Different answers for the same algorithm

2020 3 20
1 个回答
更新时间：2020 3 28
3 次查看（30 天）

0 个投票

I am getting different values of the width upon calling the findwidth function or writing its algorithm (exactly the same) in the main script. Is there a reason to justify that and know what is true?
The main script looks like this:
for i=1:10
:
for j=1:13
for p=1:length(c1)
:
P1new=[P11(concavev(i):n,:);P11(1:c1(p),:)];
width(i,j,p)=findwidth(P1new)
end
end
end
%the width function :
function width=findwidth(P11)
[k,av] = convhull(P11);
CP=[P11(k,1),P11(k,2)];
for x=1:(length(CP)-1)
z=1:(length(CP)-1);
for z=1:(length(CP)-1);
D(x,z)=Seg_point_dis(CP(z,:),CP(z+1,:),CP(x,:))
end
end
D1=max(D,[],2)
width=min(D1);
end

16 个评论
显示 14更早的评论 隐藏 14更早的评论

darova
darova 2020-3-20
What is going on in this script? It's hard to understand
farah fadel
farah fadel 2020-3-20
编辑：farah fadel 2020-3-20
I am iterating over some vertices(i), edges(j), and number of intersection(p) to define a polygon, the issue is when I want to find its width (varies across each i,j,p)using the function findwidth, I will have a value of the width different than that if i just copied the findwidth function algorithm and aded it to the script
Aditya Patil
Aditya Patil 2020-3-23
Can you provide the entire script? Most likely, the issue is due to either use of random variables somewhere, or repeated use of variables.
farah fadel
farah fadel 2020-3-23
编辑：farah fadel 2020-3-26
%The purpose of this code is to decompose a polygon into two polygones
P11=[0.45 0.75;2.37 1.49;4.15 0.32;3.63 1.48;5.58 1.78;7.45 3.21;6.12 3.58;4.75 6.15;4.32 5.94;3.75 4.55;2.45 6.44;1.55 5.45;2.51 3.67;0.45 0.75]
n=length(P11)-1;%closed polygon
[ncc,concavev]=CheckConcavity(P11);%check the number of convex polygons(ncc) and their location (concavev)
%in the above given polygon ncc=5,n=13,concavev=[2;4;7;10;13]
for i=1:ncc %iterate over the number of concave vertices
for j=1:n%iterate over all the edges
e(j,:)=P11(j,:)-P11(j+1,:);%calcualting the edges of the polygon
a(j)=e(j,2)/e(j,1);%slope of the edges
b(j)=bin(P11(concavev(i),1),P11(concavev(i),2),a(j)) ;%intercept of the edges
pp = [a(j),b(j)]; %1st line represent an edge
% 2nd line represent the drawn from the concave vertex, and parallel to the edge
for k=setdiff(1:n,[concavev(i),(concavev(i)-1)])
x2 = [P11(k) P11(k+1)];
y2 = [P11(k,2) P11(k+1,2)];
%fit linear polynomial
p2= polyfit(x2,y2,1);
%calculate intersection of the twolines
x_intersect(k) = fzero(@(x) polyval(pp-p2,x),1);
y_intersect(k) = polyval(p2,x_intersect(k));
%these intersections are the possible options to decompose the polygon at
xi(j,k)=x_intersect(k); yi(j,k)=y_intersect(k);
end
[n1,c1] =find(xi(j,:))%finding the nonzero terms, place where intersection took place
for p=1:length(c1)%number of intersection per edge
P1(p,:)=[xi(j,c1(p)),yi(j,c1(p))];
if c1(p)<concavev(i)
P1new=[P11(concavev(i):n,:);P11(1:c1(p),:);P1(p,:);P11(concavev(i),:)];%decompsed polygon
end
%now we need to calculate the width of all possible polygons we could get to choose the one with min width
[k1,av] = convhull(P1new);%calculate the convexhull of the pollygon
CP=[P1new(k1,1),P1new(k1,2)];%define the new convex polygon
for x=1:(size(CP,2)-1)%vertex
for z=1:(size(CP,2)-1) %edge
D(x,z)=Seg_point_dis(CP(z,:),CP(z+1,:),CP(x,:));
%finding all the possible distances between the vertices and the edges combination
end
end
D1=max(D,[],2) ;%The maximum dis btwn vertex and edge spans the polygon
width1=min(D1);% the width will be the maximum Span
widthmethod2(i,j,p)=width1%from script
widthmethod1(i,j,p)=findwidth(P1new)%from function
end
end
end
farah fadel
farah fadel 2020-3-23
编辑：farah fadel 2020-3-23
I tried to check for the varibles issues, but still nothing changed.(I changed the variable x in the function but still same result)
Guillaume
Guillaume 2020-3-23
You haven't provided the code for Seg_point_dis. Possibly the problem comes from there.
In any case, there are many issues with the code of findwidth:
  • No comments whatsoever. Coupled with meaningless variable names, it's impossible to know what it's meant to be doing. Even you in 6 months time won't be able to understand it.
  • You're using length on a 2D matrix. length returns either the number of rows or the number of columns (whichever is greater) Most of the times it will work fine but if your convex hull is only one point your code will error. I know it's unlikely but why leave the chance when using size(CP, 2) is guaranteed to give you the number of rows all the time.
  • You create a vector z=1:(length(CP)-1); and on the very next line throw it away and use z as a for loop iterator.
  • You're building a matrix D in a double loop. At each step of the inner loop, you calculate the maximum of D while you're still building it.
farah fadel
farah fadel 2020-3-23
Thank you for the feedback, I fixed the issues you told me about,but still the same problem.
This is the function Seg_point_dis if it would give any help
function dist=Seg_point_dis(a,b,x)
d_ab = norm(a-b);
d_ax = norm(a-x);
d_bx = norm(b-x);
if dot(a-b,x-b)*dot(b-a,x-a)>=0
A = [a,1;b,1;x,1];
dist = abs(det(A))/d_ab;
else
dist = min(d_ax, d_bx);
end
farah fadel
farah fadel 2020-3-23
编辑：farah fadel 2020-3-23
Based on the results, the answers obtained when writing the algorithm instead of calling the function seems more reasonable
Aditya Patil
Aditya Patil 2020-3-26
Can you provide both the variants of the code so that we can compare the results and find out what the issue is. Also, it would be helpful if you comment, align and rename variables to make the code more readable.
farah fadel
farah fadel 2020-3-26
I edited the scipt in the comment above, if there is anything still unclear please tell me.
Results:
The location of the nonempty var are the same but the values differ.
widthmethod2(1,:,1)=[3.4923 3.4923 3.4923 4.2720 4.0440 5.1376 5.1376 4.9211 3.4305 3.4305 4.5582 4.5582 4.8062]
widthmethod1(1;:,1)=[4.1941 ;5.1838 ; 4.4855 ; 1.4271 ;4.1062 ; 0.4544 ;4.5847 ;3.9177; 3.5832 ; 4.7346 ;4.3275 4.5915 ; 4.3693]
Aditya Patil
Aditya Patil 2020-3-27
The arguments to bin function seem incorrect. Also, can you provide the modified code with changes that you made as per above comments by me and Guillaume?
farah fadel
farah fadel 2020-3-27
the bin functions take the x,y and the slope respectively to output the intercept.
Yes sure, do I send them by email?
Aditya Patil
Aditya Patil 2020-3-27
Can you attach the files here?
Guillaume
Guillaume 2020-3-27
I've only checked the first few lines of your code and stopped at CheckConc.
You're still using length on a 2-column matrix (both in the script and in CheckConc. I recommend that you never use length. For a vector, use numel. For a matrix, use size and explicitly specify which dimension you want to know the size of, rather than assuming that length will return the size of the dimension you are hoping for. So, use size(P1, 1) and size(v, 1)
The reason I stopped at CheckConc is that the function is wrong. It only performs N-1 checks for a N sided polygon so is missing one vertex in the concavity check. I've not tried to understand the math you're doing. There are more efficient algorithm to check that concavity. See stackoverflow and math.stackexhange. If you're guaranteed that your polygon is not self-intersecting that latter one would be best. It can be implemented without any loop in matlab.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Aditya Patil
Aditya Patil 2020-3-27

0 个投票

Inside findwidth function, you have modified the variable D, which is a global variable. This creates the issue.
Instead, pass D as a parameter to the function, as follows ( parameter named E in the code below)
function width=findwidth(P11,ncc, E)
if ncc~=0
[k1,av] = convhull(P11);
CP=[P11(k1,1),P11(k1,2)];
else
CP=P11;
end
for q=1:(length(CP)-1)%:(length(CP)-1)
for z=1:(length(CP)-1)
E(q,z)=Seg_point_dis(CP(z,:),CP(z+1,:),CP(q,:));
end
end
D1=max(E,[],2) ;
width=min(D1);
end

类别

在 帮助中心 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by