What is wrong with my GCD algorithm? - c

Here is my attempt at computing the GCD of two input numbers:
int rep;
do{
system ("cls");
int a, b, gcd=2, e, d;
cin >> a >> b;
if(a % b != 0 || b % a != 0){
do{
gcd = gcd + 1;
d = a % gcd;
e = b % gcd;
} while(d==0 && e==0);
cout << gcd-1;
}else if(a == 1 || b == 1){
gcd=1;
cout << gcd;
}else if(a >= b){
gcd = a;
cout << gcd;
}else if(b >= a){
gcd = b;
cout << gcd;
}
cin >> rep;
} while(rep == 1);
If I input 8 and 24, it gives me 2 as the answer. Can anyone spot the problem in my code?

The problem is that the algorithm gives up the first time a test GCD fails. In most cases, finding the greatest means going past some values that do not work. In this case, working up to 8 means getting past 3, 5 and 7.
8 % 24 == 8. So the do loop runs at least once. gcd becomes 3 and is tested, but does not divide evenly into 8, so the while condition evaluates to false. Then 3 - 1 (2) is streamed to cout. It's not the correct GCD, though.
You could revise your algorithm to start with the lesser of the 2 inputs, and work downward until there is a success (8 here), and then a failure (7 here).

The meat of the GCD algorithm here is only 3 lines, the rest is devoted to preventing silliness.
#include <stdio.h>
unsigned GCD(unsigned x, unsigned y) {
unsigned z;
if (x < y) {
z = x; // swap
x = y;
y = z;
}
if (y == 0)
return 0;
while (z = x % y) { // perform the GCD with implicit 0 test
x = y;
y = z;
}
return y;
}
int main(void)
{
printf("GCD of %u, %u = %u\n", 1, 0, GCD(1, 0)); // billed as C
printf("GCD of %u, %u = %u\n", 0, 1, GCD(0, 1));
printf("GCD of %u, %u = %u\n", 1, 1, GCD(1, 1));
printf("GCD of %u, %u = %u\n", 8, 24, GCD(8, 24));
return 0;
}
Program output:
GCD of 1, 0 = 0
GCD of 0, 1 = 0
GCD of 1, 1 = 1
GCD of 8, 24 = 8

Related

Program to find the prime factorization

I wrote this code to find the prime factorization of a number. I just cannot figure out the last part. If x is entered as a double or float, the program should print an error message and terminate. How do I achieve this?
#include <stdio.h>
int main()
{
int x, i;
printf("Enter an integer: ");
scanf("%d", &x);
if (x <= 1)
{
return 1;
}
printf("The prime factorization of %d is ", x);
if (x > 1)
{
while (x % 2 == 0)
{
printf("2 ");
x = x / 2;
}
for (i = 3; i < 1009; i = i + 2)
{
while (x % i == 0)
{
printf("%d ", i);
x = x / i;
}
}
}
return 0;
}
Your starting point should cover all desired and undesired cases so you should take float number from a user, not int. Then, you should check whether whole decimal part of the number is 0. That is, if all of them equals 0, the user want to provide an int number instead of float.
First step is to declare a float number:
float y;
After, take its value from the user:
scanf("%f", &y);
Second step is to check whether it is int or float. There are many ways for this step. For example, I find roundf() function useful. It takes a float number and computes the nearest integer to this number. So if the nearest integer is the number itself then the number has to be int. Right?
if(roundf(y)!=y)
If you are sure it is an int, you can move onto the third step: convert float type to int type. It is called type-casting. This step is required for the remaining part of your program because in your algorithm you manipulate the number with int type so just convert it to int:
x = (int)y;
After adding the line above, you can use the rest of code which you typed. I give the whole program:
#include <stdio.h>
#include <math.h>
int main()
{
int x,i;
float y;
printf("Enter an integer: ");
scanf("%f", &y);
if(roundf(y)!=y){
printf("%f is not an integer!",y);
return 1;
}
else{
x = (int)y;
}
if (x <= 1)
{
printf("%d <= 1",x);
return 1;
}
else
{
printf("The prime factorization of %d is ", x);
while (x%2 == 0)
{
printf("2 ");
x = x / 2;
}
for ( i = 3; i < 1009; i = i + 2)
{
while (x%i == 0)
{
printf("%d ",i);
x = x / i;
}
}
}
return 0;
}
The use of scanf() is a bit tricky, I would avoid it to scan user generated input at almost all cost. But nevertheless here is a short overview for how to get the errors of scanf()
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <string.h>
int main(void)
{
int x, i, scanf_return;
printf("Enter an integer: ");
/* Reset "errno". Not necessary here, just in case. */
errno = 0;
/* scanf() returns a value in case of an error */
scanf_return = scanf("%d", &x);
/*
* scanf() returns "EOF" if it didn't find all what you wanted or
* and error happened.
* It sets "errno" to the value of the actual error. See manpage
* for all of the details.
*/
if (scanf_return == EOF) {
/*
* The error is connected to the stream, so we can differ between
* an error within scanf() and and error with the input stream
* (here: stdin)
*/
if (ferror(stdin)) {
fprintf(stderr, "Something went wrong while reading stdin: %s\n", strerror(errno));
exit(EXIT_FAILURE);
} else {
/* e.g. a conversion error, a float instead of an integer, letters
instead of a decimal number */
fprintf(stderr, "Something went wrong within scanf()\n");
exit(EXIT_FAILURE);
}
}
/*
* If no error occurred, the return holds the number of objects
* scanf() was able to read. We only need one, but it would throw an
* error if cannot find any objects, so the check is here for
* pedagogical reasons only.
*/
if (scanf_return != 1) {
fprintf(stderr, "Something went wrong within scanf(): wrong number of objects read.\n");
exit(EXIT_FAILURE);
}
if (x <= 1) {
fprintf(stderr, "Input must be larger than 1!\n");
exit(EXIT_FAILURE);
}
printf("The prime factorization of %d is ", x);
/* No need for that test, x is already larger than one at this point. */
/* if (x > 1) { */
while (x%2 == 0) {
printf("2 ");
x = x / 2;
}
for (i = 3; i < 1009; i = i + 2) {
while (x%i == 0) {
printf("%d ",i);
x = x / i;
}
}
/* } */
/* Make it pretty. */
putchar('\n');
exit(EXIT_SUCCESS);
}
Does it work?
$ ./stackoverflow_003
Enter an integer: 1234
The prime factorization of 1234 is 2 617
$ factor 1234
1234: 2 617
$ ./stackoverflow_003
Enter an integer: asd
Something went wrong within scanf(): wrong number of objects read.
$ ./stackoverflow_003
Enter an integer: 123.123
The prime factorization of 123 is 3 41
No, it does not work. Why not? If you ask scanf() to scan an integer it grabs all consecutive decimal digits (0-9) until there is no one left. The little qualifier "consecutive" is most likely the source of your problem: a floating point number with a fractional part has a decimal point and that is the point where scanf() assumes that the integer you wanted ended. Check:
$ ./stackoverflow_003
Enter an integer: .123
Something went wrong within scanf(): wrong number of objects read
How do you find out? #weather-vane gave one of many ways to do so: check if the next character after the integer is a period (or another decimal separator of your choice):
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <string.h>
int main(void)
{
int x, i, scanf_return;
char c = -1;
printf("Enter an integer: ");
/* Reset "errno". Not necessary here, just in case. */
errno = 0;
/* scanf() returns a value in case of an error */
scanf_return = scanf("%d%c", &x, &c);
/*
* scanf() returns "EOF" if it didn't find all what you wanted or
* and error happened.
* It sets "errno" to the value of the actual error. See manpage
* for all of the details.
*/
if (scanf_return == EOF) {
/*
* The error is connected to the stream, so we can differ between
* an error within scanf() and and error with the input stream
* (here: stdin)
*/
if (ferror(stdin)) {
fprintf(stderr, "Something went wrong while reading stdin: %s\n", strerror(errno));
exit(EXIT_FAILURE);
} else {
/* e.g. a conversion error, a float instead of an integer, letters
instead of a decimal number */
fprintf(stderr, "Something went wrong within scanf()\n");
exit(EXIT_FAILURE);
}
}
/*
* If no error occurred, the return holds the number of objects
* scanf() was able to read. We can use this information now.
* If there is a period (actually any character) after the integer
* it returns 2 (assuming no error happened, of course)
*/
/* If no integer given, the following character ("%c") gets ignored. */
if (scanf_return == 0) {
fprintf(stderr, "Something went wrong within scanf(): no objects read.\n");
exit(EXIT_FAILURE);
}
/* Found two objects, check second one which is the character. */
if (scanf_return == 2) {
if (c == '.') {
fprintf(stderr, "Floating point numbers are not allowed.\n");
exit(EXIT_FAILURE);
}
}
if (x <= 1) {
fprintf(stderr, "Input must be larger than 1!\n");
exit(EXIT_FAILURE);
}
printf("The prime factorization of %d is ", x);
/* No need for that test, x is already larger than one at this point. */
/* if (x > 1) { */
while (x%2 == 0) {
printf("2 ");
x = x / 2;
}
for (i = 3; i < 1009; i = i + 2) {
while (x%i == 0) {
printf("%d ",i);
x = x / i;
}
}
/* } */
/* Make it pretty. */
putchar('\n');
exit(EXIT_SUCCESS);
}
Check:
$ ./stackoverflow_003
Enter an integer: 123
The prime factorization of 123 is 3 41
$ ./stackoverflow_003
Enter an integer: 123.123
Floating point numbers are not allowed.
$ ./stackoverflow_003
Enter an integer: .123
Something went wrong within scanf(): no objects read.
Looks good enough for me. With one little bug:
$ ./stackoverflow_003
Enter an integer: 123.
Floating point numbers are not allowed
But I think I can leave that as an exercise for the dear reader.
You can try this simple C99 implementation of Pollard Rho algorithm :
// Integer factorization in C language.
// Decompose a composite number into a product of smaller integers.
unsigned long long pollard_rho(const unsigned long long N) {
// Require : N is a composite number, not a square.
// Ensure : you already performed trial division up to 23.
// Option : change the timeout, change the rand function.
static const int timeout = 18;
static unsigned long long rand_val = 2994439072U;
rand_val = (rand_val * 1025416097U + 286824428U) % 4294967291LLU;
unsigned long long gcd = 1, a, b, c, i = 0, j = 1, x = 1, y = 1 + rand_val % (N - 1);
for (; gcd == 1; ++i) {
if (i == j) {
if (j >> timeout)
break;
j <<= 1;
x = y; // "x" takes the previous value of "y" when "i" is a power of 2.
}
a = y, b = y; // computes y = f(y)
for (y = 0; a; a & 1 ? b >= N - y ? y -= N : 0, y += b : 0, a >>= 1, (c = b) >= N - b ? c -= N : 0, b += c);
y = (1 + y) % N; // function f performed f(y) = (y * y + 1) % N
for (a = y > x ? y - x : x - y, b = N; (a %= b) && (b %= a););
gcd = a | b; // the GCD(abs(y - x), N) was computed
// it continues until "gcd" is a non-trivial factor of N.
}
return gcd;
}
Usually you performed some trial division before calling the algorithm
The algorithm isn't designed to receive a prime number as input
Two consecutive calls may not result in the same answer
Alternately, there is a pure C quadratic sieve which factors numbers from 0 to 300-bit.
If in doubt about the primality of N you can use a C99 primality checker :
typedef unsigned long long int ulong;
ulong mul_mod(ulong a, ulong b, const ulong mod) {
ulong res = 0, c; // return (a * b) % mod, avoiding overflow errors while doing modular multiplication.
for (b %= mod; a; a & 1 ? b >= mod - res ? res -= mod : 0, res += b : 0, a >>= 1, (c = b) >= mod - b ? c -= mod : 0, b += c);
return res % mod;
}
ulong pow_mod(ulong n, ulong exp, const ulong mod) {
ulong res = 1; // return (n ^ exp) % mod
for (n %= mod; exp; exp & 1 ? res = mul_mod(res, n, mod) : 0, n = mul_mod(n, n, mod), exp >>= 1);
return res;
}
int is_prime(ulong N) {
// Perform a Miller-Rabin test, it should be a deterministic version.
const ulong n_primes = 9, primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
for (ulong i = 0; i < n_primes; ++i)
if (N % primes[i] == 0) return N == primes[i];
if (N < primes[n_primes - 1]) return 0;
int primality = 1, a = 0;
ulong b;
for (b = N - 1; ~b & 1; b >>= 1, ++a);
for (ulong i = 0; i < n_primes && primality; ++i) {
ulong c = pow_mod(primes[i], b, N);
if (c != 1) {
for (int j = a; j-- && (primality = c + 1 != N);)
c = mul_mod(c, c, N);
primality = !primality;
}
}
return primality;
}
To try it there is a factor function :
// return the number that was multiplied by itself to reach N.
ulong square_root(const ulong num) {
ulong res = 0, rem = num, a, b;
for (a = 1LLU << 62 ; a; a >>= 2) {
b = res + a;
res >>= 1;
if (rem >= b)
rem -= b, res += a;
}
return res;
}
ulong factor(ulong num){
const ulong root = square_root(num);
if (root * root == num) return root ;
const ulong n_primes = 9, primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
for (ulong i = 0; i < n_primes && primes[i] <= root; ++i)
if (num % primes[i] == 0) return primes[i];
if (is_prime(num))
return 1 ;
return pollard_rho(num);
}
Which is completed by the main function :
#include <assert.h>
int main(void){
for(ulong i = 2; i < 63; ++i){
ulong f = factor(i);
assert(f <= 1 || f >= i ? is_prime(i) : i % f == 0);
ulong j = (1LLU << i) - 1 ;
f = factor(j);
assert(f <= 1 || f >= j ? is_prime(j) : j % f == 0);
j = 1 | pow_mod((ulong) &main, i, -5);
f = factor(j);
assert(f <= 1 || f >= j ? is_prime(j) : j % f == 0);
}
}
There are some problems in your code:
you do not check the return value of scanf, so you cannot detect invalid or missing input and will have undefined behavior in those cases.
you only test divisors up to 1009, so composite numbers with larger prime factors do not produce any output.
prime numbers larger than 1009 do not produce any output.
you should probably output a newline after the factors.
Testing and reporting invalid input such as floating point numbers can be done more easily by reading the input as a full line and parsing it with strtol().
Here is a modified version:
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
int main() {
char input[120];
char ch;
char *p;
long x, i;
int last_errno;
printf("Enter an integer: ");
if (!fgets(input, sizeof input, stdin)) {
fprintf(stderr, "missing input\n");
return 1;
}
errno = 0;
x = strtol(input, &p, 0);
last_errno = errno;
if (p == input || sscanf(p, " %c", &ch) == 1) {
fprintf(stderr, "invalid input: %s", input);
return 1;
}
if (last_errno == ERANGE) {
fprintf(stderr, "number too large: %s", input);
return 1;
}
if (x < 0) {
fprintf(stderr, "number is negative: %ld\n", x);
return 1;
}
if (x <= 1) {
return 1;
}
printf("The prime factorization of %ld is", x);
while (x % 2 == 0) {
printf(" 2");
x = x / 2;
}
for (i = 3; x / i >= i;) {
if (x % i == 0) {
printf(" %ld", i);
x = x / i;
} else {
i = i + 2;
}
}
if (x > 1) {
printf(" %ld", x);
}
printf("\n");
return 0;
}

Understanding calling one function inside another C

I'd like to ask the following misunderstandings of C language, which I see I'm having.
I'm sorry if the code is not properly indented, I tried as much as I could but there are not so many guides on the internet.
The program asked given a starting number 'val' and a Even-Odd or Odd-Even alternating sequence (which stops whenever this rules is violated) to print the greater prime number with 'val'.
I tried with two functions and the main: one to control the GCD between two given numbers and the other to keep tracks of the greatest one, but I think I miss something in the code or in the conception of C function,
Because when compiled it returns me 0 or great number which I'm not entering.
One example to understand what I should do:
If my sequence was 10, 7, 8, 23 and my val was 3, I had to print 23, because it is the greatest integer prime with 3.
Here's the code :
#include <stdio.h>
int mcd(int a, int b)
{ // Gcd function
if (a == 0)
return b;
else
return mcd(b % a, b);
}
int valuta(int val, int h) // Valuing Max function
{
int temp = 0;
if (mcd(val, h) == 1 && h > temp)
temp = h;
return temp;
}
int main()
{
int val, d, x, y, z, t, contatore = 1;
scanf("%d", &val);
scanf("%d%d", &x, &y);
if (x > y && mcd(val, x) == 1)
{ // Two options
t = x;
}
else if (y > x && mcd(val, y) == 1)
{
t = y;
}
if ((x % 2 == 0 && y % 2 == 0) || (x % 2 == 1 && y % 2 == 1))
{ // Bad case
if (x > y && mcd(val, x) == 1)
{
t = x;
contatore = 0;
}
else if (y > x && mcd(val, y) == 1)
{
t = y;
contatore = 0;
}
}
else
{
while (contatore == 1)
{
scanf("%d", &z);
t = valuta(val, z);
if (x % 2 == 0 && z % 2 == 0)
{ // Even- Odd - Even
scanf("%d", &d);
t = valuta(val, d);
if (d % 2 == 0)
{
contatore = 0;
}
else
{
contatore = 0;
}
}
if (x % 2 == 1 && z % 2 == 1)
{ //Odd- Even- Odd
scanf("%d", &d);
t = valuta(val, d);
if (d % 2 == 1)
{
contatore = 0;
}
else
{
contatore = 0;
}
}
}
}
printf("%d\n", t);
return 0;
}
PS. Is there any way to reduce the number of lines of code or to reduce the effort in coding? I mean, a straightforward solution will be helpful.
Your valuta() function is flawed in that it needs to return the maximum qualifying value so far but has no knowledge of the previous maximum - temp is always zero. The following takes the previous maximum as an argument:
int valuta(int val, int h, int previous )
{
return ( mcd(val, h) == 1 && h > previous ) ? h : previous ;
}
And is called from main() thus:
t = valuta( val, x, t ) ;
The test mcd(val, h) == 1 is flawed, because mcd() only ever returns the value of parameter b which is not modified in the recursion, so will never return 1, unless the argument b is 1. Since I have no real idea what mcd() is intended to do, I cannot tell you how to fix it. It appear to be a broken implementation of Euclid's greatest common divisor algorithm, which correctly implemented would be:
int mcd(int a, int b)
{
if(b == 0)
return a;
else
return mcd(b, a % b);
}
But I cannot see how that relates to:
"[...] he greatest integer prime with 3 [...]
The odd/even even/odd sequence handling can be drastically simplified to the extent that it is shorter and simpler than your method (as requested) - and so that it works!
The following is a clearer starting point, but may not be a solution since it is unclear what it is it is supposed to do.
#include <stdio.h>
#include <stdbool.h>
int mcd(int a, int b)
{
if(b == 0)
return a;
else
return mcd(b, a % b);
}
int valuta(int val, int h, int previous )
{
return ( mcd(val, h) && h > previous ) ? h : previous ;
}
int main()
{
int val, x, t ;
printf( "Enter value:") ;
scanf("%d", &val);
typedef enum
{
EVEN = 0,
ODD = 1,
UNDEFINED
} eOddEven ;
eOddEven expect = UNDEFINED ;
bool sequence_valid = true ;
printf( "Enter sequence in odd/even or even/odd order (break sequence to exit):\n") ;
while( sequence_valid )
{
scanf("%d", &x);
if( expect == UNDEFINED )
{
// Sequence order determined by first value
expect = (x & 1) == 0 ? EVEN : ODD ;
}
else
{
// Switch expected odd/even
expect = (expect == ODD) ? EVEN : ODD ;
// Is new value in the expected sequence?
sequence_valid = (expect == ((x & 1) == 0 ? EVEN : ODD)) ;
}
// If the sequence is valid...
if( sequence_valid )
{
// Test if input is largest qualifying value
t = valuta( val, x, t ) ;
}
}
// Result
printf("Result: %d\n", t);
return 0;
}

Number doesn't print in expected way

This is my program on 'Sin Calculator'
#include <stdio.h>
#include <math.h>
#define PI 3.1416
void fraction(double x, int y);
void print(double sx, int x, int y, int s);
void scal();
int main(){
scal();
return 0;
}
void fraction(double x, int y){
int b = 100, a;
a = x * 100;
while ((a % 2 != 0) || (a % 5 != 0)){
if (a % 2 == 0){
a /= 2;
b /= 2;
}
else if (a % 5 == 0){
a /= 5;
b /= 5;
}
}
print(x, a, b, y);
}
void print(double sx, int x, int y, int s){
printf (" Sin(%d) = %d/%d || %.2lf\n",s,x,y,sx);
scal();
}
void scal(){
double sine, sinx;
int x, a, b;
printf ("\n Sin(X) : ");
scanf ("%d",&x);
sinx = x * (PI / 180);
sine = sin(sinx);
fraction(sine, x);
}
I don't get any errors. But when i run it, though the variable 'a' of fraction function can be divided by 5 or 2, it doesn't do it. As a result I get the whole value of 'a'. For example, the value of Sin30 degrees is 0.50, so multiplying it with 100 makes 'a' 50. 50 is dividable by 2 and 5. But in fraction function, it seems that 'a' doesn't get divided there. As a result in 'print' function, i get '50/100' instead of '1/2'. Why is this happening? And also when i enter somethin like Sin23, the program doesn't finish. It stops. What's the problem?
while ((a % 2 != 0) || (a % 5 != 0))
this test fails for the value of 50 or any integer that is multiple of both 2 an 5. Therefore the while loop will not be entered and no division will occur, so a conserves its value.
maybe your intent was the inverse of the condition:
while ((a % 2 == 0) || (a % 5 == 0))

Program Crashing and some questions about arithmetic operations in coding

Hi stackoverflow community! I'm having some problems with my code. I'm currently a student so basically I'm a beginner. Using Euclidean Algorithm, the code shown below should divide and divide two numbers until the quotient reach 0 but it just stops at the last dividing process just before the quotient turns 0. I don't know if the program crashes because of this. Hoping for a beginner-friendly reply from you guys. Thanks! (sorry if there's already an existing question like this)
Here's the code:
int quotient,quotient2,remainder,remainder2,x,y;
int foo()
{
printf("Enter a number: ");
scanf("%d", &x);
printf("Enter another number: ");
scanf("%d", &y);
if(y >= x){
quotient2 = y / x;
remainder2 = y % x;
printf("%d = %d(%d) + %d\n", y,x,quotient2,remainder2);
if(quotient2 != 0){
do{
y = x;
x = remainder2;
quotient2 = y / x;
remainder2 = y % x;
printf("%d = %d(%d) + %d\n", y,x,quotient2,remainder2);
} while(quotient2 != 0);
}
} else if(x > y){
quotient = x / y;
remainder = x % y;
printf("%d = %d(%d) + %d\n", x,y,quotient,remainder);
if(quotient != 0){
do{
x = y;
y = remainder;
quotient = x / y;
remainder = x % y;
printf("%d = %d(%d) + %d\n", x,y,quotient,remainder);
} while(quotient != 0);
}
}
system("pause");
return 0;
}
If your "Euclidean algorithm" is computing the GCD of two numbers, here is one way to do it. Note it does not divide by 0, it stops when the divisor would be 0.
So the point is, rather than guard against dividing by 0, that's the iteration's end condition anyway.
#include <stdio.h>
unsigned gcd(unsigned x, unsigned y) {
unsigned z;
if (x == 0 || y == 0) {
return 0;
}
while ((z = y % x) != 0) {
y = x;
x = z;
}
return x;
}
int main(void)
{
printf("20 ~ 20 : %u\n", gcd(20, 20));
printf("20 ~ 0 : %u\n", gcd(20, 0));
printf(" 0 ~ 20 : %u\n", gcd( 0, 20));
printf("20 ~ 16 : %u\n", gcd(20, 16));
printf("16 ~ 20 : %u\n", gcd(16, 20));
printf("20 ~ 15 : %u\n", gcd(20, 15));
printf("15 ~ 20 : %u\n", gcd(15, 20));
printf(" 1 ~ 2 : %u\n", gcd( 1, 2));
printf(" 2 ~ 1 : %u\n", gcd( 2, 1));
return 0;
}
Program output:
20 ~ 20 : 20
20 ~ 0 : 0
0 ~ 20 : 0
20 ~ 16 : 4
16 ~ 20 : 4
20 ~ 15 : 5
15 ~ 20 : 5
1 ~ 2 : 1
2 ~ 1 : 1
Note there is no need to swap the arguments. The algorithm works no matter which way round they are.
Notice that with x = remainder2;, x could take on the value of 0. Then the next quotient2 = y / x; remainder2 = y % x; both perform operations (divide by 0 and remainder 0) which are both undefined behavior. The program crash is certainly due to this.
if(y >= x){
quotient2 = y / x;
remainder2 = y % x;
printf("%d = %d(%d) + %d\n", y,x,quotient2,remainder2);
if(quotient2 != 0){
do{
y = x;
x = remainder2;
quotient2 = y / x;
remainder2 = y % x;
printf("%d = %d(%d) + %d\n", y,x,quotient2,remainder2);
} while(quotient2 != 0);
}
As code is swapping the roles of x,y with each operation, code could simplify to: (Also see What is gcd(0,a)gcd(0,a), where a is a positive integer?
unsigned gcd(unsigned a, unsigned b) {
while (b) {
a %= b;
if (a == 0) return b;
b %= a;
}
return a;
}

Point operations on an elliptic curve in a prime field in c

I am trying to write a program to perform point operations on a elliptic curve in a prime field I am using the standard formulaes for point additions and doubling in my code and these operations are performed by functions that are called but I am getting output for certain points but not all so please help me to fix the problem that are present in this code.
structure point_multiply(int x, int y, int k )
{
int xk;
int yk,m;
xk=x;
yk=y;
m=1;
int xL,yL,s,e;
e=findInverse((2*yk),211);
if((((3*(xk*xk))*e)% 211)>0)
{s = (((3*(xk*xk))*e)% 211);
}
else
s=(((3*(xk*xk))*e)% 211)+ 211;
if((((s*s)- (2*xk)) % 211)>0)
{xL=(((s*s)- (2*xk)) % 211);
}
else
xL=(((s*s)- (2*xk)) % 211) + 211;
if(((-yk+ s*(xk-xL)) % 211) > 0)
yL=(-yk+ s*(xk-xL)) % 211;
else
yL=(-yk+ s*(xk-xL)) % 211 + 211;
xk=xL;
yk=yL;
m=m+1;
while(k>m)
{
sn=point_addition(xk,yk,x,y);
xk=sn.a;
yk=sn.b;
m++;
}
s1.a=xk;
s1.b=yk;
return s1;
}
structure point_addition(int x1, int y1, int x2, int y2)
{
int s,xL,yL;
if((x1-x2)!=0)
{
if ( x1 == 0 && y1 == 0 )
{
xL = x2;
yL = y2;
s7.a=xL;
s7.b=yL;
return s7;
}
if ( x2 == 0 && y2 == 0 )
{
xL = x1;
yL = y1;
s7.a=xL;
s7.b=yL;
return s7;
}
if ( y1 == -y2 )
{
xL = yL = 0;
s7.a=xL;
s7.b=yL;
return s7;
}
l=findInverse((x1-x2),211);
if ((((y1-y2)*l) % 211)>=0)
s=((((y1-y2)*l) % 211));
else
s=(((y1-y2)*l) % 211) + 211;
if ((((s*s)-(x1+x2)) % 211)>0)
xL= (((s*s)-(x1+x2)) % 211) ;
else
xL= (((s*s)-(x1+x2)) % 211) + 211;
if(((-y1+s*(x1-xL)))>=0)
yL= ((-y1+s*(x1-xL)) % 211);
else
yL= ((-y1+s*(x1-xL)) % 211) + 211;
}
else
{
xL= 0 ;
yL= 0;
}
s7.a= xL;
s7.b= yL;
return s7 ;
}
int findInverse(int a, int b)
{
int x[3];
int y[3];
int quotient = a / b;
int remainder = a % b;
x[0] = 0;
y[0] = 1;
x[1] = 1;
y[1] = quotient * -1;
int i = 2;
for (; (b % (a%b)) != 0; i++)
{
a = b;
b = remainder;
quotient = a / b;
remainder = a % b;
x[i % 3] = (quotient * -1 * x[(i - 1) % 3]) + x[(i - 2) % 3];
y[i % 3] = (quotient * -1 * y[(i - 1) % 3]) + y[(i - 2) % 3];
}
//x[i — 1 % 3] is inverse of a
//y[i — 1 % 3] is inverse of b
if(x[(i - 1) % 3]<0)
return x[(i - 1) % 3]+211;
else
//x[i — 1 % 3] is inverse of a
//y[i — 1 % 3] is inverse of b
return x[(i - 1) % 3];
}
Edited and added main c code which uses these function to perform elliptic curve cryptography
int main()
{
int y,z=0,x=2,i[200],j[200],h=0,g,k;
while(x<200)
{
y=sqrt((x*x*x)-4);
z=modulo(y,211);
if(z!=0)
{
i[h]=x;
j[h]=z;
s[h].a=i[h];
s[h].b=j[h];
s[h+1].a=i[h];
s[h+1].b=(211 - j[h]);
printf("\nh=%d X= %d Y= %d \nh=%d X= %d Y= %d",h,s[h].a,s[h].b,h+1,s[h+1].a,s[h+1].b);
h=h+2;
}
x++;
}
printf("The total no of points we have on our elliptic curve for cryptography is %d",h-1);
x=5;
y=11;
printf("\n %d %d\n",x,y );
printf("\nEnter A number between 0 and the private key");
scanf("%d",&k);
s2=point_multiply(x,y,k);
printf("\n The public key is \n %d %d \n ",s2.a,s2.b );
printf("Enter a RANDOM number to generate the cipher texts");
scanf("\n%d",&g);
s3= point_multiply(x,y,g);
s4=point_multiply(s2.a,s2.b,g );
label:
printf("\n Enter a number to send");
scanf("%d",&h);
s6=point_addition(s4.a,s4.b,s[h].a,s[h].b);
printf("The points to be sent are X= %d Y=%d",s[h].a,s[h].b);
printf(" \n X= %d Y=%d\n X = %d Y= %d ",s3.a,s3.b,s6.a,s6.b);
//RECIEVER
s8=point_multiply(s3.a,s3.b,k);
s9=point_addition((s8.a) ,-((s8.b)%211),s6.a,s6.b);
printf(" The decrypted points are \n %d %d",s9.a,s9.b);
printf("\n If you have more no to send press 1 else press 0");
scanf("\n %d", &x1);
if(x1==1)
goto label;
else
return 0;
}
s1, s2, s3 etc are structures which hold a 2 integers which act as x and y co-ordinates
I am getting output by entering k=3,g=4, h=5 and many other cases mostly with small numbers but not for larger numbers. What could be wrong with the code?
Further edit: I guess that normal square root method is not applicable to find square roots of a modular no?.. Please tell me how to find the modular square root of a no?

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