myfunction.c

C语言开发的微粒群优化算法源程序

// ----------------------------------------------------------------------------- MYFUNCTION
float MyFunction(struct position pos,struct param param, struct model model)
{
float	a_1,a_2;
int		d,dmax,d2;
float	f_model;
int		funct;
float	hi;
int		i,j;
int		ix,ix2;
int		n1,n2;
int		used[Max_DD]={0};
float	x1,x2;

struct position	post;

float 	total;
float	x;
float	y;

// Data for model tuning example (see case 12)

static float 	data[4][2] =
{
{	10	,	19	},
{	20	,	643	},
{	30	,	8360	},
{	40	,	20687	}
};


// Data for Master Mind
static int data_M[4] =
{ 1,2,3,4};

/*
// Arbitrary table 1
static int r_w[5][5]=
{
{9	,7,	5,	3,	1},
{8  ,6,	4,	2,	0},
{7,  5,	3,	0,	0},
{6,	 4,	0,	0	,0},
{5	,0,	0	,0,	0}
};
*/

/*
// Arbitrary table 2
static int r_w[5][5]=
{
{5	,4,	3,	2,	1},
{6  ,5,	4,	3,	0},
{7,  6,	5,	0,	0},
{8,	 7,	0,	0	,0},
{9	,0,	0	,0,	0}
};
*/

/*
// Weighted Informational table
static int r_w[5][5]=
{
{94	,273,	130,	20,	1},
{249,	172,	18,	0,	0},
{224,	52,	2,	0,	0},
{56,	2,	0,	0	,0},
{3	,0,	0	,0,	0}
};

*/


// Specific Informational table for 4 different colors
static int r_w[5][5]=
{
{16	,192,	120,	20,	1},
{152,	192,	24,	0,	0},
{312,	108,	6,	0,	0},
{136,	8,	0,	0	,0},
{9	,0,	0	,0,	0}
};


switch(param.mine)
{

case 1: // ====================Solving an equation

/* parameters
dimension 1
xmin 0
xmax 10
target 0
eps 0.001
*/

x=pos.x[0];
total=x*(x*(x-5)+1)-1; // x^3 -5x^2 +x -1 = 0
return total;

case 2: //  ======================Fermat

/* Suggested parameters:
- dimension 3
- xmin 2
- xmax 100
- target 0
- granul 1
- eps 0

You will then find three integer numbers x,y,z
so that x^2 + y^2 = z^2

*/

total = 0;
dmax=pos.size-1;
for( d=0;d<dmax;d++)  //for each dimension
	{
	total=total+pos.x[d]*pos.x[d];	
	}
total=(float)fabs(total-pos.x[dmax]*pos.x[dmax]);
return total;

case 3: //  ===========================================Knapsack
/* Suggested parameters:
- dimension 10
- xmin 1
- xmax 100
- target 100
- granul 1
- eps 0
*/
strcpy(functions[99],"Knapsack");
total = 0;
	for( d=0;d<pos.size;d++) 
		total = total + pos.x[ d];
		
return total;


case 4: //  ======================================== sin wave
/*
dimension 1
*/
strcpy(functions[99],"Sin wave");
x=pos.x[0];
total=(float)sin(x);
return total;

case 5: //  ===================================== 2D Linear system
/* dimension =2, solution (5,2)
*/
x=pos.x[0];
y=pos.x[1];
total=(float)(fabs(3*x+2*y-19)+fabs(2*x-y-8));

return total;

case 6: //  =====================================  Non linear system
/* dimension =2
*/

x=pos.x[0];
y=pos.x[1];

total=(float)((x*x+y*y-1)*(x*x+y*y-1)+(sin(10*x)-y)*(sin(10*x)-y));
return total;

//-- Intersection of two circles
total=(float)(fabs(x*x+y*y-1)+fabs((x-2)*(x-2)+y*y-4));

return total;

//--
total=(float) (fabs(sin(x)-y -sqrt(3)/2+1)+fabs(x*x-log(y)-(pi/3)*(pi/3)));
return total;

case 7: //  ===================================== D linear system
// When pos.size=3, the solution is (1,1,1)
//printf("\n\n");
total=0;
for (i=0;i<pos.size;i++)
	{
	hi=0;
	for (d=0;d<pos.size;d++)
		{
		if (i!=d) {y=(float)(i-d);} else {y=(float)pos.size;}
		hi=hi+y*pos.x[d];
//printf("%f ",y);
		}
//printf("\n");
	//total=total+fabs (hi-pos.size*i);
	total=total+(hi-pos.size*i)*(hi-pos.size*i);
	}

return total;


case 8: //  ===================================== 3 linear system. Pb chinois des 100 volailles
// Several integer solutions
//printf("\n\n");
strcpy(functions[99],"100 fowls");
	total=(float)(fabs (pos.x[0]+pos.x[1]+pos.x[2]-100));
	total=(float)(total+fabs (5*pos.x[0]+3*pos.x[1]+(1/3)*pos.x[2]-100));
return total;
	x=pos.x[0]+pos.x[1]+pos.x[2]-100;
	y=5*pos.x[0]+3*pos.x[1]+(1/3)*pos.x[2]-100;
	total=x*x+y*y;

return total;

case 9: // ===================================== Magic square
strcpy(functions[99],"Magic square");
d2=(int)(sqrt(pos.size));
total=0;

for (i=0;i<d2-1;i++)  // Rows
	{
	x=0;
	for (j=0;j<d2;j++)
		{
		d=j+d2*i;
		x=x+pos.x[d];
		d=j+d2*(i+1);
		x=x-pos.x[d];
		}
	total=total+x*x;
	}
	
for (j=0;j<d2-1;j++)  // Columns
	{
	x=0;
	for (i=0;i<d2;i++)
		{
		d=j+d2*i;
		x=x+pos.x[d]-pos.x[d+1];
		}
	total=total+x*x;
	}	

return total;


case 10: //===============================Rosenbrock with homogeneous coordinates

post=homogen_to_carte(pos);

//display_position(pos);
//display_position(post);

funct=param.funct;
param.funct=4;
total=tot(post,param,model);
param.funct=funct;

return total;

case 11: //===============================Rastrigin with homogeneous coordinates

post=homogen_to_carte(pos);

//display_position(pos);
//display_position(post);

funct=param.funct;
param.funct=7;
total=tot(post,param,model);
param.funct=funct;

return total;


case 12: // ======================== Model tuning ================
/* We have some data and a D parameters model
Find the "best" set of parameters
*/

/*
Model lambda*(mu^D)
column 1: D
column 2: function value
*/
d2=4;


total=0;

for (d=0;d<d2;d++)
	{
	f_model=(float)(pos.x[0]*pow(data[d][0],pos.x[1]));
	total=total+(f_model-data[d][1])*(f_model-data[d][1]);
	}

return total; 

case 13: // =================== 4 positions 6 colors Master Mind 

/*
// Test "Distance"
total=0;
for (d=0;d<4;d++) //Right position
{
	x1=fabs(pos.x[d]-data_M[d]);
	total=total+x1*x1;
}
return total;
*/

// "Normal" estimation
n1=0; // right color, right position
n2=0; // right color, wrong position


for (d=0;d<4;d++) //Right position
{
	ix=(int)pos.x[d];
	if (data_M[d]!=ix) continue;
		 n1=n1+1;// Right position
		used[ix]=1;
}


for (d=0;d<4;d++) // Wrong position
{
	ix=(int)pos.x[d];
	if (used[ix]>0) continue;
	
	for (d2=0;d2<4;d2++) 
	{
		if (d2==d) continue;
			ix2=data_M[d2];
			if (ix2!=ix) continue;
				if (used[ix2]>0) continue;
					n2=n2+1; 
					used[ix]=1;
//printf("\n d %i =>ix %i, d2 %i => ix2 %i. n2 %i",d,ix,d2,ix2,n2);
	}
}

//total=fabs((float)r_w[n2][n1]-1); // So that the minimum is 0

total=(float)(400-100*n1 - n2);

//display_position (pos);
//printf("\n n1, n2, r_w: %i %i => %i",n1,n2, r_w[n2][n1]);
return total; 


case 14: // Dual graph coloring 
total=0;

for (i=0;i<pos.size-2;i++)
{
	x=pos.x[i]+pos.x[i+1]-pos.x[i+2];
	total=total+x*x;
}
return total;

case 15: // Catalan's conjecture  x^m - y^n = 1 has just one integer solution (3^2 - 2^3 = 1)
// dimension = 4
total=pow(pos.x[0],pos.x[1])-pow(pos.x[2],pos.x[3])-1;
if (pos.x[0]==pos.x[2]) total=total+pow(pos.x[0],pos.x[1]); // Just to avoid the local minimum x0=x2 and x1=x3
return total;


default:
	printf("\n ERROR. MyFunction: unknown param.mine");
	exit(1);
}
}