Hi all, im making a game in c++ and i have a problem.. well its not a problem but i want to know the best way to compare the colors with te row and the columns, the way im thinking is .. something like this.. :

char board[7][7];

int row , col = 1;
int srch=1;
for(row=6;row>1;row--)
{
                  if (map[row][col]== map [row][col+srch] && map[row][col]== map [row][col+srch+1] && .....                   map[row][col+srch+3]== map [row][col+srch+4]) 
                  {
                        map[row][col]=' ';
                        ...
                        map[row][col+srch+4]=' ';
                  }
                  else
               else
            else....etc
{
srch++;
}
} ....

well something like that ....establishing all the possible IF opcions :S... but there is a easier way to detect the collisions?? like in c# bounds.intersect? ... as you can see, in the board, 1 is for RED, 2 is BLUE, 3 YELLOW, 4 Green .. when there are 4 or more of the same kind intersecting they're all destroyed.. so is a better way or function ? im using c++ and allegro.. thank you all for your help.

int sequenceColor;
int startIndex;

for each row:
 for(int col = 0; col < numColumns; ++col) {
   int color = map[row, col];
   if(color != sequenceColor) {
     // new sequence begins, check if last sequence is long enough to clear
     int endIndex = col;
     if(endIndex - startIndex >= 4) {
        set map[row, i] = ' ', for i in [startIndex, ..., endIndex - 1]
     }

     startIndex = col; // start new sequence
     sequenceColor = color;
   }
 }

from your code above i assume you're only interested in finding adjacent columns of the same color
in an array. it's not really an intersection problem i think.
so

231111134
121122344

would become

23_____34
121122344

you could look for the start and end of sequences of the same color. something like this would work:

int sequenceColor;
int startIndex;

for each row:
 for(int col = 0; col < numColumns; ++col) {
   int color = map[row, col];
   if(color != sequenceColor) {
     // new sequence begins, check if last sequence is long enough to clear
     int endIndex = col;
     if(endIndex - startIndex >= 4) {
        set map[row, i] = ' ', for i in [startIndex, ..., endIndex - 1]
     }

     startIndex = col; // start new sequence
     sequenceColor = color;
   }
 }

Hi, thank you for your answer but im also looking for adjacent in the upper rows and then start again on the cols, but maybe saving the positions? in a new array? like in your example :

231111134 ---> 23_____34 saving (3,4,5,6,7) then compare in the upper row and if there is 1 delete too..

For instance:

if i have:

224133

123144

221124

2311134

would become

__4_33

1_3___

____2_

_3___3_

or maybe in the first loop will erase all the 2 then the 1 ,then all the 4 until there is no more equal adjacent

then those numbers will drop to look for combos or chains and would become:

______

______

____3_

1334233

and look again for coincidence but i dont think that would be a problem just running again the code.

Thanks again.

well something like that ....establishing all the possible IF opcions :S... but there is a easier way to detect the collisions??

1. For every unvisited tile on the board:
2.	For each unvisited neighbor of this tile:
3.		Visit each neighbor of the same color:
4.			Count and mark this neighbor tile
5.			Move to this tile and repeat step 2 for this neighbor tile
6.	If there are no more neighbors, and we have moved to this tile:
7.		Move back, and continue on step 5
8.	If we found four or more tiles of the same color neighboring this tile:
9.		Score and remove these tiles

This is important step in your growth as a programmer: When you realize what you have learned is infeasible here and that there must be an easier way. While it is certainly possible to code the logic entirely of IF-statements, it quickly becomes unmanageable and buggy. So, as a programmer, you must solve the problem in a generic way. For instance, the procedure could look something like this, in pseudo code:

1. For every unvisited tile on the board:2.	For each unvisited neighbor of this tile:3.		Visit each neighbor of the same color:4.			Count and mark this neighbor tile5.			Move to this tile and repeat step 2 for this neighbor tile6.	If there are no more neighbors, and we have moved to this tile:7.		Move back, and continue on step 58.	If we found four or more tiles of the same color neighboring this tile:9.		Score and remove these tiles

Notice steps 2, 5, and 7: Recursion might be helpful here.