Equivalence classes for cubic rotation symmetric functions

Abstract

Rotation symmetric Boolean functions, that is, Boolean functions which are invariant under any cyclic permutation of the variables, have been extensively studied in the last dozen years or so due to their importance in cryptography and coding theory. Little was known about the basic question of when two such functions are affine equivalent until very recently. The case of affine equivalence of quadratic rotation symmetric functions was solved in 2009 by Kim, Park and Hahn. In 2010, Cusick introduced the concept of patterns to study the more complicated case of cubic functions under permutations. This method made possible a detailed analysis of some special classes of cubic rotation symmetric functions in n variables where n has a special form, for example, n equal to a prime power. Here the affine equivalence classes under permutations in n variables for any n are examined. In particular, all possible sizes of these classes are identified and an exact count of the functions in the largest classes is given. It is conjectured that the classes of the largest size contain a positive proportion of all of the distinct functions in n variables.

Appendix: Examples

1.1 n = 15

First we are going to do an example demonstrating the use of the ξ function defined in Definition 3.2. Look at the case n = 15 where the number of distinct primes is 2. By definition ξ(15) = ξ(3)·ξ(5) = 1·3 = 3. Too see that this is true, let us make the matrix in the proof of Lemma 3.10 with m ₁ = 3 and m ₂ = 5:

$$ \begin{array}{ccccc} 1 & 4 &7 & 10 &13\\[3pt] 2 & 5 & 8 & 11 & 14\\[3pt] 3 & 6 & 9 & 12 & 15\\ \end{array} $$

First, we cross off the rows that are not relatively prime to 3, and then from the remaining rows, cross off the values that are not relatively prime to 5. The result is shown below:

$$ \begin{array}{ccccc} 1 & 4 &7 & {{10}{\diagup}} &13\\[3pt] 2 & {{5}{\diagup}} & 8 & 11 & 14\\[3pt] {{3}{}} &{{6}{}} & {{9}{}} & {{12}{}} & {{15}{}}\\ \end{array} $$

To get the result, count the number of “vertical” pairs remaining. These are (1, 2), (7, 8) and (13, 14). So the answer is 3, as we calculated by the formula.

Now that we have seen how the ξ function works, let us begin doing some calculations for n = 15 (see Fig. 9). Since n is odd and $5\not\equiv1\mod 6$ there will be no class of size φ(15)/3 and so we will use Fig. 3 when doing calculations.

Fig. 9

n = 15 example

We can also calculate from Lemma 2.2 that there are 3 classes of size φ(15)/2 = 4; 1 class of size φ(5)/2 = 2; and 1 class of size φ(3)/2 = 1. Therefore, from Theorem 3.1, E ₂(15) = 2 + 3 = 5. Figure 10 is a list of all of the equivalence classes when n = 15. As can be seen in this case, all of the equivalence classes we identified are actually a count for all of the equivalence classes altogether.

Fig. 10

Equivalence classes when n = 15

1.2 n = 28

The second example we look at is n = 28 (see Fig. 11). This time, since n is even and the highest power of 2 dividing n is 2, we will use Fig. 7 to help with the calculations. Notice first that φ(28) = 12 and ξ(28) = 0.

Fig. 11

n = 28 example

We can also calculate from Lemmas 2.2 and 2.4 that there are 3 classes of size φ(28)/2 = 6; 1 class of size φ(14)/2 = 3; 1 class of size φ(7)/2 = 3; 1 class of size φ(7)/3 = 2; and 1 class of size φ(4)/2. Therefore, we have identified 13 classes altogether and E ₂(28) = 6 + 3 = 9. Figure 12 is a list of all equivalence classes when n = 28. As can be seen from the figure, there are more equivalence classes than those we found. In particular, the classes of size φ(14), φ(7) and φ(4) have not been counted.

Fig. 12

Equivalence classes when n = 28