我不确定x.d <= delta 是什么意思,但我假设它应该是x <= delta。
您可以使用投影梯度法或加速投影梯度法(这只是对投影梯度法的轻微修改,“神奇地”收敛得更快)来解决这个问题。这是一些显示如何最小化 .5|| 的 python 代码Ax - b ||^2 受限于 0 manuscript 中关于近端算法的文章。
import numpy as np
import matplotlib.pyplot as plt
def fista(gradf,proxg,evalf,evalg,x0,params):
# This code does FISTA with line search
maxIter = params['maxIter']
t = params['stepSize'] # Initial step size
showTrigger = params['showTrigger']
increaseFactor = 1.25
decreaseFactor = .5
costs = np.zeros((maxIter,1))
xkm1 = np.copy(x0)
vkm1 = np.copy(x0)
for k in np.arange(1,maxIter+1,dtype = np.double):
costs[k-1] = evalf(xkm1) + evalg(xkm1)
if k % showTrigger == 0:
print "Iteration: " + str(k) + " cost: " + str(costs[k-1])
t = increaseFactor*t
acceptFlag = False
while acceptFlag == False:
if k == 1:
theta = 1
else:
a = tkm1
b = t*(thetakm1**2)
c = -t*(thetakm1**2)
theta = (-b + np.sqrt(b**2 - 4*a*c))/(2*a)
y = (1 - theta)*xkm1 + theta*vkm1
(gradf_y,fy) = gradf(y)
x = proxg(y - t*gradf_y,t)
fx = evalf(x)
if fx <= fy + np.vdot(gradf_y,x - y) + (.5/t)*np.sum((x - y)**2):
acceptFlag = True
else:
t = decreaseFactor*t
tkm1 = t
thetakm1 = theta
vkm1 = xkm1 + (1/theta)*(x - xkm1)
xkm1 = x
return (xkm1,costs)
if __name__ == '__main__':
delta = 5.0
numRows = 300
numCols = 50
A = np.random.randn(numRows,numCols)
ATrans = np.transpose(A)
xTrue = delta*np.random.rand(numCols,1)
b = np.dot(A,xTrue)
noise = .1*np.random.randn(numRows,1)
b = b + noise
def evalf(x):
AxMinusb = np.dot(A, x) - b
val = .5 * np.sum(AxMinusb ** 2)
return val
def gradf(x):
AxMinusb = np.dot(A, x) - b
grad = np.dot(ATrans, AxMinusb)
val = .5 * np.sum(AxMinusb ** 2)
return (grad, val)
def evalg(x):
return 0.0
def proxg(x,t):
return np.maximum(np.minimum(x,delta),0.0)
x0 = np.zeros((numCols,1))
params = {'maxIter': 500, 'stepSize': 1.0, 'showTrigger': 5}
(x,costs) = fista(gradf,proxg,evalf,evalg,x0,params)
plt.figure()
plt.plot(x)
plt.plot(xTrue)
plt.figure()
plt.semilogy(costs)