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

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Yihan Liu il 4 Ott 2019
Chiuso: Image Analyst il 6 Giu 2020
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
[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
##### 7 CommentiMostra 5 commenti meno recentiNascondi 5 commenti meno recenti
Walter Roberson il 6 Ott 2019
Consider column 7. Suppose you find the minimum value of the column using min. Now find the row indices where the column values are equal to the minimum. There is no point in checking any rows other than those ones, because the other rows cannot possibly have values less than the minimum. Now you can go through just that list of rows and test each one to see whether the row test is satisfied; if it is, then emit that row and column pair.
Yihan Liu il 7 Ott 2019
I rewrite my program by using max&min and this time it works. Thank you so much and I appreciate you guys’ help.

### Risposte (15)

Arafat Roney il 8 Mag 2020
[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 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
Image Analyst il 10 Mag 2020
Yes but it could be made more efficient if the min and max are computed before the loops, and a map of saddle points is preallocated in advance, like in my Answer.
Arafat Roney il 10 Mag 2020
Yep thanks, I've checked your answer. Actually I'm a beginner here. Not used to preallocating, just simply finished solving. Next time I will try to preallocate and solve in more efficient way. Thank you again.

Ductho Le il 4 Apr 2020
If you have no idea to solve this problem, you can use my code as a reference. Good luck!
[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 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Image Analyst il 8 Gen 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
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Amandeep Kaur il 9 Gen 2020
how to rectify this?
Rik il 10 Gen 2020
His code is a minimal working example (and therefore includes data to run). It is your homework, put some effort into how to adapt this to work as a function.

sri harsha juttiga il 5 Apr 2020
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 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
What is this line doing?
indices(f,:)=[ii,jj];
Walter Roberson il 26 Mag 2020
The required output is a list of locations that are saddle points, expressed as two columns, first row and then column numbers of where the saddle point was found. The line you are indicating is the one that is recording the row and column numbers into the solution.

Yaksha SJ il 10 Mag 2020
[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
end
end
end
##### 2 CommentiMostra NessunoNascondi Nessuno
Walter Roberson il 10 Mag 2020
Note: you can use min(M, [], 1) to force taking the min along the first dimension, which would return all of M if M only had one row.
Adam Casole-Buchanan il 13 Mag 2020
This was a huge help.

Amandeep Kaur il 8 Gen 2020
[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 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Vivek Jaswal il 18 Apr 2020
[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 CommentiMostra 7 commenti meno recentiNascondi 7 commenti meno recenti
Vivek Jaswal il 18 Apr 2020
Modificato: Vivek Jaswal il 18 Apr 2020
Because I am not able to submit the assignment on this question.
For input matrix Z, the indices are coming out to be correct but there is an error while submission.
Image Analyst il 18 Apr 2020
rowMaxima =
4
colMinima =
1 2 3 4 4 3 2 1
1 1
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 il 25 Apr 2020
[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
##### 2 CommentiMostra NessunoNascondi Nessuno
Walter Roberson il 25 Apr 2020
Under what condition could it be true that m(i,j)<b ?
JIGNESH PATIL il 30 Mag 2020
Great help!

Yasin Peker il 4 Mag 2020
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 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Yasin Peker il 4 Mag 2020
I always get the same error message
Error using ==
Matrix dimensions must agree.
o = (r{v,vv} == l{v,vv});
What I have to do to?
What should I do to improve my code?

Kumar Vivek il 16 Mag 2020
Modificato: Kumar Vivek il 16 Mag 2020
Try this out if you are not getting the problem.
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 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Prithvi Shams il 23 Mag 2020
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 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

David Gonzalez il 24 Mag 2020
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 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Taif Ahmed BIpul il 31 Mag 2020
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 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Nikolay Ampilogov il 2 Giu 2020
Diar Yihan Liu and other members of this topic!
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:
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.
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 CommentiMostra NessunoNascondi Nessuno
Walter Roberson il 2 Giu 2020
gourav's code returns [] if there is not saddle point, and that is an empty array. 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.
MATLAB has an indefinite number of different empty arrays, but typically when people refer to "the empty array" they are referring to [] which is 0 x 0.
Note: nothing in the assignment expects an error message for invalid parameters.
Nikolay Ampilogov il 3 Giu 2020
Modificato: Nikolay Ampilogov il 3 Giu 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 il 6 Giu 2020
Modificato: Vipin Parthan il 6 Giu 2020
[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.

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