restart:

# GAUSSIAN ELIMINATION WITH PARTIAL PIVOTING ALGORITHM 6.2

# 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 with Partial Pivoting. `):

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

for I1 from 1 to N do

NROW[I1-1] := I1:

od:

# Initialize row pointer

NN := N-1: M:=N+1:

ICHG := 0:

I1 := 1:

# Step 2

while OK = TRUE and I1 <= NN do

# Step 3

IMAX := NROW[I1-1]:

AMAX := abs(A[IMAX-1,I1-1]):

IMAX := I1:

JJ := I1+1:

for IP from JJ to N do

JP := NROW[IP-1]:

if abs(A[JP-1,I1-1]) > AMAX then

AMAX := abs(A[JP-1,I1-1]):

IMAX := IP:

fi:

od:

# Step 4

if AMAX <= 1.0e-20 then

OK := FALSE:

else

# Step 5

# Simulate row interchange

if NROW[I1-1] <> NROW[IMAX-1] then

ICHG := ICHG+1:

NCOPY := NROW[I1-1]:

NROW[I1-1] := NROW[IMAX-1]:

NROW[IMAX-1] := NCOPY:

fi:

I2 := NROW[I1-1]:

# Step 6

for J from JJ to N do

J1 := NROW[J-1]:

# Step 7

XM := A[J1-1,I1-1]/A[I2-1,I1-1]:

# Step 8

for K from JJ to M do

A[J1-1,K-1] := A[J1-1,K-1]-XM*A[I2-1,K-1]:

od:

# Multiplier XM could be saved in A[J1-1,I1-1]

A[J1-1,I1-1] := 0:

od:

fi:

I1 := I1+1:

od:

if OK = TRUE then

# Step 9

N1 := NROW[N-1]:

if abs(A[N1-1,N-1]) <= 1.0e-20 then

OK := FALSE:

# System has no unique solution

else

# Step 10

# Start backward substitution

X[N-1] := A[N1-1,M-1] / A[N1-1,N-1]:

# Step 11

for K from 1 to NN do

I1 := NN - K + 1:

JJ := I1 + 1:

N2 := NROW[I1-1]:

SUM := 0:

for KK from JJ to N do

SUM := SUM-A[N2-1,KK-1]*X[KK-1]:

od:

X[I1-1] := (A[N2-1,N] + SUM) / A[N2-1,I1-1]:

od:

# Step 12

# 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 - PARTIAL PIVOTING \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 Has solution vector:\134n `):

for I1 from 1 to N do

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

od:

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

fprintf(OUP, ` \134nThe rows have been logically re-ordered to: \134n`):

for I1 from 1 to N do

fprintf(OUP, ` %2d`, NROW[I1-1]):

od:

fprintf(OUP,`\134n `):

if OUP <> default then

fclose(OUP):

print(`Output file `,NAME,` created successfully\134n`):

fi:

if OK = FALSE then

printf(`System has no unique solution `):

fi:

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The original system - output by rows: 1.00000000 -1.00000000 2.00000000 6.00000000 4.00000000 -1.00000000 1.00000000 -2.00000000 4.00000000 1.00000000 5.00000000 1.00000000

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IiIi

GAUSSIAN ELIMINATION - PARTIAL PIVOTING The reduced system - output by rows: 0.00000000 0.00000000 3.25000000 7.62500000 4.00000000 -1.00000000 1.00000000 -2.00000000 0.00000000 2.00000000 4.00000000 3.00000000 Has solution vector: -1.88461538 -3.19230769 2.34615385 with 2 row interchange(s) The rows have been logically re-ordered to: 2 3 1

JSFH