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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 Commenti

Risposte (17)

Prasad Reddy
Prasad Reddy il 30 Mag 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 Commenti
Alexandar
Alexandar il 24 Giu 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 il 24 Giu 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 il 31 Lug 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 Commenti

Ahmed Salmi
Ahmed Salmi il 17 Lug 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 il 12 Ago 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 il 4 Gen 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 il 4 Gen 2021
Modificato: DGM il 29 Mar 2023
function [mmr,mmm] = minimax(M)
N = M';
mmr = max(N)-min(N);
mmm = max(max(N))-min(min(N));
end
  1 Commento
Rik
Rik il 4 Gen 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 il 6 Feb 2021
  2 Commenti
Stephen23
Stephen23 il 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 il 8 Giu 2021
Modificato: Stephen23 il 8 Giu 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 Commento
Fazal Hussain
Fazal Hussain il 19 Gen 2022
Modificato: DGM il 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 il 25 Giu 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 Commenti
Chappa Likhith
Chappa Likhith il 25 Giu 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 il 25 Giu 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 il 6 Giu 2022
Modificato: DGM il 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 il 27 Giu 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 il 22 Lug 2022
Modificato: Muhammad il 22 Lug 2022
function [mmr,mmm]=minimax(M)
mmr=[abs([max(M.')-min(M.')])]
mmm=abs([(max(M(:))-(min(M(:)))])
end
  1 Commento
Walter Roberson
Walter Roberson il 22 Lug 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 il 11 Ott 2022
Modificato: DGM il 29 Mar 2023
function [mmr,mmm] = minimax (m)
mmr = abs(max(m.')-min(m.'))
mmm = (max(m,[],'all')-min(m,[],'all'))
  1 Commento
Stephen23
Stephen23 il 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 il 18 Dic 2022
Spostato: DGM il 29 Mar 2023
function [mmr, mmm] = minimax(M)
mmr = max(M')-min(M');
mmm = max(max(M)) - min(min(M));

Hassan
Hassan il 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 Commenti
Stephen23
Stephen23 il 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 il 29 Mar 2023
Modificato: DGM il 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 il 27 Ott 2023
Spostato: Stephen23 il 27 Ott 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 il 5 Feb 2024
Modificato: Aramis il 5 Feb 2024
function [x, y] = minimax(M)
x = (max(M,[],2) - min(M,[],2))';
y = max(M,[], "all")- min(M,[], "all");
end
  1 Commento
DGM
DGM il 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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