how to find the element which is greater than or equal to its row and smaller or equal to its column in a matrix

328 views (last 30 days)
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
  8 Comments

Sign in to comment.

Accepted Answer

Stephen23
Stephen23 on 15 Jun 2015
Edited: Stephen23 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
  3 Comments
Akshay Padti
Akshay Padti on 7 Nov 2021
@Stephen It gives wrong result for row vectors.
Above code answer :
>> saddle([1 2 3 4 4 3 2 1])
ans =
1 1 4 5
Expected answer :
>> saddle([1 2 3 4 4 3 2 1])
ans =
1 4
1 5
Corrected code :
function indices = saddle(mat)
[row, ~] = size(mat);
row_mx = bsxfun(@ge, mat, max(mat, [], 2));
col_mn = bsxfun(@le, mat, min(mat, [], 1));
[R, C] = find(row_mx & col_mn);
if row == 1
indices = [R; C]';
else
indices = [R, C];
end
end

Sign in to comment.

More Answers (18)

vaishak p nair
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
  1 Comment
Travis Ha
Travis Ha on 30 Jul 2020
I get everything except for c and d. How does it go through each row and column to find the saddle point? Can you just thoroughly explain how c and d works please?

Sign in to comment.


Tejas Sabu
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.
  4 Comments
Mert Yalcinoz
Mert Yalcinoz on 18 Feb 2022
i didnt understand completly. when i wrote indices=[ii jj] in my code (i have extracted indices in the first row) it failed while testing row vector. why exactly??

Sign in to comment.


the cyclist
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.
  1 Comment
Muhammad Usman Saleem
Muhammad Usman Saleem on 14 Jun 2015
@the cyclist thanks for contributions... But i hope the refer link is not solution of my problem?? I am getting 1 1 for mat(5,3)?????

Sign in to comment.


Konstantinos Sofos
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,
  4 Comments

Sign in to comment.


Jaimin Motavar
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
  2 Comments

Sign in to comment.


Divya Ratna
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
  1 Comment
Garvit Kukreja
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

Sign in to comment.


SIVA SAI AKULA
SIVA SAI AKULA on 29 Jul 2020
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

Hicham Satti
Hicham Satti on 7 Sep 2020
%Hope it will help you!!
function indices = saddle(M)
M;
%[row col] = size(M);
indices=[];
ind_row_col = [];
for i=1:row
for j=1:col
if ( M(i,j) >= M(i,:) & M(i,j) <= M(:,j) )
ind_row_col = [ind_row_col M(i,j)];
indices = [indices ; i,j];
end
end
end
  4 Comments
Image Analyst
Image Analyst on 11 Jul 2021
If you say
x = [x M(a, b)]
then x must exist in advance otherwise it won't know what to concatenate M onto. Even though x is an empty array, that's enough for it to exist and allow stuff to be stitched onto it.

Sign in to comment.


charu sharma
charu sharma on 27 Aug 2015
You should use two for loops to check for each element of a row and a column. Refer this for much simpler code: http://farzicoders.blogspot.in/2015/08/write-function-called-saddle-that-finds.html

Jos (10584)
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
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

Muhammad Qaisar Ali
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
Shiladittya Debnath on 27 Jul 2020
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
Shiladittya Debnath on 27 Jul 2020
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

Abdul Quadir Khan
Abdul Quadir Khan on 6 Nov 2020
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

Mohamed El Nageeb
Mohamed El Nageeb on 17 Dec 2020
this is my answer to this problem....it works fine but i feel like I complicated it. any tips for improvement?
function indices = saddle(M)
[r, c] = size(M);
indices=[];
for ii = 1 : r
for jj = 1 : c
req=0;
for k = 1 : c
if M(ii,jj) >= M(ii,k)
req = req +1;
end
end
for d = 1 : r
if M(ii,jj) <= M(d,jj)
req = req +1;
end
end
if req == (r+c)
indices = vertcat(indices,[ii,jj]);
end
end
end
  1 Comment
Rik
Rik on 17 Dec 2020
If you're looking for code improvements: have you read the other solutions in this thread?
If not, why do you think others will read your answer and learn from it?

Sign in to comment.


VIGNESH B S
VIGNESH B S on 8 Nov 2021
function indices = saddle(Z)
indices1 = []; %Creating a temporary matrix..
[r c] = size(Z);
for i = 1:r
row_sum = sum(Z(i,:)); %To obtain the sum of row
row_max = max(Z(i,:)); %To obtain max of row
for j = 1:c
col_sum = sum(Z(:,j)); %To obtain cum of column
col_min = min(Z(:,j)); %To obtain minimum of column.
if Z(i,j)>=row_max && Z(i,j)<=col_min %The logic -> matrix element should be greatest in row or more than the sum
%and also should be least of column.
mat = [i j];%a TEMPORARY to form a matrix with row and column of the saddle element.
indices1 = [indices1;mat]; %Now we just add it to the empty matrix.
end
end
end
indices = indices1;
end

Salim Maharjan
Salim Maharjan on 3 Feb 2022
Edited: Salim Maharjan on 3 Feb 2022
This code worked for me.
Function:
function indices=saddle(M)
[row,col]=size(M);
indices=[]; % Initializing the saddle points to an empty matrix
for ii=1:row
for jj=1:col
% Check if the element is greater than or equal to every element in its row
% and return its sum
A=sum(M(ii,jj)>=M(ii,:));
% Check if the element is less than or equal to every element in its column
% and return its sum
B=sum(M(ii,jj)<=M(:,jj));
%Provided than an element is saddle point, the following condition must hold
if isequal(A,col) && isequal(B,row)
indices=[indices,[ii,jj]]; %Adding the row index and column index of saddle point in matrix indices
end
end
end
end

Products

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by