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 个评论
Sanket khullr
2019-4-3
function indices = saddle(M)
X = (max(M,[],2))'; % operation on row arranged in a column i.e maximum of each row arranged in a column and then transpose gives the row vector of the same
Y = min(M,[],1); %operation on column arranged in row i.e. minimun of each column arranged in a row
count=0;
for i=1:size(X,2)
for j=1:size(Y,2)
if X(i)==Y(j)
count = count+1;
indices(count,:) = [i,j];
end
end
end
if count==0
indices = double.empty;
end
Jack Crespo
2019-4-20
I have an issue for returning an empty matrix if there are no saddle points. I dont know where to put the statement in my code. I currently have it under the check if a point is a saddle point, but whenever a point is not, it returns the empty matrix. If i delete the else statement, the code works for matrices with saddle points.
function indices = saddle(M)
row = size(M,1);
col = size(M,2);
if row==1
maximum = max(M);
[first, second]= find(M==maximum);
indices=[first;second]';
else
maximums = max(M,[],2)';
minimums = min(M,[],1);
for i=1
for j=1:col
for x=1
for y=1:col
if maximums(i,j)==minimums(x,y) %check if a point is a saddle point
[first,second]= find(M==maximums(i,j));
indices=[first second]; %returns the indices of the saddle points
else %else statement if a matrix has no saddle points
indices=[];
break
end
end
end
end
end
end
Xenium Adil
2019-4-30
hey budies i am still stuck at sparse2matrix and have almost completed the course help me out solving that......... trying this code
function [matrix]=sparse2matrix(incell)
S=size(incell);
q=S(2)-2;
msize=incell{1};
mdef=incell{2};
matrix=repmat(mdef,msize);
while q>0
matrix(incell{q+2}(1), incell{q+2}(2)) = incell{q+2}(3);
q = q-1;
end
but failing for random cases help me out...
Sri Sai Nomula
2020-4-21
function matrix=sparse2matrix(cellvec)
matrix=cellvec{2}*ones(cellvec{1}(1),cellvec{1}(2));
for i=3:size(cellvec,2)
matrix(cellvec{i}(1),cellvec{i}(2))=cellvec{i}(3);
end
end
SIVA SAI AKULA
2020-7-29
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
2020-9-7
编辑：Hicham Satti
2020-9-7
%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
Ahmet Burhan Baglar
2020-10-12
编辑：Ahmet Burhan Baglar
2020-10-12
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
KRISH BHANDARI
2021-5-18
@Ahmet Burhan Baglar you are using the wrong logical operator . use single& not doubles .
采纳的回答
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 个评论
Ravi kumar
2020-7-2
your code is not working,but thanks for the hint.
@Ravi kumar: if you provide example data for which it does not work then I could fix my answer.
The original question defines "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". Lets test it on some random data:
>> mat = randi(9,4,4)
mat =
4 8 6 7
4 7 6 2
5 7 1 4
7 8 6 6
>> 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]
idx =
2 2
3 2
Can you tell me why this is "not working" ?
Akshay Padti
2021-11-7
编辑：Akshay Padti
2021-11-7
@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
@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])
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
更多回答（21 个）
vaishak p nair
2019-8-26
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 个评论
Tejas Sabu
2020-6-13
编辑：Tejas Sabu
2020-6-13
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 个评论
Farhan Gony
2020-8-15
can you kindly help to understand why we need to use the null matrix, indices in the 10th line. Why can't we just write :
indices= [i,j];
thank you in advance.
Rik
2020-8-20
If you do that, what would happen with the next iteration? indices will get overwritten.
THIERNO AMADOU MOUCTAR BALDE
2020-12-30
working
Mert Yalcinoz
2022-2-18
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
2015-6-14
编辑：Konstantinos Sofos
2015-6-14
2 个投票
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 个评论
Marcos Mariano
2015-6-14
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
Muhammad Usman Saleem
2015-6-15
Revant Shah
2020-4-24
@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
2021-2-18
what is saddle point?what should be the answer after we execute the code?
the cyclist
2015-6-14
1 个投票
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 个评论
Muhammad Usman Saleem
2015-6-14
Jaimin Motavar
2019-7-3
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 个评论
Faria Sultana
2020-4-30
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
2021-8-9
Prod is the keyword for product. So, here it is going to multiply the elements in that matrix.
Divya Ratna
2020-5-24
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 个评论
Garvit Kukreja
2020-5-29
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
Muhammad Qaisar Ali
2020-6-26
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
2 个评论
Avinav Ayushman
2020-8-8
Thanks man, it works and i got the main point also
Muhammad Qaisar Ali
2020-8-10
I glad that it helped someone.
SIVA SAI AKULA
2020-7-29
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
0 个评论
Hicham Satti
2020-9-7
%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 个评论
Rik
2020-9-8
Why did you post this? What does it teach? What makes it different from the other solutions in this thread?
Hicham Satti
2020-9-8
编辑：Hicham Satti
2020-9-8
This work is a explicit code using basic tools like if loops and logical tools but most of the other codes are implicit codes using advenced MATLAB tools.
Jake Armitage
2021-7-10
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
2021-7-11
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
2015-8-27
0 个投票
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 个评论
Jos (10584)
2019-4-3
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)].'
0 个评论
MADDINENI REVANTH SAI
2019-8-31
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
0 个评论
Shiladittya Debnath
2020-7-27
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
0 个评论
Shiladittya Debnath
2020-7-27
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
0 个评论
Abdul Quadir Khan
2020-11-6
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
0 个评论
Mohamed El Nageeb
2020-12-17
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 个评论
Rik
2020-12-17
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?
VIGNESH B S
2021-11-8
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
0 个评论
Salim Maharjan
2022-2-3
编辑：Salim Maharjan
2022-2-3
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
0 个评论
YUWEI LI
2022-7-10
0 个投票
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 个评论
Yifan He
2022-7-31
0 个投票
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
0 个评论
Aramis
2024-6-21
The best solution
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
1 个评论
DGM
2024-6-22
Other than being formatted, this is a verbatim duplicate of
This question is locked.
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