Neigbour array elements

I'm dealing with 2 dimensional arrays and i couldn't figure out how to do this:

for example assume we have a 4x4 matrix, and there is a predefined number k(which is between 0 and 128).well let's declare the array as a[i][j]. for example if the value of a[i][j]>k and if the neigbours of a[i][j]<=k,it will count 1.
but when a[i][j]>k , if one or more of a[i][j]'s neigbours is also bigger than k ,it will also count 1. I mean,for example if a[0][0]>k,a[0][1]>k and a[1][1]>k this whole 3 cell count as 1.
how can i make this?

Quote:

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Originally Posted by ashley90

I mean,for example if a[0][0]>k,a[0][1]>k and a[1][1]>k...

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Not sure about your requirements and by chance did you mean

Code:

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a[0][0]>k, a[0][1]>k and a[1][0]>k

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Since a[1][0] comes before a[1][1] in relation to a[0][0].

|0 |0 |0 |1 |0 |1 |
|0 |1 |1 |1 |1 |0 |
|0 |0 |1 |0 |0 |0 |
|0 |0 |1 |1 |0 |0 |
|0 |0 |0 |0 |0 |0 |

for example in here the count will be 2. a[0][5] counts 1 because it has no neigbors with 1( bigger than k)(diagonal not counts). but from a[0][3] to other 1's below it, there is a path without entering a 0 cell. So all of it counts 1. and totally count=2.

well i couldn't figure out how to find that kind of paths.

Okay so I am confused about your requirements. State them as simply and clearly as possible and more than likely someone from the cboard will help out.

Quote:

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Originally Posted by itCbitC

Okay so I am confused about your requirements. State them as simply and clearly as possible and more than likely someone from the cboard will help out.

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Look, for example consider this 2x2 matrix:

when

1 0
0 1 in here count=2 because we can't find a path between cell of 1's without entering a cell of 0

1 1
0 0 but in here count=1 because there is path between 1's without entering 0 cell and it counts 1

it's obvious i guess. the same thing goes for the above matrix i gave.

I probably won't be one of the ones helping with code, but...

EDIT: Oh, never mind that post if you read it. You're counting the number of "continents," or contiguous blocks of cells >k, in the matrix.

Do it out on paper. How would you count continents without the computer. Then, translate the algorithim you come up with into C.

Quote:

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Originally Posted by King Mir

Do it out on paper. How would you count continents without the computer. Then, translate the algorithim you come up with into C.

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Yeah i did it on paper but couldn't convert it to a code. For example i wrote first a function which gives 1 or 0 to cell according to their value and k. Then i tried to check every cell one by one. If it has no neigbor with 1's , its easy to convert to code but i couldn't do the second part which if a cell has a neigbor with 1 and also if this neigbor cell of 1 has neigbors with 1...

Quote:

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Originally Posted by ashley90

Look, for example consider this 2x2 matrix:

when

1 0
0 1 in here count=2 because we can't find a path between cell of 1's without entering a cell of 0

1 1
0 0 but in here count=1 because there is path between 1's without entering 0 cell and it counts 1

it's obvious i guess. the same thing goes for the above matrix i gave.

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In the
1 0
0 1 matrix, the count is 2 because you have two 1's that have no path to other cells, right?

And in this matrix:
1 1
0 0
the count is 1 because it has no path to the last row.

If the matrix were:
1 1
0 0
0 0
then would the count be 2 (meaning we're counting the number of rows with no path to them), or would it be one, because we're counting the link(s) between two cells.

What would these counts be?
1 1
0 1
0 1

1 1
0 0
1 1

As we loop (iterate) through the loop, what *exactly* triggers a count increase? That's all we need to know. Your prior description was more of a pronouncement. "Count is N!" :(

If you can "nail down" the description, of just exactly what we're counting, the problem will be solved in a jiffy. Then it's a trivial problem to code up.

Quote:

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Originally Posted by Adak

In the
1 0
0 1 matrix, the count is 2 because you have two 1's that have no path to other cells, right?

And in this matrix:
1 1
0 0
the count is 1 because it has no path to the last row.

If the matrix were:
1 1
0 0
0 0
then would the count be 2 (meaning we're counting the number of rows with no path to them), or would it be one, because we're counting the link(s) between two cells.

What would these counts be?
1 1
0 1
0 1

1 1
0 0
1 1

As we loop (iterate) through the loop, what *exactly* triggers a count increase? That's all we need to know. Your prior description was more of a pronouncement. "Count is N!" :(

If you can "nail down" the description, of just exactly what we're counting, the problem will be solved in a jiffy. Then it's a trivial problem to code up.

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Ok, i'll try to make it clear,

1 1
0 0
0 0 in here count is 1. why? because there is path between them without entering 0 cell.

1 1
0 1
0 1 in here count is again 1 since you can see the path through the 1's.

1 1
0 0
1 1 in here count is 2 since 1 from the first row and 1 from the 3rd.

another example

1 1 1
0 1 0
1 1 1 in here count is again 1 because you can go through 1 cell's without entering 0's.

i hope it makes sense now

Ok, so the count is counting up every group of 1's in the matrix.

Two isolated groups - count is 2
One group all connected - count is 1

The size of the group(s) doesn't matter, just how many distinct, separate groups there are.

Now I see what you're saying! <light comes on> :)

This is what I had in mind. For ease of programming, I surrounded the "real" matrix, with a row of zero's on the top and bottom, and with a column of zero's, on either side.

That's just to make the programming easier to avoid array out of bounds, and not necessary at all.

Code:

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#include <stdio.h>
#define Rows 5
#define Cols 7

int main(void) {
  int r, c, count, links, dirr; //dirr = direction
  int a[Rows][Cols] = {
  { 0,0,0,0,0,0,0 },
  { 0,1,1,1,0,1,0 },
  { 0,0,1,0,0,0,0 },
  { 0,0,0,0,1,1,0 },
  { 0,0,0,0,0,0,0 }, };

  count = links = 0;
  for(r=1;r<Rows-1;r++)  {
      putchar('\n');
      for(c= 1; c<Cols-1;c++)  {
        printf("%d ", a[r][c]);
        if(links == 0 && a[r][c])
            count++;
        if(a[r][c] == 1)  {
            for(dirr = 0, links = 0; dirr < 4; dirr++) {
              if(a[r-1][c] == 1) {  //12 o'clock
                  links++; break; //you have a link, so quit the testing
              }

              //continue like the above, for the other 3 directions: 3, 6, and 9 o'clock
            }
            if(links == 0)
              count++;

        }
      }
  }
  printf("\n This matrix has %d distinct groups ", count);
  printf("\n\n\t\t\t    press enter when ready ");
  r = getchar();
  return 0;
}

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This wasn't as easy as I thought it would be. The problem is with this configuration:

Code:

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0 0 0 1
0 1 1 1
0 0 0 0

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The elements in red, will be counted as a new group, unless the program "walks" over the links, and finds the blue '1' in the row above them, knows that it's connected to it, and makes no increment of the group count.

The earlier algorithm worked in 3 tested configurations, simply because the array was very small, which kept the configurations, relatively simple.

This is my attempt to make this, more capable:

Code:

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/* groups.c Finds distinct contiguous groups in a matrix.
Status: ok
*/

#include <stdio.h>

#define Rows 6
#define Cols 9

int trails(int a[Rows][Cols], int r, int c);

int main(void) {
  int r, c, group, links, ready, hit;

  int a[Rows][Cols] = {
  { 0,0,0,0,0,0,0,0,0 },
  { 0,1,1,1,1,1,0,0,0 },
  { 0,0,1,0,0,0,0,1,0 },
  { 0,1,0,1,0,1,1,1,0 },   
  { 0,0,1,1,1,0,1,1,0 },   
  { 0,0,0,0,0,0,0,0,0 },
  };

  printf("\n\n");
  group = links = 0;
  ready = 1;

  for(r=1, hit= 0;r<Rows-1;r++)  {
      putchar('\n');
      for(c= 1; c<Cols-1;c++)  {
        printf("%d ", a[r][c]);
        if(a[r][c]) {
            if(a[r-1][c] == 1) {  //12 o'clock
              links++; ready = 0;
            }
            if(a[r][c+1] == 1) {  //3 o'clock
              links++;
              hit = trails(a, r, c);
              if(hit) {
                  ready = 0; 
              }
            }
            if(a[r][c-1] == 1) {  //9 o'clock
              links++; 
            }
        }

        if(!links) ready = 1;
        if(hit == 0 && ready == 1 && a[r][c]) {
            group++;
            ready = 0;
        }
        links = 0; hit = 0;
      }
  }
  printf("\n This matrix has 4 distinct groups. I found: %d groups ", group);
  printf("\n\n\t\t\t    press enter when ready ");
  r = getchar();
  return 0;
}
/* recursively calls itself to follow the trail of links upward and to the
  right, to see if it hits (is linked with), a previous sqr in an already
  counted group.
*/
int trails(int a[Rows][Cols], int r, int c) {
  //hit = 1 means you've hit a previous groups sqr
  static hit = 0;
  int link = 1;

  while(link > 0)  {
      if(a[r-1][c] == 1) {  //12 o'clock
        hit = 1; break;
      }
      else
        link = 0;

      if(a[r][c+1] == 1) {  //3 o'clock
        trails(a, r, c+1);
      }
      else
        link = 0;
  }
  return hit;
}

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