is that the dilation of an object f subtracts the object itself. 
The other method is that the object substracts ils erosion. 
dg(f)-(f6B)-f 
Or 
dg(f)-f-(fOB) 
where B is the structuring element. 
For the grey level f, there is the summability. Supposc 
11, f(m,n)>=e, 
fg(m,n)= | 
10, f(m,n)<g 
where grey level g70,1,2,...,N (usually N=225 ). Thus, 
N 
f(m,n)- 2 fg(m,n) = max{g! fg(m,n)=1} 
=1 
and, : 
N N 
eg(f) -eg( X fg(m.n)- X eg(fg(m,n)) 
g=1 g=1 
It is interest that the region can be obtained by the inverse 
procedure. That is the region filling can be completed 
with mathematical morphology. If X is edges, and P is a 
point of a region R. Repeat 
Sn-(p*nB)flXc 
until the result is the same as previous one. 
32.2 Thinning. The edge exíracted by above 
processing is nol one pixel thinkness, even though it is 
Skeletized progressively. How can the connected edge 
with one pixel thinkness be captured? the feasible way is 
that the pixels in out layer are removed gradually on the 
condition of connectedness, until no pixel can be 
removed. 
The thinning operator is difined by Hitmiss operation as 
XOT=X/X@T where X is the edge, and T is the structure 
element. For structure element sequence D={D1, D2, D3, 
D4 }, the m thinning is 
{XOD}m=(..((((XOD1)OD2)OD3)OD4)...) 
for m times 
Algorithm 1: 
(1). X'={XOD}y where D={D1,D2,D3,D4}, 
00- -00 -00 Li 
Dis011 D2-110 D3=110 D4=011 
s EON .00 00- 
(2). X"={X'OE}y. where E={E1,E2,E3,E4} 
-0- i. af 4 41. 
Fl=111 E2=110 E3-11 E4=01 
A. =} + (0 4 
(3). X-X" and repeat. 
If the orders of structure elements are different, the results 
are different also. The improved algorithm is 
Algorithm 2: 
where 
X={(XODi) (XODi+1) (XOEi)}m 
where 
i=1,2,3,4 and i=i(mod4) when i>4. 
The result with the constant length of the branch includes 
some noisy branch which should be cut off. The 
corrected method is 
Algorithm 3: 
(1) Before (or after) cach iteration 
X=X0Gi, 1=1,2,...,8 
where 
MS 001 00. 000 
Gi-010 G2=010 363-011 G4=010 
000 000 00. 001 
000 000 .00 1 00 
65-010 G6=010 G7=110 G8=010 
2j. 100 .00 0 00 
8 
(2) X1-(XODJOE, X2-(X0G)2. X3-U(XOGi) 
i=1 
X=X2U{(X3®2M) fX1} 
The result can obviously be improved with the possibility 
of one pixel less in length. 
3.2.3 Node detection 
(1) End-point set 
8 
end(x)= U (X@Gi) 
i=1 
(2) 3-intersection set 
4 4 4 
cross3(X)- U (XGTi) U U (x8Fi) U U (X8Bi) 
i=l i=l íi 
where 
.01 101 101 10. 
T1=010 T2=010 T3=010 T4=010 
101 10. 401. 101 
1,1 01 +. 10. 
F1-010 F2-11. F3-010 F4-.11 
d. .01 1.1 10. 
10. .01 A. d. 
Bi-011 B2=110 B3=110 B4=011 
I 1. .01 19. 
(3) 4-intersection set 
cross4(X)-(X8M1)(X8M2) 
101 010 
MI-010 M2-111 
101 010 
32.44. . 
each curve 
is carried « 
where 
and the m. 
the straigl 
thresholds 
according 
repeated. ; 
3.3 Regk 
Region d 
objects(re 
satisfies: 
then (X1. 
decompos 
(1) concis 
(2) invari: 
(3) repres 
(4) unique 
331 A 
with symi 
processing 
where ni i 
1) in step i 
-.. is a deci 
as morc px 
3.32 
elements F 
where Sr 
Step 1: Re 
Step 2: In 
according 
least. In th 
333 A