106.

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Page 193

third co-ordinate, z, the image can then be viewed as a three-dimension intensity surface, as illustrated in Figure 5.3b.

Keller and Chen (1989) propose a modified version of a procedure described by Voss (1986). Let a cube with side length L, where L is an odd, positive integer centre on an arbitrary point on the image intensity surface as shown in Figure 5.3b, and let P(m, L) be the probability that m filled cells are contained by the box. For any permissible value of L, one obtains:

(5.6)

In this equation, n(L)=L³−((L−1)/2) is the maximum number of non-empty cells that a box of size L can contain. For instance, n(3)=3³−1 =26, n(5)=123. A box contains at least (L+1)/2 non-empty cells. The idea of empty and non-empty cells is explained by reference to Figure 5.4a, which shows a box of size L=3 centred on the point x=2, y=2, z=16 in Figure 5.3a. Consider the intensity surface corresponding to Figure 5.3a as represented by Figure 5.3b. The centre cell of the box is located at x= 2, y=2, z=16 and covers cells x±1, y±1. The z limits of the box are 16±1. Taking the matrix fragment shown on the left of Figure 5.4a as a guide, and working from row 1 to row 3, the value at x=1, y=3 is z=30. The intensity surface at this point, shown in Figure 5.3b, passes through the box cells z=15, z=16, and z=17. All three values of 30 on row 1 can be visualised in this way. At x=1, y=2 the value of z is 78, and the intensity surface shown in Figure 5.3b also passes through cells z=15 to z=17 in the box. The same is true for values on row 3, column 1 to 3. However, at row 2 column 3 the height of the intensity surface is 15, so that surface overlaps the box at z=15 but does not reach the box cells at z=16 or z=17. At row 3, column 3 the intensity surface value z=12 does not reach the bottom of the box, so all three box cells are empty. The centre of the matrix fragment is x=2, y=2, z=16, which is the centre cell of the box. The cells at x=2, y=2, z=15, and x=2, y =2, z=16 are filled, but the cell at x=2, y=2, z=17 is empty. The 3 ×3 box therefore contains m=21 filled cells and six empty cells, as shown on the right hand side of Figure 5.4a.

Figure 5.4b shows the situation when the box is moved one cell to the right, to the point x=3, y=2, z=15. The reader should follows the logic of the previous paragraph to show that m=20 cells of the 3×3×3 box are filled.

The probability term P(m, L) in Equation (5.6) can be estimated using a histogram technique. To illustrate the methods, imagine a box with side length L=3 and an image of size 12×12. The number m of non-empty

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