如何将高斯消除lua程序应用于任何矩阵？ Lua中的数组

如何解决如何将高斯消除lua程序应用于任何矩阵？ Lua中的数组

我只是在进行高斯消去实验，以尝试在这种用于计算梯形形式的特定方法中加快工作速度。

当我尝试运行它时，它不起作用。

我想在这些矩阵上运行该程序。 您能在下面的矩阵上应用程序，然后告诉我它们给您什么 ？

--[[

Gaussian Elimination in Lua

--]]

-- print matrix
local function printmatrix (m)
    for _,row in ipairs (m) do
        io.write (table.concat (row,',') .. '\n')
   end
end

-- read matrix in CSV format
local function readcsv (file)
    io.input (file)
    local m = {columns = 0,rectangular = true}
    for line in io.lines () do
        local row = {}
        -- the next line is tricky and goes over all entries in the row
        for w in line:gmatch '[^,]+' do
            row [#row + 1] = tonumber (w) or 0
        end
        m [#m + 1] = row
        -- Update matrix dimensions
        m.rectangular = m.rectangular and (#row == m.columns or #m == 1)
        m.columns = #row > m.columns and #row or m.columns
    end
    return m
end

-- if m[r][c] is zero swap row r with some row i>r to make m[r][c] nonzero,if possible
local function swap (m,r,c)
    local nrows,ncols = #m,m.columns
    if r <= 0 or r > nrows or c <= 0 or c > ncols then error 'Position out of range' end
    if m [r] [c] ~= 0 then
        -- nothing to do
        return
    end
    -- find a suitable row
    local i = r + 1
    while i <= nrows and m [i] [c] == 0 do
        i = i + 1
    end
    if i <= nrows then
        m [r],m [i] = m [i],m [r]
    end
end

-- if m[r][c] is nonzero apply row operations to make each m[i][c]==0 for i>r
local function clear (m,m.columns
    if r <= 0 or r > nrows or c <= 0 or c > ncols then error 'Position out of range' end
    if m [r] [c] == 0 then
        -- nothing to do
        return
    end
    for i = r + 1,nrows do
        local f = m [i] [c] / m [r] [c]
        for j = 1,#m [i] do
            m [i] [j] = m [i] [j] - f * m [r] [j]
        end
    end
end

-- apply Gaussian elimination to m to get it into echelon form
function echelon (m)
    local nrows,m.columns
    local r,c = 1,1 -- current position
    while r <= nrows and c <= ncols do
        -- try to get a nonzero value at this position
        swap (m,c)
        if m [r] [c] == 0 then
            -- can't,so move right
            c = c + 1
        else
            clear (m,c)
            -- done,so move diagonally
            r = r + 1
            c = c + 1
        end
    end
end

local m = readcsv (arg [1])
print 'Original:'
printmatrix (m)
if m.rectangular then
    echelon (m)
    print 'Echelon form:'
    printmatrix (m)
else
    error 'Matrix not rectangular!'
end

在以下矩阵上运行该程序会得到什么：

1,3,5,7
2,1,-1,0
3,4,7
5,7

1,1
2,2,1
3,1
4,0

 1,6,7,8,9
-1,-2,-3,-4,-5,-6,-7,-8,-8
 1,3
 1,3
 0,-6
 9,1

解决方法

一段熟悉的代码。

在for line in io.lines () do中将if line ~= '' then ... end循环的内部包裹起来，以安全地处理空行。

之后：

alexander@maicube:~$ lua t.lua t1.csv
Original:
1,3,5,7
2,1,-1,0
3,4,7
5,7
Echelon form:
1,7
0,-5,-11,-14
0,0
0,0
alexander@maicube:~$ lua t.lua t2.csv
Original:
1,1
2,2,1
3,1
4,0
Echelon form:
1,1
0,-2,-1
0,-2
0,-1
alexander@maicube:~$ lua t.lua t3.csv
Original:
1,6,7,8,9
-1,-3,-4,-6,-7,-8,-8
1,3
1,3
0,-6
9,1
Echelon form:
1,9
0,-6
0,0