Reordering the rows in a matrix in a specific order - c

Question

I am successfully storing the calculated subsets in a 2-D array matrix in C language.Now I want to print the subsets in an order desired.
For eg.
2-D array matrix is
10 7 3 2 1
10 7 5 1
7 6 5 3 2
10 6 5 2
10 7 6
Desired Output
10 7 6
10 7 5 1
10 7 3 2 1
10 6 5 2
7 6 5 3 2
How quick sort can be applied to sort/order these rows?

Answer 1

As #chqrlie noted, this can be easily solved with qsort.
Depending on the way the matrix is declared (is it an array of pointers to arrays of ints? do all arrays have the same length? is it a global array of fixed size?) the code will have to do slightly different things.
So, assuming the array is a global variable and all rows have same length (padded with 0s):
MWE:
#include <stdio.h>
#include <stdlib.h>
/*
Compare 2 integers
returns:
-1 if *i1 < *i2
+1 if *i1 > *i2
0 if *i1 == *i2
*/
int intcmp(const int *i1, const int *i2)
{
return (*i2 < *i1) - (*i1 < *i2);
}
#define ROWS 5
#define COLS 5
/*
Assumes rows already sorted in descending order
NOTE: qsort calls the comparison function with pointers to elements
so this function has to be tweaked in case the matrix is an array of
pointers. In that case the function's declaration would be:
int rowcmp(int **pr1, int **pr2)
{
const int *r1 = *pr1;
const int *r2 = *pr2;
// the rest is the same
}
*/
int rowcmp(const int *r1, const int *r2)
{
int i = 0, cmp;
do {
cmp = intcmp(&r1[i], &r2[i]);
i++;
} while (i < COLS && cmp == 0);
return -cmp; /* return -cmp to sort in descending order */
}
int data[5][5] = {
{10,7,3,2,1},
{10,7,5,1,0},
{ 7,6,5,3,2},
{10,6,5,2,0},
{10,7,6,0,0}
};
void printmatrix()
{
int i, j;
for (i = 0; i < ROWS; i++) {
for (j = 0; j < COLS; j++) {
printf("%d ", data[i][j]); /* leaves a trailing space in each row */
}
printf("\n");
}
}
int main()
{
printmatrix();
qsort(data, 5, sizeof(data[0]), (int (*)(const void *, const void *))rowcmp);
printf("\n");
printmatrix();
return 0;
}
For the most flexible solution, I would define
struct row {
size_t len;
int *elems;
};
struct matrix {
struct row *rows;
size_t nrows;
};
and change the code accordingly.
NOTE: code not thoroughly tested, use with caution ;)

Answer 2

First of all, are you sure that the 1 on row 3,col 5 should be there and not on the last line?
Anyway, an efficient way to achieve what you want is:
compute the frequency array
declare a new matrix
go from the highest element (10 in your case) from frequency array and put in your matrix using your desired format.
It is time-efficient because you don't use any sorting algorithm, thus you don't waste time there.
It is NOT space-efficient because you use 2 matrices and 1 array, instead of only 1 matrix as suggested in other posts, but this should not be a problem, unless you use matrices of millions of rows and columns
C code for frequency array:
int freq[11] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
for(int i=0; i<NO_ROWS; i++) {
for(int j=0; j<NO_COLS; j++) {
if(MATRIX[i][j]!=null && MATRIX[i][j]>0 && MATRIX[i][j]<11) {
freq[MATRIX[i][j]]++;
}
}
}
C code for computing the new matrix dimensions
(assuming you want to keep the number of rows)
OUTPUT_MATRIX[100][100] /*I declared it statically, but I would advise to make it dinamically */
/* first, compute the number columns.
To do so, we need the number of elements
(we get them by simply summing up frequency array's elements) */
int s=0;
for(int i=0; i<11; i++) {
s+=frequency[i];
}
int addOne = 0 /* boolean value to check if we will have to add one extra column for safety */
if(s % NO_ROWS) {
addOne = 1; /* division is not even, so we will have to add extra column */
}
NO_COLS = s/NO_ROWS + addOne;
Now, final part, assigning the values from frequency array to the OUTPUT_MATRIX
int k=0;
int currentNumber = 10; /* assigning starts from 10 */
for(int i=0; i<NO_ROWS; i++) {
for(int j=0; j<NO_COLS; j++) {
if(currentNumber>0) {
if(frequency[currentNumber]==0 || k>=frequency[currentNumber]) {
currentNumber--;
k=0;
}
OUTPUT_MATRIX[i][j] = frequency[currentNumber];
k++;
} else {/*here, you can assign the rest of the value with whatever you want
I will just put 0's */
OUTPUTMATRIX[i][j] = 0;
}
}
}
Hope this helps!

Answer 3

This is what I do in C++ to reorder a matrix:
// b is the matrix and p is an array of integer containing the desired order of rows
for(i=0; i<n; i++){
if( p[i]==i )
continue;
b[i].swap(b[p[i]]);
j = p[i]; // New row i position
// Update row i position to new one
for(int k=i+1; k<n; k++){
if( p[k] == i )
p[k] = j;
}
printRow( b[i] );
}

Answer 4

You need to define an array of pointers of the data type you use and then you can reorder your matrix.
for example your matrix is: arr[5][10], and you want to print line 4 before line 3:
int *[5] arr2;
arr2[0] = &arr[0][0];
arr2[1] = &arr[1][0];
arr2[2] = &arr[2][0];
arr2[3] = &arr[4][0];
arr2[4] = &arr[3][0];
in regard to how will the ordering algorithm work, i would suggest placing a header in the start of each array in the matrix which will tell you how many elements it has(basically the first element of each array can be a counter of the total elements) afterwards you can order the strings by comparing the header, and if it is equal comparing the first element and so on. this can be done in a loop that iterates as many times as there are elements in the array, when the elements are not equal, break out of the loop.
hope this helps.