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new_fem1d.c
1896 lines (1541 loc) · 49.4 KB
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new_fem1d.c
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#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <omp.h>
#include <string.h>
#include <math.h>
//FUNCTON DECLARTIONS
/******************************************************************************/
int main ( int argc, char **argv);
double run(int version);
double Average(double ary[]);
int check();
void assemble (int version);
double ff ( double x );
void geometry (int version);
void init ();
void output (int version);
void phi ( int il, double x, double *phii, double *phiix, double xleft,
double xrite );
double pp ( double x );
void prsys ();
double qq ( double x );
void solve ();
void timestamp ( void );
//GLOBALS
/******************************************************************************/
long int NSUB;
//# define NL 20
long int NL;
FILE *fp_out;
FILE *fp_sol;
double *adiag;
double *aleft;
double *arite;
double *f;
double *h;
int ibc;
int *indx;
int *node;
int nquad;
int nu;
double ul;
double ur;
double xl;
double *xn;
double *xquad;
double xr;
/******************************************************************************/
/** main proccess user arguments, runs version trails and
* provides analytical information about performance and
* code correctness.
*
*/
int main(int argc, char **argv)
{
int i;
int threads;
int trials;
//get NSUB, threads and trails from argument
if(argc != 5){
printf("Invalid number of arguments.\n");
printf("Usage: ./fem1d [SUB_SIZE] [NL] [NUM_THREADS] [TRIALS]\n");
exit(EXIT_FAILURE);
}
else if((NSUB = atoi(argv[1])) == 0) {
printf("Invalid subdivison size.\n");
printf("Usage: ./fem1d [SUB_SIZE] [NL] [NUM_THREADS] [TRIALS]\n");
exit(EXIT_FAILURE);
} else if ((NL = atoi(argv[2])) == 0){
printf("Invalid number of threads.\n");
printf("Usage: ./fem1d [SUB_SIZE] [NL] [NUM_THREADS] [TRIALS]\n");
exit(EXIT_FAILURE);
} else if ((threads = atoi(argv[3])) == 0){
printf("Invalid number of threads.\n");
printf("Usage: ./fem1d [SUB_SIZE] [NL] [NUM_THREADS] [TRIALS]\n");
exit(EXIT_FAILURE);
} else if ((trials = atoi(argv[4])) == 0){
printf("Invalid number of trails.\n");
printf("Usage: ./fem1d [SUB_SIZE] [NL] [NUM_THREADS] [TRIALS]\n");
exit(EXIT_FAILURE);
}
printf("NSUB = %ld, NL = %ld, threads = %d, trails = %d\n",
NSUB,NL,threads,trials);
//set number of threads
omp_set_num_threads(threads);
//wipe solutions files
fp_sol = fopen("old_sol.txt","w");
fclose(fp_sol);
fp_sol = fopen("new_sol.txt","w");
fclose(fp_sol);
//time spent on execution for each trail
double time_spent[trials];
int ver = 0;
double old_time[trials];
double new_time[trials];
int oc = 0;
int nc = 0;
//toggle between versions and perform trials
for(i=0;i<(trials*2);i++){
if(ver == 0){
printf("Running old version: trail %d...\n",oc+1);
old_time[oc]= run(0);
printf("Succesfully complete: Time taken: %fsec\n",old_time[oc]);
oc++;
ver++;
} else if (ver == 1){
printf("Running new version trail %d...\n",nc+1);
new_time[nc]= run(1);
printf("Succesfully complete: Time taken: %fsec\n",new_time[nc]);
nc++;
ver = ver - 1;
}
}
//print results
printf("********************************** RESULTS ************************************\n");
double o_time = Average(old_time);
double n_time = Average(new_time);
printf("NSUB = %ld, NL = %ld, threads = %d, trails = %d\n",NSUB,NL,threads,trials);
printf("New version Average Time: %fsec\n",n_time);
printf("Old version Average Time: %fsec\n",o_time);
printf("Speed up: %fsec\n", (o_time - n_time));
/*
//RUN OLD VERSION TRAILS
printf("**************************** RUNNING OLD VERSION ******************************\n");
for(i=0;i<trials;i++){
printf("Running old version, trail %d...\n",i+1);
time_spent[i] = run(0);
printf("Old version, trail %d succesfully complete.\n",i+1);
printf("Time taken: %fsec\n",time_spent[i]);
}
//store old time for later display
double old_time = Average(time_spent);
printf("**************************** RUNNING NEW VERSION ******************************\n");
//RUN NEW VERSION TRAILS
for(i=0;i<trials;i++){
printf("Running new version, trail %d...\n",i+1);
time_spent[i] = run(1);
printf("New version, trail %d succesfully complete.\n", i+1);
printf("Time taken: %fsec\n",time_spent[i]);
}
printf("********************************** RESULTS ************************************\n");
double new_time = Average(time_spent);
printf("New version Average Time: %fsec\n",new_time);
printf("Old version Average Time: %fsec\n",old_time);
printf("Speed up: %fsec\n", (old_time - new_time));
*/
//CHECK FOR CORRCTNESS
if(check()==0){
printf("CORRECT: Outputs are identical.\n");
} else {
printf("INCORRECT: Outputs are not identical.\n");
}
return 0;
}
/******************************************************************************/
/**
* Computes and returns average of an array of doubles
*
*/
double Average(double ary[]){
double ave = 0;
int i;
int length = (int)(sizeof(ary)/sizeof(double));
for(i=0;i < length;i++){
ave += ary[i];
}
return ave / (double)length;
}
/******************************************************************************/
//TODO take print statments out of parallel for loops
/******************************************************************************/
/*
Purpose:
RUN is the main program for FEM1D.
Discussion:
FEM1D solves a one dimensional ODE using the finite element method.
The differential equation solved is
- d/dX (P dU/dX) + Q U = F
The finite-element method uses piecewise linear basis functions.
Here U is an unknown scalar function of X defined on the
interval [XL,XR], and P, Q and F are given functions of X.
The values of U or U' at XL and XR are also specified.
The interval [XL,XR] is "meshed" with NSUB+1 points,
XN(0) = XL, XN(1)=XL+H, XN(2)=XL+2*H, ..., XN(NSUB)=XR.
This creates NSUB subintervals, with interval number 1
having endpoints XN(0) and XN(1), and so on up to interval
NSUB, which has endpoints XN(NSUB-1) and XN(NSUB).
Licensing:
This code is distributed under the GNU LGPL license.
Modified:
29 May 2009
Author:
C version by John Burkardt
Parameters:
double ADIAG(NU), the "diagonal" coefficients. That is, ADIAG(I) is
the coefficient of the I-th unknown in the I-th equation.
double ALEFT(NU), the "left hand" coefficients. That is, ALEFT(I)
is the coefficient of the (I-1)-th unknown in the I-th equation.
There is no value in ALEFT(1), since the first equation
does not refer to a "0-th" unknown.
double ARITE(NU).
ARITE(I) is the "right hand" coefficient of the I-th
equation in the linear system. ARITE(I) is the coefficient
of the (I+1)-th unknown in the I-th equation. There is
no value in ARITE(NU) because the NU-th equation does not
refer to an "NU+1"-th unknown.
double F(NU).
ASSEMBLE stores into F the right hand side of the linear
equations.
SOLVE replaces those values of F by the solution of the
linear equations.
double H(NSUB)
H(I) is the length of subinterval I. This code uses
equal spacing for all the subintervals.
int IBC.
IBC declares what the boundary conditions are.
1, at the left endpoint, U has the value UL,
at the right endpoint, U' has the value UR.
2, at the left endpoint, U' has the value UL,
at the right endpoint, U has the value UR.
3, at the left endpoint, U has the value UL,
and at the right endpoint, U has the value UR.
4, at the left endpoint, U' has the value UL,
at the right endpoint U' has the value UR.
int INDX[NSUB+1].
For a node I, INDX(I) is the index of the unknown
associated with node I.
If INDX(I) is equal to -1, then no unknown is associated
with the node, because a boundary condition fixing the
value of U has been applied at the node instead.
Unknowns are numbered beginning with 1.
If IBC is 2 or 4, then there is an unknown value of U
at node 0, which will be unknown number 1. Otherwise,
unknown number 1 will be associated with node 1.
If IBC is 1 or 4, then there is an unknown value of U
at node NSUB, which will be unknown NSUB or NSUB+1,
depending on whether there was an unknown at node 0.
int NL.
The number of basis functions used in a single
subinterval. (NL-1) is the degree of the polynomials
used. For this code, NL is fixed at 2, meaning that
piecewise linear functions are used as the basis.
int NODE[NL*NSUB].
For each subinterval I:
NODE[0+I*2] is the number of the left node, and
NODE[1+I*2] is the number of the right node.
int NQUAD.
The number of quadrature points used in a subinterval.
This code uses NQUAD = 1.
int NSUB.
The number of subintervals into which the interval [XL,XR] is broken.
int NU.
NU is the number of unknowns in the linear system.
Depending on the value of IBC, there will be NSUB-1,
NSUB, or NSUB+1 unknown values, which are the coefficients
of basis functions.
double UL.
If IBC is 1 or 3, UL is the value that U is required
to have at X = XL.
If IBC is 2 or 4, UL is the value that U' is required
to have at X = XL.
double UR.
If IBC is 2 or 3, UR is the value that U is required
to have at X = XR.
If IBC is 1 or 4, UR is the value that U' is required
to have at X = XR.
double XL.
XL is the left endpoint of the interval over which the
differential equation is being solved.
double XN(0:NSUB).
XN(I) is the location of the I-th node. XN(0) is XL,
and XN(NSUB) is XR.
double XQUAD(NSUB)
XQUAD(I) is the location of the single quadrature point
in interval I.
double XR.
XR is the right endpoint of the interval over which the
differential equation is being solved.
*/
double run(int version){
if(version == 0){
if((fp_out = fopen("old_out.txt", "w+")) == NULL ||
(fp_sol = fopen("old_sol.txt", "a")) == NULL){
printf("Old Version files not found.\n");
exit(EXIT_FAILURE);
}
}
else if (version == 1){
if((fp_out = fopen("new_out.txt", "w+")) == NULL ||
(fp_sol = fopen("new_sol.txt", "a")) == NULL){
printf("New Version files not found.\n");
exit(EXIT_FAILURE);
}
}
//Allocate array memory
adiag = (double *)malloc(sizeof(double)*(double)(NSUB+1));
aleft = (double *)malloc(sizeof(double)*(double)(NSUB+1));
arite = (double *)malloc(sizeof(double)*(double)(NSUB+1));
f = (double *)malloc(sizeof(double)*(double)(NSUB+1));
h = (double *)malloc(sizeof(double)*(double)(NSUB));
indx = (int *)malloc(sizeof(int)*(int)(NSUB+1));
node = (int *)malloc(sizeof(int)*((int)NL*(int)NSUB));
xn = (double *)malloc(sizeof(double)*(double)(NSUB+1));
xquad = (double *)malloc(sizeof(double)*(double)(NSUB));
//START TIMER//
double begin, end, time_spent;
begin = omp_get_wtime();
timestamp ();
fprintf (fp_out, "\n" );
fprintf (fp_out, "FEM1D\n" );
fprintf (fp_out, " C version\n" );
fprintf (fp_out, "\n" );
fprintf (fp_out, " Solve the two-point boundary value problem\n" );
fprintf (fp_out, "\n" );
fprintf (fp_out, " - d/dX (P dU/dX) + Q U = F\n" );
fprintf (fp_out, "\n" );
fprintf (fp_out, " on the interval [XL,XR], specifying\n" );
fprintf (fp_out," the value of U or U' at each end.\n" );
fprintf (fp_out, "\n" );
fprintf (fp_out," The interval [XL,XR] is broken into NSUB = %ld subintervals\n", NSUB );
fprintf (fp_out, " Number of basis functions per element is NL = %ld\n", NL );
/*
Initialize the data.
*/
init ();
/*
Compute the geometric quantities.
*/
geometry (version);
/*
Assemble the linear system.
*/
assemble (version);
/*
Print out the linear system.
*/
prsys ();
/*
Solve the linear system.
*/
solve ();
/*
Print out the solution.
*/
output (version);
/*
Terminate.
*/
fprintf (fp_out, "\n" );
fprintf (fp_out,"FEM1D:\n" );
fprintf (fp_out, " Normal end of execution.\n" );
fprintf ( fp_out,"\n" );
//END TIMER//
end = omp_get_wtime();
time_spent = end - begin;
timestamp ( );
//CLOSE STREAMS
fclose(fp_out);
fclose(fp_sol);
//FREE MEMORY
free(adiag);
free(aleft);
free(arite);
free(f);
free(h);
free(indx);
free(node);
free(xn);
free(xquad);
return time_spent;
}
/******************************************************************************/
/**
* checks output files for equality to determind correctness
* of new version.
*/
int check(){
FILE *fp1 = fopen("old_sol.txt","r");
FILE *fp2 = fopen("new_sol.txt","r");
int ch1, ch2;
if (fp1 == NULL) {
printf("Cannot open for reading ld_out.txt");
exit(1);
} else if (fp2 == NULL) {
printf("Cannot open for reading new_out.txt");
exit(1);
} else {
ch1 = getc(fp1);
ch2 = getc(fp2);
//printf("%d, %d\n",ch1,ch2);
while ((ch1 != EOF) && (ch2 != EOF) && (ch1 == ch2)) {
ch1 = getc(fp1);
ch2 = getc(fp2);
}
if (ch1 == ch2){
fclose(fp1);
fclose(fp2);
return 0;
} else if (ch1 != ch2) {
fclose(fp1);
fclose(fp2);
return 1;
}
}
return 1;
}
/******************************************************************************/
/*
Purpose:
ASSEMBLE assembles the matrix and right-hand-side of the linear system.
Discussion:
The linear system has the form:
K * C = F
that is to be solved for the coefficients C.
Numerical integration is used to compute the entries of K and F.
Note that a 1 point quadrature rule, which is sometimes used to
assemble the matrix and right hand side, is just barely accurate
enough for simple problems. If you want better results, you
should use a quadrature rule that is more accurate.
Licensing:
This code is distributed under the GNU LGPL license.
Modified:
29 May 2009
Author:
C version by John Burkardt
Parameters:
Output, double ADIAG(NU), the "diagonal" coefficients. That is,
ADIAG(I) is the coefficient of the I-th unknown in the I-th equation.
Output, double ALEFT(NU), the "left hand" coefficients. That is,
ALEFT(I) is the coefficient of the (I-1)-th unknown in the I-th equation.
There is no value in ALEFT(1), since the first equation
does not refer to a "0-th" unknown.
Output, double ARITE(NU).
ARITE(I) is the "right hand" coefficient of the I-th
equation in the linear system. ARITE(I) is the coefficient
of the (I+1)-th unknown in the I-th equation. There is
no value in ARITE(NU) because the NU-th equation does not
refer to an "NU+1"-th unknown.
Output, double F(NU).
ASSEMBLE stores into F the right hand side of the linear
equations.
SOLVE replaces those values of F by the solution of the
linear equations.
Input, double H(NSUB)
H(I) is the length of subinterval I. This code uses
equal spacing for all the subintervals.
Input, int INDX[NSUB+1].
For a node I, INDX(I) is the index of the unknown
associated with node I.
If INDX(I) is equal to -1, then no unknown is associated
with the node, because a boundary condition fixing the
value of U has been applied at the node instead.
Unknowns are numbered beginning with 1.
If IBC is 2 or 4, then there is an unknown value of U
at node 0, which will be unknown number 1. Otherwise,
unknown number 1 will be associated with node 1.
If IBC is 1 or 4, then there is an unknown value of U
at node NSUB, which will be unknown NSUB or NSUB+1,
depending on whether there was an unknown at node 0.
Input, int NL.
The number of basis functions used in a single
subinterval. (NL-1) is the degree of the polynomials
used. For this code, NL is fixed at 2, meaning that
piecewise linear functions are used as the basis.
Input, int NODE[NL*NSUB].
For each subinterval I:
NODE[0+I*2] is the number of the left node, and
NODE[1+I*2] is the number of the right node.
Input, int NU.
NU is the number of unknowns in the linear system.
Depending on the value of IBC, there will be NSUB-1,
NSUB, or NSUB+1 unknown values, which are the coefficients
of basis functions.
Input, int NQUAD.
The number of quadrature points used in a subinterval.
This code uses NQUAD = 1.
Input, int NSUB.
The number of subintervals into which the interval [XL,XR] is broken.
Input, double UL.
If IBC is 1 or 3, UL is the value that U is required
to have at X = XL.
If IBC is 2 or 4, UL is the value that U' is required
to have at X = XL.
Input, double UR.
If IBC is 2 or 3, UR is the value that U is required
to have at X = XR.
If IBC is 1 or 4, UR is the value that U' is required
to have at X = XR.
Input, double XL.
XL is the left endpoint of the interval over which the
differential equation is being solved.
Input, double XR.
XR is the right endpoint of the interval over which the
differential equation is being solved.
*/
void assemble (int version){
if(version == 0){
double aij;
double he;
int i;
int ie;
int ig;
int il;
int iq;
int iu;
int jg;
int jl;
int ju;
double phii;
double phiix;
double phij;
double phijx;
double x;
double xleft;
double xquade;
double xrite;
/*
Zero out the arrays that hold the coefficients of the matrix
and the right hand side.
*/
for ( i = 0; i < nu; i++ )
{
f[i] = 0.0;
}
for ( i = 0; i < nu; i++ )
{
adiag[i] = 0.0;
}
for ( i = 0; i < nu; i++ )
{
aleft[i] = 0.0;
}
for ( i = 0; i < nu; i++ )
{
arite[i] = 0.0;
}
/*
For interval number IE,
*/
for ( ie = 0; ie < NSUB; ie++ )
{
he = h[ie];
xleft = xn[node[0+ie*2]];
xrite = xn[node[1+ie*2]];
/*
consider each quadrature point IQ,
*/
for ( iq = 0; iq < nquad; iq++ )
{
xquade = xquad[ie];
/*
and evaluate the integrals associated with the basis functions
for the left, and for the right nodes.
*/
for ( il = 1; il <= NL; il++ )
{
ig = node[il-1+ie*2];
iu = indx[ig] - 1;
if ( 0 <= iu )
{
phi ( il, xquade, &phii, &phiix, xleft, xrite );
f[iu] = f[iu] + he * ff ( xquade ) * phii;
/*
Take care of boundary nodes at which U' was specified.
*/
if ( ig == 0 )
{
x = 0.0;
f[iu] = f[iu] - pp ( x ) * ul;
}
else if ( ig == NSUB )
{
x = 1.0;
f[iu] = f[iu] + pp ( x ) * ur;
}
/*
Evaluate the integrals that take a product of the basis
function times itself, or times the other basis function
that is nonzero in this interval.
*/
for ( jl = 1; jl <= NL; jl++ )
{
jg = node[jl-1+ie*2];
ju = indx[jg] - 1;
phi ( jl, xquade, &phij, &phijx, xleft, xrite );
aij = he * ( pp ( xquade ) * phiix * phijx
+ qq ( xquade ) * phii * phij );
/*
If there is no variable associated with the node, then it's
a specified boundary value, so we multiply the coefficient
times the specified boundary value and subtract it from the
right hand side.
*/
if ( ju < 0 )
{
if ( jg == 0 )
{
f[iu] = f[iu] - aij * ul;
}
else if ( jg == NSUB )
{
f[iu] = f[iu] - aij * ur;
}
}
/*
Otherwise, we add the coefficient we've just computed to the
diagonal, or left or right entries of row IU of the matrix.
*/
else
{
if ( iu == ju )
{
adiag[iu] = adiag[iu] + aij;
}
else if ( ju < iu )
{
aleft[iu] = aleft[iu] + aij;
}
else
{
arite[iu] = arite[iu] + aij;
}
}
}
// printf("%d\n",jl );
}
}
}
}
return;
//****************************** PARALLELIZED CODE **********************************
} else {
double aij;
double he;
int i;
int ie;
int ig;
int il;
int iq;
int iu;
int jg;
int jl;
int ju;
double phii;
double phiix;
double phij;
double phijx;
double x;
double xleft;
double xquade;
double xrite;
/*
Zero out the arrays that hold the coefficients of the matrix
and the right hand side.
*/
#pragma omp parallel for
for ( i = 0; i < nu; i++ )
{
f[i] = 0.0;
adiag[i] = 0.0;
aleft[i] = 0.0;
arite[i] = 0.0;
}
/*
For interval number IE,
*/
for ( ie = 0; ie < NSUB; ie++ )
{
he = h[ie];
xleft = xn[node[0+ie*2]];
xrite = xn[node[1+ie*2]];
/*
consider each quadrature point IQ,
*/
for ( iq = 0; iq < nquad; iq++ )
{
xquade = xquad[ie];
/*
and evaluate the integrals associated with the basis functions
for the left, and for the right nodes.
*/
for ( il = 1; il <= NL; il++ )
{
ig = node[il-1+ie*2];
iu = indx[ig] - 1;
if ( 0 <= iu )
{
phi ( il, xquade, &phii, &phiix, xleft, xrite );
f[iu] = f[iu] + he * ff ( xquade ) * phii;
/*
Take care of boundary nodes at which U' was specified.
*/
if ( ig == 0 )
{
x = 0.0;
f[iu] = f[iu] - pp ( x ) * ul;
}
else if ( ig == NSUB )
{
x = 1.0;
f[iu] = f[iu] + pp ( x ) * ur;
}
/*
Evaluate the integrals that take a product of the basis
function times itself, or times the other basis function
that is nonzero in this interval.
*/
for ( jl = 1; jl <= NL; jl++ )
{
jg = node[jl-1+ie*2];
ju = indx[jg] - 1;
phi ( jl, xquade, &phij, &phijx, xleft, xrite );
aij = he * ( pp ( xquade ) * phiix * phijx
+ qq ( xquade ) * phii * phij );
/*
If there is no variable associated with the node, then it's
a specified boundary value, so we multiply the coefficient
times the specified boundary value and subtract it from the
right hand side.
*/
if ( ju < 0 )
{
if ( jg == 0 )
{
f[iu] = f[iu] - aij * ul;
}
else if ( jg == NSUB )
{
f[iu] = f[iu] - aij * ur;
}
}
/*
Otherwise, we add the coefficient we've just computed to the
diagonal, or left or right entries of row IU of the matrix.
*/
else
{
if ( iu == ju )
{
adiag[iu] = adiag[iu] + aij;
}
else if ( ju < iu )
{
aleft[iu] = aleft[iu] + aij;
}
else
{
arite[iu] = arite[iu] + aij;
}
}
}
// printf("%d\n",jl );
}
}
}
}
return;
}
}
/******************************************************************************/
/*
Purpose:
FF evaluates the right hand side function.
Discussion:
This routine evaluates the function F(X) in the differential equation.
-d/dx (p du/dx) + q u = f
at the point X.
Licensing:
This code is distributed under the GNU LGPL license.
Modified:
29 May 2009
Author:
John Burkardt
Parameters:
Input, double X, the argument of the function.
Output, double FF, the value of the function.
*/
double ff ( double x ){
double value;
value = 0.0;
return value;
}
/******************************************************************************/
/*
Purpose:
GEOMETRY sets up the geometry for the interval [XL,XR].
Modified:
29 May 2009
Author:
C version by John Burkardt
Parameters:
Output, double H(NSUB)
H(I) is the length of subinterval I. This code uses
equal spacing for all the subintervals.
Input, int IBC.
IBC declares what the boundary conditions are.
1, at the left endpoint, U has the value UL,
at the right endpoint, U' has the value UR.
2, at the left endpoint, U' has the value UL,
at the right endpoint, U has the value UR.
3, at the left endpoint, U has the value UL,
and at the right endpoint, U has the value UR.
4, at the left endpoint, U' has the value UL,
at the right endpoint U' has the value UR.
Output, int INDX[NSUB+1].
For a node I, INDX(I) is the index of the unknown
associated with node I.
If INDX(I) is equal to -1, then no unknown is associated
with the node, because a boundary condition fixing the
value of U has been applied at the node instead.
Unknowns are numbered beginning with 1.
If IBC is 2 or 4, then there is an unknown value of U
at node 0, which will be unknown number 1. Otherwise,
unknown number 1 will be associated with node 1.
If IBC is 1 or 4, then there is an unknown value of U
at node NSUB, which will be unknown NSUB or NSUB+1,
depending on whether there was an unknown at node 0.
Input, int NL.
The number of basis functions used in a single
subinterval. (NL-1) is the degree of the polynomials
used. For this code, NL is fixed at 2, meaning that
piecewise linear functions are used as the basis.
Output, int NODE[NL*NSUB].
For each subinterval I:
NODE[0+I*2] is the number of the left node, and
NODE[1+I*2] is the number of the right node.
Input, int NSUB.