# matrix related matlab query

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Siddharth Vidyarthi on 22 Mar 2019
Edited: Ronald Grant on 5 Aug 2020 at 12:19
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

Stephan on 22 Mar 2019
This appears to be homework - to be exactly: your homework.
Sahil Deshpande on 30 May 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

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
Thank you!
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

Rose Potter 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

Guillaume on 4 Apr 2019
What is the error message you get?
Kailash Ramasubramaniam on 8 May 2020
function [mmr,mmm]=minimax(r)
mmr= [max(r(1,[1:end]))- min(r(1,[1:end])),max(r(2,[1:end]))- min(r(2,[1:end])),...
max(r(3,[1:end]))- min(r(3,[1:end]))];
mmm=max(r(:))-min(r(:));
end
This the code which I wrote for this question. This works fine for matrices till 3 rows,after which it fails. I am new to matlab. Can someone help me to correct this code for random matrices please?
Ronald Grant on 5 Aug 2020 at 12:18
function [mmr, mmm] = minimax(M)
mmq = max(M,[],2)-min(M,[],2);
mmr = transpose(mmq);
mmm = max(M(:))-min (M(:));
end
try to modify few things in your code, and here you go. nice code btw really help me alot :D

Rose Potter on 4 Apr 2019 Durga Charan on 13 Apr 2019
you needs to transpose the final results. your function calls in colum. but it is asked as a row vector.
Gregorio Giorgi on 6 Jul 2019
Thank you Durga, I was getting the same as Rose, but then I transposed the final results to a row vector like you suggested.
This is how I solved it in the end (there are probably other ways to do it):
function [mmr, mmm] = minimax (M)
mmr_1 = max(M, [], 2) - min(M, [], 2);
mmr = mmr_1'
mmm = max(M, [], 'all') - min(M, [], 'all');
end
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

AYUSH GURTU on 28 May 2019
function [mmr, mmm] = minimax(M)
mmr = (max(M,[],2)-min(M,[],2))';
mmm = max(M(:))-min(M(:));
end

Muhammad Najeeb khan on 19 Jun 2019
can you please explain the mmr code.
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.

Harsha Vyshalli on 22 May 2020
function [mmr, mmm] = minimax(A);
B = A';
maxi= max(B);
mini = min(B);
mmr = max(B) - min(B);
mmm = max(maxi) - min(mini);
end

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

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.

#### 1 Comment

Dorbala sankarshana on 23 Jun 2020