In the one point-system, we suppose a optimum triangular area and this area should be contain maximum points in the selected point-system. Extarly, this triangular area's of boundary conditions should be depend only three-node in point-system.
Also, this sub-program running of similarly cyclic-permutation technique than not more-speedly. This program's cyclic-permutation run-time is depend matlab main function as -nchoosek-. Plainly, If you selected more 100 point than solution time possible be few minute. This program's low-order-level of run-time not depent is my program's base-algorithm.
for example:try run-time.
tic;nchoosek(500,3);toc <----|
Syntax:
Syntax.input
random=nodes = [x(i) y(i)]
Syntax.outputs
random_nodes_in = maximum node computed triangular area
random_nodes_out = other node
Example:
random_node = rand(20,2); <---|
random_nodes=
node value x(i) y(i)
3.0000 0.6104 0.2033
4.0000 0.9886 0.8193
9.0000 0.6655 0.2923
12.0000 0.4604 0.8841
13.0000 0.8000 0.6156
17.0000 0.7132 0.8267
18.0000 0.7204 0.6114
19.0000 0.7333 0.8473
20.0000 0.6223 0.1141
random_nodes_out=
node value x(i) y(i)
1.0000 0.5503 0.9898
2.0000 0.7772 0.1524
5.0000 0.0483 0.0584
6.0000 0.9854 0.5385
7.0000 0.2047 0.1902
8.0000 0.9125 0.5995
10.0000 0.4623 0.0913
11.0000 0.0483 0.5068
14.0000 0.2894 0.0464
15.0000 0.6951 0.9519
16.0000 0.2593 0.1690
Run-times:
tic;[A,B] = maxnodetrn(rand(5,2)),toc Elapsed time : 0.018329 s.
tic;[A,B] = maxnodetrn(rand(10,2)),toc Elapsed time : 0.062679 s.
tic;[A,B] = maxnodetrn(rand(20,2)),toc Elapsed time : 0.115247 s.
tic;[A,B] = maxnodetrn(rand(30,2)),toc Elapsed time : 0.193706 s.
tic;[A,B] = maxnodetrn(rand(40,2)),toc Elapsed time : 0.447564 s.
tic;[A,B] = maxnodetrn(rand(50,2)),toc Elapsed time : 0.935133 s.
tic;[A,B] = maxnodetrn(rand(100,2)),toc Elapsed time : 10.459298 s.
tic;[A,B] = maxnodetrn(rand(200,2)),toc Elapsed time : 140.301166 s.
Notice:
This sub-function is running the cyclic-permutation algorithm. If you selected more 100 point than solution time be possible a few minute. |
