Info

# 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,

1 view (last 30 days)
Debaditya Chakraborty on 27 May 2019
Closed: Rik on 24 Jul 2020
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])
Is my logic correct?
my approach
function [a,b]= minimax(M)
m=M([1:end,0);
a= [abs(max(M(m))-min(M(m)))];
b= max(M(:)) - min(M(:));
end
Matrika Shukla on 18 Jul 2020
What does (M.') do exactly?
Walter Roberson on 18 Jul 2020
M.' is transpose (not conjugate transpose, just plain transpose)

mayank ghugretkar on 5 Jun 2019
here's my function....
went a little descriptive for good understanding to readers.
function [a,b]=minimax(M)
row_max=max(M');
overall_max=max(row_max);
row_min=min(M');
overall_min=min(row_min);
a=row_max - row_min;
b=overall_max-overall_min;
[mmr, mmm] = minimax([1:4;5:8;9:12])
Purushottam Shrestha on 8 Jun 2020
We need to transpose because max(M.') gives a row vector of maximum elements of each row. I want you to try by giving command >>max(A.') Then you can see clearly.
Stephen23 on 17 Jul 2020
"We need to transpose because max(M.') gives a row vector of maximum elements of each row."
In some specific cases it will, but in general it does not.
"I want you to try by giving command >>max(A.') Then you can see clearly."
Okay, lets take a look:
>> A = [1;2;3]
A =
1
2
3
>> max(A.')
ans = 3
I can clearly see that this does NOT give the maximum of each row of A.

Arooba Ijaz on 1 May 2020
function [mmr,mmm] =minimax (M)
%finding mmr
a=M'
b=max(a)
c=min(a)
mmr=b-c
%finding mmm
d=max(M)
e=max(d)
f=min(M)
g=min(f)
mmm=e-g
##### 3 CommentsShow 1 older commentHide 1 older comment
Walter Roberson on 9 Jun 2020
M is two dimensional. When you take max() of a two-dimensional matrix, then by default the maximum is taken for each column, so you would go from an m x n matrix to a 1 x n matrix of output. Then max() applied to that 1 x n matrix would take the maximum of those values, giving you a 1 x 1 result.
Rik on 9 Jun 2020
This is done, because max only operates on a single dimension. Starting from R2018b you can specify a vector of dimensions, or use the 'all' keyword, see the documentation. In this answer they probably should have written max(M(:)) instead. I don't know who upvoted this function, as it is undocumented and takes a strange path to an answer.

Nisheeth Ranjan on 28 May 2020
function [mmr,mmm]=minimax(A)
mmt=[max(A,[],2)-min(A,[],2)];
mmr=mmt'
mmm=max(max(A))-min(min(A))
This is the easiest code you cold ever find. Thank me later.
Jessica Avellaneda on 22 Jul 2020
Walter Roberson on 22 Jul 2020

Geoff Hayes on 27 May 2019
Edited: Geoff Hayes on 27 May 2019
Is my logic correct?
I'm not clear on why you need the m. In fact, doesn't the line of code
m=M([1:end,0);
fail since there is no closing square bracket? What is the intent of this line?
Take a look at max and min and in particular the "dimension to operate along" parameter and see how that can be used to find the minimum and maximum value in each row (as opposed to in each column).
RAHUL KUMAR on 8 May 2020
function [mmr mmm] = minimax(M);
mmr = (max(M,[],2) - min(M,[],2))';
mmm = max(M(:))-min(M(:));
end
Sahil Deshpande on 30 May 2020
function [mmr,mmm] = minimax(M)
mmr = abs(max(M.')-min(M.'));
mmm = max(max(M)) - min(min(M));
I did it this way

pradeep kumar on 26 Feb 2020
function [mmr,mmm]=minimax(M)
mmr=abs(max(M')-min(M'));
mmm=(max(max(M'))-min(min(M')))
end
Rik on 26 Feb 2020
Edited: Stephen23 on 17 Jul 2020
Why would you use the transpose if you can also simply use the third input argument for min?
Also, max(max(M')) is equivalent to max(max(M)) and max(M(:)) (and also to max(M,[],'all'), so you could even use that).

Rohan Singla on 17 Apr 2020
function [mmr,mmm] = minimax(M)
a=M';
mmr=max(a,[],1)-min(a,[],1);
mmm= max(M(:)) - min(M(:));
end
Walter Roberson on 12 May 2020
M' is conjugate transpose. Unless you are doing specialized linear algebra, it is recommended that you use .' instead of ' as .' is regular (non-conjugate) transpose.
Walter Roberson on 12 May 2020

AYUSH MISHRA on 26 May 2020
function [mmr,mmm]=minimax(M)
mmr=max(M')-min(M');
mmm=max(max(M'))-min(min(M'));
end
% here M' is use because when we are using M than mmr generate column matrix
SOLUTION
[mmr, mmm] = minimax([1:4;5:8;9:12])
mmr =
3 3 3
mmm =
11
saurav Tiwari on 11 Jun 2020
whatttt, it's so easy code omg and i make it very difficult. Same on me

Anurag Verma on 26 May 2020
function [mmr,mmm]=minimax(M)
a = max(M(1,:))-min(M(1,:));
b = max(M(2,:))- min(M(2,:));
c = max(M(3,:))- min(M(3,:));
mmr = [a,b,c];
mmm = max(M(:))-min(M(:));
what's wrong with this code. can anyone explain please it gives an error with the random matrix question?
Rik on 26 May 2020
Your code will only consider the first 3 rows. It will error for arrays that don't have 3 rows, and will return an incorrect result for arrays that have more than 3 rows.
You should read the documentation for max and min, and look through the other solutions on this thread for other possible strategies to solve this assignment.
saurav Tiwari on 11 Jun 2020
yaa, RIK is right. your code can only work for 3 rows matrix but random matrix contain a matrix of rows>1 . ok so, you should have to make a code that can work for any type of matrix

Md Naim on 30 May 2020
function [mmr, mmm]= minimax(M)
mmr = max(M')-min(M')
mmm = max(max(M'))-min(min(M'))
end

ROHAN SUTRADHAR on 6 Jun 2020
function [mmr,mmm] = minimax(A)
X = A';
mmr = max(X([1:end],[1:end]))- min(X([1:end],[1:end]));
mmm = max(X(:))-min(X(:));
end

saurav Tiwari on 11 Jun 2020
function [a,b]=minimax(M)
[m,n]=size(M);
x=1:m;
a=max(M(x,:)')-min(M(x,:)');
v=M(:);
b=max(v)-min(v);
end
saurav Tiwari on 11 Jun 2020
most easiest code of the world

A.H.M.Shahidul Islam on 21 Jul 2020
Edited: A.H.M.Shahidul Islam on 21 Jul 2020
function [mmr,mmm]=minimax(M)
m=M';
mmr=abs(max(m)-min(m));
mmm=max(M(:))-min(M(:));
%works like a charm
Stephen23 on 21 Jul 2020
"works like a charm"
Does not work:
>> M = [1;2;4]
M =
1
2
4
>> minimax(M)
ans =
3

Akinola Tomiwa on 23 Jul 2020
Function [mmr, mmm] = minmax(x)
mmr = (max(x, [], 2) - min(x, [], 2)';
%the prime converts it to a row matrix
mmm = (max(x(:)) - min(x(:));
end
Walter Roberson on 23 Jul 2020
mmm = (max(x(:)) - min(x(:)) ;
1 2 3 21 2 3 21
The number indicates the bracket nesting level in effect "after" the corresponding character. You can see that you have one open bracket in effect at the end of the line.
youssef boudhaouia on 24 Jul 2020
function [mmr,mmm]=minimax(M)
a=M';
ma=max(a);
mi=min(a);
mmr = ma - mi ;
mmm=max(max(M)) - min(min(M));
end
Here's my answer, as simple as possible and it works.

youssef boudhaouia on 24 Jul 2020
function [mmr,mmm]=minimax(M)
a=M';
ma=max(a);
mi=min(a);
mmr = ma - mi ;
mmm=max(max(M)) - min(min(M));
end
here's my answer as simple as possible , it works!

R2018b

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!