On the algebraic immunity of multiplexer Boolean functions

Theorem 1

A Boolean function

where k ≥ 1 represents a 2 k : 1 multiplexer Boolean function if and only if it satisfies the recurrence relation:

where

and

Proof

We apply induction on k to show that for k ≥ 1 , f k ( y 0 , y 1 , … , y k − 1 , Z 1 ) represents a 2 k : 1 multiplexer Boolean function. For k = 1 we have f 1 ( y 0 , z 0 , z 1 ) = y 0 z 0 + y 0 z 1 + z 0 , which is clearly a 2 : 1 multiplexer Boolean function. Now we assume that the theorem is true for k ∈ N and let t = 2 k y k + 2 k − 1 y k − 1 + … + y 0 . Now from the definition of multiplexer Boolean function and the value of t ,

and

Using (7) and (8) and from induction hypothesis we have,

There are two choices of y k ∈ F 2 and using them separately on (9), we obtain two following cases:

Case 1: ( y k = 0 )

Case 2: ( y k = 1 )

Finally from both of the above cases, the multiplexer Boolean function is

Thus, induction completes the proof.□

Theorem 2

Let X = ( Y , Z ) , where Y ∈ F 2 k , Z ∈ F 2 2 k , and f k ( X ) = f k ( ( Y , Z ) ) is a multiplexer Boolean function. Then there exists a one-to-one correspondence between annihilators of f k and f k + 1 , such that each annihilator of f k is mapped onto some annihilator of f k + 1 of the same degree.

Proof

Let 1 ∈ F 2 2 k denote the vector ( 1 , 1 , … , 1 ) . Now from the definition of multiplexer,

and

Now (10) and (11) imply that

Now from (12), it is clear that if g k ( Y , Z ) is any annihilator of f k ( Y , Z ) of degree d , then g k ( Y , Z + 1 ) is an annihilator of degree d of 1 + f k ( Y , Z ) . This ensures one-to-one correspondence of f k ( Y , Z ) and 1 + f k ( Y , Z ) .□

Lemma 1

For k ≥ 1 , let f k be a multiplexer Boolean function. Then

Proof

We apply induction on k . For k = 1 we have f 1 ( y 0 , z 0 , z 1 ) = y 0 z 0 + y 0 z 1 + z 0 . To find its annihilator, we see that

which implies that y 0 z 0 + y 0 annihilates f 1 . Therefore, Aℐ ( f 1 ) ≤ 2 . Now we assume the theorem to be true for some k = m , m ∈ N . Then from the hypothesis,

Let g m ( Y , Z ) = g m ( y 0 , y 1 , … , y m − 1 , z 0 , z 1 , … , z 2 m − 1 ) be an annihilator of the least degree, where Y = ( y 0 , y 1 , … , y m − 1 ) and Z = ( z 0 , z 1 , … , z 2 m − 1 ) , then deg g m ≤ m + 1 . We define function g m + 1 as

It is easy to verify that g m + 1 annihilates f m + 1 . As

we conclude that

This completes the proof of induction as well as that of lemma.□

Theorem 3

For k ≥ 1 , let f k be a multiplexer Boolean function. Then,

Proof

We apply induction on k . For k = 1 we have f 1 ( y 0 , z 0 , z 1 ) = y 0 z 0 + y 0 z 1 + z 0 . Let L be the set of all linear and affine functions in variables y 0 , z 0 , and z 1 , therefore, L = { y 0 + z 0 + z 1 , y 0 + z 0 + z 1 + 1 , y 0 , z 0 , z 1 , y 0 + 1 , z 0 + 1 , z 1 + 1 , y 0 + z 0 , y 0 + z 0 + 1 , z 0 + z 1 , z 0 + z 1 + 1 , y 0 + z 1 , y 0 + z 1 + 1 } . It is verified exhaustively that no one from L annihilates f 1 . In other words, there is no one degree annihilator of f 1 , while y 0 z 0 + y 0 annihilates f 1 . Therefore,

Now we assume the theorem to be true for some k = m , m ∈ N . Then from the induction hypothesis

Let g m + 1 ( y 0 , y 1 , … , y m , z 0 , z 1 , … , z 2 m + 1 − 1 ) be a least positive degree annihilator of f m + 1 . Without loss of generality, we may assume that

where g m 1 and g m 2 are two polynomials in y 0 , y 1 , … , y m − 1 , z 0 , z 1 , … , z 2 m + 1 − 1 . We have

Now (6) and (14) imply that

and

From (15) and the induction hypothesis, we have either g m 1 = 0 or deg g m 1 ≥ m + 1 . Similarly from (16) and induction hypothesis, either g m 2 = 0 or deg g m 2 ≥ m + 1 . Observe that both g m 1 and g m 2 cannot be simultaneously zero as g m + 1 is of positive degree. In view of Lemma 1,

Further if any of the g m 1 or g m 2 is zero, other must have degree m + 1 . In this case, we have Aℐ ( f m + 1 ) = m + 2 . Now we have two cases:

Case 1: { deg g m 1 = m + 2 }

Either this is an impossible case (when deg ( g m 1 + g m 2 ) ≥ m + 2 ) or Aℐ ( f m + 1 ) = m + 2 .

Case 2: { deg g m 1 = m + 1 }

In this case, deg g m 2 must be equal to m + 1 . For deg g m 2 = m + 2 , deg g m + 1 = m + 3 . Here we observe two things. First, g m 1 must contain a term involving any of the variables y 0 , y 1 , … , y k − 1 , z 0 , z 1 , … , z 2 m − 1 as a function of z 2 m , z 2 m + 1 , … , z 2 m + 1 − 1 and it cannot annihilate

The second thing we observe is that g m 1 ( y 0 , y 1 , … , y m − 1 , z 0 , z 1 , … , z 2 m + 1 − 1 ) is a family of annihilators of f m ( y 0 , y 1 , … , y m − 1 , z 0 , z 1 , … , z 2 m − 1 ) given by different values of variables z 2 m , z 2 m + 1 , … , z 2 m + 1 − 1 . We claim that in the ANF of g m 1 , there exists a term of degree m + 1 which does not involve any of the variables out of z 2 m , z 2 m + 1 , … , z 2 m + 1 − 1 . This holds true because in such case putting z 2 m = z 2 m + 1 = … = z 2 m + 1 − 1 = 1 will give an annihilator of f m ( y 0 , y 1 , … , y m − 1 , z 0 , z 1 , … , z 2 m − 1 ) of a positive degree less than m + 1 . Similarly, we can prove that in the ANF of g m 2 , there exists a term of degree m + 1 , which does not involve any of the variables out of z 0 , z 1 , … , z 2 m − 1 . Because of this fact, the degree of g m 1 ( y 0 , y 1 , … , y m − 1 , z 0 , z 1 , … , z 2 m + 1 − 1 ) + g m 2 ( y 0 , y 1 , … , y m − 1 , z 0 , z 1 , … , z 2 m + 1 − 1 ) will not be less than m + 1 . Consequently, Aℐ ( f m + 1 ) = m + 2 . Thus, from induction the theorem is proved.□