How can I use a nested for loop to find multiple minimum values in a matrix array?

I am looking to write a script that will determine what the minimum value in a matrix is, and where it is located (it occurs more than once). I have written a code that will find a single minimum, but I'm struggling to adapt it for more than one minimum. This is a homework so any hints would be greatly appreciated. here is my code so far for matrix A.

A=[12,44,7,32,3;66,32,3,55,9;14,53,28,3,5];

MinVal=A(1,1); 
RowVal=1; 
ColVal=1;

for i=1:3
    for j=1:3
    if A(i,j)<MinVal; 
        MinVal=A(i,j);
        RowVal=i;
        ColVal=j; 
    end
    end
end

formatspec='The minimum value is %0.2f in row %0.0f and column %0.0f.\n';
fprintf(formatspec,MinVal,RowVal,ColVal)

Answer 1

[Guillaume]

One possible way is to have RowVal and ColVal as vectors instead of scalars. You then have two if conditions:

• A(i,j) == MinVal : add current row and column to the end of the RowVal and ColVal vectors
• A(i,j) < MinVal : update MinVal and reset RowVal and ColVal

That should be enough for you to write the code.

As an aside, avoid hardcoding the end values of your loop. Use

for i = 1:size(A, 1) %similar for j

That way the code works even if you change the size of A.

[Max Campbell]

Thanks for you help. I'm still struggling to find the correct answer, The vector that I'm creating seems to only record every new minimum. Here is my current script:

A=[12,44,7,32,3;66,32,3,55,9;14,53,28,3,5];

MinVal=A(1,1);
RowVal=1;
ColVal=1;

for i=1:size(A,1)
    for j=1:size(A,2)
        if A(i,j)<MinVal;
            MinVal=A(i,j);
            RowVal(i)=i;
            ColVal(j)=j;
            if A(i,j)==MinVal
                RowVal(i)=i;
                ColVal(j)=j;
            end
        end
    end
end

[Thorsten]

Guillaume told you to add current row and column to the end of the RowVal and ColVal vectors if A(i,j) == MinVal. You have just updated RowVal and ColVal. See my answer on how to add values.

[Guillaume]

There are several issues with your code:

• you're indexing RowVal and ColVal with the row and column. That makes no sense.
• the second if is never going to be true as it is embedded in the first if. That is you can only ever reach if A(i,j)==MinVal iif the first if is true and that's only when A(i,j)<MinVal.
• Even if that second if could be executed it doesn't do what it should: "reset RowVal and ColVal"

To fix the first one, create an additional variable that you only increase when you've found a minimum and use that variable as index. Alternatively, use end+1.

To fix the second one, use elseif

To reset a variable that was previously a vector, either:

%var = [1 2 3 4 5]; %vector to reset
var = [];
var(1) = value

or:

%var = [1 2 3 4 5]; %vector to reset
var = value

Answer 2

[Thorsten]

Simplify your code by using a linear index

ind = 1;
minval = A(ind);
for i =1:numel(A)
  if A(i) < minval
    minval = A(i);
    ind = i;
  elseif A(i) == minval
    ind = [ind i]; % add current index
  end
end

Convert linear index to sub scripts:

[i j] =ind2sub(size(A), ind)

Answer 3

[Stephen23]

It is easy to get the row and column indices of the minimum without any loops using code vectorization. This is faster and much more efficient than using nested loops:

>> A = [12,44,7,32,3;66,32,3,55,9;14,53,28,3,5];
>> X = min(A(:));
>> [R,C] = find(A==X)
R =
   2
   3
   1
C =
   3
   4
   5

One can also display the text without any loops:

>> fmt = 'The minimum value is %0.2f in row %0.0f and column %0.0f.\n';
>> fprintf(fmt,[X*ones(size(R)),R,C].')
The minimum value is 3.00 in row 2 and column 3.
The minimum value is 3.00 in row 3 and column 4.
The minimum value is 3.00 in row 1 and column 5.

[Thorsten]

It's homework, so min is probably not allowed.

[Guillaume]

As it's homework I assume the OP has to use a loop. Otherwise, yes, a loop is pointless.

[Stephen23]

I realize this is probably homework and restricted to loops, but in case someone is browsing Answers or using an internet search engine then there needs to be atleast one answer that gives a more efficient solution to this problem, and links to the reasons why using loops is a poor solution.