restart:

# SYMMETRIC POWER METHOD ALGORITHM 9.2

# To approximate the dominant eigenvalue and an associated

# eigenvector of the n by n symmetric matrix A given a nonzero vector x:

# INPUT: Dimension n: matrix A: vector x: tolerance TOL:

# maximum number of iterations N.

# OUTPUT: Approximate eigenvalue MU: approximate eigenvector x or

# a message that the maximum number of iterations was

# exceeded.

print(`This is the Symmetric Power Method.\134n`):

OK := FALSE:

print(`Choice of input method`):

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

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), A(2,1), A(2,2), ...,

A(2,N)`):

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

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

print(`entries with`):

print(`at least one blank.\134n`):

print(`The initial approximation should follow in same format.\134n`):

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 dimension n - 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 do

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

od:

for I1 from 1 to N do

Y[I1-1] := 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 dimension n - an integer.`):

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

if N > 0 then

print(`Input Matrix `):

for I1 from 1 to N do

for J from 1 to N 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:

print(`Input initial approximation vector `):

for I1 from 1 to N do

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

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

od:

OK := TRUE:

else

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

fi:

od:

fi:

if OK = TRUE then

OK := FALSE:

while OK = FALSE do

print(`Input the tolerance.\134n`):

TOL := scanf(`%f`)[1]:print(`Tolerance = `):print(TOL):

if TOL > 0 then

OK := TRUE:

else

print(`Tolerance must be positive number.\134n`):

fi:

od:

OK := FALSE:

while OK = FALSE do

print(`Input maximum number of iterations `):

print(`- integer.\134n`):

NN := scanf(`%d`)[1]:print(`Maximum number of iterations = `):print(NN):

# Use NN in place of N

if NN > 0 then

OK := TRUE:

else

print(`Number must be positive integer.\134n`):

fi:

od:

fi:

if OK = TRUE then

OUP := default:

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

for I1 from 1 to N do

for J from 1 to N do

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

od:

fprintf(OUP, `\134n`):

od:

fprintf(OUP, `The initial approximation vector is :\134n`):

for I1 from 1 to N do

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

od:

fprintf(OUP, `\134n`):

fi:

if OK = TRUE then

for I1 from 1 to N do

X[I1-1] := 0:

od:

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(`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(`Output file is `):print(NAME):

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

else

OUP := default:

fi:

fprintf(OUP, `SYMMETRIC POWER METHOD\134n\134n`):

fprintf(OUP, `iter approx approx eigenvector\134n`):

fprintf(OUP, ` eigenvalue\134n`):

# Step 1

K := 1:

XL := 0:

for I1 from 1 to N do

XL := XL+Y[I1-1]*Y[I1-1]:

od:

# 2-Norm of Y

XL := sqrt(XL):

ERR := 0:

if XL > 0 then

for I1 from 1 to N do

T := Y[I1-1]/XL:

ERR := ERR+(X[I1-1]-T)*(X[I1-1]-T):

X[I1-1] := T:

od:

# X has a 2-Norm of 1.0

ERR := sqrt(ERR):

else

print(`A has a zero eigenvalue - select new vector and begin again\134n`):

OK := FALSE:

fi:

if OK = TRUE then

# Step 2

while K <= NN and OK = TRUE do

# Steps 3 and 4

YMU := 0:

for I1 from 1 to N do

Y[I1-1] := 0:

for J from 1 to N do

Y[I1-1] := Y[I1-1]+A[I1-1,J-1]*X[J-1]:

od:

YMU := YMU+X[I1-1]*Y[I1-1]:

od:

# Steps 5 and 6

XL := 0:

for I1 from 1 to N do

XL := XL+Y[I1-1]*Y[I1-1]:

od:

# 2-Norm of Y

XL := sqrt(XL):

ERR := 0:

if XL > 0 then

for I1 from 1 to N do

T := Y[I1-1]/XL:

ERR := ERR+(X[I1-1]-T)*(X[I1-1]-T):

X[I1-1] := T:

od:

# X has a 2-Norm of 1.0

ERR := sqrt(ERR):

else

printf(`A has a zero eigenvalue - select new vector and begin again\134n`):

OK := FALSE:

fi:

fprintf(OUP, `%d %12.8f`, K, YMU):

for I1 from 1 to N do

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

od:

fprintf(OUP, `\134n`):

if OK = TRUE then

# Step 7

if ERR < TOL then

# Procedure completed successfully.

fprintf(OUP, `\134n\134nThe eigenvalue = %12.8f`,YMU):

fprintf(OUP, ` to tolerance = %.10e\134n`, TOL):

fprintf(OUP, `obtained on iteration number = %d\134n\134n`, K):

fprintf(OUP, `Unit eigenvector is :\134n\134n`):

for I1 from 1 to N do

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

od:

fprintf(OUP, `\134n`):

OK := FALSE:

else

# Step 8

K := K+1:

fi:

od:

# Step 9

if K > NN then

print(`No convergence within `,NN,` iterations\134n`):

fi:

if OUP <> default then

fclose(OUP):

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

fi:

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IiNE

The original matrix - output by rows: 4.00000000 -1.00000000 1.00000000 -1.00000000 3.00000000 -2.00000000 1.00000000 -2.00000000 3.00000000 The initial approximation vector is : 1.00000000 0.00000000 0.00000000

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IiIi

SYMMETRIC POWER METHOD iter approx approx eigenvector eigenvalue 1 4.00000000 0.94280904 -0.23570226 0.23570226 2 5.00000000 0.81649658 -0.40824829 0.40824829 3 5.66666666 0.71066905 -0.49746834 0.49746834 4 5.90909091 0.64699664 -0.53916387 0.53916387 5 5.97674418 0.61283648 -0.55876267 0.55876267 6 5.99415204 0.59524716 -0.56819047 0.56819047 7 5.99853587 0.58633558 -0.57280476 0.57280476 8 5.99963383 0.58185194 -0.57508622 0.57508622 9 5.99990845 0.57960333 -0.57622043 0.57622043 10 5.99997711 0.57847736 -0.57678590 0.57678590 11 5.99999428 0.57791395 -0.57706822 0.57706822 12 5.99999857 0.57763214 -0.57720928 0.57720928 13 5.99999964 0.57749122 -0.57727978 0.57727978 14 5.99999991 0.57742074 -0.57731503 0.57731503 15 5.99999998 0.57738551 -0.57733265 0.57733265 16 5.99999999 0.57736789 -0.57734146 0.57734146 17 6.00000000 0.57735908 -0.57734586 0.57734586 18 6.00000000 0.57735467 -0.57734807 0.57734807 The eigenvalue = 6.00000000 to tolerance = 1.0000000000e-05 obtained on iteration number = 18 Unit eigenvector is : 0.57735467 -0.57734807 0.57734807

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