1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
|
/************************************************************************/
/* */
/* Cluster line segments. */
/* */
/************************************************************************/
# include <appDebugon.h>
# if 0
================================================================
Find the line that minimises the sum of squared distances of the
end points of two segments to itself.
================================================================
We have two line segments:
R: (r_x1,r_y1) -> (r_x2,r_y2)
S: (s_x1,s_y1) -> (s_x2,s_y2)
Distance to a line with normal formula sin( a )*x+ cos( a )*y+ c= 0 of
each of the four end points. Where q is the angle of the line with the
x axis.
d= sin( q )* r_x1+ cos( q )* r_y1+ c. Etc
Squares of the distances are:
( sin( q )* r_x1+ cos( q )* r_y1+ c )^2. Etc
Sum of the squared distances is: ...
T_a_a= r_x1*r_x1+ r_x2*r_x2+ s_x1*s_x1+ s_x2*s_x2;
T_a_b= r_x1*r_y1+ r_x2*r_y2+ s_x1*s_y1+ s_x2*s_y2;
T_a_c= r_x1 + r_x2 + s_x1 + s_x2;
T_b_b= r_y1*r_y1+ r_y2*r_y2+ s_y1*s_y1+ s_y2*s_y2;
T_b_c= r_y1 + r_y2 + s_y1 + s_y2;
T_c_c= 4;
DS= T_a_a* sin( q )* sin( q )+
2* T_a_b* sin( q )* cos( q )+
2* T_a_c* sin( q )* c+
T_b_b* cos( q )* cos( q )+
2* T_b_c* cos( q )* c+
T_c_c* c* c;
Should be minimal. Partial derivatives to q,c respectively.
to q: T_a_a* [ sin( q )* cos( q )+ cos( q )* sin( q ) ]+
2* T_a_b* [ sin( q )* -sin( q )+ cos( q )* cos( q ) ]+
2* T_a_c* [ cos( q )* c ]+
T_b_b* [ cos( q )* -sin( q )+ -sin( q )* cos( q ) ]+
2* T_b_c* [ -sin( q )* c ]+
T_c_c* 0.
= 0
to c: cos( q )- sin( q )+ 2* c.
= 0;
or T_a_a* [ sin( q )* cos( q ) ]+
T_a_b* [ cos^2( q )- sin^2( q ) ]+
T_a_c* [ c* cos( q ) ]+
T_b_b* [ cos( q )* -sin( q ) ]+
T_b_c* [ c* -sin( q ) ]
= 0
===
To avoid gonio use the line a* x+ b* y+ c= 0 with a+ b= 1. In stead of
a*a+ b*b= 1.
d= a* r_x1+ ( 1- a )* r_y1+ c. Etc
d= a* r_x1+ r_y1- a* r_y1+ c. Etc
d= ( r_x1- r_y1 )* a+ r_y1+ c. Etc
Squares:
d*d= ( r_x1- r_y1 )^2* a*a+
2*( r_x1- r_y1 )* r_y1* a+
2*( r_x1- r_y1 )* a* c+
r_y1* r_y1=
2* r_y1* c+
c* c;
Be V_a_a= sum( ( r_x1- r_y1 )^2 )
V_a= sum( ( r_x1- r_y1 )* r_y1 );
V_a_c= sum( ( r_x1- r_y1 ) );
V_c= r_y1;
V_c_c= 4;
Sum of the squared distances is: ...
DS= V_a_a* a*a+
2* V_a* a+
2* V_a_c* a*c+
2* V_c* c+
V_c_c* c* c;
Should be minimal. Partial derivatives to a,c respectively.
to a: 2* V_a_a* a+ 2* V_a+ 2* V_a_c* c,
to c: 2* V_a_c* a+ 2* V_c+ 2* V_c_c* c,
Divide by 2, should be 0.
(a.1) V_a_a* a+ V_a_c* c+ V_a= 0.
(c.1) V_a_c* a+ V_c_c* c+ V_c= 0.
(a.2) V_a_c* a+
( V_a_c^2 )/V_a_a* c+
( V_a* V_a_c )/V_a_a= 0
(c.2) ( V_a_c^2 )/V_c_c* a+
V_a_c* c+
( V_a_c* V_c )/ V_c_c= 0;
(a.2-c.1) [ ( V_a_c^2 )/V_a_a- V_c_c ]* c+
[ ( V_a* V_a_c )/V_a_a- V_c ]= 0.
(c.2-a.1) [ ( V_a_c^2 )/V_c_c- V_a_a ]* a+
[ ( V_a_c* V_c )/V_c_c- V_a ]= 0.
a= [ V_a- ( V_a_c* V_c )/V_c_c ]/ [ ( V_a_c^2 )/V_c_c- V_a_a ].
c= [ V_c- ( V_a* V_a_c )/V_a_a ]/ [ ( V_a_c^2 )/V_a_a- V_c_c ].
# endif
|
