Intersecting and non-intersecting box regions

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Elysi Cochin
Elysi Cochin am 10 Jul. 2021
Kommentiert: Matt J am 12 Jul. 2021
Having a set of bounding box values [x y width height] , how can i find the number of bounding box that gets intersected and that do not gets intersected when plotted
From the above example, there are 5 intersecting boxes and 2 non-intersecting boxes
How can i do so with the attached bounding box values

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Matt J
Matt J am 10 Jul. 2021
Bearbeitet: Matt J am 10 Jul. 2021
Using rectint(), you can straightforwardly obtain a binary mask A such that A(i,j)=1 if rectangle bbx(i,:) and bbx(j,:) intersect.
load bbx
A=rectint(bbx,bbx)>0
A = 20×20 logical array
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
and therefore the number of rectangles that have an intersection another rectangle would be,
N=sum(tril(A,-1),'all')
  6 Kommentare
Elysi Cochin
Elysi Cochin am 12 Jul. 2021
Bearbeitet: Elysi Cochin am 12 Jul. 2021
load bbx
A = rectint(bbx,bbx)>0
Sir there are total 20 boxes
and non-intersecting box = 13, but there are more zeros, in the matrix A
So Sir, if I use A(i,j), how will i get 13?
Matt J
Matt J am 12 Jul. 2021
>> nnz(sum(A,1)==1)
ans =
13

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Weitere Antworten (2)

Simon Chan
Simon Chan am 10 Jul. 2021
If viusal inspection is allowed, then the number can be counted by plotting the boxes in a figure:
rawdata=load('bbx.mat');
figure
for k=1:length(rawdata.bbx)
rectangle('Position',rawdata.bbx(k,:))
end
  4 Kommentare
Elysi Cochin
Elysi Cochin am 10 Jul. 2021
when i increase the number of rows in bbx, i get error
Matrix dimensions must agree.
Error in demo
B = B+A;
Simon Chan
Simon Chan am 10 Jul. 2021
Would you please run the following commands before the above script:
clear;
clc;
If the error happens again, would you please run the following:
size(A) % Check the dimension of Matrix A
size(B) % Check the dimension of Matrix B

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KSSV
KSSV am 10 Jul. 2021
Run a loop for each box and find the intersection points. Use this to get the intersection points.
If your output is empty, it means there is no intersection.
  3 Kommentare
KSSV
KSSV am 10 Jul. 2021
You have origin, length and width of bounding box. So you can get four vertices of the box. You need to use four coordinates for each bounding box to get the intersection points.
Elysi Cochin
Elysi Cochin am 10 Jul. 2021
i didnt get that Sir. Please can you show me an example?

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