Numerical Optimization 拟牛顿BFGS+精确线搜索 源码实现 python

BFGS求解凸二次规划

$$\begin{gathered} f(x)=\frac{1}{2}{x}^{T}Ax+b^{T}x \\ g=Ax+b \end{gathered}$$

def f(x):
    return 0.5*np.dot(np.dot(x,A),x)+np.dot(b,x)+c
def  g(x):
    return np.dot(A,x)+b

采用精确线搜索求步长：

alpha_k = -np.dot(pk.T, g(xk))/np.dot(pk.T,np.dot(A,pk))

样正HK 采用如下公式

from scipy.optimize import minimize
from numpy import *
import numpy
import numpy as np

def minimize_bfgs(f, x0, jac=None,gtol=1e-5,disp=False):
    # 第一步，给出初始点x0, Ho, 计算go
    gfk = g(x0)
    k = 0
    N = len(x0)
    I = numpy.eye(N, dtype=int)
    Hk = I
    # 初始化Hessian 矩阵
    xk = x0
    gnorm = numpy.amax(numpy.abs(gfk))
    while (gnorm > gtol):
        # 第二步 ： 求下降方向
        pk = -numpy.dot(Hk, gfk)
        # 第三步： 这里使用了精确线搜索确定步长，并更新xk+1
        alpha_k = -np.dot(pk.T, g(xk))/np.dot(pk.T,np.dot(A,pk))
        xkp1 = xk + alpha_k * pk
        # 第四步：  校正Hk
        sk = xkp1 - xk
        xk = xkp1
        gfkp1 = g(xkp1)
        yk = gfkp1 - gfk
        gfk = gfkp1
        k += 1
        gnorm = numpy.amax(numpy.abs(gfk))

        if (gnorm <= gtol):
            break
        rhok = 1.0 / (numpy.dot(yk, sk))
        # A1 = I - np.dot(sk[:,None],yk[:,None].T)* rhok
        A1= I - sk[:, None] * yk[None, :] * rhok
        A2 = I - yk[:, None] * sk[None, :] * rhok

        Hk = np.dot(A1, np.dot(Hk, A2)) + (rhok * np.dot(sk[:,None] , sk[:,None].T))

        # Hk = numpy.dot(A1, numpy.dot(Hk, A2)) + (rhok * sk[:, None] *
        #                                          sk[None, :])
        # numpy.newaxis是None
        # 都是二维np.dot 相当于矩阵乘



    print("         Current function value xk:[%f %f]"% (xk[0],xk[1]))
    print("         Current function value: %f" % f(xk))

测试如下

if __name__ == '__main__':
    x0 = [1, 1]
    A = np.array([[1, -1], [-1, 2]])
    b = [-2, 0]
    g0 = g(x0)
    c = 0
    print(minimize_bfgs(f, x0, jac=g,gtol=1e-6, disp= True))
    res = minimize(f, x0, method='BFGS', jac=g, options={'gtol': 1e-6, 'disp': True})
    print(res)

超线性收敛，次于牛顿方法

    # sk=np.array([-0.00835751 ,-0.00901759 ,-0.02819486, -0.1701776 ])
    # yk=np.array([ -5.30212743   ,3.92364059 , 46.13550229 ,-11.46023375])
    # (fun, x0, args=(), jac = None, gtol = 1e-5, disp = False):
    # res = minimize(rosen, x0, method='BFGS', jac=rosen_der, options={'gtol': 1e-6, 'disp': True})
    # res = minimize(f, x0, method='BFGS', jac=g, options={'gtol': 1e-6, 'disp': True})
    # print(res)
    # I = numpy.eye(4, dtype=int)
    # A1 = I - sk[:, None] * yk[None, :]
    # 不能真接用sk*yk.T,先得将数组变为二维的才行，所以等价于
    # a=sk[:,None]
    # b=yk[:,None]
    # print(a)
    # print(I-np.dot(a,b.T)==A1)
    # *用于array 表示对应元素相乘，np.dot()点乘