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

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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
  8 comentarios
Ahmet Burhan Baglar
Ahmet Burhan Baglar el 12 de Oct. de 2020
Editada: Ahmet Burhan Baglar el 12 de Oct. de 2020
hello, can I ask why did u use ind_row_col array ?
Actually I used smilar code but I got an error message :( can you pls explain?
I generally got error at if line statement
function indices = saddle(M)
%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
indices = [];
[row, col] = size(M)
for ii = 1:row
for jj = 1:col
if M(ii,jj) >= M(ii,:) && M(ii,jj) <= M(:,jj)
indices = [indices ; ii,jj];
end
end
end

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Respuesta aceptada

Stephen23
Stephen23 el 15 de Jun. de 2015
Editada: Stephen23 el 15 de Jun. de 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
  4 comentarios
Akshay Padti
Akshay Padti el 7 de Nov. de 2021
Editada: Akshay Padti el 7 de Nov. de 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
Stephen23
Stephen23 el 31 de Jul. de 2022
Editada: Stephen23 el 31 de Jul. de 2022
@Akshay Padti: a simpler approach is to just replace the last line with this:
idx = [R(:),C(:)];
For example, using your test data:
saddle([1,2,3,4,4,3,2,1])
ans = 2×2
1 4 1 5
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

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Más respuestas (20)

vaishak p nair
vaishak p nair el 26 de Ag. de 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 comentario
Travis Ha
Travis Ha el 30 de Jul. de 2020
Editada: Travis Ha el 30 de Jul. de 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?

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Tejas Sabu
Tejas Sabu el 13 de Jun. de 2020
Editada: Tejas Sabu el 13 de Jun. de 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 comentarios
Mert Yalcinoz
Mert Yalcinoz el 18 de Feb. de 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??

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Konstantinos Sofos
Konstantinos Sofos el 14 de Jun. de 2015
Editada: Konstantinos Sofos el 14 de Jun. de 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 comentarios
Revant Shah
Revant Shah el 24 de Abr. de 2020
@Marco I tried this man, but somehow the code is not working. Its showing an error which says that && operator should be convertible to scalar logical value.

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the cyclist
the cyclist el 14 de Jun. de 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 comentario
Muhammad Usman Saleem
Muhammad Usman Saleem el 14 de Jun. de 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)?????

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Jaimin Motavar
Jaimin Motavar el 3 de Jul. de 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 comentarios
Faria Sultana
Faria Sultana el 30 de Abr. de 2020
Hello, I'm learning MATLAB from the very beginning nowadays. So, I didn't understand the approach using prod function. Would you please tell me what is going on inside the built-in function 'prod'?
Naga
Naga el 9 de Ag. de 2021
Prod is the keyword for product. So, here it is going to multiply the elements in that matrix.

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Divya Ratna
Divya Ratna el 24 de Mayo de 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 comentario
Garvit Kukreja
Garvit Kukreja el 29 de Mayo de 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

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Muhammad Qaisar Ali
Muhammad Qaisar Ali el 26 de Jun. de 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

SIVA SAI AKULA
SIVA SAI AKULA el 29 de Jul. de 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 el 7 de Sept. de 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 comentarios
Jake Armitage
Jake Armitage el 10 de Jul. de 2021
This is very close to what I was working on. Can you explain the empty array and the use of them in the "if" statement please?
I'm trying to place why the necessary syntax is "x = [x M(a, b)]" and "y = [y ; a,b]".
Image Analyst
Image Analyst el 11 de Jul. de 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.

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charu sharma
charu sharma el 27 de Ag. de 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) el 3 de Abr. de 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 el 31 de Ag. de 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

Shiladittya Debnath
Shiladittya Debnath el 27 de Jul. de 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 el 27 de Jul. de 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 el 6 de Nov. de 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 el 17 de Dic. de 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 comentario
Rik
Rik el 17 de Dic. de 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?

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VIGNESH B S
VIGNESH B S el 8 de Nov. de 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 el 3 de Feb. de 2022
Editada: Salim Maharjan el 3 de Feb. de 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

YUWEI LI
YUWEI LI el 10 de Jul. de 2022
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

Yifan He
Yifan He el 31 de Jul. de 2022
function indices = saddle(M)
m = size(M,1);
n = size(M,2);
indices = [];
for i = 1:m
for j = 1:n
if (sum(M(i,j) >= M(i,1:end)) == n) & (sum(M(i,j) <= M(1:end,j)) == m)
indices = [indices;[i,j]];
end
end
end

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