This is a program to calculate a definite integral using numerical quadrature method (I don't know if this is the right translation):
#include <math.h>
#include <stdio.h>
float f(float x){
float y;
y = 4/(+x*x);
return y;
}
void quadra(float a, float b, float *Q, float *E, float f(float)){
float q1, q2, m, h, fa, fb;
h = b - a;
fa = f(a);
fb = f(b);
q1 = (fa+fb)*h/2.;
m = (a+b)/2.;
q2 = ( (fa+2*f(m)+fb) ) *h/4;
*Q = q2;
*E = fabs(q2-q1)/3;
}
void scambia(float *x, float *y) {
float z;
z = *x;
*x = *y;
*y = z;
return;
}
void sort(float x[], int n) {
int flag=1, k=n-1, i;
while (flag == 1 && k > 0) {
flag = 0;
for (i=0; i<k; i++) {
if (x[i]>x[i+1]) {
scambia(&x[i], &x[i+1]);
flag = 1;
}
}
k = k-1;
}
return;
}
int intautri(float A, float B, float TOL, int MAXFUN, float *Q, float *E, int *N, float FUN(float)){
void sort(float [], int);
void quadra(float, float, float*, float*, float(float));
float Q1,Q2,c,d,Iold,Eold,E0,E1,E2,alist[100],blist[100],qlist[100],elist[100];
int n, flag;
quadra(A, B, &Q1, &E0,FUN);
*N = 3;
n = 1;
alist[n] = A;
blist[n] = B;
qlist[n] = Q1;
elist[n] = E0;
*Q = Q1;
*E = E0;
if (*E<=TOL || *N>=MAXFUN){
flag = 1;
}else{
c = alist[n];
d = blist[n];
Iold = qlist[n];
Eold = elist[n];
n = n-1;
quadra(c, (c+d)/2, &Q1, &E1, FUN);
quadra((c+d)/2, d, &Q2, &E2, FUN);
*Q = *Q - Iold + Q1 + Q2;
*E = *E - Eold + E1 + E2;
*N = *N + 6;
intautri(A , B, TOL, MAXFUN, Q, E, N, FUN);
alist[n+1] = c;
blist[n+1] = (c+d)/2;
qlist[n+1] = Q1;
elist[n+1] = E1;
alist[n+2] = (c+d)/2;
blist[n+2] = d;
qlist[n+2] = Q2;
elist[n+2] = E2;
n = n+2;
sort(alist, n);
sort(blist, n);
sort(qlist, n);
sort(elist, n);
flag = 0;
}
;
return flag;
}
int main(){
int intautri(float, float, float, int, float *, float *, int*, float(float));
float TOL, MAXFUN, A, B,Q,E;
int N, J;
float f(float);
A = 0;
B = 1;
TOL = 0.0001;
MAXFUN = 200;
J = intautri(A, B, TOL, MAXFUN, &Q, &E, &N, f);
printf("%d\n", J);
printf("%d\n", N);
printf("%f\n", Q);
return 0;
}
Anyway I compile without problems that code, but when I run it "segmentation fault (core dumped)" appears on terminal. I know that error could depend by memory allocation but I don't understand what is wrong...can you help me??
You have an exit condition in intautri:
if (*E<=TOL || *N>=MAXFUN){
but in the beginning of the function you also have:
*N = 3;
This means that your second exit condition will never fire. Without attempting to understand the logic of the code, I believe that you should:
Initialize N in main().
Remove the *N = 3 assignment in intautri().
Related
i have a C code that finds the solution of a function using the methode of bisection and Newton Raphson, and i want to compare the results using graphs ( i am asked to do so in matlab as it's a school project ), but i have no idea how.
here's my code :
`
#include <stdio.h>
#include <math.h>
#include<stdio.h>
#include<math.h>
float F0 (float V)
{
float J0 = 1e-15;
float n = 0.68;
float V0 = 0.025;
int E = 1;
int R = 100;
return (E-V-R*J0*(exp(n*V/V0)-1));
}
float F1 (float V)
{
float J0 = 1e-15;
float n = 0.68;
float V0 = 0.025;
int E = 1;
int R = 100;
return (-1-n*R*J0/V0*exp(n*V/V0));
}
void Dichotomie (float *V, float a, float b, int *itr)
{
*V=(a+b)/2;
++(*itr);
printf("Iteration no. %3d V = %7.5f\n", *itr, *V);
}
void Newton(int itr, int maxmitr, float h, float V0, float V1, float err)
{
for (itr=1; itr<=maxmitr; itr++)
{
h=F0(V0)/F1(V0);
V1=V0-h;
printf("Iteration no. %3d, V = %9.6f\n", itr, V1);
if (fabs(h) < 2*err)
{
printf("After %3d iterations, root = %8.6f\n", itr, V1);
return;
}
V0=V1;
}
printf(" The required solution does not converge or iterations are insufficient\n");
return;
}
int main ()
{
float J0 = 1e-15;
float n = 0.68;
float V0 = 0.025;
int E = 1;
int R = 100;
int itr = 0, maxmitr;
float V, a, b, err, V1;
float h;
a = 0;
b = 1;
err = 0.000001;
maxmitr = 100;
Newton(itr, maxmitr,h,V0,V1,err);
Dichotomie (&V, a, b, &itr);
do
{
if (F0(a)*F0(V) < 0)
b=V;
else
a=V;
Dichotomie (&V1, a, b, &itr);
if (fabs(V1-V) < 2*err)
{
printf("After %d iterations, root = %6.6f\n", itr, V1);
return 0;
}
V=V1;
}
while (itr < maxmitr);
printf("The solution does not converge or iterations are not sufficient");
return 1;
}
`
I read some documentations about this in the Matlab website, and i found that there is a function block in Simulink to be used, but i have no idea how Simulink works.
I'm trying to parallelize my code, but i got errors. I need to calc a Cauchy problem (it's already done) but than i need to parallelize it using OpenMP lib.
I've tried to write some code with OpenMP, but it's not working.
I've created a struct to collect result.
struct Dots {
double par;
double x;
double y;
};
This is my target function with parameter.
int ode_func (double x, const double y[], double f[], void *params)
{
double mu = *(int *)params;
f[0] = x + 2 * y[0] / (1 + mu * mu);
return GSL_SUCCESS;
}
This is the main function. I currently didn't find a way how to create a array of arrays of struct, but this is not the main problem.
void calc_cauchy_problem(struct Dots ArrayOfDots[], double x_start, double x_end, double y_start,
int count) {
int dim = 1;
double x = x_start;
double y[1] = {y_start};
int mu = 5;
int param = 0;
gsl_odeiv2_system sys = {ode_func, NULL, dim, ¶m};
gsl_odeiv2_driver * d = gsl_odeiv2_driver_alloc_y_new (&sys,
gsl_odeiv2_step_rkf45, 1e-6, 1e-6, 0.0);
int status = 0;
#pragma omp parallel for shared(ArrayOfDots) private(sys, param, d, status)
for (int param = 1; param < mu; param++) {
gsl_odeiv2_system sys = {ode_func, NULL, dim, ¶m};
gsl_odeiv2_driver * d = gsl_odeiv2_driver_alloc_y_new (&sys,
gsl_odeiv2_step_rkf45, 1e-6, 1e-6, 0.0);
for (int i = 1; i <= count; i++)
{
double xi = x_start + i * (x_end - x_start) / count;
int status = gsl_odeiv2_driver_apply(d, &x, xi, y);
if (status != GSL_SUCCESS)
{
printf ("error, return value=%d\n", status);
break;
}
// ArrayOfDots[i].par = mu;
// ArrayOfDots[i].x = xi;
// ArrayOfDots[i].y = y[0];
}
gsl_odeiv2_driver_free (d);
}
}
The main
int main() {
double x_start = 0;
double x_end = 10;
double y_start = 0;
int count = 10;
struct Dots ArrayOfDots[count];
calc_cauchy_problem(ArrayOfDots, x_start, x_end, y_start, count);
return 0;
}
It's compiled successfully with this gcc main.c -o main -fopenmp -lgsl -std=gnu11 but when i launch it i got error
gsl: driver.c:354: ERROR: integration limits and/or step direction not consistent
Default GSL error handler invoked.
I think that the main problem with this #pragma omp parallel for shared(ArrayOfDots) private(sys, param, d, status) but i have no idea how to rewrite this in the other way.
Thanks for your responses.
UPD:
With Kaveh Vahedipour help my code partially start to work. It means that half of my for cycle start to work.
UPD UPD:
After another investigations i had the following code:
It's compile and run, but i got Process finished with exit code 4 and printf("Elapsed time = %f\n", omp_get_wtime() - start_time); don't print anything.
struct Dots {
double par;
double x;
double y;
};
int ode_func (double x, const double y[], double f[], void *params)
{
double mu = *(int *)params;
f[0] = (x + 2 * y[0]) / (1 + mu * mu);
return GSL_SUCCESS;
}
void calc_cauchy_problem(double x_start, double x_end, double y_start,
int count, int param1, int param2) {
int dim = 1;
double x = x_start;
double y[1] = {y_start};
int param = param1;
int j = 0;
int status = 0;
char filename[10];
#pragma omp parallel for private(param, status, x, y)
for (param = param1; param <= param2; param++) {
struct Dots ArrayOfDots[count];
gsl_odeiv2_system sys = {ode_func, NULL, dim, ¶m};
gsl_odeiv2_driver * d =
gsl_odeiv2_driver_alloc_y_new (&sys, gsl_odeiv2_step_rkf45, 1e-6, 1e-6, 0.0);
for (int i = 1; i <= count; i++) {
double xi = x_start + i * (x_end - x_start) / count;
int status = gsl_odeiv2_driver_apply(d, &x, xi, y);
if (status != GSL_SUCCESS)
{
printf ("error, return value=%d\n", status);
break;
}
ArrayOfDots[i].par = param;
ArrayOfDots[i].x = xi;
ArrayOfDots[i].y = y[0];
}
gsl_odeiv2_driver_free (d);
}
}
int main() {
double start_time = omp_get_wtime();
double x_start = 0;
double x_end = 10;
double y_start = 0;
const int count = 500;
int param1 = 1;
int param2 = 10;
calc_cauchy_problem(x_start, x_end, y_start, count, param1, param2);
printf("Elapsed time = %f\n", omp_get_wtime() - start_time);
return 0;
}
Add x to private loop vars: private(sys, param, d, status, x). Please get back to me, if you still experience issues.
void calc_cauchy_problem(double x_start, double x_end, double y_start,
int count, int param1, int param2) {
int dim = 1;
double x = x_start;
double y[1] = {y_start};
int param = param1;
int j = 0;
int status = 0;
char filename[10];
#pragma omp parallel for private(param, status, x, y)
for (param = param1; param <= param2; param++) {
struct Dots ArrayOfDots[count];
gsl_odeiv2_system sys = {ode_func, NULL, dim, ¶m};
gsl_odeiv2_driver * d =
gsl_odeiv2_driver_alloc_y_new (&sys, gsl_odeiv2_step_rkf45, 1e-6, 1e-6, 0.0);
for (int i = 1; i <= count; i++) {
double xi = x_start + i * (x_end - x_start) / count;
int status = gsl_odeiv2_driver_apply(d, &x, xi, y);
if (status != GSL_SUCCESS)
{
printf ("error, return value=%d\n", status);
break;
}
ArrayOfDots[i].par = param;
ArrayOfDots[i].x = xi;
ArrayOfDots[i].y = y[0];
}
//write_data_to_file(param, count, ArrayOfDots);
for (int i = 0; i < count; ++i) {
printf ("%d: %f, %f, %f\n", omp_get_thread_num(),
ArrayOfDots[i].par, ArrayOfDots[i].x, ArrayOfDots[i].y);
}
gsl_odeiv2_driver_free (d);
}
}
Seems like this version works fine. I think problem was with this struct Dots ArrayOfDots[count]; and when i try to push values to this struct.
ArrayOfDots[i].par = param;
ArrayOfDots[i].x = xi;
ArrayOfDots[i].y = y[0];
Here is the full code.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <omp.h>
// GSL lib includes
#include <gsl/gsl_sf_bessel.h>
#include <gsl/gsl_errno.h>
#include <gsl/gsl_matrix.h>
#include <gsl/gsl_odeiv2.h>
int ode_func (double x, const double y[], double f[], void *params)
{
double mu = *(int *)params;
f[0] = (x + 2 * y[0]) / (1 + mu * mu);
return GSL_SUCCESS;
}
void calc_cauchy_problem(double x_start, double x_end, double y_start,
int count, int param1, int param2) {
#pragma omp parallel for
for(int param = param1; param < param2; param++) {
gsl_odeiv2_system sys = {ode_func, NULL, 1, ¶m};
gsl_odeiv2_driver * d =
gsl_odeiv2_driver_alloc_y_new (&sys, gsl_odeiv2_step_rk8pd,
1e-6, 1e-6, 0.0);
int i;
double x = x_start, x1 = x_end;
double y[1] = { y_start };
for (i = 1; i <= count; i++)
{
double xi = i * x1 / count;
int status = gsl_odeiv2_driver_apply (d, &x, xi, y);
if (status != GSL_SUCCESS)
{
printf ("error, return value=%d\n", status);
break;
}
// printf ("%d %d %.5e %.5e\n", omp_get_thread_num(), param, x, y[0]);
}
gsl_odeiv2_driver_free (d);
}
}
int main() {
double start_time = omp_get_wtime();
double x_start = 0;
double x_end = 10;
double y_start = 0;
const int count = 100000;
int param1 = 1;
int param2 = 20;
calc_cauchy_problem(x_start, x_end, y_start, count, param1, param2);
printf("Elapsed time = %f\n", omp_get_wtime() - start_time);
return 0;
}
Really thanks to Kaveh Vahedipour.
I want to create a polynomial with only one term: 6/5 and display it.
It should print this: 6/5X^0
The structure of polynomial is :
typedef struct __poly_struct_t *poly_t;
struct __poly_struct_t{
unsigned int deg;
ratio_t *coeffs;
};
Where ratio_t is an array of rational numbers, it's structure is:
typedef struct __ratio_struct_t{
int64_t num;
int64_t den;
}ratio_t;
I used two functions to construct this polynomial. polyFromRatioArray works: it prints 6/5X^0
poly_t polyFromRatioArray(ratio_t *c, unsigned int degree){
poly_t p = (struct __poly_struct_t*)malloc(sizeof(struct __poly_struct_t));
p->deg = degree;
p->coeffs = c;
return p;
}
The other one made the denominator overflowed: polyFromRatio prints 6/140218959144480X^0
poly_t polyFromRatio(ratio_t c){
return polyFromRatioArray(&c, 0);
}
Main function:
int main(){
ratio_t ra = createRatio((int64_t)6,(int64_t)5);
poly_t p1 = polyFromRatioArray(&ra, 0); // one that works
polyPrint(p1);
poly_t p2 = polyFromRatio(ra); // this doesn't
polyPrint(p2);
free(p1);
free(p2);
return 0;
}
Other fonctions involved:
ratio_t createRatio(int64_t a, int64_t b){
if(b == 0){
printf("Error : a divise by 0 \n");
exit(1);
}
ratio_t r;
int64_t pgcd = gcd(a, b); // gcd(int64_t a, int64_t b) is a function that finds pgcd using Euclid.
r.num = a/pgcd;
r.den = b/pgcd;
return r;
}
int64_t gcd(int64_t a, int64_t b){
int64_t u, v, g;
ext_eucl_div(&u, &v, &g, llabs(a), llabs(b));
return g;
}
void ext_eucl_div(int64_t *u, int64_t *v, int64_t *g, int64_t a, int64_t b){ // this function stocks pgcd of a and b in g
int64_t u1, u2, u3 , v1, v2, v3, q, t1, t2, t3;
int tour = 0;
do{
if(tour == 0){
u1 = 1; u2 = 0; u3 = a; v1 = 0; v2 = 1; v3 = b;
}
else{
u1 = v1; u2 = v2; u3 = v3; v1 = t1; v2 = t2; v3 = t3;
}
q = u3/v3;
t1 = u1 - q*v1;
t2 = u2 - q*v2;
t3 = u3%v3;
tour++;
} while(t3>=1);
*u = v1;
*v = v2;
*g = v3;
}
void polyPrint(poly_t p){
unsigned int i;
for(i=0; i<= p->deg; i++){
if(p->coeffs[i].num != 0){
printRatio(p->coeffs[i]);
if(i != p->deg) printf("X^%u + ", i);
else printf("X^%u\n", i);
}else printf("0\n");
}
}
void printRatio(ratio_t a){
printf("%" PRId64, a.num);
printf("/%" PRId64, a.den);
}
This is very strange, polyFromRatioArray and polyFromRatio seem like doing the same thing but nope.
I have a general idea of how to do this, as you can see by the code I did below. The only problem I am having is finishing the interchange part. Basically, what I am trying to do is move the value of the lowest variable into the 1st variable, the second middle value to the 2nd variable, and the biggest to the 3rd variable.
I know I do this with the interchange and using temp somehow, but how would I complete that, because with three values temp would get overridden somehow. What am I missing? So basically a = 4.0, b = 7.0, c = 1.0, c (1.0) needs to go into a, a (4.0) needs to go into b, and b (7.0) needs to go into c.
Thanks!
#include <stdio.h>
void interchange(double * x, double * y, double * z);
int main(void)
{
double a = 4.0, b = 7.0, c = 1.0;
printf_s("Originally a = %d, b = %d, and c = %d.\n", a, b, c);
interchange(&a, &b, &c);
printf_s("Now, a = %d, b = %d, and c = %d.\n", a, b, c);
return 0;
}
void interchange(double * x, double * y, double * z)
{
double temp;
temp = *z;
*y = *z;
* = temp
// what am I missing here? I cant get my head around this above ^^^
}
Thanks for the guidance!
Something like:
void swap(double* first, double* last){
double temp = *first;
*first = *last;
*last = temp;
}
void interchange(double * x, double * y, double * z){
if(*x > *y) swap(x, y);
if(*y > *z) swap(y, z);
if(*x > *y) swap(x, y);
}
the simplest way is:
if (*x > *y) {
temp = *x; // store value of x
*x = *y; // overwrite x with y
*y = temp; // overwrite y with x (which is in temp)
}
// now we sure that x <= y
if (*y > *z) {
temp = *z;
*z = *y;
*y = temp;
}
// now we sure that x <= z and y <= z, but we don't know x/y relationships
if (*x > *y) {
temp = *x;
*x = *y;
*y = temp;
}
// now x <= y <= z
I am tyring to make velocity Verlet method, by using C language.
I thought I made it good. However, there pops up 'Segmentation fault(core dumped)' whenever, I increase the size of the vector or array, x and y.
For the size n equal and less than 1e3, it's fine, but at the point of n = 1e4, the program gets error.
Please anybody help me on this.
Thank you.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double verlet(double t, double x)
{
double E = 0.252;
double B = 0.052;
double a = M_PI/2;
return -sin(x) + E*cos(t) + B*cos(2*t+a);
}
double pverlet(double(*f)(double, double), double dt, double t, double x, double y)
{
return x + dt*( y + (dt/2)*f(t, x));
}
double vverlet(double(*g)(double, double), double dt, double t, double x, double y)
{
return y + (dt/2) * g(t, x);
}
int main(void)
{
int i;
double t;
int n = 1e4;
double ti = 0, tf = 1e5, dt = (tf-ti)/n;
double *x = (double *) malloc(sizeof(double)*n);
double *y = (double *) malloc(sizeof(double)*2*n);
if (x == NULL)
{
printf("error allocating memory!\n");
return 1;
}
if (y == NULL)
{
printf("error allocating memory!\n");
return 1;
}
for (y[0] = 0, i = 1; i <2*n; i++)
{
y[i] = vverlet(verlet, dt, ti + dt*(i-1), x[i-1], y[i-1]);
}
for (x[0] = 0, i = 1; i < n; i++)
{
x[i] = pverlet(verlet, dt, ti + dt*(i-1), x[i-1], y[2*(i-1)]);
}
for (i = 0; i < n; i++)
{
t = ti + dt * i;
printf("%e %e %e\n", t, x[i], y[2*i]);
}
return 0;
free(x);
free(y);
}
for (y[0] = 0, i = 1; i <2*n; i++)
{
y[i] = vverlet(verlet, dt, ti + dt*(i-1), x[i-1], y[i-1]);
}
x is defined from 0 to n-1.