A code for the detection and correction of errors was implemented using the Hamming method, this code was made in c ++. Make the code in Matlab, in such a way that the c ++ code is a base, to be able to develop the algorithm or the error detection and correction program using the Hamming method. The algorithm made in c ++ is shown below so that you can develop it in matlab. #include #include int main() { int a[20], b[20], c[20], d[20], i, k, m, f, n, j, r, p, x, y, z, ch, key, q, v, sum = 0; system("clear"); printf("\n Ingresa la longitud de la palabra de datos :"); scanf("%d", &k); printf("\n Ingresa la palabra de datos: \n"); for (i = 1; i <= k; i++) { scanf("%d", &a[i]); } m = 1; while ((k + m + 1) >= pow(2, m)) { m++; } printf("\n El valor de m es : %d", m); n = k + m; j = 1; r = 0; for (i = 1; i <= n; i++) { p = pow(2, r); if (i == p) { b[i] = 0; r++; } else { b[i] = a[j]; j++; } } printf("\n Palabra código intermedia es: \n"); for (i = 1; i <= n; i++) printf("%d", b[i]); p = 0; for (i = 1; i <= m; i++) { x = pow(2, p); r = 1; for (j = x; j <= n; j = j + (x * 2)) { for (y = j; y < (j + x); y++) { c[r] = b[y]; r++; } } z = 0; for (y = 1; y <= (r - 1); y++) { if (c[y] == 1) z++; } if (z % 2 == 0) b[x] = 0; else b[x] = 1; for (y = 1; y <= 20; y++) c[y] = 0; p++; } printf("\n\n El código hamming es: \n"); for (i = 1; i <= n; i++) printf("%d", b[i]); while (1) { printf ("\n\n Presiona 1 para cambiar un bit o 0 para salir \n\n"); scanf("%d", &ch); if (ch == 1) { printf("\n Ingresa el bit que quieras cambiar: \n"); scanf("%d", &key); for (i = 1; i <= n; i++) { if (i == key) { if (b[key] == 1) b[key] = 0; else b[key] = 1; break; } } printf("\n El nuevo código es: \n\n"); for (i = 1; i <= n; i++) printf("%d", b[i]); } else break; } p = 0; q = 0; for (i = 1; i <= m; i++) { x = pow(2, p); r = 1; for (j = x; j <= n; j = j + (x * 2)) { for (y = j; y < (j + x); y++) { c[r] = b[y]; r++; } } z = 0; for (y = 1; y <= (r - 1); y++) { if (c[y] == 1) z++; } v = z % 2; d[q] = v; sum = sum + (v * pow(2, q)); q++; for (y = 1; y <= 20; y++) c[y] = 0; p++; } if (sum == 0) printf("\n\n Ningún error encontrado....... \n\n"); else printf("\n\n Error en la posición %d", sum); printf("\n"); return 0;
A code for the detection and correction of errors was implemented using the Hamming method, this code was made in c ++. Make the code in Matlab, in such a way that the c ++ code is a base, to be able to develop the
The algorithm made in c ++ is shown below so that you can develop it in matlab.
#include <stdio.h>
#include <math.h>
int main()
{
int a[20], b[20], c[20], d[20], i, k, m, f, n, j, r, p, x, y, z, ch,
key, q, v, sum = 0;
system("clear");
printf("\n Ingresa la longitud de la palabra de datos :");
scanf("%d", &k);
printf("\n Ingresa la palabra de datos: \n");
for (i = 1; i <= k; i++) {
scanf("%d", &a[i]);
}
m = 1;
while ((k + m + 1) >= pow(2, m)) {
m++;
}
printf("\n El valor de m es : %d", m);
n = k + m;
j = 1;
r = 0;
for (i = 1; i <= n; i++) {
p = pow(2, r);
if (i == p) {
b[i] = 0;
r++;
} else {
b[i] = a[j];
j++;
}
}
printf("\n Palabra código intermedia es: \n");
for (i = 1; i <= n; i++)
printf("%d", b[i]);
p = 0;
for (i = 1; i <= m; i++) {
x = pow(2, p);
r = 1;
for (j = x; j <= n; j = j + (x * 2)) {
for (y = j; y < (j + x); y++) {
c[r] = b[y];
r++;
}
}
z = 0;
for (y = 1; y <= (r - 1); y++) {
if (c[y] == 1)
z++;
}
if (z % 2 == 0)
b[x] = 0;
else
b[x] = 1;
for (y = 1; y <= 20; y++)
c[y] = 0;
p++;
}
printf("\n\n El código hamming es: \n");
for (i = 1; i <= n; i++)
printf("%d", b[i]);
while (1) {
printf
("\n\n Presiona 1 para cambiar un bit o 0 para salir \n\n");
scanf("%d", &ch);
if (ch == 1) {
printf("\n Ingresa el bit que quieras cambiar: \n");
scanf("%d", &key);
for (i = 1; i <= n; i++) {
if (i == key) {
if (b[key] == 1)
b[key] = 0;
else
b[key] = 1;
break;
}
}
printf("\n El nuevo código es: \n\n");
for (i = 1; i <= n; i++)
printf("%d", b[i]);
} else
break;
}
p = 0;
q = 0;
for (i = 1; i <= m; i++) {
x = pow(2, p);
r = 1;
for (j = x; j <= n; j = j + (x * 2)) {
for (y = j; y < (j + x); y++) {
c[r] = b[y];
r++;
}
}
z = 0;
for (y = 1; y <= (r - 1); y++) {
if (c[y] == 1)
z++;
}
v = z % 2;
d[q] = v;
sum = sum + (v * pow(2, q));
q++;
for (y = 1; y <= 20; y++)
c[y] = 0;
p++;
}
if (sum == 0)
printf("\n\n Ningún error encontrado....... \n\n");
else
printf("\n\n Error en la posición %d", sum);
printf("\n");
return 0;
}
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