= Piecewise-constant segmentation contrast threshold
{title2=$\beta L<\frac{a(A-a)}A d^2$}
Suppose two constant data intensities differ by $d$, have areas $a$ and $A-a$, and share an internal boundary of length $L$. Keeping the boundary costs $\beta L$ with zero fidelity; merging the two regions costs $a(A-a)d^2/A$ after fitting their common mean. The displayed inequality makes splitting cheaper among these two candidate geometries. This comparison does not prove either geometry is globally optimal among all partitions.
Back to article page
