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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])
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
  15 comentarios

Respuestas (14)

mayank ghugretkar
mayank ghugretkar el 5 de Jun. de 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;
Code to call your function
[mmr, mmm] = minimax([1:4;5:8;9:12])
  5 comentarios
Purushottam Shrestha
Purushottam Shrestha el 8 de Jun. de 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
Stephen23 el 17 de Jul. de 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
Arooba Ijaz el 1 de Mayo de 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 comentarios
Walter Roberson
Walter Roberson el 9 de Jun. de 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
Rik el 9 de Jun. de 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
Nisheeth Ranjan el 28 de Mayo de 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.
  5 comentarios

Geoff Hayes
Geoff Hayes el 27 de Mayo de 2019
Editada: Geoff Hayes el 27 de Mayo de 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).
  4 comentarios
RAHUL KUMAR
RAHUL KUMAR el 8 de Mayo de 2020
function [mmr mmm] = minimax(M);
mmr = (max(M,[],2) - min(M,[],2))';
mmm = max(M(:))-min(M(:));
end
Sahil Deshpande
Sahil Deshpande el 30 de Mayo de 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
pradeep kumar el 26 de Feb. de 2020
function [mmr,mmm]=minimax(M)
mmr=abs(max(M')-min(M'));
mmm=(max(max(M'))-min(min(M')))
end
  1 comentario
Rik
Rik el 26 de Feb. de 2020
Editada: Stephen23 el 17 de Jul. de 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
Rohan Singla el 17 de Abr. de 2020
function [mmr,mmm] = minimax(M)
a=M';
mmr=max(a,[],1)-min(a,[],1);
mmm= max(M(:)) - min(M(:));
end
  5 comentarios
Walter Roberson
Walter Roberson el 12 de Mayo de 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.

AYUSH MISHRA
AYUSH MISHRA el 26 de Mayo de 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

Anurag Verma
Anurag Verma el 26 de Mayo de 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?
  2 comentarios
Rik
Rik el 26 de Mayo de 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
saurav Tiwari el 11 de Jun. de 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
Md Naim el 30 de Mayo de 2020
function [mmr, mmm]= minimax(M)
mmr = max(M')-min(M')
mmm = max(max(M'))-min(min(M'))
end

ROHAN SUTRADHAR
ROHAN SUTRADHAR el 6 de Jun. de 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
saurav Tiwari el 11 de Jun. de 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

A.H.M.Shahidul Islam
A.H.M.Shahidul Islam el 21 de Jul. de 2020
Editada: A.H.M.Shahidul Islam el 21 de Jul. de 2020
function [mmr,mmm]=minimax(M)
m=M';
mmr=abs(max(m)-min(m));
mmm=max(M(:))-min(M(:));
%works like a charm

Akinola Tomiwa
Akinola Tomiwa el 23 de Jul. de 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
  4 comentarios
Walter Roberson
Walter Roberson el 23 de Jul. de 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
youssef boudhaouia el 24 de Jul. de 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
youssef boudhaouia el 24 de Jul. de 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!

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