gauss-seidel程序修改-技术交流区-MoHub

在gauss-Seidel迭代中
function Gauss(A, B)
(m, n) = size(A)
z = [convert(Matrix{Float64}, A) convert(Vector{Float64}, B)]
x = zeros(Float64, n)

for i = 1:n - 1
    for j = i + 1:n
        if z[i, i]!= 0
            m = -z[j, i] / z[i, i]
            z[j, :] = z[j, :] + m * z[i, :]
        else
            println("不能用高斯消去")
        end
    end
end

B = z[:, n + 1]
x[n] = B[n] / z[n, n]

for k = (n - 1):-1:1
    s = 0.0
    for t = k + 1:n
        s = s + z[k, t] * x[t]
    end
    x[k] = (B[k] - s) / z[k, k]
    # println("解x[%.0f]的值为%.1f", k, x[k])
end

return x

end
function GaussSeidelMatrix(A, B, tol = 1e-6)
(m, n) = size(A)
# 确认系数矩阵A是方阵（Gauss-Seidel迭代法要求系数矩阵为方阵，这里添加检查更严谨）
if m!= n
error("系数矩阵A必须是方阵")
end
# 确保B是列向量形式，维度为 (n, 1)，如果不是则进行转换
if size(B, 2)!= 1
B = reshape(B, n, 1)
elseif size(B, 1)!= n
error("常数项向量B的维度与系数矩阵A的行数不匹配")
end
# 初始化迭代向量x，明确初始化为列向量形式（确保维度正确）
x = zeros(Float64, n, 1)
xGS = zeros(Float64, n, 1)
L = tril(A)
U = triu(A, 1)

max_iterations = 1000
iteration_count = 0
while true
    xGS = (B - U * x) / L
    error = norm(xGS - x)
    if error < tol || iteration_count >= max_iterations
        return xGS[:, 1]
    end
    x = xGS
    iteration_count += 1
end

end
A = [10 -1 2 0; -1 11 -1 3; 2 -1 10 -1; 0 3 -1 8]
B = [6; 25; -11; 15]
solution_gauss = Gauss(A, B)
println("The solution of Gauss is")
for i in 1:length(solution_gauss)
println("x$i = ", solution_gauss[i])
end

solution_gauss_seidel = GaussSeidel(A, B)
println("The solution of Gauss-Seidel is")
for i in 1:length(solution_gauss_seidel)
println("x$i = ", solution_gauss_seidel[i])
end
Julia报错DimensionMismatch: Both inputs should have the same number of columns，怎么修改代码？

您好，您的gauss_seidel迭代里面的函数可能写的有点问题，一直报输入参数必须有相同的列。可以尝试使用如下代码实现：
function GaussSeidelMatrix(A, b, x0; tol=1e-6, max_iter=1000)
# A: 系数矩阵
# b: 常数向量
# x0: 初始猜测值
# tol: 误差容限（默认1e-6）
# max_iter: 最大迭代次数（默认1000）
# 获取矩阵的维度
n = length(b)
# 创建解向量 x，初始化为 x0
x = copy(x0)
# 开始迭代
for k in 1:max_iter
# 记录当前解向量的旧值，方便计算误差
x_old = copy(x)
for i in 1:n
sum1 = dot(A[i, 1:i-1],x[1:i-1]) # 左边的已知项
sum2 = dot(A[i, i+1:n] ,x_old[i+1:n]) # 右边的已知项
x[i] = (b[i] - sum1 - sum2) / A[i, i]
end
err = norm(x - x_old)
# 如果误差小于容忍值，停止迭代
if err < tol
println("Converged after $k iterations")
return x
end
end
println("Reached maximum iterations")
return x
end

全部回答 1