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I have a 2D array with 2 types of values : '-1' and non - '-1'. -1 implies pixel off else on. After each loop my array changes.
Hoa can i make a video from these images ? any idea , tools , any help :X

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Last Post by suho
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Can't tell you without seeing the code.

Actually, i have a DLA simulation program that moves particles in a 2D grid. So i wanted to make a video of these movements. I have not written the code for visualization. Please give me any suggestion for the visualization of the data .... :(

int main()

{

	int a=0,s_num=0,side_length=10;

	mass_count=new int[N/side_length];  

	

	for(a=0;a<N/side_length;a++)

	    mass_count[a]=0;

		

	printf("\n Enter number of simulations to be done");

	scanf("%d",&s_num);

	for(a=0;a<s_num;a++)

	{

     int i,j,k,l,v,d,direction,p,steps=100,cluster_id=0,c_id=0,px,py,n;

	 map<int,cluster*>::iterator it;

	 particles= new particle[num];

	 cluster_list = new cluster[num];



     //printf("\nEnter number of steps ... ");

     //scanf("%d",&steps);

	 int seed=time(0);// System time is taken as the seed

     srand(seed);//seeding the random number generator

     

     

	 for( i=0;i<num;i++)

	 {

		 clusters[i]=&cluster_list[i];

	 }



     for(i=0;i<N;i++)

     {

        for(j=0;j<N;j++)

            lattice[i][j]=-1;

     }



	 

     

     //Generating atoms randomly



     for(i=0;i<num;i++)

     {

        for(j=0;j< D;j++)

        {

            particles[i].x[j]=int(   (   (double)rand() / ((double)(RAND_MAX)+(double)(1)) )*N);

        }

        

        if( lattice[particles[i].x[0]][particles[i].x[1]] == -1) // if site is unoccupied

        {

            lattice[particles[i].x[0]][particles[i].x[1]] = i;

        }

        else

        {

            i--;

        }



     //printf("%d %d %d \n",i,particles[i].x[0],particles[i].x[1]);

     }

     

     for( i=0;i<num;i++)

     {

          if( particles[i].ID == -1)

          {

              

              identity(i,cluster_id);           

              cluster_id++;

          }

     }

     n_clusters=cluster_id-1;        

     

     ofstream data;


//...
     // a lot of code for simulation
	//.....

	 int steps_taken = j;

	      	 

     data.open("final.txt");

     for(i=0;i<num;i++)

     {

         for( j=0;j<D;j++)

              data<<particles[i].x[j]<<" ";

              

         data<<"\n";

     }

     data.close();



	 cout<<"\n Number of Clusters = "<<n_clusters+1<<"\nSimulation Complete........";

	 cout<<"\n Time taken = "<<time(0)-seed;



     data.open("info.txt");

     data<<N<<" "<<fraction<<" "<<time(0)-seed<<" "<<steps_taken;

	 data.close();

	 

	 

	 /*data.open("N_L.txt");

	 data<<N<<"\n";

	 

     for( i=2;i<N;i*=2)  // assuming that lattice side length = 2^n

     {

		  box b;

          b.setbox_length(i);// boxes of side length 'i'

		  b.setboxes();

		  b.fill_boxes();

          b.box_counting();

                                

          data<<i<<" "<<b.getbox_count()<<"\n";

     }

     data.close();*/

     mass_fractal(side_length);

	 find_length();

//     getch();

     //for(i=num-1;i>=0;i--)

	 //for(i=0;i<num;i++)

     //{

		 delete []particles;

		 //delete (cluster_list+i);

	 //}

   }

   write_length(s_num);

   write_mass_fractal(side_length,s_num);

   delete []mass_count;

   delete []cluster_list;

     return(0);

}

Edited by suho: n/a

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Hi there, I'm trying to do the same thing. Did you get any answers anywhere else?

cheers

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Hi there, I'm trying to do the same thing. Did you get any answers anywhere else?

cheers

Not yet... i was on a vacation. But i will definitely do it. Give me some contact info, i'll tell you when its done.

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