راهنمایی در مورد حرکت اسب در صفحه شطرنج

کسی می تونه دونه دونه کدهایی رو که در زیر می نویسم رو توضیح بده.
همون برنامه حرکت اسب در صفحه شطرنج هست. بخوبی هم کار می کنه.

#include <iostream.h>
#include <iomanip.h>
#include <conio.h>
int min,npos,l,m,counter,
board[9][9]={{0},{0}} ,nexti[9]={0},
exits[11][2] ,nextj[9]={0}, //main
ktmov1[9]={0,-2,-1,1,2,2,1,-1,-2}, // variables
ktmov2[9]={0,1,2,2,1,-1,-2,-2,-1};
int i,j,t,p,q,r,temp; //temporary required elements
void move(int min)
{ //This function
i = nexti[min]; //move horse place
j = nextj[min]; //if needed
board[i][j] = m;
}
void prompt(){
cout<<'\n'<<setw(22)<<"The chessboard is: "<<'\n'<<'\n';
for(p=1 ; p<=8 ; p++) {
for(q=1 ; q<=8 ; q++)
cout<<setw(4)<< board[p][q] ;
cout<< endl;
} // end_first for
} //end_prompt
void main(){
cout<<'\n'<<"Enter Number Of Row And Column You Want To Start From ? ";
cout<<'\n'<<" ( 0<row<9 and 0<column<9 ) >> ";
cin>>i>>j;
board[i][j] = 1;
for(m=2 ; m<=64 ; m++) //main loop
{
for(p=0 ; p<=10 ;p++) //giving
for(q=0 ; q<=1 ;q++) //zero to
exits[p][q]=0; //variables
for(p=0 ; p<=8 ;p++) //and arrays
nexti[r]=nextj[r]=0; //needed
l=1 ;npos=0;
for (r=1 ; r<=8 ; r++) //This loop
{ //find rows and columns of
p =i + ktmov1[r]; //places the horse
q =j + ktmov2[r]; //can go and save
if( p<=8 && p>0 //rows in 'nexti' array and
&& q<=8 && q>0 //column in 'nextj' array
&& board [p][q]==0)
{
nexti[l] = p;
nextj[l] = q;
++l;
++npos; //number of positions
} // end_if
} //end_for
if(npos == 0){ //This condition check
prompt(); //the next place .
return; }
if(npos == 1) { // This condition check
min = 1; // if there is only one
move(min); // place to go, move horse
continue; } // to it.
t=0;
for( l=1 ; l<=npos ; l++) // This loop find and
{ // save numbers of exits
counter=0; // from every next place
for( r=1 ; r<=8 ; r++)
{
p=nexti[l] + ktmov1[r];
q=nextj[l] + ktmov2[r];
if( p<=8 && p>0 // checking next
&& q<=8 && q>0 // move is in board
&& board [p][q]==0) // and is empty
++counter;} // or not
exits[t][0]=l;
exits[t][1]=counter;
t++;
} //end for l
t=0;
min=exits[t][0];
temp=exits[t][1];
t++;
while(exits[t][0]!=0){ // finding minimum exit
if(temp>exits[t][1]){ // in next place
temp=exits[t][1];
min=exits[t][0];} // end_if
++t;
} //end_while
move(min);

}
prompt(); //showing board
getch();
cin.get();
}