For n = 3, we have n(3,0): = R3(0) + ${f}_{4}^{\ast }$ (3) = 0 + 3 = 3 f3(S3,X3) = R3(X3) + ${f}_{4}^{\ast }$ (S3 − X3) n(3,1): = R3(1) + ${f}_{4}^{\ast }$ (2) = 3 + 3 = 6 n(0,0): f3(0,0) = R3(0) + ${f}_{4}^{\ast }$ (0 − 0) n(3,2): = R3(2) + ${f}_{4}^{\ast }$ (1) = 3 + 3 = 6 = R3(0) + ${f}_{4}^{\ast }$ (0) = 0 n (3,3): = R3(3) + ${f}_{4}^{\ast }$ (0) = 2 + 0 = 2 n(1,0): f3(1, 0) = R3(0) + ${f}_{4}^{\ast }$ (1 − 0) n(4,0): = R3(0) + ${f}_{4}^{\ast }$ (4) = 0 + 3 = 3 = R3 (0) + ${f}_{4}^{\ast }$ (1) = 0 + 3 = 3 n(4,1): = R3(1) + ${f}_{4}^{\ast }$ (3) = 3 + 3 = 6 n(1,1): f3(1,1) = R3(1) + ${f}_{4}^{\ast }$ (0) = 3 + 0 = 3 n(4,2): = R3(2) + ${f}_{4}^{\ast }$ (2) = 3 + 3 = 6 n(1,1): f3(1,1) = R3(1) + ${f}_{4}^{\ast }$ (0) = 3 + 0 = 3 n(4,3): = R3(3) + ${f}_{4}^{\ast }$ (1) = 2+3 = 5 n (2,0): = R3(0) + ${f}_{4}^{\ast }$ (2) = 0 + 3 = 3 n(4,4): = R3(4) + ${f}_{4}^{\ast }$ (0) = 3 + 0 = 3 n(2,1): = R3(1) + ${f}_{4}^{\ast }$ (1) = 3 + 3 = 6 n(5,1): = R3(1) + ${f}_{4}^{\ast }$ (4) = 3 + 3 = 6 n(2,2): = R3(2) + ${f}_{4}^{\ast }$ (0) = 3 + 0 = 3 n (5,2): = R3(2) + ${f}_{4}^{\ast }$ (3) = 3 + 3 = 6 n(5,0): = R3(0) + ${f}_{4}^{\ast }$ (5) = 0 + 3 = 3 n(5,3): = R3(3) + ${f}_{4}^{\ast }$ (2) = 2 + 3 = 5 n(6,0): = R3(0) + ${f}_{4}^{\ast }$ (6) = 3 n(5,4): = R3(4) + ${f}_{4}^{\ast }$ (1) = 3 + 3 = 6 n(6,1): = R3(1) + ${f}_{4}^{\ast }$ (5) = 6 n(5,5): = R3(5) + ${f}_{4}^{\ast }$ (0) = 3 + 0 = 3 n(6,2): = R3(2) + ${f}_{4}^{\ast }$ (4) = 6 n(6,3): = R3(3) + ${f}_{4}^{\ast }$ (3) = 5 n(6,4): = R3(4) + ${f}_{4}^{\ast }$ (2) = 6 n(6,5): = R3(5) + ${f}_{4}^{\ast }$ (1) = 6 n (6,6): = R3(6) + ${f}_{4}^{\ast }$ (0) = 3