Can anyone explain me why M' is taken in the following code?

%{ 
Question : Write a function called minimax that takes M, a matrix input argument and 
           returns mmr, a row vector containing the absolute values of the difference 
           between the maximum and minimum valued elements in each row. As a second 
           output argument called mmm, it provides the difference between the maximum 
           and minimum element in the entire matrix 
%}

%Code :
            function [mmr, mmm] = minimax(M)
            mmr = max(M') - min(M');
            mmm = max(M, [], 'all') - min(M, [], 'all');

[John D'Errico]

Didn't you write the code? Oh, I guess not. Should you ask the person who did this homework assignment for you?

What does the ' (prime) operator do? Hint:

help transpose

[Walter Roberson]

help ctranspose

would be more correct for determining what the function does.

See also https://www.mathworks.com/help/matlab/ref/max.html#bubhhu5-5

• If A is complex, then max(A) returns the complex number with the largest magnitude. If magnitudes are equal, then max(A) returns the value with the largest magnitude and the largest phase angle.

[Aniket Bordekar]

Yes I didn't wrote the code but I want to know why was that transpose taken.

@JohnD'Errico

[Aniket Bordekar]

Thanks @WalterRoberson

[Stephen23]

Detailed discussion on this homework task, including explanations of why the shown approach is buggy:

https://www.mathworks.com/matlabcentral/answers/538171-write-a-function-called-minimax-that-takes-m-a-matrix-input-argument-and-returns-mmr-a-row-vector

Answer 1

[Walter Roberson]

By default, min() and max() operate along the first non-scalar dimension. If you have a 2D array, m x n, with m and n both not 1, then that means that min() or max() of the array would produce a 1 x n output -- it has operated along columns, producing one result for each column.

Now suppose you transpose the m x n array to become n x m, with the rows becoming columns and the columns becoming rows, and you min() or max that. You will get a 1 x m result -- one result for each of what were originally rows.

Thus, min(M') is one way of producing a minimum for each row (but it has a problem if the data is complex-valued.)

More clear and robust is to use the syntax min(M,[],2) to process dimension #2 specifically, producing one result for each row.

Answer 2

[Shubham Shah]

function [mmr,mmm]= minimax(M)
mmt = [max(M,[],2)-min(M,[],2)]
mmr = mmt'
mmm = max(M, [], 'all') - min(M, [], 'all')
end

[Shubham Shah]

Question : Write a function called minimax that takes M, a matrix input argument and

returns mmr, a row vector containing the absolute values of the difference

between the maximum and minimum valued elements in each row. As a second

output argument called mmm, it provides the difference between the maximum

and minimum element in the entire matrix