Solving 2nd part of minimax question

Good day, Please I've been stuck on this assignment for awhile now, i got the second part but the first part can't seem to make it work with random numbers. Had an eureka with 'mmm' and it was thrilling but my assignment is 'over' overdue and i really need a guide to solve 'mmr' properly. i'll greatly appreciate any help. Thank you

Here's what i've achieved so far.

mmr works but not with random numbers and it seems cumbersome.

function [mmr,mmm] = minimaxTrial(M)
    A = ([M]);
    A_max = max(A,[],'all');
    A_min = min (A,[],'all');
    mmm = (A_max - A_min);
    [row,col] = size(A);
  C_1 = (A(1,1:col)); 
  C_2 = (A(2,1:col)); 
  C_3 = (A(3,1:col));
  C_4 = (A(n,1:col));
    M1 = max(C_1) - min(C_1);
    M2 = max(C_2) - min(C_2);
    M3 = max(C_3) - min(C_3);
    M4 = max(c_4) - min(C_4);
    mmr =[M1, M2, M3, M4];
end 

[Turlough Hughes]

It's not clear what you want to do. Anyway, the min and max function will return the min/max for each column or for each row of your input matrix if you have the third input as 1 or 2 respectively. Also instead of A(1,1:col) you can simply write A(1,:), even though you don't actually need those lines.

[Sarah Majin]

Thank you for your response. I'm trying to solve the first part of this question

"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"

When you say I do not need those lines, do you mean I can try use the max, min function directly as in the case of mmr??

[John D'Errico]

Remember that when you ask for help because something does not work properly, if you show us ONLY the code, then we cannot know what you really need to get as a result. Just telling us this is the second part of a minimax question does not help, because we don't see the question, and we are not taking your course. Showing the code you wrote is hugely important, just not quite enough.

I will add that the line:

A = ([M]);

is comepletely irrelevant to anything. It does nothing except use CPU cycles.

It looks like the question in respect to mmm is to return the difference between the overall largest and smallest elements in the matrix.

Is mmr supposed to be somehow related to that, perhaps applying only to the rows of the matrix?

[Turlough Hughes]

Yes, the last 9 lines can be boiled down to one. Try min(A,[],2), which gives the minimum of each row.

[Sarah Majin]

Thank you John I'll note that for next time. Here's the question

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

I am having difficulties with the highlighted part.

I used

A = ([M]); to represent the matrix input. I dunno, is another way of expressing that?

[Sarah Majin]

Thank you Turlough, I'll try that ?

[Turlough Hughes]

Also, you should look at the Matlab Onramp courses (the first one) to get to grips with the basics.

[Madumathi Krishnan]

I got the output for that question

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

function [mmr,mmm]=minimax (M)
mmr=max(M,[],2)'-min(M,[],2)';
c=max(M,[],2);
d=min(M,[],2);
mmm=max(c)-min(d);
end
[mmr, mmm] = minimax([1:4;5:8;9:12])

Answer 1

[John D'Errico]

I see you did answer the question in my comment.

"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"

This seems a little confusing as a question, because the difference between the max and min will ALWAYS be a positive number. So I'm not at all sure why it seemed important to indicate the absolute value. I guess that just means those writing homework assignments are human too?

Anyway, the global max minus the min is easy to write, as is the row-wise result.

function [mmr,mmm] = minimaxTrial(M)
  mmm = max(A,[],'all') - min(A,[],'all');
  mmr = max(A,[],2) - min(A,[],2);
end

One thing you don't want to do, is to extract each row of the matrix as you did.

C_1 = (A(1,1:col)); 
C_2 = (A(2,1:col)); 
C_3 = (A(3,1:col));
C_4 = (A(n,1:col));

That never works, because what happens if tomorrow your matrix has 3 or 5 rows? That little n at the end does not tell MATLAB to extend the pattern for n rows. (I suppose it should, but computers are not that smart. Well, not yet.)

Instead, you either needed to use the correct form for the max and min functions as I did, or you could have used a loop. Far better to write it in one line as I did though. using a loop, I might write it as:

function [mmr,mmm] = minimaxTrial(M)
  mmm = max(A,[],'all') - min(A,[],'all');
  [Ar,Ac] = size(A);
  mmr = NaN(Ar,1);
  for ir = 1:Ar
    mmr(ir) = max(A(ir,:)) - min(A(ir,:));
  end
end

See why this works for any number of rows.

Also see that I pre-allocated the vector mmr in the looped version. That is something you will need to do as your code becomes more sophisticated, because you don't want to dynamically adjust the size of vectors of arrays in a loop.

[Sarah Majin]

Thank so much for your help John. This is my final code

function [mmr,mmm] = minimaxTrial(M) A = ([M]); A_max = max(A,[],'all'); A_min = min (A,[],'all'); mmm = (A_max - A_min); M1 = max (A,[],2); M2 = min (A,[],2); M3 = M1 - M2; mmr= M3'; end I believe it will work, I'm not too familiar with loops yet. I appreciate

[Sarah Majin]

 function [mmr,mmm] = minimaxTrial(M)
    A = ([M]);
    A_max = max(A,[],'all');
    A_min = min (A,[],'all');
    mmm = (A_max - A_min);
    M1 = max (A,[],2);
    M2 = min (A,[],2);
    M3 = M1 - M2;
    mmr= M3';
  end

[Prathibha v]

function [mmr, mmm]= minimax (A)

mmr = [max(A,[],2)- min(A,[],2)]';

mmm = max(A,[],'all')-min(A,[],'all');

end

Answer 2

[ERTIZA HOSSAIN SHOPNIL]

function [mmr,mmm]= minimax(x)
    A=x';
    mmr=max(A)-min(A);
    mmm=(max(A,[],'all')-min(A,[],'all'));
end

Answer 3

[Naveen Gehlot]

function [mmr, mmm]= minimax (A)
mmr = [max(A,[],2)- min(A,[],2)]'; 
mmm = max(A,[],'all')-min(A,[],'all');


OUTPUT
mmr =

     3     3     3


mmm =

    11