Finding the minimum value of a line in a 3D array

[learner999]

Hello everyone,

I need a small help. I have a 3D array like X[a][b][c]. Each cell of this array holds a value. I need the b indice which corresponds to the minimum value. Let me explain it with an example
X[0][0][0]=2
X[0][1][0]=4
X[0][2][0]=5
X[0][3][0]=1

I need to find 3.

X[0][0][1]=0
X[0][1][1]=2
X[0][2][1]=5
X[0][3][1]=7 I need to find 0 here. I need to do that for all a and c.

and if there are more than one indice which corresponds the same minimum value, I should stock them in an array. Like

X[2][0][5]=10
X[2][1][5]=2
X[2][2][5]=2
X[2][3][5]=7

I should save 1 and 2.Thank you so much in advance

[Ganado]

What have you tried so far? Do you know how to iterate through a multi-dimensional array with nested for loops?

[learner999]

Yes, I tried Following:

	
	for (int p = 0; p <a; ++p) {
		for (int t = 0; t < c; ++t) {
                      int min2 = 99999;
			for (int m = 0; m < b; ++m) {
				
				if (X[a][b][c] < min2) {
					min2 = X[a][b][c];

				}
			}


			//here we stock the b 
			
			vector<int>k;
			for (int m = 0; m < b; ++m) {
				if (min2 ==X[a][b][c])
				{
					k.push_back(b);
				}

			}
}
}

[keskiverto]

for (int p = 0; p <a; ++p) {

    if (X[a] < min2) {
        min2 = X[a];
    }

If first dimension has a elements, then X[a] is out of range error.

Furthermore, why have a loop, if you always dereference the same element: X[a][b][c]

[Ganado]

In other words: You should be using your iteration variables (p, t, m) as indices.
It's more "idiomatic" to use (i, j, k) as for-loop variables if you don't have better names for them. Otherwise, usually you should avoid one-letter variable names.

[keskiverto]

There is more. You want minimum for each "line", don't you?

Since you have array int X[A][B][C]; you will have A*C "lines".
If you would keep only one index per line, then you would need array int min_index[A][C]; to store them.

You want potentially many indices per line. You would need, for example:
vector<int> min_indices[A][C];

For your example, the min_indices[2][5] would then be a vector<int> that contains values 1 and 2.

[learner999]

Thank you so much @keskiverto and @Ganado,

I just need to store one indice which is B.

I guess I did not understand your comments. For using the p,t,m. I did it

	int min2 = 99999;
	for (int p = 0; p < nbProducts; ++p) {
		for (int t = 0; t < nbPeriods; ++t) {
			for (int m = 0; m < nbMachines; ++m) {
				
				if (X[p][m][t] < min2) {
					min2 = lagpart2[p][m][t];

				}
			}
}
}

here , I try to find the minimum value for each X[p][t]. It should iterate on m and find the minimum of it. So, the compared X[p][m][t] should be like

X[2][0][5]
X[2][1][5]
X[2][2][5]
X[2][3][5]

Thank you so much for your help.

[keskiverto]

Those are better names. You code, however, looks for the smallest value of the entire array.

This is the part that iterates every possible pair of (p,t):

for (int p = 0; p < nbProducts; ++p) {
  for (int t = 0; t < nbPeriods; ++t) {
    // code X
  }
}

The code X should find the answer for one pair. (2,5) is one such pair.

To get minumum "cell values" for each pair:

int results[nbProducts][nbPeriods] {};
for (int p = 0; p < nbProducts; ++p) {
  for (int t = 0; t < nbPeriods; ++t) {

    int min2 = 99999;
    for (int m = 0; m < nbMachines; ++m) {
      if ( X[p][m][t] < min2 ) {
        min2 = lagpart2[p][m][t];
      }
    }
    results[p][t] = min2;

  }
}

[learner999]

Thank you so much @keskiverto. However, I need the middle indice of the minimum value. I need the "m" indices. Your code holds p and t , does not it ? I solved the problem thank you