. x,fmin = search(f,a,b,tol=1.0e-6) golden section method for determining x that minimizes the user-supplied scalar function f(x). the minimum must be bracketed in (a,b).''' from math import log, ceildef bracket(f,x1,h): c = 1.618033989 f1 = f(x1) x2 = x1 + h; f2 = f(x2) # determine downhill direction and change sign of h if needed if f2 > f1: h = -h x2 = x1 + h; f2 = f(x2) # check if minimum between x1 - h and x1 + h if f2 > f1: return x2,x1 - h # search loop for i in range (100): h = c*h x3 = x2 + h; f3 = f(x3) if f3 > f2: return x1,x3 x1 = x2; x2 = x3 f1 = f2; f2 = f3 print bracket did not find a mimimum def search(f,a,b,tol=1.0e-9): niter = int(ceil(-2.078087*log(tol/abs(b-a)))) # eq. (10.4) r = 0.618033989 c = 1.0 - r # first telescoping x1 = r*a + c*b; x2 = c*a + r*b f1 = f(x1); f2 = f(x2) # main loop for i in range(niter): if f1 > f2: a = x1 x1 = x2; f1 = f2 x2 = c*a + r*b; f2 = f(x2) else: b = x2 x2 = x1; f2 = f1 x1 = r*a + c*b; f1 = f(x1) if f1 < f2: return x1,f1 else: return x2,f2
希望本文所述对大家的python程序设计有所帮助。