probl01.h

此代码是文化算法的一个例子

/* Write in this file your function and constraints */

/* Modify the following parameters */
#define VARIABLES 13
/* Total number of constraints */
#define NUMCONSTR 9
/* Number of equiality constraints (must be the last) */
#define NUMEQCONSTR 0
/* Number of tests for each node expansion
        PRUEBAS_ARBOL = VARIABLES or more */
#define PRUEBAS_ARBOL 13
/* Maximum depth of the tree */
#define PROFUNDIDAD_MAX 5
/* n for the 2^n-tree (must be less or equal than the
   number of variables, but we suggest not more than 4 */
#define TREEDIMS 3
/*      TREENODES = 2^TREEDIMS */
#define TREENODES 8

struct individuo {
	float variable[VARIABLES];
	float aptitud;
	float g[NUMCONSTR];
	float viol;
	char factible;
	struct celda *celda;
	unsigned char victorias;
};

struct celda {
	char clase;
	char profundidad;
	char d[TREEDIMS];
	int factibles, noFactibles;
	float lnodo[VARIABLES], unodo[VARIABLES];
	struct celda *hijo, *padre;
};

struct creencias {
	float l[VARIABLES], u[VARIABLES], L[VARIABLES], U[VARIABLES], lp[VARIABLES], up[VARIABLES];
	struct celda *raiz;
};

void limites(float[], float[]);
void evalua(struct individuo *, float, float, int);

/* Give here the lower and upper bounds for the variables */
void limites(float l[], float u[]) {
	int i;
	for (i = 0; i < 9; i++) {
		l[i] = 0.0;
		u[i] = 1.0;
	}
	for (i = 9; i < 12; i++) {
		l[i] = 0.0;
		u[i] = 100.0;
	}
	l[12] = 0.0;
	u[12] = 1.0;
}

/* Evaluaci髇 de la aptitud y de las restricciones.
   Funci髇 que cambia con el problema */
void evalua(struct individuo *ind, float deltai, float deltaf, int Gmax) {
	int i;
	float s1 = 0.0, s2 = 0.0, s3 = 0.0;
	float delta = 0.001; /* Para las restricciones de igualdad */


	/* Evaluate the objective function, and store its value in ind->aptitud.
	   Evaluate the constraints and store 1 in ind->factible if the point is feasible,
	   or 0 if it is infeasible.
	   Use the values stored in ind->variable[i]. */


	for (i = 0; i < 4; i++) {
		s1 += ind->variable[i];
		s2 += ind->variable[i]*ind->variable[i];
	}
	for (i = 4; i < 13; i++) {
		s3 += ind->variable[i];
	}

	ind->aptitud = 5*s1 - 5*s2 - s3;

	ind->g[0] = 2*ind->variable[0] + 2*ind->variable[1] + ind->variable[9] + ind->variable[10] - 10;
	ind->g[1] = 2*ind->variable[0] + 2*ind->variable[2] + ind->variable[9] + ind->variable[11] - 10;
	ind->g[2] = 2*ind->variable[1] + 2*ind->variable[2] + ind->variable[10] + ind->variable[11] - 10;
	ind->g[3] = -8*ind->variable[0] + ind->variable[9];
	ind->g[4] = -8*ind->variable[1] + ind->variable[10];
	ind->g[5] = -8*ind->variable[2] + ind->variable[11];
	ind->g[6] = -2*ind->variable[3] - ind->variable[4] + ind->variable[9];
	ind->g[7] = -2*ind->variable[5] - ind->variable[6] + ind->variable[10];
	ind->g[8] = -2*ind->variable[7] - ind->variable[8] + ind->variable[11];

	ind->factible = 1;
	for (i = 0; i < NUMCONSTR; i++) {
		if (i < NUMCONSTR - NUMEQCONSTR) {
			if (ind->g[i] > 0) {
				ind->factible = 0;
				break;
			}
		}
		else {
			if (fabs(ind->g[i]) - delta > 0) {
				ind->factible = 0;
				break;
			}
		}
	}
}