# matrix related matlab query

16 visualizzazioni (ultimi 30 giorni)
Siddharth Vidyarthi il 22 Mar 2019
Modificato: DGM il 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 il 30 Mag 2020
Modificato: Walter Roberson il 8 Giu 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 il 26 Feb 2023
Modificato: DGM il 26 Feb 2023

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### Risposte (9)

KETAN PATEL il 11 Giu 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
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KETAN PATEL il 14 Giu 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 il 17 Dic 2019

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Saurabh Bhardwaj il 8 Giu 2020
function [a,b]=minimax(M)
A= min(M,[],2);
B= max(M,[],2);
a=(B-A)';
b=max(B)-min(A);
end
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
DGM il 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.

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RP il 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
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Crystal Judd Unson il 25 Apr 2021
Modificato: Crystal Judd Unson il 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 il 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.

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RP il 4 Apr 2019
##### 4 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
sneha sharma il 10 Set 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 il 13 Ago 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'

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AYUSH GURTU il 28 Mag 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 il 2 Mar 2020
Can you please explain the syntax for mmr?
Ashitha Nair il 15 Giu 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.

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Ashitha Nair il 15 Giu 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.
##### 2 CommentiMostra NessunoNascondi Nessuno
Dorbala sankarshana il 23 Giu 2020
DGM il 27 Feb 2023
Why would you take ceil()? That will give you the wrong result for non-integer inputs.

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

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Amit Jain il 24 Ott 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
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
DGM il 27 Feb 2023
A(1:end,1:end)
is the same as
A

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ANDIE MEDDAUGH il 7 Lug 2021
Modificato: DGM il 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.
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

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