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From: ImageAnalyst on 17 Jun 2010 22:19
On Jun 17, 9:44 pm, "someone" <some...(a)somewhere.net> wrote:
> "Manal " <be....(a)live.com> wrote in message <hvegm6$qn...(a)fred.mathworks.com>...
> > I'll draw a graph from these points (depending on their index)
> > they have to be ordered sequentially starting from the head for example and finish at the same point ( the same head point)
>
> > I don't know how this can be happen :(
>
> If I understand your situation, you might try a procedure something like this:
>
> 1) Start with A = M(x1,y1) % an initial point
> 2) Set M(x1,y1) = []   % remove that point
> 3) Calculate the minimum distance (or distance squared) to every other point using
>       d = min(sqrt((M-x1).2 + (M-y1).2))  % don't really need sqrt
> 4) Use find to get the index of the minimum distance.
>       (You will have to decide what to do in case of ties.)
> 5) Set A = [A;M(xd,yd)]
> 6) Set M(xd,yd) = []
> 7) Repeat steps 3 - 6 until M is empty
>
> A should be a matrix with your desired order.
>
> I'm sure there is lots of optimizing and degugging required with the above procedure, but you should get the idea.

---------------------------------------------------------------
Yeah I thought of something similar except that you need to find the
TWO closest other points since every point will have two lines leaving
it, not just one. And like I said before (and you hinted at) even
that will break down for certain points (like that point in the middle
which will likely have two strange lines coming from somewhere on the
boundary to snag it). I'd just tackle the problem at an earlier stage
to avoid the mess. For example, he may have got the points from a
binary image and could have just gotten the boundary with
bwboundaries() instead of find() or whatever he did.
From: Manal on 18 Jun 2010 06:53
ImageAnalyst <imageanalyst(a)mailinator.com> wrote in message <2f3f77cc-5532-4603-a02e-952f30d66ce4(a)z25g2000vbk.googlegroups.com>...
> I don't have any idea which of the dozens of "boundary" points should
> be attached to the single point in the center, near the centroid. And
> I don't know which point of the "main" boundary should be connected to
> which, and how many, of the points in the small cluster near the
> bottom of the image.
>
> Why don't you take a step back and look at the situation so that you
> don't have to solve that problem. So . . . how did you get the points
> in the first place? Maybe you can get the boundary at that time, say
> with bwboundaries(), and not have to solve the problem that you
> initially posed.

thanks for your replay

the centroid is not a problem since its drawn in another stage
the problem is only in the boundary

I was thinking to solve it from earlier stage as you have suggested but I did not found a solution
this boundary has been extracted from the image using Sobel edge detection
B = edge(im1,'sobel'); %finding edges

after that I extracted the none zero values from the B to be the boundary points
if B(i,j)>0
s(x,1)=i;
s(x,2)=j;
x=x+1;
end

I know that the problem start from here
but I don't have any idea how can I solve it
From: Manal on 18 Jun 2010 06:57
"someone" <someone(a)somewhere.net> wrote in message <hvej13$loi$1(a)fred.mathworks.com>...
> "Manal " <be.san(a)live.com> wrote in message <hvegm6$qn7$1(a)fred.mathworks.com>...
> > I'll draw a graph from these points (depending on their index)
> > they have to be ordered sequentially starting from the head for example and finish at the same point ( the same head point)
> >
> > I don't know how this can be happen :(
>
> If I understand your situation, you might try a procedure something like this:
>
> 1) Start with A = M(x1,y1) % an initial point
> 2) Set M(x1,y1) = [] % remove that point
> 3) Calculate the minimum distance (or distance squared) to every other point using
> d = min(sqrt((M-x1).2 + (M-y1).2)) % don't really need sqrt
> 4) Use find to get the index of the minimum distance.
> (You will have to decide what to do in case of ties.)
> 5) Set A = [A;M(xd,yd)]
> 6) Set M(xd,yd) = []
> 7) Repeat steps 3 - 6 until M is empty
>
> A should be a matrix with your desired order.
>
> I'm sure there is lots of optimizing and degugging required with the above procedure, but you should get the idea.


thanks a lot
I'll try your suggestion
it seems reasonable for me
but do I have to calculate the distance from the current point that I am testing point to each other points??
I'll try your code and comeback with the results

thanks a lot
From: Manal on 18 Jun 2010 09:15
"someone" <someone(a)somewhere.net> wrote in message <hvej13$loi$1(a)fred.mathworks.com>...
> "Manal " <be.san(a)live.com> wrote in message <hvegm6$qn7$1(a)fred.mathworks.com>...
> > I'll draw a graph from these points (depending on their index)
> > they have to be ordered sequentially starting from the head for example and finish at the same point ( the same head point)
> >
> > I don't know how this can be happen :(
>
> If I understand your situation, you might try a procedure something like this:
>
> 1) Start with A = M(x1,y1) % an initial point
> 2) Set M(x1,y1) = [] % remove that point
> 3) Calculate the minimum distance (or distance squared) to every other point using
> d = min(sqrt((M-x1).2 + (M-y1).2)) % don't really need sqrt
> 4) Use find to get the index of the minimum distance.
> (You will have to decide what to do in case of ties.)
> 5) Set A = [A;M(xd,yd)]
> 6) Set M(xd,yd) = []
> 7) Repeat steps 3 - 6 until M is empty
>
> A should be a matrix with your desired order.
>
> I'm sure there is lots of optimizing and degugging required with the above procedure, but you should get the idea.



% Reorder the Matrix s
n=x-1;
u=1;
for k=1:n
A(u,1)=s(k,1);
A(u,2)=s(k,2);
s(k,:)=0;
i=1;
dt=0;
for o=1:n
xt=s(o,2);
yt=s(o,1);
dt(i,1) = sqrt((xt-A(u,2))^2 + (yt-A(u,1))^2);
dt(i,2) = A(u,1);
dt(i,3) = A(u,2);
i=i+1;
end
u=u+1;
[r,c,v]=find(min(dt(:,1)));
end

is this true so far??
I feel there is something wrong
how can I find the values that have the minimum distance ??
From: someone on 18 Jun 2010 09:52
"Manal " <be.san(a)live.com> wrote in message <hvfrh9$qpj$1(a)fred.mathworks.com>...
> "someone" <someone(a)somewhere.net> wrote in message <hvej13$loi$1(a)fred.mathworks.com>...
> > "Manal " <be.san(a)live.com> wrote in message <hvegm6$qn7$1(a)fred.mathworks.com>...
> > > I'll draw a graph from these points (depending on their index)
> > > they have to be ordered sequentially starting from the head for example and finish at the same point ( the same head point)
> > >
> > > I don't know how this can be happen :(
> >
> > If I understand your situation, you might try a procedure something like this:
> >
> > 1) Start with A = M(x1,y1) % an initial point
> > 2) Set M(x1,y1) = [] % remove that point
> > 3) Calculate the minimum distance (or distance squared) to every other point using
> > d = min(sqrt((M-x1).2 + (M-y1).2)) % don't really need sqrt
> > 4) Use find to get the index of the minimum distance.
> > (You will have to decide what to do in case of ties.)
> > 5) Set A = [A;M(xd,yd)]
> > 6) Set M(xd,yd) = []
> > 7) Repeat steps 3 - 6 until M is empty
> >
> > A should be a matrix with your desired order.
> >
> > I'm sure there is lots of optimizing and degugging required with the above procedure, but you should get the idea.
>
>
>
> % Reorder the Matrix s
> n=x-1;
> u=1;
> for k=1:n
> A(u,1)=s(k,1);
> A(u,2)=s(k,2);
> s(k,:)=0;
> i=1;
> dt=0;
> for o=1:n
> xt=s(o,2);
> yt=s(o,1);
> dt(i,1) = sqrt((xt-A(u,2))^2 + (yt-A(u,1))^2);
> dt(i,2) = A(u,1);
> dt(i,3) = A(u,2);
> i=i+1;
> end
> u=u+1;
> [r,c,v]=find(min(dt(:,1)));
> end
>
> is this true so far??
> I feel there is something wrong
> how can I find the values that have the minimum distance ??

% To find the index (idx) of values that have the minimum distance:
mind = min(dt(:,1));
idx = find(dt(:,1) - mind)
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