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Title
Mapping without the sun
Author
Zhang, Jixian

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

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*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