write a function called saddle in the input matrix M , the function should return with exactly two column vector and each row of output vector represent dimensions of saddle point in the input matrix M

i have written this code , but its giving me error please help to correct my code ,Thanks in advance

function indices = saddle(M)
indice=[];
r=0;
for n1=1:size(M,1)
    maximum=max(M(n1,:));
    point=find(M(n1,:)==maximum);
    lowest=min(M(:,point));
    if maximum == lowest
        condition=1;  
    else
        condition=0;
    end
    if condition==1 
        r=r+1;
        indice(r,1)=n1;
        indice(r,2)=point;
    end
    indices=indice;
end

[Rik]

You can find guidelines for posting homework on this forum here. Have you tried putting a break point at the start of your function and checked when your variable started containing incorrect values?

You can even find this specific question on this forum, with a lot of advice (and sadly even complete working solutions). I'm tempted to close this question now, before it starts attracting the same behavior.

[Amrita Pritam]

function s = saddle(M)
% Create logical vector that are true for each saddle condition separately 
     minLocs = M <= min(M, [], 1);   
     maxLocs = M >= max(M, [], 2); 
% Find the indices where both conditions are true!     
     [row, col] = find(minLocs & maxLocs);
% If the input is a row vector, row and col returned from the find
% function need to be transposed to fit the output format 
    if isrow(M) 
         s = [row', col'];
    else
         s = [row, col]; 
    end
end

[Rik]

I would suggest removing the code and leaving the code comments. That way students can fill in a framework solution and learn from you.

Answer 1

[Duong Nguyen]

Hi, I got my answer for this question as shown below. I am wondering that if there is any test cases fail with my answer? Thank you!

function indices = saddle(M)
indices = [];
[row,~] = size(M);
for i = 1:row
    Max = strfind(M(i,:),max(M(i,:))); [~, indMax] = size(Max);
    for ii = 1:indMax
        if M(i,Max(ii)) == min(M(:,Max(ii)))
            indices = [indices; [i, Max(ii)]];
        end
    end
end
end 

[Rik]

You don't have any comments. How is anyone supposed to learn from your answer?

Also, mlint is warning you that your indices array is growing every iteration.

To answer your question: where is 4,4 in the output?

M= [10 12 7 3 12;
    3 10 6 2 8;
    12 24 17 6 10;
    15 21 10 8 12;
    1 18 22 4 15];
indices = saddle(M)

indices = 1×2

2 2

[Luis Eduardo Pacheco González]

If i'm right, you are using strfind to compare each element of the matrix in the for loop so you can evaluate if it's greater or equal to...right?

On the other hand, can you explain the line: indices = [indices; [i, Max(ii)]];

Why do you write indices inside the argument?

Answer 2

[Chandan Kumar]

function s = saddle(M) 
[r, c] = size(M);
% Initialize the saddle points to an empty array
% Check the dimensions to see if input is a row or column vector
% find the min value in each column if more than 1 row
% min would give a single value
% find the max value in each row
% max would give a single value
% visit each column
% and each row, that is, each element of M
if M(jj,ii) == cols(ii) && M(jj,ii) == rows(jj) % if both conditions hold
    s = [s; jj ii]; % saddle point! Let's add it!
end

[Rik]

Please use the tools explained on this page to format your post.

Also, why did you answer this question with a complete solution? That only encourages cheating.

Answer 3

[Ankit Sharma]

%  I have write this code in matlab 2015 
%  creating a function named as saddle
function indices = saddle(M)  % input matrix is M
s = size(M); % finding the size of the M matrix
[max_per_row] = max(M , [] ,2); % finding the maximum in row by max command, here 2 reads the row
[min_per_col] = min(M , [] ,1); % finding the minimum in column by min command, here 1 reads the column 
indices = []; % assining th =e indices to empty matrix if there is no saddle point present then it returns empty matrix there
% creating a loop to check the requireties like, whose value is 
% greater then or equal to every element in row, and less then  or 
% equal to every element in column

for j=1:s(2) 
    for i=1:s(1)
        if M(i,j) >= max_per_row(i) && M(i,j) <= min_per_col(j) % here we are applying the condition to check requireties
            indices = [indices; [i, j]] % if any indices find then assigning the value of indices
        end
    end
end
end

Answer 4

[Vishnu V]

function indices = saddle(M)
[m,n]=size(M);
indices = [];
for ii = 1 : m    %accessing each elements of M
    for jj = 1 : n
        if sum(M(ii,jj)>=M(ii,:))==n  % checking whether the element is greater than or equal to all the other elements in the row 
            if sum([M(ii,jj)<=M(:,jj)]')==m % checking whether the element is lesserr than or equal to all the other elements in the column 
               indices = [ indices; ii jj]; % append the indices of saddle element 
            end
        end
    end
end

Answer 5

[David]

function s = saddle(M)
[r, c] = size(M);
% Initialize the saddle points to an empty array
s = [];
% Check the dimensions to see if input is a row or column vector
if r > 1
    cols = min(M);          % find the min value in each column if more than 1 row
else
    cols = M;               % vector is a special case, min would give a single value
end
if c > 1
    rows = max(M');         % find the max value in each row
else
    rows = M;               % vector is a special case, max would give a single value
end
for ii = 1:c                % visit each column
    for jj = 1:r            % and each row, that is, each element of M
        if M(jj,ii) == cols(ii) && M(jj,ii) == rows(jj) % if both conditions hold
            s = [s; jj ii];                             % saddle point! Let's add it!
        end
    end
end

[Vishal]

what is the code to genreate the following function

[DGM]

Oh that's easy:

Answer 6

[youssef]

function indices=saddle(M)
indices=[];
[a,b]=size(M);
rows=cell(a,1);
%making the ceLLs
for i=1:a
    rows{i,1}=M(i,:);
end
%finding max of each row
maxindx=[];
for i=1:a
    tempcellrow=rows{i,1};
    c1=find(tempcellrow==max(tempcellrow));
    for q=1:length(c1)
    maxindx=[maxindx;i c1(q)];
    end
end
%maxindex now holds the indices of all elements which are the max of its
%row and with respect to the repetition if exist

%Lets check these element by its indecies if they are the min of their cols
count=0;
[L,~]=size(maxindx);
for c=1:L
    check=M(maxindx(c,1),maxindx(c,2));
    for i=1:a
        if check<=M(i,maxindx(c,2))
            count=count+1;
        end
    end
    if count==a
        indices=[indices;maxindx(c,:)];
    end
    count=0;
end
end