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,

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. See the code below for an example:

>> A = randi(100,3,4) %EXAMPLE
A =
    66    94    75    18
     4    68    40    71
    85    76    66     4
>> [x, y] = minimax(A)
x =
    76    67    81
y =
    90
%end example
%calling code: [mmr, mmm] = minimax([1:4;5:8;9:12])

My answer to this:

function [mmr,mmm] = minimax(M)
mmr = abs(max(M.')-min(M.')); 
mmm = max(max(M)) - min(min(M)); 

This is shortest code I could write. What do you guys think of this?

[Stephen23]

Fails when there are multiple rows but each row contains only one element.

This was already discussed here: https://www.mathworks.com/matlabcentral/answers/464240-write-a-function-called-minimax-that-takes-m-a-matrix-input-argument-and-returns-mmr-a-row-vector#comment_713890

https://www.mathworks.com/matlabcentral/answers/464240-write-a-function-called-minimax-that-takes-m-a-matrix-input-argument-and-returns-mmr-a-row-vector#comment_802432

Simply put: your assumption that min and max always operate over the row dimension is incorrect: it actually depends on the size of the input array. The robust solution is thus not to rely on behavior which changes with array size, but to simply specify the dimension argument.

[Sahil Deshpande]

Hey Stephen, Thanks for the reply!

How do I do this?

[Ritika Mehra]

What is the meaning of max(M.'). I didn't get that command.

[Rik]

https://www.mathworks.com/help/matlab/ref/transpose.html

[Emirhan Bilgiç]

Nice job but abs is arbitrary, you already subtract min from max. So conclusion will be always positive.

[Keerthi M S]

this code works thank you so much.

[Chandan Kumar]

this code dont work for me so i did some changes and now its like this and now its good for me.

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

[Hritik Chowdhary]

why are we taking transpose of M here while calculating mmr?

[henok gashaw]

because max(M) gives the maximum value of the column vector.

[Rik]

You can also use the optional third input to change the default direction that max operates on: max(M,[],2)

[Stephen23]

"because max(M) gives the maximum value of the column vector."

It takes the maximum along the first non-scalar dimension.

Which is the cause of the bug in many of the answers on this page.

[M KK]

Hello ,

why do .' in max(M.') does why you have used it ?

[khaula]

not working still, can any one code it rightly please??

[DGM]

See also:

https://www.mathworks.com/matlabcentral/answers/451695-matrix-related-matlab-query

https://www.mathworks.com/matlabcentral/answers/496687-solving-2nd-part-of-minimax-question

https://www.mathworks.com/matlabcentral/answers/530008-i-had-write-the-function-for-find-the-difference-of-maximum-and-minimum-from-each-martix-row-and-ex

https://www.mathworks.com/matlabcentral/answers/507190-making-a-function-for-matrix

https://www.mathworks.com/matlabcentral/answers/1607910-write-a-function-called-minmax

Answer 1

[Prasad Reddy]

function [mmr,mmm] = minimax(M)
a=max(M');
b=min(M');
mmr=a-b;
c=max(a);
d=min(b);
mmm=c-d;
end
% This is what i came up with. Please give a upthumb if it works.

[Stephen23]

Incorrect output:

>> M = [1;2;3];
>> minimax(M)
ans =  2

The correct output would be:

[0;0;0]

https://www.mathworks.com/matlabcentral/answers/464240-write-a-function-called-minimax-that-takes-m-a-matrix-input-argument-and-returns-mmr-a-row-vector

[Alexandar]

How come you put a single apostrophe for this: M'. I am having trouble understanding that portion since I am new to coding.

[Rik]

The apostrophe is the operator to determine the conjugate. In the case of non-complex numbers that means swapping the rows with columns.

Answer 2

[Rushi Auti]

function [mmr,mmm] = minimax(M)
a = max(M,[],2);
b = min(M,[],2);
c= a-b;
d = c';
mmr = c'
e = max(M,[],'all');
f = min(M,[],'all');
mmm = e-f

[Rushi Auti]

I wrote this code.

[Aniket Bordekar]

Can you pelese explain why is c' taken?

[Rushi Auti]

bcz we want the difference between the maximum and minimum numbers.

actually we can reduce it like this. This will also run better.

function [mmr,mmm] = minimax(M)
    a = max(M,[],2);
    b = min(M,[],2);
    mmr = (a-b)';
    e = max(M,[],'all');
    f = min(M,[],'all');
    mmm = e-f;
end

[Aniket Bordekar]

Yes, but actually my question was why that transpose is needed?

[ali yasser]

yes ,cuz in example he want answer as vector not 3 by 1 matrix i guess

[Nur Ameera Nabila Abdul Rahim]

may i know what is the 2 at the a = max(M,[],2) means ?

may i know also what is the [] in the code means ?

[Rik]

@Nur, did you read the documentation for the max function? It will explain this syntax to you.

[Samson Ihuoma]

Pls guys I've been using the free 1month trial for matlab and it has currently expired... I need a license no, can someone help me out?

[Rik]

@Samson Ihuoma:

If you are a student: ask your school/university IT for a license or buy the discounted student license. If you are not a student: ask your company's IT for a license.

Asking for this on a public forum may not be the smartest move. I doubt anyone here will even help you find illegal options.

[Wenceslao Jr Sevilla]

@rushi I got my student account with the help of coursera

[Renz Reven Mariveles]

Thank youu. omg I have been searching for so longg. I HAVE FINALLY FIND THE ANSWER.

[Shanu]

Thanku so much😊

Answer 3

[Ahmed Salmi]

function [mmr,mmm]=minimax(m)
mmr=max(m')-min(m');
mmm=max(m,[],'all')-min(m,[],'all');
end

or

function [mmr,mmm]=minimax(m)
a=max(m');
b=min(m');
mmr=a-b;
c=max(m,[],'all');
d=min(m,[],'all');
mmm=c-d;
end

[Stephen23]

Incorrect output:

>> M = [1;2;3]
M =
   1
   2
   3
>> minimax(M)
ans =   2

Answer 4

[Harry Virani]

function [mmr, mmm] = minimax(input)

matrix = [input];

maxr = max(matrix.');
minr = min(matrix.');
mmr = maxr - minr;

maxm = max(maxr);
minm = min(minr);
mmm = maxm - minm;
end

[Stephen23]

Fails for any matrix with only one column:

>> minimax([1;2;3])
ans =  2

Answer 5

[durgesh patel]

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

[Stephen23]

Fails for any matrix with only one column:

minimax([1;2;3])

ans = 2

Answer 6

[Shamith Raj Shetty]

function [mmr,mmm] = minimax(M)
    N = M';
    mmr = max(N)-min(N);
    mmm = max(max(N))-min(min(N));
end

[Rik]

Your function fails for column vectors.

M = [1;2;3];
minimax(M) % ans =  [0,0,0]

ans = 2

M=[1:4;5:8;9:12];
minimax(M) % ans = [3,3,3]

ans = 1×3

3 3 3

●

Also, what is the point of posting this answer? What does it teach? Why should it not be deleted?

Answer 7

[Francisco Moto]

[Francisco Moto]

My function works but it failed the random question . Need help

[Stephen23]

@Francisco Moto: your function does not do what your assignment requires. In particular:

• Your function accepts one input. It then ignores this input completely.
• You have hard-coded values for one specific matrix. The assignment requests a general solution.

Most of the operations in your function are not used for anything.

Answer 8

[Balakrishna Peram]

on a General sense this should be the answer

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

[Fazal Hussain]

There is some mistake in second line but now it will give you output okay.

thanks

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

Answer 9

[Chappa Likhith]

In editor window:

function [mmr,mmm]=minimax(M)
    mmr=difference(M') %M' is a tranpose of M. If you want to know why this.. go to COMPUTER PROGRAMMING WITH MATLAB book of author J. MICHAEL FITZPATRICK AND ÁKOS LÉDECZI... go to page 90 tabel 2.7
    mmm=difference(M(:));
function a=difference(v)
    a=max(v)-min(v);

In comand window:

    >>>[mmr, mmm] = minimax([1:4;5:8;9:12])
    % you can write any other matrix too

[Walter Roberson]

This code is incorrect for the case where M has complex values. The meaning of minimum and maximum of complex numbers is not clear, but this code will get it wrong.

If it does not have complex values, then it is better style to use M.' rather than M'

[Chappa Likhith]

May be you are correct. I'm not that much familiar with matlab and I don't know for what M.' is used for... This is the question in coursera assignment of vanderbilt university. This question appears after completion of few topics where the topic of M.' is not covered.... My answer is for them who are facing the same situation like me. Because I too didn't got the answer for a long time and I saw your solution(I think so) I didn't understand what's going on in your code. I hope you understand my situation...

[Walter Roberson]

I have not posted a solution for this, as it is a homework question, and I avoid posting complete answers to homework questions.

The difference between M' and M.' is that M.' is plain transpose, but M' is conjugate transpose.

M = [1+2i 2-3i 4]

M = 
   1.0000 + 2.0000i   2.0000 - 3.0000i   4.0000 + 0.0000i

M'

ans = 
   1.0000 - 2.0000i
   2.0000 + 3.0000i
   4.0000 + 0.0000i

M.'

ans = 
   1.0000 + 2.0000i
   2.0000 - 3.0000i
   4.0000 + 0.0000i

Notice that in the M' that the signs of the complex part have changed but in the M.' version they do not change. You can see from the final entry that the result is the same for values that have no complex part.

As a matter of style, I recommend that you always use .' unless you specifically need conjugate transpose: using .' will save people having to think a lot about your code to figure out whether you should have used .' instead of '

Answer 10

[Vetrimurasu Baskaran]

function [mmr,mmm] = minimax(M)
    r = size(M);
    val = r(1);
    mmr = inf;
    for i = 1:val
        mmr(i) = max(M(i,1:end)) - min(M(i,1:end));
    end
    A = M(:);
    mmm = max(A)-min(A);
end

Answer 11

[Adwaith G]

I am new to Matlab , so i am explaining what i learned here.

Initially i solved it by using the code

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

# Both M' and M.' gives the transpose of a matrix. However, M' gives the conjugate transpose. So, I suggest that u only use M.'

However, this code fails if the matrix has only 1 column. So, i used the code

function [mmr,mmm] = minimax(M)
mmk = max(M,[],2)-min(M,[],2);
mmr = mmk.';
mmm = max(M(:))-min(M(:));

#max(M,[],2) computes the max value of each row and returns a column vector and in order to get a row vector, we take the transpose.

Answer 12

[Muhammad]

function [mmr,mmm]=minimax(M)
mmr=[abs([max(M.')-min(M.')])]
mmm=abs([(max(M(:))-(min(M(:)))])
end

[Walter Roberson]

@Muhammad

What purpose do those [ ] serve in the body of the code?

mmm=abs([(max(M(:))-(min(M(:)))])
       123   4 5 43 4   5 6 54321

You have one more open bracket than you have close brackets

Answer 13

[Arah Cristal]

function [mmr,mmm] = minimax (m)
mmr = abs(max(m.')-min(m.'))

mmm = (max(m,[],'all')-min(m,[],'all'))

[Stephen23]

Fails for any matrix with only one column:

minimax([1;2;3])

ans = 2

function [mmr,mmm] = minimax (m)
mmr = abs(max(m.')-min(m.'));
mmm = (max(m,[],'all')-min(m,[],'all'));
end

Answer 14

[Muhammad Faizan Ahmed]

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

Answer 15

[Hassan]

function [mmr,mmm]=minimax(M)
mmr=[abs(max(M(1:end,:).')-min(M(1:end,:).'))];
mmm=max(M(:))-min(M(:));
end

[Stephen23]

Fails for any matrix with only one column:

minimax([1;2;3])

ans = 2

function [mmr,mmm]=minimax(M)
mmr=[abs(max(M(1:end,:).')-min(M(1:end,:).'))];
mmm=max(M(:))-min(M(:));
end

[DGM]

• There's no point in doing abs(max(X)-min(X)). Think about why.
• There's no point in doing X(1:end,:). Think about why.
• There's no point in doing [X]. Think about why.
• Why reshape/transpose the array multiple times instead of just once?

How do so many people keep writing these same nonsense things unless:

• they're just building collages of code based on other bad code
• people somehow gravitate to these superfluous decorations when they want to make their code appear superficially unique for some bizarre purpose

This isn't where you turn in your assignment. There's little merit in posting something unless it correctly answers the question or provides the reader with new information. It should then stand to reason that there's little merit in repeating prior examples which have been demonstrated to be incorrect.

If you're going to post an answer in a thread full of junk answers, try to break that trend. Post an answer which is tested and documented. Explain why your answer is different than others (both strengths and weaknesses are important to know). Since you can run your code in the editor, you have the opportunity to demonstrate that it does what you say it does.

Answer 16

[JASON]

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

% this is what i did

Answer 17

[Aramis]

function [x, y] = minimax(M)
    x = (max(M,[],2) - min(M,[],2))';
    y = max(M,[], "all")- min(M,[], "all");
end

[DGM]

While this is correct, it's not really any different than the answer above it. The only difference is the change in output variable names, which are (I assume) dictated by the assignment. If the grader actually requires the outputs to be mmr, mmm respectively, then this would be a problem. Fixing the variable names would make this a duplicate answer.