162 
I\ — TJlO + Tj02 
12 = (7720-7702) 2 +4 77^ 
/3 = (?730 — 377l2) 2 + (377l2 — 37703) 2 
IA = (7J30 + Tj\2) 2 + (rjn + TJoi) 2 (5) 
/5 = (7730 - 37712X7730 + 77u)[(7730 + 7712) 2 - 3(7712 + 7703) 2 ] + (7703 - 37721X7703 + 772>)[(77o3 + 7721) 2 - 3(7712 + 7730) 2 ] 
/6 = (7720 — 7702)[(7730 + 7712) 2 — (7721 + 7703) ] + 4771,[(7730 + 7712) (7721 + 7703)] 
/ 7 = (37721 - 7703X7730 + 7712)1X7730 + 7712) 2 - 3(772. + 7703) 2 ] + (37712 - 7730X7703 + 7721)1X7703 + 7721) 2 - 3(7712 + 7730) 2 ] 
d(A,B) 
7 
I 
* = 1 
X (IT-ITJ 
(6) 
4. GLOBAL RELAXATION MATCHING 
The basic thought of the strategy is: m, n control point in image 
A and B respectively, that control point series are 
{A 1 ,A 2 ,---, A m } and ‘-,B n } .We had known k pairs of 
the homologous (k>=6, and k«m, n).If matching points pair 
(Ah ,Br) is corresponding image points, then random Ai point 
in the image A might has Bj point in the image B satisfying 
conditional distance | A j , A h ] = \b j , B r J with Bj and Aj are 
homologous. In other word, all homologous in image A, B can 
corresponding support the registering point. 
Given a 777 X 77 matrix P — [P- \ , Py is the matching 
probability between ( A;, B j ) . 
1) When / = 0 , Py is the initial probability of P. We choose 
the similarity measurement for image invariant moments of (A is 
Bj) in the matching window as the initial value 
p°=p(A,,B,). 
2) When / >0, define the support probability of ( A h , B r ) for (Aj, 
Bj) as: 
S'{i, j;h,r)= p l hr min 
d» 
djr 
dih 
(7) 
Where p hr is the probability for (A h , B r )being homologous 
when the 1 times iteration is proceeded,. d ih , dj r are the distances 
of(Ai, A h )and( B J5 B r ), S‘ (/, j; h, r) E [0,1] . 
3) When iteration number is 1+1, the matching probability 
between Aj and Bj is 
./+1 
Pf=P\jSy 
s\j= 2X 
h=l 
h*i 
maxS 1 (i,j;h,r) 
l<r<n 
r*j 
(8) 
Where Sy is the support of other pair of homologous for (Aj, 
Bj), W h is weight function of point h for point i, it is a distance 
function of the two points, the shorter distance is, the bigger 
weight is. 
4) If p[j > T , stop iteration and determine a pair of 
homologous, which is taken as the support for the matching of 
other undetermined registering point. T is given threshold and 
its value usually approach 1. 
After iteration, up to one element of every column or every row 
in matrix P whose value is near to 1, which indicates that the 
corresponding points are homologous. If the value of the 
column i in matrix P are near to 0, which indicates that there is 
no homologous in A, and Bj. 
5. ELIMINATION OF MISMATCHING POINTS 
After getting pairs of homologous used the global relaxation 
matching algorithm, there would be some mismatching points. 
Then we used the quadratic polynomial model [8] to eliminate 
the error points. 
Given a point (X, Y) in reference image A, its coordinate in 
undetermined registering image B is (X, y) , the relationship 
of (X, Y) and (x,y) is 
\x — Clo + Cl\X + Q2Y + ChXY + CIaX + CI5 Y 
^ (9) 
[y = b» + b>X + b,Y + biXY + b.X 2 + bsY 2 
Since image A and B have more than 6 accurate homologous 
before registration, we can compute the coefficients of the 
quadratic polynomial using these pair of homologous by 
adjustment^] method. 
For points (Ax, Ay) in image A, we can compute the 
corresponding coordinate (x,, y<) in image B according to 
equation (9). The coordinate of the homologous in image B 
is (Bx, By) . The difference between (Xt,y t ) and 
(Bx , By) are Ax, = Bx-Xr , Ay, = By - y, . The mean 
square error could compute as below: 
(10) 
where N is the count of homologous, 
l N j N 
If = —^ Ax,, LI y = —^ Ay, and threshold Tx = 2(7* . 
Nji Nti