Splitting a 2D array to an array of smaller 2D arrays in C

Question

Given:

1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8 
1 2 3 4 5 6 7 8

I want to split the 2d array (struct MATRIX) into the an array of struct MATRIX given a chunksize CS: assume cs to be 2, the answer would be

Seg[0]:
1 2 
1 2 
1 2
Seg[1]:
3 4 
3 4 
3 4
....
Seg[3]:
7 8
7 8
7 8

Here is my Matrix Struct:

typedef struct MATRIX {
    int nrow;
    int ncol;
    int **element;
} MATRIX;

and here is the function the seperates them:

void SegmentMatrix(MATRIX input,MATRIX* segs,int Chunksize, int p) {
    int i,j,r;

    //Allocate segs
    for (i = 0; i<p;i++)
    {
        CreateMatrix(&(segs[i]),input.nrow ,Chunksize,0);
    }

    //Now Copy the elements from input to the segs
    //where seg0 takes from 0 to cs cols of a, and all their rows, and seg1 takes from cs to 2cs ...
    printf("Stats:\n\t P: %d\t CS: %d\n",p,Chunksize);
    for (r = 0; r<p; r++) {
        for (i = 0; i<input.nrow;i++) {
            for (j = r*Chunksize; j<r*Chunksize+Chunksize-1; j++) {
                 //I tried (&(segs[r]))->element... Doesn't work, produces wrong data
                 segs[r].element[i][j] = input.element[i][j];

        }
    }
    PRINTM(segs[r]);
    }


}

Note that PRINTM basically prints the matrix, it knows the limits by checking segs[r].nrow and ncol and CreateMatrix takes the following inputs (&matrix, number of rows, number of colums, filltype) and mallocs from within.

filltype: 
0- generates zeroth matrix
1- generates identity
else A[i][j] = j; for simplicity

The problem is that the if i print the matrices Segs[i], they all come down with their default value given by CreateMatrix, and not the newly added values.

CLARIFICATION: Okay, so if you guys check that last PRINTM in SegmentMatrix function, it outputs the matrices as if the for loops didn't happen, aka, i can delete the for loops and would get the same output.. did i do something wrong in this line (taken from the SegmentMatrix)

Segs[r].element[i][j] = input.element[i][j];

Where would you place your call to PRINTM to have the erroneous input appearing? I would like to see that called out in your above code to make sure it is not a cart before the horse type problem. — Michael Dorgan
– Michael Dorgan, Commented May 13, 2013 at 19:03
If you look at the last statement in SegmentMatrix you will see the PRINTM, it shows the default values of the segs, as if the whole for loops didn't happen — zellwwf
– zellwwf, Commented May 13, 2013 at 21:11

score 6 · Accepted Answer · 2013-05-13 19:23:28Z

6

I don't see why and what you are manipulating with multiplication by ChunkSize and r (which is uninitialized anyway), I'd suggest simplifying the code (rule of thumb: if it seems messy, it's too complex). All you need is a 3-dimensional array to store the array of chunks, and modulo arithmetic plus integer division to insert into the appropriate column of the appropriate chunk:

/* the variable-sized dimension of the `chunks' argument is w / chsz elements big
 * (it's the number of chunks)
 */
void split(int h, int w, int mat[h][w], int chsz, int chunks[][h][chsz])
{
    /* go through each row */
    for (int i = 0; i < h; i++) {
        /* and in each row, go through each column */
        for (int j = 0; j < w; j++) {
            /* and for each column, find which chunk it goes in
             * (that's j / chsz), and put it into the proper row
             * (which is j % chsz)
             */
            chunks[j / chsz][i][j % chsz] = mat[i][j];
        }
    }
}

Demonstration, a. k. a. how to call it:

int main(int agrc, char *argv[])
{
    const size_t w = 8;
    const size_t h = 3;
    const size_t c = 2;

    int mat[h][w] = {
        { 1, 2, 3, 4, 5, 6, 7, 8 },
        { 1, 2, 3, 4, 5, 6, 7, 8 },
        { 1, 2, 3, 4, 5, 6, 7, 8 }
    };

    int chunks[w / c][h][c];

    split(h, w, mat, c, chunks);

    for (int i = 0; i < w / c; i++) {
        for (int j = 0; j < h; j++) {
            for (int k = 0; k < c; k++) {
                printf("%3d ", chunks[i][j][k]);
            }
            printf("\n");
        }
        printf("\n\n");
    }

    return 0;
}

edited May 13, 2013 at 19:23

answered May 13, 2013 at 19:14

user529758

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3 Comments

qwr Over a year ago

+1 think he should accept your answer . i was too tired to check his code . so simple wrote my pseudo code

zellwwf Over a year ago

Thank you guys... i will look at it and respond ASAP, but the structs are there to keep count of the size of the matrix just in case

zellwwf Over a year ago

This is part of a program that is supposed to read matrices from files, and operates on them using MPI (MPICH2) This part (the one i have posted) is about multiplying them. I have done the serial one easily, using the struct i have above. Chunksize is literally the number of columns i would send from matrix A to the nodes, it's calculated in process 0 and it's just passed to the function (i don't see why u think it's uninit) and P is, the number of processes. The last for loop is to make sure that no matter how many processes i've got i can summon this loop correctly, @H2CO3

qwr · Accepted Answer · 2013-05-13 19:32:06Z

2

Question was unclear . so i thought he wanted just to know how to achieve this. So i wrote this simple Pseudo code . Otherwise accept my apologize :

matrix[i] matrix
//matrixes total column size should be bigger big 2d array column size
first condition check: sum(matrix[i].colsize)>=big2d.colsize
//in this simple code raw sizes must be equal
second condition: for all i matrix[i].rawsize=big2d.rawsize
//if columns sizes will be equal the algorithm could be simplified , does not mean optimized
 //splitting big2d into matrixes
for (int br=0;br<big2d.rawsize;br++){
i=0;//store matrix index
int previndex=0;//store offset for next matrix
  for(int bc=0;bc<big2d.colsize;bc++){

      matrix[i].val[bc-previndex][br]=big2d.val[bc][br]; //assign (bc,br) 

      if(bc-previndex==matrix[i].colsize-1){
             i++; //move to next matrix;//if we not have next matrix then break;
            previndex=bc+1; 
          }
     /*if it be for equal chunks matrixes offset can be calculated this way too
         matrix[bc/chunk].val[bc%chunk][br]=big2d.val[bc][br];
      */
  }//loop columns
}//loop raws

edited May 13, 2013 at 19:32

answered May 13, 2013 at 18:59

qwr

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1 Comment

Michael Dorgan Over a year ago

Please, add some comments on exactly what your code is doing. It is not clear by itself...

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