For n = 1, we have n(4,3) = R1(3) + ${f}_{2}^{\ast }$ (1) = 1 + 3 = 4 f1(S1,X1) = R1(X1) + ${f}_{2}^{\ast }$ (S1 ? X1) n(4,4) = R1(4) + ${f}_{2}^{\ast }$ (0) = 3 + 0 = 3 f1(0,0) = R1(0) + ${f}_{2}^{\ast }$ (0) = 0 n(5,0) = R1(0) + ${f}_{2}^{\ast }$ (5) = 9 n(1,0) = R1(0) + ${f}_{2}^{\ast }$ (1) = 3 n(5,1) = R1(1) + ${f}_{2}^{\ast }$ (4) = 3 + 9 = 12 n(1,1) = R1(1) + ${f}_{2}^{\ast }$ (0) = 3 + 0 = 3 n(5, 2) = R1(2) + ${f}_{2}^{\ast }$ (3) = 3 + 9 = 12 n(2,0) = R1(1) + ${f}_{2}^{\ast }$ (2) = 6 n(5, 3) = R1(3) + ${f}_{2}^{\ast }$ (2) = 1 + 6 = 7 n(2,1) = R1(1) + ${f}_{2}^{\ast }$ (1) = 3 + 3 = 6 n(5, 4) = R1(4) + ${f}_{2}^{\ast }$ (1) = 3 + 3 = 6 n (2, 2) = R1(2) + ${f}_{2}^{\ast }$ (0) = 3 + 0 = 3 n(5, 5) = R1(5) + ${f}_{2}^{\ast }$ (0) = 3 + 0 = 3 n (3, 0) = R1(0) + ${f}_{2}^{\ast }$ (3) = 9 n(6,0) = R1(0) + ${f}_{2}^{\ast }$ (6) = 9 n(3,1) = R1(1) + ${f}_{2}^{\ast }$ (2) = 3 + 6 =9 n(6,1) = R1(1) + ${f}_{2}^{\ast }$ (5) = 3 + 9 = 12 n(3,2) = R1(2) + ${f}_{2}^{\ast }$ (1) = 3 + 3 = 9 n(6, 2) = R1(2) + ${f}_{2}^{\ast }$ (4) = 3 + 9 = 12 n(3,3) = R1(3) + ${f}_{2}^{\ast }$ (0) = 1 n(6,3) = R1(3) + ${f}_{2}^{\ast }$ (3) = 1 + 9 = 10 n(4,0) = R1(0) + ${f}_{2}^{\ast }$ (4) = 9 n (6, 4) = R1(4) + ${f}_{2}^{\ast }$ (2) = 3 + 9 = 9 n(4,1) = R1(1) + ${f}_{2}^{\ast }$ (3) = 3 + 6 = 9 n(6,5) = R1(5) + ${f}_{2}^{\ast }$ (1) = 3 + 3 = 6 n(4,2) = R1(2) + ${f}_{2}^{\ast }$ (2) = 3 + 6 = 9 n(6,6) = R1(6) + ${f}_{2}^{\ast }$ (0) = 1 + 0 = 1