restart:

# GAUSSIAN ELIMINATION WITH BACKWARD SUBSTITUTION ALGOTITHM 6.1

# To solve the n by n linear system

# E1: A[1,1] X[1] + A[1,2] X[2] +...+ A[1,n] X[n] = A[1,n+1]

# E2: A[2,1] X[1] + A[2,2] X[2] +...+ A[2,n] X[n] = A[2,n+1]

# :

# .

# EN: A[n,1] X[1] + A[n,2] X[2] +...+ A[n,n] X[n] = A[n,n+1]

# INPUT: number of unknowns and equations n: augmented

# matrix A = (A(I,J)) where 1<=I<=n and 1<=J<=n+1.

# OUTPUT: solution x(1), x(2),...,x(n) or a message that the

# linear system has no unique solution.

print(`This is Gaussian Elimination to solve a linear system.`):

print(`Choice of input method`):

print(`1. input from keyboard - not recommended for large systems`):

print(`2. input from a text file`):

print(`Please enter 1 or 2.`):

FLAG := scanf(`%d`)[1]: print(`Your input is`): print(FLAG):

if FLAG = 2 then

print(`The array will be input from a text file in the order`):

print(`A(1,1), A(1,2), ..., A(1,N+1), A(2,1), A(2,2), ...,

A(2,N+1)`):

print(`..., A(N,1), A(N,2), ..., A(N,N+1)\134n`):

print(`Place as many entries as desired on each line, but separate `):

print(`entries with`):

print(`at least one blank.`):

print(`Has the input file been created? - enter 1 for yes or 2 for no.`):

AA := scanf(`%d`)[1]: print(`Your response is`): print(AA):

if AA = 1 then

print(`Input the file name in the form - drive:\134\134name.ext`):

print(`for example: A:\134\134DATA.DTA`):

NAME := scanf(`%s`)[1]: print(`The file name is`): print(NAME):

INP := fopen(NAME,READ,TEXT):

OK := FALSE:

while OK = FALSE do

print(`Input the number of equations - an integer.`):

N := scanf(`%d`)[1]: print(`N is`): print(N):

if N > 0 then

for I1 from 1 to N do

for J from 1 to N+1 do

A[I1-1,J-1] := fscanf(INP, `%f`)[1]:

od:

OK := TRUE:

fclose(INP):

else print(`The number must be a positive integer.\134n`):

fi:

od:

else

print(`The program will end so the input file can be created.\134n`):

fi:

else

OK := FALSE:

while OK = FALSE do

print(`Input the number of equations - an integer.`):

N := scanf(`%d`)[1]: print(`N= `): print(N):

if N > 0 then

for I1 from 1 to N do

for J from 1 to N+1 do

print(`input entry in position `,I1,J):

A[I1-1,J-1] := scanf(`%f`)[1]:print(`Data is `):print(A[I1-1,J-1]):

od:

OK := TRUE:

else print(`The number must be a positive integer.\134n`):

fi:

od:

fi:

if OK = TRUE then

OUP := default:

fprintf(OUP, `The original system - output by rows:\134n`):

for I1 from 1 to N do

for J from 1 to N+1 do

fprintf(OUP, ` %11.8f`, A[I1-1,J-1]):

od:

fprintf(OUP, `\134n`):

od:

# Step 1

NN := N-1:

M := N+1:

ICHG := 0:

I1 := 1:

while OK = TRUE and I1 <= NN do

# Step 2

# Use IP in place of P

IP := I1:

while abs(A[IP-1,I1-1]) <= 1.0e-20 and IP <= N do

IP := IP+1:

od:

if IP = M then

OK := FALSE:

else

# Step 3

if IP <> I1 then

for JJ from 1 to M do

C := A[I1-1,JJ-1]:

A[I1-1,JJ-1] := A[IP-1,JJ-1]:

A[IP-1,JJ-1] := C:

od:

ICHG := ICHG+1:

fi:

# Step 4

JJ := I1+1:

for J from JJ to N do

# Step 5

# XM is used in place of m(J,I)

XM := A[J-1,I1-1]/A[I1-1,I1-1]:

# Step 6

for K from JJ to M do

A[J-1,K-1] := A[J-1,K-1] - XM * A[I1-1,K-1]:

od:

# Multiplier XM could be saved in A[J,I]

A[J-1,I1-1] := 0:

od:

fi:

I1 := I1+1:

od:

if OK = TRUE

then

# Step 7

if abs(A[N-1,N-1]) <= 1.0e-20 then

OK := FALSE:

else

# Step 8

# Start backward substitution

X[N-1] := A[N-1,M-1] / A[N-1,N-1]:

# Step 9

for K from 1 to NN do

I1 := NN-K+1:

JJ := I1+1:

SUM := 0:

for KK from JJ to N do

SUM := SUM - A[I1-1,KK-1] * X[KK-1]:

od:

X[I1-1] := (A[I1-1,M-1]+SUM) / A[I1-1,I1-1]:

od:

# Step 10

# Process is complete

print(`Choice of output method`):

print(`1. Output to screen`):

print(`2. Output to text file`):

print(`Please enter 1 or 2.`):

FLAG := scanf(`%d`)[1]: print(`Your input is`): print(FLAG):

if FLAG = 2 then

print(`Input the file name in the form - drive:\134\134name.ext`):

print(`for example: A:\134\134OUTPUT.DTA`):

NAME := scanf(`%s`)[1]: print(`The output file is`): print(NAME):

OUP := fopen(NAME, WRITE,TEXT):

else

OUP := default:

fi:

fprintf(OUP, `GAUSSIAN ELIMINATION\134n\134n`):

fprintf(OUP, `The reduced system - output by rows:\134n`):

for I1 from 1 to N do

for J from 1 to M do

fprintf(OUP, ` %11.8f`, A[I1-1,J-1]):

od:

fprintf(OUP, `\134n`):

od:

fprintf(OUP, `\134n\134nHas solution vector:\134n`):

for I1 from 1 to N do

fprintf(OUP, ` %12.8f`, X[I1-1]):

od:

fprintf (OUP, `\134n\134nwith %d row interchange(s)\134n`, ICHG):

if OUP <> default then

fclose(OUP):

print(`Output file `,NAME,` created successfully`):

fi:

if OK = FALSE then

print(`System has no unique solution\134n`):

fi:

SVdUaGlzfmlzfkdhdXNzaWFufkVsaW1pbmF0aW9ufnRvfnNvbHZlfmF+bGluZWFyfnN5c3RlbS5HNiI=

STdDaG9pY2V+b2Z+aW5wdXR+bWV0aG9kRzYi

SWVuMS5+aW5wdXR+ZnJvbX5rZXlib2FyZH4tfm5vdH5yZWNvbW1lbmRlZH5mb3J+bGFyZ2V+c3lzdGVtc0c2Ig==

SToyLn5pbnB1dH5mcm9tfmF+dGV4dH5maWxlRzYi

STVQbGVhc2V+ZW50ZXJ+MX5vcn4yLkc2Ig==

SS5Zb3VyfmlucHV0fmlzRzYi

IiIi

SUxJbnB1dH50aGV+bnVtYmVyfm9mfmVxdWF0aW9uc34tfmFufmludGVnZXIuRzYi

SSROPX5HNiI=

IiIj

NiVJOWlucHV0fmVudHJ5fmlufnBvc2l0aW9ufkc2IiIiIkYl

SSlEYXRhfmlzfkc2Ig==

JCIiIyIiIQ==

NiVJOWlucHV0fmVudHJ5fmlufnBvc2l0aW9ufkc2IiIiIiIiIw==

SSlEYXRhfmlzfkc2Ig==

JCIiJCIiIQ==

NiVJOWlucHV0fmVudHJ5fmlufnBvc2l0aW9ufkc2IiIiIiIiJA==

SSlEYXRhfmlzfkc2Ig==

JCEiIiIiIQ==

NiVJOWlucHV0fmVudHJ5fmlufnBvc2l0aW9ufkc2IiIiIyIiIg==

SSlEYXRhfmlzfkc2Ig==

JCIiJSIiIQ==

NiVJOWlucHV0fmVudHJ5fmlufnBvc2l0aW9ufkc2IiIiI0Yl

SSlEYXRhfmlzfkc2Ig==

JCIiIyIiIQ==

NiVJOWlucHV0fmVudHJ5fmlufnBvc2l0aW9ufkc2IiIiIyIiJA==

SSlEYXRhfmlzfkc2Ig==

JCIiJyIiIQ==

The original system - output by rows: 2.00000000 3.00000000 -1.00000000 4.00000000 2.00000000 6.00000000

SThDaG9pY2V+b2Z+b3V0cHV0fm1ldGhvZEc2Ig==

STQxLn5PdXRwdXR+dG9+c2NyZWVuRzYi

STcyLn5PdXRwdXR+dG9+dGV4dH5maWxlRzYi

STVQbGVhc2V+ZW50ZXJ+MX5vcn4yLkc2Ig==

SS5Zb3VyfmlucHV0fmlzRzYi

IiIi

GAUSSIAN ELIMINATION The reduced system - output by rows: 2.00000000 3.00000000 -1.00000000 0.00000000 -4.00000000 8.00000000 Has solution vector: 2.50000000 -2.00000000 with 0 row interchange(s)

JSFH