Subtracting a value from lines of a matrix - c

Question

I have the following matrix:
1 4 5
5 7 6
5 8 8
I want to find the minimum value of line 1 and subtract from all values of the same line. The same thing for line 2 and 3. The minimum value of line 1 is 1, line 2 is 5 and line 3 is 5. So I subtract 1 from all values of line 1, subtract 5 from all values in line 2 and subtract 5 from all values in line 3.
0 3 4
0 2 1
0 3 3
My matrix is called "a":
min = a[0][0] \\\ min = minimum value
for (k = 0; k < 3; k++) {
for (l = 1; l < 3; l++) {
if (a[k][l] < min)
min = a[k][l - 1];
}
for (l = 0; l < 3; l++) {
a[k][l] = a[k][l] - min;
}
min = a[k+1][0];
}
For k = 0, the value a[k+1][0] = 5 is changing to 4. Why is that?
EDIT: I declared the array as:
a[0][0] = 1;
a[0][1] = 4;
a[0][2] = 5;
a[1][0] = 5;
a[1][1] = 7;
a[1][2] = 6;
a[2][0] = 5;
a[2][1] = 8;
a[2][2] = 8;
Following Kresimir I changed the code to:
for (k = 0; k < 3; k++) {
min = 10000;
for (l = 0; l < 3; l++) {
if (a[k][l] < min)
min = a[k][l];
}
for (l = 0; l < 3; l++) {
a[k][l] = a[k][l] - min;
}
}
To print the matrix:
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {
printf("a[%d][%d] = %d\n", i, j, a[i][j]);
}
}
But the output is:
a[0][0] = 0
a[0][1] = 3
a[0][2] = 0
a[1][0] = 0
a[1][1] = 3
a[1][2] = 0
a[2][0] = 0
a[2][1] = 0
a[2][2] = 3
When k = 0, the first line is changed correctly and the rest is kept the same as they should, when k = 1, all lines are changed wrong to the above.

Answer 1

When approaching any bit of coding, it generally helps to break your coding tasks down into a series of individual steps -- that then provide a road-map to follow as you begin actual coding. You can type the steps out in a separate editor window, (or what I find just as helpful is an 8 1/2 x 11 sheet of paper and pencil).
Think through what your code must do and write it down, e.g.
loop over all rows in the matrix (a/k/a the 2D array of int);
loop over all column values to find the minimum in each row; and finally
loop over all column values (again) subtracting the minimum from each value.
You don't have to get it perfect the first time, now look again at the steps you have written and determine if there are any constraints you must impose.
(Here, yes, you must determine the minimum by examining each value in a row before you can begin subtracting the minimum from each value -- this necessitates at least two loops over the column values. Further, you must also reset or re-initialize your min variable so that it holds the minimum for the current row -- not the minimum from the last row that just happened to be less than this one. Use the re-initialization requirement to make a logical choice for the scope within which each variable should be declared)
Now with your steps refined with any constraints you must impose, you can logically lay out your code (keeping in mind that you must always protect against reading or writing beyond your array bounds, etc.) With the benefit of a good outline, you know you will need one outer loop that loops over all rows, and then two inner loops that (a) find the minimum, and (b) subtract that from all values in that row. You could do something similar to:
#include <stdio.h>
#include <limits.h> /* for INT_MAX, INT_MIN */
#define ASZ 3 /* if you need a constant, define one (or more), a size */
int main (void) {
int a[][ASZ] = {{ 1, 4, 5 }, { 5, 7, 6 }, { 5, 8, 8 }};
puts ("Original matrix:"); /* output original matrix */
for (int row = 0; row < ASZ; row++) {
for (int col = 0; col < ASZ; col++)
printf (" %2d", a[row][col]);
putchar ('\n');
}
puts ("\nModified matrix:"); /* subtract row-min from each element */
for (int row = 0; row < ASZ; row++) { /* loop over rows */
int min = INT_MAX; /* declare min = INT_MAX */
for (int col = 0; col < ASZ; col++) /* loop over column vals */
if (a[row][col] < min) /* find row-min value */
min = a[row][col];
for (int col = 0; col < ASZ; col++) { /* loop over column vals */
a[row][col] -= min; /* subtract row-min value */
printf (" %2d", a[row][col]); /* output new value */
}
putchar ('\n');
}
}
Example Use/Output
$ ./bin/mtrx_subtract_rowmin
Original matrix:
1 4 5
5 7 6
5 8 8
Modified matrix:
0 3 4
0 2 1
0 3 3
There is no magic to it, it just takes approaching each problem in a systematic way. Doing it often enough, it gets easier each time. Pay attention to where each variable was declared (or constant defined) and understand why. Let me know if you have any further questions.

Answer 2

If you want to find the minimum for each line, you need to initialise it for each line, not only once.
for (k = 0; k < 3; k++) {
min = a[k][0];
...
You were only searching for the minimum so far.

Answer 3

Whenever I'm looking for a minimum, I set min to positive infinity (nothing can be larger than that - you don't have to do it, but it makes for a nicer code, in my opinion).
Alternatively, you can use min = Math.min(...a[k]); instead of that entire first for loop (will not work on IE, though).
Also, keep in mind that indices go from 0 to 2. Also, check the order of k and l (depending on how you implement the rows and columns of the matrix).
let k, l, min;
let a=[[1, 4, 5],
[5, 7, 6],
[5, 8, 8]];
for (k = 0; k < 3; k++) {
min = +Infinity;
for (l = 0; l < 3; l++) {
if (a[k][l] < min)
min = a[k][l];
}
for (l = 0; l < 3; l++) {
a[k][l] = a[k][l] - min;
}
}
console.log(a);