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# [Assignment]Write a function called saddle that finds saddle points in the input matrix M.

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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.

I am working on this assignment...but i got error feedback all the time,i'm still a noob,so can anyone help me to figure out what is wrong in my program? how to improve it?

Here is my program

function indices = saddle(M)

[a,~] = size(M);

j = 1;i = 1;

[max_M,~] = max(M,[],2);

count = 0;

for ii = 1:a

[c,d] = find(M==max_M(ii));

[e,f] = size(c);

mi = min(M(:,d));

if(max_M(ii) == mi)

while i<=f

while i<=e

output(j,1) = c(i);

output(j,2) = d(i);

j = j+1;

i = i+1;

count = count+1;

end

end

end

if count>0

indices = output;

else

indices = [];

end

end

end

### Answers (15)

Arafat Roney
on 8 May 2020

function in=saddle(M)

[m,n]=size(M); %%SIZE CALCULATED

in=[]; %%'in' IS INITIALIZED AS AN EMPTY MATRIX

for ii=1:m

for jj=1:n

if (M(ii,jj)==max(M(ii,:))&& M(ii,jj)==min(M(:,jj)))

in=[in; ii,jj]; %%INDICES CALCULATION AND STORING TO 'in'

end

end

end

indices=in; %%FINAL INDICES AS OUTPUT ARGUMENT

end

##### 4 Comments

Arafat Roney
on 10 May 2020

Ductho Le
on 4 Apr 2020

If you have no idea to solve this problem, you can use my code as a reference. Good luck!

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

##### 0 Comments

Image Analyst
on 8 Jan 2020

Try this:

numPoints = 7;

M = randi(9, numPoints, numPoints)

% M = [1,2,3,4,4,3,2,1] % Sample data

rowMaxima = max(M, [], 2)

colMinima = min(M, [], 1)

[rows, columns] = size(M);

output = false(size(M));

for col = 1 : columns

for row = 1 : rows

if M(row, col) >= rowMaxima(row) && M(row, col) <= colMinima(col)

output(row, col) = true;

end

end

end

[saddleX, saddleY] = find(output)

output = [saddleX(:), saddleY(:)] % Make into N by 2 array

##### 3 Comments

Rik
on 10 Jan 2020

sri harsha juttiga
on 5 Apr 2020

function [indices,j] = saddle(M)

indices=[];

[m,n]=size(M);

f=0;

for ii=1:m

for jj=1:n

a=M(ii,jj);

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

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

if (a>=x && a<=y)

f=f+1;

indices(f,:)=[ii,jj];

end

end

end

if isempty(indices)

indices=[];

else

indices=indices(1:f,:);

end

end

##### 3 Comments

Walter Roberson
on 26 May 2020

Yaksha SJ
on 10 May 2020

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

##### 2 Comments

Amandeep Kaur
on 8 Jan 2020

function [indices] = saddle(M)

[a,~] = size(M);

j = 1;i = 1;

[max_M,~] = max(M,[],2);

count = 0;

for ii = 1:a

[c,d] = find(M==max_M(ii));

mi = min(M(:,d));

if(max_M(ii) == mi)

output(j,1) = c(i);

output(j,2) = d(i);

j = j+1;

i = i+1;

count = count+1;

end

end

if count>0

indices = output;

else

indices = [];

end

end

I am writing this code, it is giving following message

##### 0 Comments

Vivek Jaswal
on 18 Apr 2020

Someone please help me out where am I going wrong in this code?

function x=saddle(M)

[b,c]=size(M);

for i=1:b

for j=1:c

if M(i,j)==max(M(i,:))

[d,e]=max(M(i,:));

if d==min(M(:,e))

a(i,1)=i; a(i,2)=e;

else

a(i,1)=0; a(i,2)=0;

end

end

j=j+1;

end

i=i+1;

end

x=a(any(a,2),:);

if x==0

x=[];

end

end

##### 9 Comments

Image Analyst
on 18 Apr 2020

The code in my answer https://www.mathworks.com/matlabcentral/answers/483543-assignment-write-a-function-called-saddle-that-finds-saddle-points-in-the-input-matrix-m#answer_409282

gives the correct answer:

rowMaxima =

4

colMinima =

1 2 3 4 4 3 2 1

saddleX =

1 1

saddleY =

4 5

output =

1 4

1 5

You can see that the 4's are saddle points and there are two of the 4's at locations (1,4) and (1,5).

gourav naik
on 25 Apr 2020

function out=saddle(m)

[x,y]=size(m);

ind=[];

for i=1:x

for j=1:y

a=max(m(i,:));

b=min(m(:,j));

if m(i,j)<=b && m(i,j)>=a

ind=[ind ;i,j];

end

end

end

out=ind;

end

try these simple and short

Yasin Peker
on 4 May 2020

function indices = saddle(M)

k = size(M);

r = cell(k(1),k(2));

l = cell(k(1),k(2));

for i = 1:k(2)

for ii = 1:k(1)

t = (M(ii,i) <= M(:,i));

if sum(t) == k(1)

r{ii,i} = [ii,i];

end

end

end

for y = 1:k(1)

for yy = 1:k(2)

p = M(y,yy) >= M(y,:);

if sum(p) == k(2)

l{y,yy} = [y,yy];

end

end

end

for v = 1:k(1)

for vv = 1:k(2)

o = (r{v,vv} == l{v,vv});

indices=[];

if sum(o) == 2

indices = r{v,vv};

end

end

end

end

##### 1 Comment

Yasin Peker
on 4 May 2020

I always get the same error message

Error using ==

Matrix dimensions must agree.

Error in saddle (line 25)

o = (r{v,vv} == l{v,vv});

What I have to do to?

What should I do to improve my code?

Kumar Vivek
on 16 May 2020

Edited: Kumar Vivek
on 16 May 2020

Try this out if you are not getting the problem.

function indices = saddle(M)

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

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

d = b==l; %converting the common elements to logical 1

if isempty(d)

indices = [];

else

[rd,cd] = find(d); %finding the location of that common element

indices = [rd(:),cd(:)];

end

end

##### 0 Comments

Prithvi Shams
on 23 May 2020

function indices = saddle(M)

B = size(M);

idx = 0;

A = [];

%%

for i = 1:B(1)

for j = 1:B(2)

if (M(i,j) == max(M(i,:))) && (M(i,j) == min(M(:,j)))

idx = idx + 1;

A(idx) = M(i,j); %Capture the saddle points in array A

R(idx) = i; %Capture row indices of saddle points in vector R

C(idx) = j; %Capture column indices of saddle points in vector C

end

end

end

%%

if isempty(A)

indices = [];

else

for k = 1:numel(A)

indices(k, 1) = R(k);

indices(k, 2) = C(k);

end

end

end

##### 0 Comments

David Gonzalez
on 24 May 2020

function indices = saddle(M)

indices = [];

for jj = 1:size(M,2)

for ii = 1:size(M,1)

if M(ii,jj) == min(M(:,jj)) && M(ii,jj) == max(M(ii,:))

indices = [indices; ii jj];

end

end

end

##### 0 Comments

Taif Ahmed BIpul
on 31 May 2020

function indices=saddle2(M)

k=0;

indices1=zeros(k);

indices2=zeros(k);

rowIndices=zeros(k);

colIndices=zeros(k);

for i=1:size(M,1)

for j=1:size(M,2)

if M(i,j)==max(M(i,:)) && M(i,j)==min(M(:,j))

k=k+1;

rowIndices(k)=i;

colIndices(k)=j;

indices1=[rowIndices' colIndices'];

else

indices2=[];

end

end

end

if isempty(indices1)

indices=indices2;

else

indices=indices1;

end

##### 0 Comments

Nikolay Ampilogov
on 2 Jun 2020

I think that there is a mistake in the verification pricedure or discription of this exercise [https://www.coursera.org/ ; coourse Introduction to Programming with MATLAB > Week 9 > Assignment: Saddle Points].

There is a frase at the end of the discription: "If there is no saddle point in M, then indices is the empty array." It means that in this case your result should back somethink like:

indices =

0×2 empty double matrix

My code wich passed by the test calling (Code to call your function) is:

function indices = saddle(M)

indices = double.empty(0,2); % To return empty matrix in the mistake case

if (size(M, 1) < 2) || (size(M, 2) < 2) || ~isnumeric(M) % Check the argument

fprintf('The argument is not a numeric matrix with proper size.');

return

end

% Find the indices of saddle points defined as points wich are the local

% minimums for columns and the local maximums for rows of M simultaneously.

[indLocalSaddleRow, indLocalSaddleCol] = find(islocalmin(M, 1) & islocalmax(M, 2));

indices = [indLocalSaddleRow indLocalSaddleCol];

end

To check it for multi saddle surface, please place this before calling:

Z = repmat(Z, 2);

But my code is not acceptable by Assessment after Submit.

The code by gourav naik (on 25 Apr 2020; here) is not sattisfy to requarement "'If there is no saddle point in M, then indices is the empty array." But his code pass by Assessment after Submit succssesfully.

##### 2 Comments

Nikolay Ampilogov
on 3 Jun 2020

Edited: Nikolay Ampilogov
on 3 Jun 2020

Dear Walter, thanks for your attention.

1. It does not have exactly two columns, but it is empty, and that is valid to interpret the bit about empty as over-riding the part about two columns.

Agreed, rather better to use [] instead double.empty(0,2).

And to add this before the end:

if isempty(indices)

indices = [];

end

2. Note: nothing in the assignment expects an error message for invalid parameters.

Agreed, this user-friendly polite here is not necessary:

fprintf('The argument is not a numeric matrix with proper size.');

3. What do you think?

How many saddle points an angulated plane has? Z = [1 2 3; 4 5 6; 7 8 9];

How many saddle points a not angulated plane has? Z = ones(3,3);

Thanks a lot!

Vipin Parthan
on 6 Jun 2020

Edited: Vipin Parthan
on 6 Jun 2020

function indices=saddle(M)

[r,c]=size(M);

%Compute largest row element and the least column element.

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

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

% Initialize empty matrix

indices=[];

% Conditional to find saddle point

for j=1:c

for i=1:r

if M(i,j)==rMax(i) && M(i,j)==cMin(j)

indices=[indices; i,j]; % Store into output argument

end

end

end

end

Hope this code helps. Please suggest if any modifications can be made to make the code more efficient.

##### 0 Comments

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