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Chapter 6 ■ Thinning 233

higher resolutions until no change is seen; then the growth process continues
at the original resolution in the original way (minimal force path).
This certainly approximates the set of skeletal pixels S for a digital band.
For example, assume an infinitely long, straight band along the x axis, having
width 2w. Then the boundaries of the band are the lines y = w and y =−w.
Then the force acting on the point (x,y)would be:
∞ ∞

L 1 L 2
F(x, y) = 3 dl x + 3 dl x (EQ 6.7)
−∞ |L 1 | −∞ |L 2 |
where L 1 = (x − l, y − w), L 2 = (x − l, w + y), and l is the length along the
boundary. This becomes:
4y

F(x, y) = 0, (EQ 6.8)
(w + y)(y − w)
Now, any of the dot products referred to previously can be written as:
16y(y + dy)

d i = (EQ 6.9)
(w + y)(y − w)(w + y + dy)(y + dy − w)
All that is needed is to know in what circumstances this expression is negative.
Since −w + dy < y < w − dy it is known that y − w and y + dy − w are negative
and that w + y and w + y + dy are positive, the sign of the dot product is the
sign of y(y + dy). Solving this quadratic reveals that it is negative only between
0and −dy.Thus,
)
1 if − dy < y < 0
C(x, y, dx, dy) = (EQ 6.10)
0 otherwise
As dy approaches 0 this becomes:

)
1 y = 0
C(x, y) = (EQ 6.11)
0 otherwise
which means that the x axis is the skeleton, as was suspected. This demonstra-
tion holds for infinitely long straight lines in any orientation and having any
width.
The application of this method to real figures is based on three assumptions:
What is true for infinitely long lines is approximately true for shorter (and
curved) ones;
A figure can be considered to be a collection of concatenated digital band
segments.

Intersections can be represented by multiple bands, one for each crossing
line.