前一段時間學習了無約束最優化方法，今天用Matlab實現了求解無約束最優化問題的FR共軛梯度法。關于共軛梯度法的理論介紹，請參考我的另一篇文章無約束最優化方法學習筆記。

文件testConjungateGradient.m用于測試共軛梯度法函數。測試文件需要定義函數f和自變量x，給定迭代初值x0和允許誤差?。函數設置了show_detail變量用于控制是否顯示每一步的迭代信息。

% test conjungate gradient method

% by TomHeaven, hanlin_tan@nudt.edu.cn, 2015.08.25

%% define function and variable

syms x1 x2;

%f = xs^2+2*ys^2-2*xs*ys + 2*ys + 2;

f = (x1-1)^4 + (x1 - x2)^2;

%f = (1-x1)^2 + 2*(x2 - x1^2)^2;

x = {x1, x2};

% initial value

x0 = [0 0];

% tolerance

epsilon = 1e-1;

%% call conjungate gradient method

show_detail = true;

[bestf, bestx, count] = conjungate_gradient(f, x, x0, epsilon, show_detail);

% print result

fprintf('bestx = %s, bestf = %f, count = %d\n', num2str(bestx), bestf, count);

文件conjungate_gradient.m是共軛梯度法的實現函數。變量nf表示函數f的梯度?f(梯度的希臘字母是nabla，故用nf)。

function [fv, bestx, iter_num] = conjungate_gradient(f, x, x0, epsilon, show_detail)

%% conjungate gradient method

% by TomHeaven, hanlin_tan@nudt.edu.cn, 2015.08.25

% Input:

% f - syms function

% x - row cell arrow for input syms variables

% $x_0$ - init point

% epsilon - tolerance

% show_detail - a boolean value for wether to print details

% Output:

% fv - minimum f value

% bestx - mimimum point

% iter_num - iteration count

%% init

syms lambdas % suffix s indicates this is a symbol variable

% n is the dimension

n = length(x);

% compute differential of function f stored in cell nf

nf = cell(1, n); % using row cells, column cells will result in error

for i = 1 : n

nf{i} = diff(f, x{i});

end

% $\nabla f(x_0)$

nfv = subs(nf, x, x0);

% init $\nabla f(x_k)$

nfv_pre = nfv;

% init count, k and xv for x value.

count = 0;

k = 0;

xv = x0;

% initial search direction

d = - nfv;

% show initial info

if show_detail

fprintf('Initial:\n');

fprintf('f = %s, x0 = %s, epsilon = %f\n\n', char(f), num2str(x0), epsilon);

end

%% loop

while (norm(nfv) > epsilon)

%% one-dimensional search

% define $x_{k+1} = x_{k} + \lambda d$

xv = xv+lambdas*d;

% define $\phi$ and do 1-dim search

phi = subs(f, x, xv);

nphi = diff(phi); % $\nabla \phi$

lambda = solve(nphi);

% get rid of complex and minus solution

lambda = double(lambda);

if length(lambda) > 1

lambda = lambda(abs(imag(lambda)) < 1e-5);

lambda = lambda(lambda > 0);

lambda = lambda(1);

end

% if $\lambda$ is too small, stop iteration

if lambda < 1e-5

break;

end

%% update

% update $x_{k+1｝ = x_{k} + \lambda d$

xv = subs(xv, lambdas, lambda);

% convert sym to double

xv = double(xv);

% compute the differential

nfv = subs(nf, x, xv);

% update counters

count = count + 1;

k = k + 1;

% compute alpha based on FR formula

alpha = sumsqr(nfv) / sumsqr(nfv_pre);

% show iteration info

if show_detail

fprintf('Iteration: %d\n', count);

fprintf('x(%d) = %s, lambda = %f\n', count, num2str(xv), lambda);

fprintf('nf(x) = %s, norm(nf) = %f\n', num2str(double(nfv)), norm(double(nfv)));

fprintf('d = %s, alpha = %f\n', num2str(double(d)), double(alpha));

fprintf('\n');

end

% update conjungate direction

d = -nfv + alpha * d;

% save the previous $$\nabla f(x_k)$$

nfv_pre = nfv;

% reset the conjungate direction and k if k >= n

if k >= n

k = 0;

d = - nfv;

end

end % while

%% output

fv = double(subs(f, x, xv));

bestx = double(xv);

iter_num = count;

end

運行testConjungateGradient后輸出結果如下：

>> testConjungateGradient

Initial:

f = (x1 - x2)^2 + (x1 - 1)^4, x0 = 0 0, epsilon = 0.100000

Iteration: 1

x(1) = 0.41025 0, lambda = 0.102561

nf(x) = 1.08e-16 -0.82049, norm(nf) = 0.820491

d = 4 0, alpha = 0.042075

Iteration: 2

x(2) = 0.52994 0.58355, lambda = 0.711218

nf(x) = -0.52265 0.10721, norm(nf) = 0.533528

d = 0.1683 0.82049, alpha = 0.422831

Iteration: 3

x(3) = 0.63914 0.56115, lambda = 0.208923

nf(x) = -0.031994 -0.15597, norm(nf) = 0.159223

d = 0.52265 -0.10721, alpha = 0.089062

Iteration: 4

x(4) = 0.76439 0.79465, lambda = 1.594673

nf(x) = -0.11285 0.060533, norm(nf) = 0.128062

d = 0.078542 0.14643, alpha = 0.646892

Iteration: 5

x(5) = 0.79174 0.77998, lambda = 0.242379

nf(x) = -0.012614 -0.023517, norm(nf) = 0.026686

d = 0.11285 -0.060533, alpha = 0.043425

bestx = 0.79174 0.77998, bestf = 0.002019, count = 5

修改允許誤差為

epsilon=1e-8;

則可以得到更加精確的結果：

Iteration: 6

x(6) = 0.9026 0.9122, lambda = 6.329707

nf(x) = -0.022884 0.019188, norm(nf) = 0.029864

d = 0.017515 0.020888, alpha = 1.252319

Iteration: 7

x(7) = 0.90828 0.90744, lambda = 0.247992

nf(x) = -0.0014077 -0.0016788, norm(nf) = 0.002191

d = 0.022884 -0.019188, alpha = 0.005382

Iteration: 8

x(8) = 0.97476 0.97586, lambda = 43.429293

nf(x) = -0.0022668 0.0022025, norm(nf) = 0.003161

d = 0.0015309 0.0015756, alpha = 2.080989

Iteration: 9

x(9) = 0.97533 0.97531, lambda = 0.249812

nf(x) = -2.9597e-05 -3.0461e-05, norm(nf) = 0.000042

d = 0.0022668 -0.0022025, alpha = 0.000181

Iteration: 10

x(10) = 0.99709 0.99712, lambda = 725.188481

nf(x) = -5.2106e-05 5.2008e-05, norm(nf) = 0.000074

d = 3.0006e-05 3.0063e-05, alpha = 3.004594

Iteration: 11

x(11) = 0.9971 0.9971, lambda = 0.249997

nf(x) = -4.8571e-08 -4.8663e-08, norm(nf) = 0.000000

d = 5.2106e-05 -5.2008e-05, alpha = 0.000001

Iteration: 12

x(12) = 0.99992 0.99992, lambda = 57856.826721

nf(x) = -9.3751e-08 9.3748e-08, norm(nf) = 0.000000

d = 4.8616e-08 4.8617e-08, alpha = 3.718503

Iteration: 13

x(13) = 0.99992 0.99992, lambda = 0.250000

nf(x) = -1.1858e-12 -1.1855e-12, norm(nf) = 0.000000

d = 9.3751e-08 -9.3748e-08, alpha = 0.000000

bestx = 0.99992 0.99992, bestf = 0.000000, count = 13

這與問題的最優解(1,1)T已經非常接近了。

算法實現沒有經過大量測試，實際使用可能會有BUG。這里只是用于說明基本實現原理，有興趣的讀者可以在此基礎上改進。

總結

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