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,

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

[Geoff Hayes]

Debaditya - what is your question?

[Debaditya Chakraborty]

How can make this work for any given matrix??

especially function a

[Geoff Hayes]

The code that you have posted will work for any given matrix though you probably want to remove the m as it is unclear why it is needed. Have you looked at the documentation to see how to use min and max to get the minimum or maximum across each row?

[JEET BANERJEE]

I think the code (as shown below) generated by me is quite time saving and could solve it in two lines.

function [mmr,mmm] = minimax(M)
row_vector = abs(max(M,[],2)- min(M,[],2];
Overall = max(M,[],’all’)- min(M,[],’all’);
mmr = (row_vector)’;
mmm = Overall;
end

[vignesh suresh]

function [mmr, mmm]= minimax(M)
mmr=max(M')-min(M');
mmm=[max(M(:))-min(M(:))];

[Rik]

Comment posted as answer by Bece:

Hello everyone,

I am using Mathlab Online R2020a for this course.

I was able to pass this assignment with the following code:

function [mmr, mmm]= minimax(M)
mmr=max(M')-min(M');
mmm=[max(M(:))-min(M(:))]; 

Then calling it by the given [mmr, mmm] = minimax([1:4;5:8;9:12]) function.

BUT when i try to run this on Mathlab Online, i am getting the Unrecognized function or variable 'minimax' error. Does anyone know why? thanks.

P.s: i created a new script 'assignment' and wrote the code in there, in EDITOR then saved it. Then i tried to call the funtion by [mmr, mmm] = minimax([1:4;5:8;9:12]) in command window and it doesn't work. The other solutions, which are given here by other participants, don't work as well.

Regards

[Sai Sandeep Kajuluri]

I got the same code of your's executed. kindly, name the file as minimax and try again

[Zarin Rafa Shaitee]

Will you kindly describe why transpose of M is important? Thank you

[Jessika Lisboa]

can you explain me the string? what does the string do and what it means for this function?

[Walter Roberson]

I do not see anyone using string() for this topic ?

[Sahil Deshpande]

function [mmr,mmm] = minimax(M)
mmr = abs(max(M.')-min(M.')); 
mmm = max(max(M)) - min(min(M)); 

Shortest code I could write

[abhishek kansal]

it works

[Rik]

My reason for closing this question: people are just posting their answers to this homework question, without substantial discussion happening. Once there is an option to disallow new answers while allowing comments that option should be considered for this thread.

[Matrika Shukla]

What does (M.') do exactly?

[Walter Roberson]

M.' is transpose (not conjugate transpose, just plain transpose)

Answer 1

[mayank ghugretkar]

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])

[Rik]

In general completely working solutions to homework questions are discouraged. The point of homework is to learn, not to outsource work to strangers on the internet. Sometimes helping someone understand a problem and help them solving it on their own can take more time, but it is a lot more satifying than solving someone else's homework.

A broader view on the topic of homework can be found here.

[Rik]

Comment by Jagmeet Singh mistakenly posted as answer:

In the answer why do we need to transpose the matrix M?

[Rik]

Because the min function operates only in one direction. If you want it to operate along a different direction, you can either transpose the matrix (as Mayank has done), or use the third input, as described in the documentation (which I would suggest).

[Purushottam Shrestha]

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]

"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.

Answer 2

[Arooba Ijaz]

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

[vinod suthar]

can please give the reason behind finding e as max(d) which is max(M).

[Walter Roberson]

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]

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.

Answer 3

[Nisheeth Ranjan]

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.

[sandeep kumar singh]

yeah its working

[Sahil Deshpande]

function [mmr,mmm] = minimax(M)
mmr = abs(max(M.')-min(M.')); 
mmm = max(max(M)) - min(min(M)); 

I did it this way.

[Shabarish Muralidharan]

Bro can anyone explain the logic behind this code

[Jessica Avellaneda]

Can you explain please

[Walter Roberson]

Are you asking about Nisheeth's code or about Sahil's code?

Answer 4

[Geoff Hayes]

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).

[Debaditya Chakraborty]

how can I assign each row to this function one after the other?

[Geoff Hayes]

how can I assign each row to this function one after the other?

I don't understand your question. Why do you need to assign each row to this function? If you do

>> min(A, [], 2)

then that will return a vector/array with the minimum value in each row of A...

[RAHUL KUMAR]

function [mmr mmm] = minimax(M);
mmr = (max(M,[],2) - min(M,[],2))';
mmm = max(M(:))-min(M(:));
end

[Sahil Deshpande]

function [mmr,mmm] = minimax(M)
mmr = abs(max(M.')-min(M.')); 
mmm = max(max(M)) - min(min(M)); 

I did it this way

Answer 5

[pradeep kumar]

function [mmr,mmm]=minimax(M)
mmr=abs(max(M')-min(M'));
mmm=(max(max(M'))-min(min(M')))
end

[Rik]

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).

Answer 6

[Rohan Singla]

function [mmr,mmm] = minimax(M)
a=M';
mmr=max(a,[],1)-min(a,[],1);
mmm= max(M(:)) - min(M(:));
end

[Rohan Singla]

you can try this code as it genrated a successfull result .

[RAHUL KUMAR]

function [mmr mmm] = minimax(M);
mmr = (max(M,[],2) - min(M,[],2))';
mmm = max(M(:))-min(M(:));
end

[wikhyath Tirumala]

can i please know what does M' indicate???

[Walter Roberson]

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]

M' is conjugate transpose, https://www.mathworks.com/help/matlab/ref/ctranspose.html

M' is not transpose, https://www.mathworks.com/help/matlab/ref/transpose.html which would be M.'

Answer 7

[AYUSH MISHRA]

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]

whatttt, it's so easy code omg and i make it very difficult. Same on me

Answer 8

[Anurag Verma]

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]

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]

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

Answer 9

[Md Naim]

function [mmr, mmm]= minimax(M)
mmr = max(M')-min(M')
mmm = max(max(M'))-min(min(M'))
end

Answer 10

[ROHAN SUTRADHAR]

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

Answer 11

[saurav Tiwari]

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]

most easiest code of the world

Answer 12

[A.H.M.Shahidul Islam]

function [mmr,mmm]=minimax(M)
m=M';
mmr=abs(max(m)-min(m));
mmm=max(M(:))-min(M(:));
%works like a charm

[Stephen23]

"works like a charm"

Does not work:

>> M = [1;2;4]
M =
     1
     2
     4
>> minimax(M)
ans =
     3

Answer 13

[Akinola Tomiwa]

    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]

you have too many ( on two lines

[Akinola Tomiwa]

Oh, alright, thanks I'll do well to write codes in smaller piece. Thanks a lot

Function [mmr, mmm] = minimax(x)
min_value = min(x,[], 2); % this takes the row minimum values of the supplied matrix  
max_value = max(x, [],2); % this takes the row maximum values
mmr = (max_value-min_value)'; % the prime converts the colon to a row 
mmm = (max(x(:)) - min(x(:)) ;
%x(:) converts the matrix to a single colon matrix so max(x(:)) returns a single value the maximum value 
end

How is this ?

[Walter Roberson]

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]

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.

Answer 14

[youssef boudhaouia]

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!