Solving ill-posed problems with artificial neural networks

Doughnut Method We will calculate the dose deposited at the point P in the middle of the concentric circles. It can be calculated as the difference between the cases : first, the dose deposition by full big circle exposition, and second, the dose deposited when only full small circle is exposed : () () E E E Q f S dS Q f S dS p b s b b b S s s s S b s = − = − ∫ ∫ (A.1.1) where Eb and Es are absorbed exposures in the middle point in case of big circle and small circle is exposed exclusively, Qb and Qs are doses applied to big and small circles patterning, Sb and Ss are fields of the big and small circles respectively. With S r dS rdr Q Q Q i i i i i b s e = = = = π π 2 2 , , (A.1.2) where Qe is the common dose applied to both circles, it is easy to find that () () () E Q rf r dr Q rf r dr Q rf r dr p b b e s s e r r s b = − = ∫ ∫ ∫ 2 2 2 3 π π π (A.1.3)