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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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Debaditya Chakraborty il 27 Mag 2019
Chiuso: Rik il 24 Lug 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
##### 15 CommentiMostra 13 commenti meno recentiNascondi 13 commenti meno recenti
Matrika Shukla il 18 Lug 2020
What does (M.') do exactly?
Walter Roberson il 18 Lug 2020
M.' is transpose (not conjugate transpose, just plain transpose)

### Risposte (14)

mayank ghugretkar il 5 Giu 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])
##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
Purushottam Shrestha il 8 Giu 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 il 17 Lug 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 il 1 Mag 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 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Walter Roberson il 9 Giu 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 il 9 Giu 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 il 28 Mag 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 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
Jessica Avellaneda il 22 Lug 2020
Walter Roberson il 22 Lug 2020

Geoff Hayes il 27 Mag 2019
Modificato: Geoff Hayes il 27 Mag 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 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
RAHUL KUMAR il 8 Mag 2020
function [mmr mmm] = minimax(M);
mmr = (max(M,[],2) - min(M,[],2))';
mmm = max(M(:))-min(M(:));
end
Sahil Deshpande il 30 Mag 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 il 26 Feb 2020
function [mmr,mmm]=minimax(M)
mmr=abs(max(M')-min(M'));
mmm=(max(max(M'))-min(min(M')))
end
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Rik il 26 Feb 2020
Modificato: Stephen23 il 17 Lug 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 il 17 Apr 2020
function [mmr,mmm] = minimax(M)
a=M';
mmr=max(a,[],1)-min(a,[],1);
mmm= max(M(:)) - min(M(:));
end
##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
Walter Roberson il 12 Mag 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 il 12 Mag 2020

AYUSH MISHRA il 26 Mag 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
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
saurav Tiwari il 11 Giu 2020
whatttt, it's so easy code omg and i make it very difficult. Same on me

Anurag Verma il 26 Mag 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 CommentiMostra NessunoNascondi Nessuno
Rik il 26 Mag 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 il 11 Giu 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 il 30 Mag 2020
function [mmr, mmm]= minimax(M)
mmr = max(M')-min(M')
mmm = max(max(M'))-min(min(M'))
end
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

ROHAN SUTRADHAR il 6 Giu 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
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

saurav Tiwari il 11 Giu 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
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
saurav Tiwari il 11 Giu 2020
most easiest code of the world

A.H.M.Shahidul Islam il 21 Lug 2020
Modificato: A.H.M.Shahidul Islam il 21 Lug 2020
function [mmr,mmm]=minimax(M)
m=M';
mmr=abs(max(m)-min(m));
mmm=max(M(:))-min(M(:));
%works like a charm
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Stephen23 il 21 Lug 2020
"works like a charm"
Does not work:
>> M = [1;2;4]
M =
1
2
4
>> minimax(M)
ans =
3

Akinola Tomiwa il 23 Lug 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 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
Walter Roberson il 23 Lug 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 il 24 Lug 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 il 24 Lug 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!
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Questa domanda è chiusa.

R2018b

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