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Hi everyone; I am going to find the saddle points of a matrix M. The question is given below...

Write a function called saddle that finds saddle points in the input matrix M. For the purposes of this problem, a saddle point is defined as an element whose value is greater than or equal to every element in its row, and less than or equal to every element in its column. Note that there may be more than one saddle point in M. Return a matrix indices that has exactly two columns. Each row of indices corresponds to one saddle point with the first element of the row containing the row index of the saddle point and the second column containing the column index. The saddle points are provided in indices in the same order they are located in M according to column-major ordering. If there is no saddle point in M, then indices is the empty array.

I am trying that code:

function indices = saddle(M)

[ rows,cols ] = size(M);

[valR,posR] = max(M,[],2);

[valC,posC] = min(M,[],1);

indices= [];

for i = 1:length(posR)

if i == posC(posR(i))

indices= [indices; i, posR(i)];

end

end

end

It is running fine. But when i test my code for

>> mat=zeros(5,3)

mat =

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

I am getting wrong output :

saddle(mat)

ans =

1 1

the correct output must be

saddle(mat)

ans =

1 1

2 1

3 1

4 1

5 1

1 2

2 2

3 2

4 2

5 2

1 3

2 3

3 3

4 3

5 3

What i am doing wrong?? Thanks in advance

Stephen Cobeldick
on 15 Jun 2015

Edited: Stephen Cobeldick
on 15 Jun 2015

Copying code from the internet is not always a good way to learn best-practice coding: Solving this problem using loops misses using MATLAB's excellent code vectorization abilities. It would be much neater and faster to use bsxfun instead, like this:

function idx = saddle(mat)

row_mx = bsxfun(@ge,mat,max(mat,[],2));

col_mn = bsxfun(@le,mat,min(mat,[],1));

[R,C] = find(row_mx & col_mn);

idx = [R,C];

end

Which gives this:

>> saddle(zeros(5,3))

ans =

1 1

2 1

3 1

4 1

5 1

1 2

2 2

3 2

4 2

5 2

1 3

2 3

3 3

4 3

5 3

vaishak p nair
on 26 Aug 2019

Write a function called saddle that finds saddle points in the input matrix M. For the purposes of this problem, a saddle point is defined as an element whose value is greater than or equal to every element in its row, and less than or equal to every element in its column. Note that there may be more than one saddle point in M. Return a matrix called indices that has exactly two columns. Each row of indices corresponds to one saddle point with the first element of the row containing the row index of the saddle point and the second element containing the column index. If there is no saddle point in M, then indices is the empty array.

solution :

function indices=saddle(M)

indices=[];

[a b]=size(M);

q=1;

for i=1:a

for j=1:b

x=M(i,:);

y=M(:,j);

c=M(i,j)>=x;

d=M(i,j)<=y;

if ~ismember(0,c) && ~ismember(0,d)

indices(q,1)=i;

indices(q,2)=j;

q=q+1;

end

end

end

end

Travis Ha
on 30 Jul 2020 at 23:14

the cyclist
on 14 Jun 2015

The reason your code doesn't give your expected result can be summarized by this sentence from the documentation for max: If the maximum value occurs more than once, then max returns the index corresponding to the first occurrence.

I think you were expecting it to return the indices of all the maxima.

Konstantinos Sofos
on 14 Jun 2015

Edited: Konstantinos Sofos
on 14 Jun 2015

Dear Muhammad,

You know it's very unfair continuously to ask the forum to solve your exercises/homework. I can understand you because also I was student and I wanted to solve my exercise to proceed but just as a friendly recommendation "Try to understand your exercises!". It's the only way to go one step further without cheating most of all yourself. The code that you posted has been posted before 4 days also to another programming forum Find saddle points in Matlab.

Now to your exercise. Your problem is very simple.

1. Take a piece of white paper and write down a matrix

2. Try to write down an algorithm

3. Write your own function

In my opinion this the way to learn programming. Good luck!

Regards,

Marcos Mariano
on 14 Jun 2015

I am totally agree with Konstantinos Sofos. Look at the forum of your course, you may find the help you need. But I am giving you a hint, try this:

etc..

saddle_mat = [ ];

for jj =

for ii =

if ...

saddle_mat = %add ii jj to the saddle points list as a new row

end

end

end

Revant Shah
on 24 Apr 2020

Jaimin Motavar
on 3 Jul 2019

I hope this answer is helpful to you.

function indices = saddle(M)

[m,n]=size(M);

a=[];

for i=1:m

for j=1:n

if prod(M(i,j)>=M(i,:))==1 && prod(M(i,j)<=M(:,j))==1

a=[i,j;a];

end

end

end

indices=a;

end

Faria Sultana
on 30 Apr 2020

charu sharma
on 27 Aug 2015

Jos (10584)
on 3 Apr 2019

function out = saddle(M)

[r, c] = ind2sub(size(M), 1:numel(M)) ;

tf = arrayfun(@(r, c) all(M(r, c) >= A(:, c)) && all(M(r, c) <= M(r, :)), r, c)

out = [r(tf) ; c(tf)].'

MADDINENI REVANTH SAI
on 31 Aug 2019

function s = saddle(M)

[r, c] = size(M);

s = [];

if r > 1

cols = min(M);

else

cols = M;

end

if c > 1

rows = max(M');

else

rows = M;

end

for ii = 1:c

for jj = 1:r

ifM(jj,ii) = cols(ii)&&M(jj)==rows(jj)

s = [s;jj ii];

end

end

end

Divya Ratna
on 24 May 2020

i think anyone should try their own first rather than looking for answers in the community.

my attempt was this.

this passes all the test cases...

function indices = saddle (M)

s = size (M);

indices = [];

for ii = 1 : s(1)

maxy = max ( M(ii,:) );

for jj = 1 : s(2)

if M(ii,jj) == maxy;

miny = min (M(:,jj));

if M(ii,jj) == miny;

indices = [indices; ii jj];

end

end

end

end

end

Garvit Kukreja
on 29 May 2020

can you help me with this.

Thankyou

function [indices] = saddle(z)

[ii jj ]= size(z)

indices = [];

for i=1:ii

for j=1:jj

x(i ,j)= [ z(i,j)]

end

[p,q]= max(x(i,:)) %max value in a row. p give max value, q gives column

for k=1:ii

y(k,q)= [z(k,q)]

end

[m,n]= min(y(:,q)) %min value in a row. m give min value, n gives column

if p==m

indices = [indices; i q]

end

end

end

Garvit Kukreja
on 30 May 2020

Can anyone Help me with this. Thankyou

function [indices] = saddle(z)

[ii jj ]= size(z)

indices = [];

for i=1:ii

for j=1:jj

x(i ,j)= [ z(i,j)]

end

[p,q]= max(x(i,:)) %max value in a row. p give max value, q gives column

for k=1:ii

y(k,q)= [z(k,q)]

end

[m,n]= min(y(:,q)) %min value in a row. m give min value, n gives column

if p==m

indices = [indices; i q]

end

end

end

Tejas Sabu
on 13 Jun 2020

Edited: Tejas Sabu
on 13 Jun 2020

function indices=saddle(M)

[m,n] = size (M);

indices=[];%we want an empty matrix if there r no saddle points

for i=1:m; %going thru all the rows and each element of the row.

maxi=max(M(i,:));% finding the max of the elements of the specific row.

for j=1:n;% running thru all the coloumns and each element of the column .

mini=min(M(:,j));% finding the min of the elements of each column.

if maxi==mini%checking if the max of a row is same as the min of a column, if yes then

indices=[indices;i j];% indices will give null matrix in first column and i and j in the next row

end

end

end

hope this helps...try to understand the code instead of copying.

Muhammad Qaisar Ali
on 26 Jun 2020

another approch

function indices = saddle(Z)

indices=[];

for r=1:size((Z),1) % going through Rows

for c=1:size((Z),2) % going through Cols

if sum((Z(r,c)>=(Z(r,:))))>=size((Z),2) && sum((Z(r,c)<=(Z(:,c))))>=size((Z),1) % then saddle point

indices=[indices;[r,c]];

end

end

end

end

Shiladittya Debnath
on 27 Jul 2020 at 7:53

For Function :

function id = saddle(M)

[a,b]=size(M);

id = zeros(a+b,2);

count = 0;

for i = 1:a

mah = max(M(i,:));

[c1,c2] = find(M(i,:) == mah);

for k = 1:length(c1)

c1k = c1(k); c2k = c2(k);

mic = min(M(:,c2k));

if M(i,c2k)==mic

count = count+1;

id(count,:) = [i,c2k];

end

end

end

id = id(1:count,:);

end

Shiladittya Debnath
on 27 Jul 2020 at 7:54

And for Code to Call your Function :

% create an interesting surface

[X,Y] = meshgrid(-15:0.5:10,-10:0.5:10);

Z = (X.^2-Y.^2)';

% find saddle points

indices = saddle(Z)

% plot surface

surf(Z);

hold on

% mark saddle points with red dots in the same figure

for ii = 1:size(indices,1)

h = scatter3(indices(ii,2),indices(ii,1),Z(indices(ii,1),indices(ii,2)),'red','filled');

h.SizeData = 120;

end

% adjust viewpoint

view(-115,14);

hold off

SIVA SAI AKULA
on 29 Jul 2020 at 3:04

function indices = saddle(M)

row_max = max(M,[],2);

col_min = min(M,[],1);

[row,col]=find((M == row_max).*(M == col_min));

if isempty(col) || isempty(row)

indices=[]

else

for i=1:length(row)

indices(i,:)=[row(i),col(i)];

end

end

end

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