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