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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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Muhammad Usman Saleem el 14 de Jun. de 2015
Locked: Stephen23 el 14 de Jul. de 2024
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:
[ 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 :
ans =
1 1
the correct output must be
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 comentariosMostrar 6 comentarios más antiguosOcultar 6 comentarios más antiguos
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
%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
KRISH BHANDARI el 18 de Mayo de 2021
@Ahmet Burhan Baglar you are using the wrong logical operator . use single& not doubles .

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:
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:
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 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
Akshay Padti el 7 de Nov. de 2021
@Stephen It gives wrong result for row vectors.
>> saddle([1 2 3 4 4 3 2 1])
ans =
1 1 4 5
>> saddle([1 2 3 4 4 3 2 1])
ans =
1 4
1 5
Corrected code :
[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 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:
ans = 2×2
1 4 1 5
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

### Más respuestas (21)

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 :
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 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
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?

Tejas Sabu el 13 de Jun. de 2020
Editada: Tejas Sabu el 13 de Jun. de 2020
[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 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
THIERNO AMADOU MOUCTAR BALDE el 30 de Dic. de 2020
working
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??

Konstantinos Sofos el 14 de Jun. de 2015
Editada: Konstantinos Sofos el 14 de Jun. de 2015
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.
1. Take a piece of white paper and write down a matrix
2. Try to write down an algorithm
In my opinion this the way to learn programming. Good luck!
Regards,
##### 4 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
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.
Jobin Geevarghese Thampi el 18 de Feb. de 2021
what is saddle point?what should be the answer after we execute the code?

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 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
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)?????

Jaimin Motavar el 3 de Jul. de 2019
[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 comentariosMostrar NingunoOcultar Ninguno
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 el 9 de Ag. de 2021
Prod is the keyword for product. So, here it is going to multiply the elements in that matrix.

Divya Ratna el 24 de Mayo de 2020
i think anyone should try their own first rather than looking for answers in the community.
this passes all the test cases...
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 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Garvit Kukreja el 29 de Mayo de 2020
can you help me with this.
Thankyou
[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

Muhammad Qaisar Ali el 26 de Jun. de 2020
another approch
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
##### 2 comentariosMostrar NingunoOcultar Ninguno
Avinav Ayushman el 8 de Ag. de 2020
Thanks man, it works and i got the main point also
Muhammad Qaisar Ali el 10 de Ag. de 2020
I glad that it helped someone.

SIVA SAI AKULA el 29 de Jul. de 2020
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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Hicham Satti el 7 de Sept. de 2020
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 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
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 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.

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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Jos (10584) el 3 de Abr. de 2019
[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)].'
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

MADDINENI REVANTH SAI el 31 de Ag. de 2019
[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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Shiladittya Debnath el 27 de Jul. de 2020
For Function :
[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 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

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)';
% 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
view(-115,14);
hold off
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Abdul Quadir Khan el 6 de Nov. de 2020
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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

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?
[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 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Rik el 17 de Dic. de 2020
If you're looking for code improvements: have you read the other solutions in this thread?

VIGNESH B S el 8 de Nov. de 2021
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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Salim Maharjan el 3 de Feb. de 2022
Editada: Salim Maharjan el 3 de Feb. de 2022
This code worked for me.
Function:
[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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Yifan He el 31 de Jul. de 2022
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
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Aramis el 21 de Jun. de 2024
The best solution
% 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
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
DGM el 22 de Jun. de 2024

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