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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,

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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?
  14 Comments

Answers (17)

Prasad Reddy
Prasad Reddy on 30 May 2020
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.
  3 Comments
Alexandar
Alexandar on 24 Jun 2022
How come you put a single apostrophe for this: M'. I am having trouble understanding that portion since I am new to coding.
Rik
Rik on 24 Jun 2022
The apostrophe is the operator to determine the conjugate. In the case of non-complex numbers that means swapping the rows with columns.

Rushi Auti
Rushi Auti on 31 Jul 2020
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
  12 Comments

Ahmed Salmi
Ahmed Salmi on 17 Jul 2020
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

Harry Virani
Harry Virani on 12 Aug 2020
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

durgesh patel
durgesh patel on 4 Jan 2021
function [mmr , mmm] = minimax(M)
mmr = max(M') - min(M');
mmm = max(M,[],'all')- min(M,[],'all');
end

Shamith Raj Shetty
Shamith Raj Shetty on 4 Jan 2021
Edited: DGM on 29 Mar 2023
function [mmr,mmm] = minimax(M)
N = M';
mmr = max(N)-min(N);
mmm = max(max(N))-min(min(N));
end
  1 Comment
Rik
Rik on 4 Jan 2021
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?

Francisco Moto
Francisco Moto on 6 Feb 2021
  2 Comments
Stephen23
Stephen23 on 6 Feb 2021
@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.

Balakrishna Peram
Balakrishna Peram on 8 Jun 2021
Edited: Stephen23 on 8 Jun 2021
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
  1 Comment
Fazal Hussain
Fazal Hussain on 19 Jan 2022
Edited: DGM on 29 Mar 2023
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

Chappa Likhith
Chappa Likhith on 25 Jun 2021
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
  3 Comments
Chappa Likhith
Chappa Likhith on 25 Jun 2021
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
Walter Roberson on 25 Jun 2021
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 '

Vetrimurasu Baskaran
Vetrimurasu Baskaran on 6 Jun 2022
Edited: DGM on 26 Feb 2023
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

Adwaith G
Adwaith G on 27 Jun 2022
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.

Muhammad
Muhammad on 22 Jul 2022
Edited: Muhammad on 22 Jul 2022
function [mmr,mmm]=minimax(M)
mmr=[abs([max(M.')-min(M.')])]
mmm=abs([(max(M(:))-(min(M(:)))])
end
  1 Comment
Walter Roberson
Walter Roberson on 22 Jul 2022
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

Arah Cristal
Arah Cristal on 11 Oct 2022
Edited: DGM on 29 Mar 2023
function [mmr,mmm] = minimax (m)
mmr = abs(max(m.')-min(m.'))
mmm = (max(m,[],'all')-min(m,[],'all'))
  1 Comment
Stephen23
Stephen23 on 7 Nov 2022
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

Muhammad Faizan Ahmed
Muhammad Faizan Ahmed on 18 Dec 2022
Moved: DGM on 29 Mar 2023
function [mmr, mmm] = minimax(M)
mmr = max(M')-min(M');
mmm = max(max(M)) - min(min(M));

Hassan
Hassan on 29 Mar 2023
function [mmr,mmm]=minimax(M)
mmr=[abs(max(M(1:end,:).')-min(M(1:end,:).'))];
mmm=max(M(:))-min(M(:));
end
  2 Comments
Stephen23
Stephen23 on 29 Mar 2023
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
DGM on 29 Mar 2023
Edited: DGM on 29 Mar 2023
  • 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.

JASON
JASON on 27 Oct 2023
Moved: Stephen23 on 27 Oct 2023
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

Aramis
Aramis on 5 Feb 2024
Edited: Aramis on 5 Feb 2024
function [x, y] = minimax(M)
x = (max(M,[],2) - min(M,[],2))';
y = max(M,[], "all")- min(M,[], "all");
end
  1 Comment
DGM
DGM on 5 Feb 2024
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.

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