Symmetry and Computational Complexity of Boolean Functions: Problems and Results By Evangelos Kranakis (Carleton U, Ottawa)

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1 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 1 Symmetry and Computational Complexity of Boolean Functions: Problems and Results By Evangelos Kranakis (Carleton U, Ottawa)

2 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 2 Symmetry and complexity Problem 1 How does the symmetry of a boolean function f : {0, 1} n {0, 1} affect its computational complexity? Must also answer the following questions: Question 1 1. Why is symmetry relevant? 2. How do you measure symmetry? 3. How do you measure complexity? 4. What is the model of computation?

3 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 3 The circuit model Input Gates Gates Gates Output Gate

4 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 4 What does symmetry have to do with it? Theorem 1 For each n there is a fan-in 2 boolean circuit of size O(n) computing the function x = (x 1, x 2,..., x n ) x 1 := n i=1 x i. W.l.o.g. n = 2 m. Use induction on m. Compute the weight of input (x 1,..., x n ) by computing the weights of (x 1,..., x n/2 ) and (x n/2+1,..., x n ) and then add the resulting weights. Symmetric functions depend only on the weight of the input, rather than on the input itself. Hence a circuit for f will consist of a circuit for the function x x 1, which outputs the binary representation of x 1 for any input x, and a circuit with log n inputs for determining whether or not f is 1 on inputs with a given weight.

5 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 5 Let s agree on something: some notation Definition 1 Aut(f) : the automorhism (or invariance) group of f B n is the set of permutations σ S n such that for all x 1, x 2,..., x n {0, 1} f(x 1, x 2,..., x n ) = f(x σ(1), x σ(2),..., x σ(n) ) B n,k : the set of functions f : {0, 1} n {0, 1,..., k 1} B n := B n,2 : is the set of boolean functions S n : the set of permutations σ : {1, 2,..., n} {1, 2,..., n}. (x 1, x 2,..., x n ) σ = (x σ(1), x σ(2),..., x σ(n) ) f σ : {0, 1} n {0, 1} : x f σ (x) := f(x σ ) Aut(f) = {σ S n : x 2 n (f(x σ ) = f(x))} S n is a permutation group! A Thought: Can Aut(f) measure the symmetry of f?

6 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 6 So, we have the following: The more symmetric the easier! Theorem 2 If Aut(f) = S n then f can be computed by a fan-in 2 boolean circuit of size O(n). What would be more natural than the following conjecture? Conjecture 1 The more symmetric a boolean function f is, the easier it should be to compute f Of course, f is more symmetric invariance group Aut(f) is larger Question 2 When is Aut(f) large?

7 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 7 Some permutation groups will be missed! Question 3 Is every permutation group G S n k-represantable (i.e., the invariance group of a function f B n,k )? Theorem 3 For n 3, the alternating group A n is not 2-representable. Assume A n Aut(f), for some f B n. For any x 2 n, there exist 1 i < j n, such that x i = x j. It follows that the alternating group A n, as well as the transposition (i, j) fix f on the input x. Consequently, every permutation in S n must also fix f on x. As this holds for every x 2 n, it follows that Aut(f) = S n. Hence, A n Aut(f), for all f B n.

8 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 8 Up to isomorphism nothing is missed! Theorem 4 ([4]) Every permutation group G S n is isomorphic to the invariance group of a boolean function f B n( log n+1 ). Idea of Proof. View words x 2 n( log n+1 ) as n blocks of length log n + 1 each. n( log n+1 ) From each permutaion σ G extract a word x(σ) 2 such that σ(i) := value of i-th block of x(σ) in binary representation. Essentially, the boolean function f is 1 exactly on strings of the form x(σ). Open Problem 1 Given n, what s the smallest k such that every permutation group G S n is isomorphic to the invariance group of a boolean function f B k?

9 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 9 Graphs and boolean functions Theorem 5 ([5]) The automorphism group of an n node, undirected graph is the invariance group of a boolean function. Idea of Proof. For each two-element set e = {i, j}, consider the n-tuple x e = (x e 1,..., x e n) {0, 1} n such that x e i = xe j = 1 and xe k all k i, j. = 0, for Let the graph G = (V, E) have vertex set V = {1, 2,..., n} and edge set E. Define the boolean function 1 if x = x e, for some e E f(x) = 0 otherwise.

10 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 10 Orbits and more Open Problem 2 Given G S n, what s the smallest k (if any) such that G = Aut(f) for some function f B n,k? The orbit of 1 i n in the group G S n is the set i G = {σ(i) : σ G}. The number of orbits of G is equal to the average number of fixed points of a permutation σ G; i.e., ω n (G) = 1 G σ G The group G is also acting on 2 n. {i : σ(i) = i}, The number Θ(G) := {x G : x 2 n } is called the cycle index of the permutation group G.

11 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 11 Maximality and representability Theorem 6 (Maximality Theorem, [4, 5]) 1. A permutation group G S n is representable if and only if it is a maximal subgroup of S n among those having the same number Θ(G) of orbits in {0, 1} n. In such a case G is representable by a function f B n,k with k ( n n/2 ). 2. All maximal subgroups of S n are 2-representable, the only exceptions being: (a) the alternating group A n, for all n 3, and the conjugates of the following three types of groups: (b) the 1-dimensional, linear, affine group AGL 1 (5) over the field of 5 elements, for n = 5; (c) the group of linear transformations P GL 2 (5) of the projective line over the field of 5 elements, for n = 6; (d) the group of semi-linear transformations P ΓL 2 (8) of the projective line over the field of 8 elements, for n = 9.

12 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 12 Is there a hierarchy? Definition 2 Let R n k denote the class of k-representable permutation groups on n letters. Clearly R n k Rn k+1. Open Problem 3 Does R n k However, the following can be proved. form a proper hierarchy. Theorem 7 ([5]) R n 2 R n 3, i.e., there exist 3-representable groups which are not 2-representable. Theorem 8 ([4]) There is a logspace algorithm, which, when given as input a cyclic group G S n, decides whether the group is 2-representable, in which case it outputs a function f B n such that G = Aut(f). Open Problem 4 Prove similar results for other groups! Some results obtained by [4] and [5]..

13 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 13 How hard it is: some asymptotics Representability obeys a 0 1 law. Theorem 9 ([4]) For any family G n : n 1 of permutations groups such that each G n S n lim n {f B n : Aut(f) = {id n }} 2 2n = lim n Moreover, if lim inf G n > 1 then lim n {f B n : Aut(f) G n } 2 2n = lim n {f B n : Aut(f) G n } 2 2n = 1. {f B n : Aut(f) = G n } 2 2n = 0. This indicates, that such boolean functions are hard to find, if they exist.

14 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 14 Back to complexity A language L {0, 1} can be represented as a sequence L n of boolean functions, where L n is the characteristic function of L {0, 1} n. Conjecture 2 If size of Aut(L n ) is any indication of the complexity of L then it must be related to the index of Aut(L n ), S i.e., the quantity S n : Aut(L n ) := n. Aut(L n ) In fact we can prove Theorem 10 ([4]) For any language L {0, 1}, if S n : Aut(L n ) = n O(1) then L can be computed by a family of fan-in 2 boolean circuits of polylogarithmic depth and polynomial size. Proof is a bit complicated, and uses the O Nan-Scott characterization of permutation subgroups of S n that do not contain the alternating group A n.

15 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 15 Can go further: From orbits to complexity Conjecture 3 ([4]) The size of S n : Aut(L n ) is related to the complexity of L. Can we show that if Θ(Aut(L n )) is small then L is easily computable? We can prove the following: Theorem 11 ([2]) Every language L with transitive automorphism groups Aut(L n ) and polynomially many cycles, i.e., Θ(Aut(L n )) = n O(1), is computable with a constant depth, polynomial size family of circuits with unbounded fan-in threshold gates. Proof is quite complicated to explain here.

16 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 16 Anonymous networks

17 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 17 Computing in anonymous networks The cyclic group C n is generated by the cycle (1, 2,..., n) and the dihedral group D n is generated by the cycle (1, 2,..., n) and the reflection ρ n = 1 2 n n n 1 1 Let R n denote the ring of n processors. Theorem 12 ([1]) Let f be a boolean function in B n. Then. 1. f is computable in the oriented ring R n if and only if Aut(f) C n. 2. f is computable in the unoriented ring R n if and only if Aut(f) D n. This generalizes to Cayley networks [6].

18 Evangelos Kranakis, School of Computer Science, Carleton University, Ottawa 18 References [1] H. Attiya and M. Snir and M. K. Warmuth, Computing on an Anonymous Ring, JACM, 35 (4), , Oct., [2] L. Babai and R. Beals and P. Takacsi-Nagy, Symmetry and Complexity, , STOC92, [3] M. Clausen, Almost all boolean functions have no linear symmetries, IPL, 41 (6), , [4] P. Clote and E. Kranakis, Boolean Functions Invariance Groups and Parallel Complexity, SIAM J. Comp., 20 (3), , [5] A. Kisielewicz, Symmetry Groups of Boolean Functions and Constructions of Permutation Groups, J. Algebra, 199 (2), , [6] E. Kranakis and D. Krizanc, Computing Boolean Functions on Cayley Networks, Proceedings of the 4th IEEE Symposium on Parallel and Distributed Processing, , 1992.