Info
Diese Frage ist geschlossen. Öffnen Sie sie erneut, um sie zu bearbeiten oder zu beantworten.
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,
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Antworten (14)
mayank ghugretkar
am 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;
Code to call your function
[mmr, mmm] = minimax([1:4;5:8;9:12])
5 Kommentare
Purushottam Shrestha
am 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
am 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
am 1 Mai 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 Kommentare
Walter Roberson
am 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
am 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
am 28 Mai 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 Kommentare
Geoff Hayes
am 27 Mai 2019
Bearbeitet: Geoff Hayes
am 27 Mai 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?
4 Kommentare
RAHUL KUMAR
am 8 Mai 2020
function [mmr mmm] = minimax(M);
mmr = (max(M,[],2) - min(M,[],2))';
mmm = max(M(:))-min(M(:));
end
Sahil Deshpande
am 30 Mai 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
am 26 Feb. 2020
function [mmr,mmm]=minimax(M)
mmr=abs(max(M')-min(M'));
mmm=(max(max(M'))-min(min(M')))
end
Rohan Singla
am 17 Apr. 2020
function [mmr,mmm] = minimax(M)
a=M';
mmr=max(a,[],1)-min(a,[],1);
mmm= max(M(:)) - min(M(:));
end
5 Kommentare
Walter Roberson
am 12 Mai 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
am 12 Mai 2020
AYUSH MISHRA
am 26 Mai 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 Kommentar
saurav Tiwari
am 11 Jun. 2020
whatttt, it's so easy code omg and i make it very difficult. Same on me
Anurag Verma
am 26 Mai 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 Kommentare
Rik
am 26 Mai 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
am 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
am 30 Mai 2020
function [mmr, mmm]= minimax(M)
mmr = max(M')-min(M')
mmm = max(max(M'))-min(min(M'))
end
0 Kommentare
ROHAN SUTRADHAR
am 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
0 Kommentare
saurav Tiwari
am 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
1 Kommentar
A.H.M.Shahidul Islam
am 21 Jul. 2020
Bearbeitet: A.H.M.Shahidul Islam
am 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
1 Kommentar
Stephen23
am 21 Jul. 2020
"works like a charm"
Does not work:
>> M = [1;2;4]
M =
1
2
4
>> minimax(M)
ans =
3
Akinola Tomiwa
am 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
4 Kommentare
Walter Roberson
am 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
am 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
am 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!
0 Kommentare
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!