A simple gaussian elimination implemented in C.

To simplify, I hard coded the linear system 10 x1 + 2 x2 + 3 x3 + 4 x4 = 5 6 x1 + 17 x2 + 8 x3 + 9 x4 = 10 11 x1 + 12 x2 + 23 x3 + 14 x4 = 15 16 x1 + 17 x2 + 18 x3 + 29 x4 = 20

into the AB float matrix.

```/* * Description: Solve a hard coded linear system by gaussian elimination * Author: Silveira Neto * License: Public Domain */   #include <stdio.h> #include <stdlib.h>   #define ROWS 4 #define COLS 5   /** * Linear System, Ax = B * * 10*x1 + 2*x2 + 3*x3 + 4*x4 = 5 * 6*x1 + 17*x2 + 8*x3 + 9*x4 = 10 * 11*x1 + 12*x2 + 23*x3 + 14*x4 = 15 * 16*x1 + 17*x2 + 18*x3 + 29*x4 = 20 */ float AB[ROWS][COLS] = { {10, 2, 3, 4, 5}, { 6, 17, 8, 9, 10}, {11, 12, 23, 14, 15}, {16, 17, 18, 29, 20} };   /* Answer x from Ax=B */ float X[ROWS] = {0,0,0,0};   int main(int argc, char** argv) { int row, col, i;   /* gaussian elimination */ for (col=0; col<COLS-1; col++) { for (row=0; row<ROWS; line++){ float pivot = AB[row][col]/AB[col][col]; if(row!=col) { for(i=0; i<COLS; i++) { AB[row][i] = AB[row][i] - pivot * AB[col][i]; }   } } }   /* X = B/A and show X */ for(row=0; row<ROWS; line++) { X[row] = AB[row][ROWS] / AB[row][row]; printf("%3.5f ", X[row]); } printf("n");   return (EXIT_SUCCESS); }```

Before the gaugassian elimination, AB is

```10  2  3  4  5
6 17  8  9 10
11 12 23 14 15
16 17 18 29 20```

and after it is

```10.00000 0.00000 0.00000 0.00000 2.82486
0.00000 15.80000 0.00000 0.00000 3.92768
0.00000 0.00000 15.85443 0.00000 3.85164
0.00000 0.00000 0.00000 14.13174 3.35329 ```

that corresponds to 10 x1 = 2.82486 15.80000 x2 = 3.92768 15.85443 x3 = 3.85164 14.13174 x4 = 3.35329

The solution vector is X = (x1, x2, x3, x4). We get it by X=B/A.

The program output, X, is

`0.28249 0.24859 0.24294 0.23729`

Benchmarking:
I’m this serial implementation over one node of our cluster, a machine with 4 processors (Intel Xeon 1.8 Ghz) and 1Gb RAM memory. I tried random systems from 1000 to 5000 variables and got the average time. Published inenglish

1. sombriks

nice, i remember what is it now, heh.

where will you apply it??

2. Silveira

sombriks, I also wrote a parallel version of gaussian elimination using MPI. My idea now is to benchmark them.

3. otavio

Hi, nice work. Just for language war, some time ago I’ve done a python version:
http://pastebin.com/f17960864

• Silveira

Cool! I’ve pasted here:

```		import sys

#############

# Buggy Gauss Elimination

# (Zero division is not checked)

#

def det(rows):

v = None

if len(rows) == 2:

r1 = rows

r2 = rows

v = r1 * r2 - r1 * r2

else:

firstRow = rows

aboveRows = rows[1:]

subDets = []

# At time I din't know the existence of enumerate

for c in range(0, len(firstRow)):

subMatrix = []

for ar in aboveRows:

subRow = []

for c2 in range(0, len(ar)):

if c != c2:

subRow.append(ar[c2])

subMatrix.append(subRow)

subDets.append(det(subMatrix) * firstRow[c])

evens = [subDets[e] for e in range(0, len(subDets), 2)]

odds = [subDets[e] for e in range(1, len(subDets), 2)]

v = reduce(lambda x, y: x+y, evens) - reduce(lambda x, y: x+y, odds)

return v

#non-recursive

def solveSystem(rows):

if det(rows) == 0:

return None

zerorColumnNth = 0

solvedSystem = rows

for workRowNth in range(0, len(rows)-1):

for prodRowNth in range(workRowNth+1, len(rows)):

workRow = solvedSystem[workRowNth]

prodRow = solvedSystem[prodRowNth]

mul = -prodRow[zerorColumnNth] / workRow[zerorColumnNth]

newProdRow = map(lambda a, b: a + b, map(lambda x: x*mul, workRow), prodRow)

solvedSystem[prodRowNth] = newProdRow

zerorColumnNth += 1

return solvedSystem

#recursive

def solveSystem2(rows):

def solveLoop(rows, workNth, prodNth, zeroColumn, nrows):

if prodNth < nrows:

workRow = rows[workNth]

prodRow = rows[prodNth]

mul = -prodRow[zeroColumn] / workRow[zeroColumn]

rows[prodNth] = map(lambda a, b: a + b, map(lambda x: x*mul, workRow), prodRow)

return solveLoop(rows, workNth, prodNth+1, zeroColumn, nrows)

else:

if workNth = 0:

row = rs[nth]

vt = [vals[v]*row[v] for v in range(nth+1, len(row)-1)]

if len(vt) > 0:

coff = -reduce(lambda a, b: a+b,  vt)

else:

coff = 0

vals[nth] = (coff + row[len(row)-1])/row[nth]

return retroLoop(rs, nth-1, vals)

return vals

nrows = len(rows)

return retroLoop(rows, nrows-1, [0 for i in range(0, nrows)])

if __name__ == '__main__':

try:

f = open(sys.argv, 'r')

except IOError, msg:

print(msg)

sys.exit(1)

rows = []

for line in f:

if line != '\n':

atoms = line.split(' ')

r = []

for a in atoms:

r.append(float(a))

rows.append(r)

f.close();

maxColumn = 0

for r in rows:

cs = len(r)

if cs > maxColumn:

maxColumn = cs

for r in rows:

cs = len(r)

if cs != maxColumn:

sys.exit(1)

ss = solveSystem2(rows)

r = retroSub(ss)

print(r)

##############

# Sample input file:

#

# 10 2 3 4 5

# 6 17 8 9 10

# 11 12 23 14 15

# 16 17 18 29 20

# Should output:

# [0.28248587570621464, 0.24858757062146891, 0.24293785310734467, 0.23728813559322035]

```
4. A

I’ll do just a observation: if A and B are matrixes, the operation B/A doesn’t exist. You have to do B*inv(A), in fact if C = inv(A) then C(i,j) = 1/A(i,j) because A is a digonal matrix.

Merry Christmas for everyone!!!
God will help us more in next year!

5. T

I’m not the only one who finds amusement in the irony of the C version being shorter and clearer than the Python one, am I?

6. Done

have you declared the variable ‘line’?

for (row=0; row<ROWS; line++) // causes error.

7. Done

full of error

8. Yigit

For those who think the code above is full of errors: why don’t you just replace line with row :evil: !

9. richard

this is neat, thanks.