我正在尝试编写一个代码来求解 A*x = b,线性系统。
我使用高斯消去过程编写了下面的代码,如果 A 中没有任何 0,它每次都会起作用。如果 A 中有零,则有时有效,有时无效。基本上我正在尝试 MATLAB 中“A\b”的替代方案。
有更好/更简单的方法吗?
A = randn(5,5);
b = randn(5,1);
nn = size(A);
n = nn(1,1);
U = A;
u = b;
for c = 1:1:n
k = U(:,c);
for r = n:-1:c
if k(r,1) == 0
continue;
else
U(r,:) = U(r,:)/k(r,1);
u(r,1) = u(r,1)/k(r,1);
end
end
for r = n:-1:(c+1)
if k(r,1) == 0
continue;
else
U(r,:) = U(r,:) - U(r-1,:);
u(r,1) = u(r,1) - u(r-1,1);
end
end
end
x = zeros(size(b));
for r = n:-1:1
if r == n
x(r,1) = u(r,1);
else
x(r,1) = u(r,1);
x(r,1) = x(r,1) - U(r,r+1:n)*x(r+1:n,1);
end
end
error = A*x - b;
for i = 1:1:n
if abs(error(i)) > 0.001
disp('ERROR!');
break;
else
continue;
end
end
disp('x:');
disp(x);
使用 0 的工作示例:
A = [1, 3, 1, 3;
3, 4, 4, 1;
3, 0, 3, 9;
0, 4, 0, 1];
b = [3;
4;
5;
6];
失败的示例(A*x-b 不是 [0])
A = [1, 3, 1, 3;
3, 4, 4, 1;
0, 0, 3, 9;
0, 4, 0, 1];
b = [3;
4;
5;
6];
我的算法的解释: 假设我有以下 A 矩阵:
|4, 1, 9|
|3, 4, 5|
|1, 3, 5|
对于第一列,我将每一行除以该行中的第一个数字,因此每行都以 1 开头
|1, 1/4, 9/4|
|1, 4/3, 5/3|
|1, 3, 5|
然后我用最后一行减去它上面的一行,然后我将对上面的行执行相同的操作,依此类推。
|1, 1/4, 9/4|
|0, 4/3-1/4, 5/3-9/4|
|0, 3-4/3, 5-5/3|
|1, 0.25, 2.250|
|0, 1.083, -0.5833|
|0, 1.667, 3.333|
然后我对其余列重复相同的操作。
|1, 0.25, 2.250|
|0, 1, -0.5385|
|0, 1, 1.999|
|1, 0.25, 2.250|
|0, 1, -0.5385|
|0, 0, -8.7700|
|1, 0.25, 2.250|
|0, 1, -0.5385|
|0, 0, 1|
我在 A 中执行的操作与在 B 中执行的操作相同,因此系统保持等效。
重新更新:
我在“for c = 1:1:n”之后添加了此内容
因此,在执行任何操作之前,它会对 A(和 b)的行进行排序,以使“c”列的条目递减(0 将保留在 A 的底行上)。现在它似乎适用于任何可逆方阵,尽管我不确定它是否有效。
r = c;
a = r + 1;
while r <= n
if r == n
r = r + 1;
elseif a <= n
while a <= n
if abs(U(r,c)) < abs(U(a,c))
UU = U(r,:);
U(r,:) = U(a,:);
U(a,:) = UU;
uu = u(r,1);
u(r,1) = u(a,1);
u(a,1) = uu;
else
a = a+1;
end
end
else
r = r+1;
a = r+1;
end
end
最佳答案
带旋转的高斯消除如下。
function [L,U,P] = my_lu_piv(A)
n = size(A,1);
I = eye(n);
O = zeros(n);
L = I;
U = O;
P = I;
function change_rows(k,p)
x = P(k,:); P(k,:) = P(p,:); P(p,:) = x;
x = A(k,:); A(k,:) = A(p,:); A(p,:) = x;
x = v(k); v(k) = v(p); v(p) = x;
end
function change_L(k,p)
x = L(k,1:k-1); L(k,1:k-1) = L(p,1:k-1);
L(p,1:k-1) = x;
end
for k = 1:n
if k == 1, v(k:n) = A(k:n,k);
else
z = L(1:k-1,1:k -1)\ A(1:k-1,k);
U(1:k-1,k) = z;
v(k:n) = A(k:n,k)-L(k:n,1:k-1)*z;
end
if k<n
x = v(k:n); p = (k-1)+find(abs(x) == max(abs(x))); % find index p
change_rows(k,p);
L(k+1:n,k) = v(k+1:n)/v(k);
if k > 1, change_L(k,p); end
end
U(k,k) = v(k);
end
end
为了解决系统..
% Ax = b (1) original system % LU = PA
(2) factorization of PA or A(p,:) into the product LU % PAx = Pb (3) multiply both sides of (1) by P % LUx = Pb
(4) substitute (2) into (3) % let y = Ux (5) define y as Ux % let c = Pb (6) define c as Pb % Ly = c
(7) subsitute (5) and (6) into (4) % U*x = y (8) a rewrite of (5)
要做到这一点..
% [L U p] = lu (A) ; % factorize % y = L \ (P*b) ; % forward solve of (7), a lower triangular system % x = U \ y ; % backsolve of (8), an upper triangular system
关于matlab - 高斯消元法求解 A*x = b 线性系统 (MATLAB),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51199586/