# matrix related matlab query

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Siddharth Vidyarthi on 22 Mar 2019
Edited: DGM on 27 Feb 2023
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)
A =
66 94 75 18
4 68 40 71
85 76 66 4
>> [x, y] = minimax(A)
x =
76 67 81
y =
90
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Sahil Deshpande on 30 May 2020
Edited: Walter Roberson on 8 Jun 2020
What do you guys think of this?
function [mmr,mmm] = minimax(M)
T = M.'; %Transposed matrix M
S = max(T) - min(T); %S will return a row vector of max - min values of each column of T, which is transpose of S.
%So S returns max - min of each row of M, which is required
mmr = abs(S); %gives the absolute value
mmm = max(max(M)) - min(min(M)); %max(M) and min (M) return a row vector, I used the function twice.
end
DGM on 26 Feb 2023
Edited: DGM on 26 Feb 2023

KETAN PATEL on 11 Jun 2019
function [mmr, mmm] = minimax(A);
B = A';
maxi= max(B);
mini = min(B);
mmr = max(B) - min(B);
mmm = max(maxi) - min(mini);
end
KETAN PATEL on 14 Jun 2019
I have another problem and I just posted it in the community. It is titled as "if-statement with conditions". Could you please take a look if you have time? It's really easy but I don't where I am going wrong.
Ammara Haider on 17 Dec 2019

Saurabh Bhardwaj on 8 Jun 2020
function [a,b]=minimax(M)
A= min(M,[],2);
B= max(M,[],2);
a=(B-A)';
b=max(B)-min(A);
end
DGM on 27 Feb 2023
.' is the regular transpose
' is the complex conjugate transpose
It could use some commentary too. Otherwise, this is more thoughtful than most of the solutions on these threads.

RP on 4 Apr 2019
I saw this exercise on Coursera and seemed to have solved it, anyway when I ran the code it worked, but when I submit the answer and it is evaluated with random input, I get an error message every time. When I try to run it with the random numbers that were used for the evaluation, I get the correct results. Does anyone have the same problem? This is my code:
function [mmr, mmm] = minimax(M)
mmr = (max(M,[],2)-min(M,[],2))'
mmm = max(M(:))
end
Crystal Judd Unson on 25 Apr 2021
Edited: Crystal Judd Unson on 25 Apr 2021
Hi, I'm new to MATLAB so I'm a little confused on max(A,[],dim). How does this code instruct mmr to be a row vector and not a column vector? Why do I get an error message when
function [mmr, mmm] = test(M)
mmr = (max(M,[],0)-min(M,[],0))';
mmm = max(M(:))-min(M(:));
end
Thanks!
Steven Lord on 25 Apr 2021
x = magic(4);
max(x, [], 0)
Error using max
Dimension argument must be a positive integer scalar, a vector of unique positive integers, or 'all'.
Arrays in MATLAB do not have a dimension 0 so it does not make sense to ask for the maximum along that dimension.

RP on 4 Apr 2019
sneha sharma on 10 Sep 2019
function [mmr,mmm]=minimax(A)
a=max(A(1,:))-min(A(1,:));
b=max(A(1,:))-min(A(1,:));
c=max(A(3,:))-min(A(3,:));
d=max(A(end,:))-min(A(end,:));
mmr=[a b c];
mmm=max(A(:))-min(A(:));
end
%this is my program it is not working for random matrices , can you define an error
VIJAY VIKAS MANGENA on 13 Aug 2020
What if the random matrix has more than 3 rows?
1)You have fixed the no.of outputs using this code.You get only 4 values ( if you meant ,b=max(A(2,:))-min(A(2,:));)
2)You have assumed that mmr can have only three outputs which is not always true..it depends on the matrix chosen and your code is supposed to work for any random matrix (the reason you got this error 'not working for random matrices'

AYUSH GURTU on 28 May 2019
function [mmr, mmm] = minimax(M)
mmr = (max(M,[],2)-min(M,[],2))';
mmm = max(M(:))-min(M(:));
end
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RUSHI SHAH on 2 Mar 2020
Can you please explain the syntax for mmr?
Ashitha Nair on 15 Jun 2020
M = max(A,[],dim) returns the maximum element along dimension dim. For example, if A is a matrix, then max(A,[],2) is a column vector containing the maximum value of each row.

Ashitha Nair on 15 Jun 2020
function [mmr,mmm]=minimax(M)
a=ceil(max(M.'));
b=ceil(min(M.'));
x=a-b;
mmr=x';
y=max(M(:));
z=min(M(:));
mmm=y-z;
end
This is how I've written it.
Dorbala sankarshana on 23 Jun 2020
DGM on 27 Feb 2023
Why would you take ceil()? That will give you the wrong result for non-integer inputs.

anuj petkar on 13 Sep 2020
function [mmr,mmm]=minimax(M)
A=(M(:,:))';
mmr=max(A(:,:))-min(A(:,:));
mmm=max(max(A))-min(min(A));
end
DGM on 27 Feb 2023
A(:,:)
is the same as
A

Amit Jain on 24 Oct 2020
function [mmr,mmm] = minimax(A)
T = A';
mmr = max(T)-min(T);
p= max(max(A(1:end,1:end)));
q = min(min(A(1:end,1:end)));
mmm= p-q;
end
DGM on 27 Feb 2023
A(1:end,1:end)
is the same as
A

ANDIE MEDDAUGH on 7 Jul 2021
Edited: DGM on 27 Feb 2023
Here's the code I used:
function [mmr, mmm] = minimax(M)
B = M';
maxie = max(B);
minnie = min(B);
mmr = abs(maxie - minnie)
mmm = abs(max(maxie) - min(minnie));
end
The max and min functions read columns, not rows. So the M' switches columns to rows, so that issue is resolved. Abs() is used to ensure absolute value and no negative numbers.