The Complexity of Solving Equations over Finite Groups A collecion - PowerPoint PPT Presentation

The Complexity of Solving Equations over Finite Groups A collecion of results by Goldmann and Russell from 1999 Philipp Nuspl in the seminar Universal Algebra und Computational Complexity Johannes Kepler University Linz March 26, 2019

Introduction

Problem What is a polynomial over G ? Each polynomial p over G is of the form 1 Defjnition (Horváth and Szabó 2006) We will assume that ( G , · ) is a fjnite group. Given polynomials p 1 , . . . , p r , q 1 , . . . , q r over G we want to decide if there is an x = ( x 1 , . . . , x n ) ∈ G n such that p i ( x ) = q i ( x ) , for all i = 1 , . . . , r . We write P OL S YS S AT ( G ) for short. If r = 1 we write P OL S AT ( G ) . p = w 1 · w 2 · · · w s where w j ∈ G ∪ { x 1 , . . . , x n } ∪ { x − 1 1 , . . . , x − 1 n } . Hence we can assume that q i ( x ) = 1, i.e. our system is given as p i ( x ) = 1 , for all i = 1 , . . . , r . We ask: For which groups G is P OL S YS S AT ( G ) ∈ P and for which P OL S AT ( G ) ∈ P ?

Examples 1 1 3 2 5 3 1 5 1 3 Some examples of polynomial equations include: 1 3 2 2 2 1 2 3 3 2 • ( Z 8 , +) : 2 + 3 x 1 + 5 x 2 + 7 x 3 = 0 , • ( D 4 , · ) with a 4 = b 2 = 1 (so | D 4 | = 8): a · a · x 1 · x 1 · b · x − 1 · b · a = x − 1 · b , • ( S 3 , ◦ ) : ( ) ( ) ( ) ◦ x − 1 ◦ x ◦ = , • ( S 5 , ◦ ) : ( ) ( ) ( ) ( ) x 1 ◦ ◦ x 2 ◦ ◦ = .

Goal of today Goldmann and Russell proved two important theorems: Theorem 1 (Goldmann and Russell 1999, Thm. 1+2) Theorem 2 (Goldmann and Russell 1999, Thm. 10 + Cor. 12) 3 If G is an abelian group, then P OL S YS S AT ( G ) ∈ P and P OL S YS S AT ( G ) ∈ NPC otherwise. If G is a nilpotent group, then P OL S AT ( G ) ∈ P and if G is not solvable then P OL S AT ( G ) ∈ NPC .

System of Equations

Solving systems over abelian groups As a fjrst step we will show: Theorem 1 (part 1, (Goldmann and Russell 1999, Thm. 1)) 4 If G is an abelian group, then P OL S YS S AT ( G ) ∈ P . Proof: Every fjnite abelian group G can be written as G ∼ = Z n 1 ⊕ · · · ⊕ Z n l . Want to solve system p i ( x 1 , . . . , x n ) = 0 for i = 1 , . . . , r with polynomials p i over G . Instead of solving the system over G we can rewrite it as l individual systems over Z n k . Hence we only consider the case Z m . Over Z m we can solve a system using (essentially) Gaussian elimination.

5 i r r x 1 x n 1 r We do not change the satisfjablity of the system if we: 1 i • Interchange rows of A : Reordering equations. i • Interchange columns of A : Reordering variables. • Adding multiple of row to difgerent row. • Adding multiple of column to difgerent column. 1 Solving systems over Z m For a polynomial ˜ p i over Z m we can write: p i ( x 1 , . . . , x n ) = p ( 1 ) x 1 + · · · + p ( n ) x n − p ( 0 ) ˜ . Hence the system ˜ p i ( x 1 , . . . , x n ) = 0 is equivalent to       p ( 1 ) p ( n ) p ( 0 ) . . . . . . .       . . . . ( a ij ) r , n i , j = 1 x := Ax :=  =  =: b =: ( b i ) r i = 1 .       . . . .     p ( 1 ) p ( n ) p ( 0 ) . . .

Algorithm For computing a diagonal form of the matrix using these operations do: 2. Reduce all entries in row i and column j . 3. If all entries in row i and column j (except a ij ) are zero, then swap row i with row 1 and column j with column 1 and proceed with step 1 with the submatrix arising by removing the fjrst row and fjrst column. 4. Otherwise we have created an element which is smaller than a ij . Again proceed with step 1 with the whole matrix. The elements in the matrix get strictly smaller, so the algorithm groups G . 6 1. Find a nonzero minimal entry a ij of A . terminates. It has polynomial complexity O ( rn min( r , n )) . Hence in total P OL S YS S AT ( Z n ) ∈ P , so P OL S YS S AT ( G ) ∈ P for abelian

NP-completeness The more diffjcult part of Theorem 1 will be: Theorem 1 (part 2) How can one show NP -completeness? (Polynomially) reduce a problem which is known to be NP -complete to the problem for which we want to 7 If G is an not abelian, then P OL S YS S AT ( G ) is NP complete. show NP -completeness. Here: Graph-Colorability.

Graph-colorability Theorem (Karp 1972) there is a color for each vertex of G such that two vertices which are connected by an edge do not have the same color is NP -complete. Figure 1: Source: Wikimedia Commons (David Eppstein), https://commons.wikimedia.org/wiki/File:Triangulation_3-coloring.svg 8 Given a graph G and k ≥ 3 difgerent colors. The problem of deciding if

Small groups 7 Table 1: Hungerford 2003 15 D 7 14 13 12 11 D 5 10 9 order 8 9 3 abelian groups 6 non-abelian groups 5 1 4 2 Z 1 Z 2 Z 3 Z 4 , Z 2 × Z 2 Z 5 D 3 ∼ Z 6 = S 3 Z 7 Z 8 , Z 2 × Z 4 , Z 2 × Z 2 × Z 2 D 4 , Q 8 Z 9 , Z 3 × Z 3 Z 10 Z 11 Z 12 , Z 2 × Z 6 D 6 , A 4 , T Z 13 Z 14 Z 15

10 i we use induction on order of the groups. Smallest non-abelian group is S 3 . Lemma (Goldmann and Russell 1999, Thm. 3) 3 Proof: We will show that coloring a graph with 6 colors can be reduced 2 total). With each vertex i in the graph we associate a variable x i . For 1 equation ij j ij P OL S YS S AT ( S 3 ) To prove that P OL S YS S AT ( G ) is NP -complete for non-abelian groups G P OL S YS S AT ( S 3 ) is NP -complete. to P OL S YS S AT ( S 3 ) . Every element in S 3 corresponds to a color (6 colors each edge ( i , j ) in the graph we introduce two variables y ij , z ij and the ( ) y ij x i x − 1 z ij x j x − 1 z − 1 y − 1 = .

11 3 3 1 2 1 1 1 2 3 2 z ij 1 1 3 1 2 3 1 3 2 y ij 2 2 1 solving a system of equations over S 3 . If we can solve the system of j Hence we have reduced the problem of coloring a graph to the problem of 1 1 3 . The equation 2 ij ij y ij 3 2 1 2 3 1 3 2 z ij 3 P OL S YS S AT ( S 3 ) If the coloring is legal, then for every edge ( i , j ) we have ( ) α := x i x − 1 ̸ = ( ) y ij α z ij α − 1 z − 1 y − 1 = has a solution if and only if α is not the identity: α α ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) equations over S 3 we can color the graph. Therefore P OL S YS S AT ( S 3 ) ∈ NPC .

Inducible subgroups before we will prove the general result. Defjnition (Goldmann and Russell 1999, Def. 1) such that Inducible subgroups have the nice property that NP completeness carries over to the larger group. Namely: Lemma (Goldmann and Russell 1999, Lemma 4) Let H be an inducible subgroup of G . 12 Having the base-case S 3 settled we will introduce some more concepts A subset H ⊆ G is called inducible if there is a polynomial p over G H = Im ( p ) = { p ( g 1 , . . . , g n ) : g 1 , . . . , g n ∈ G } . 1. If P OL S YS S AT ( H ) ∈ NPC , then P OL S YS S AT ( G ) ∈ NPC . 2. If H is a normal subgroup of G and P OL S YS S AT ( G / H ) ∈ NPC , then P OL S YS S AT ( G ) ∈ NPC .

Proof complexity of inducible subgroups j the original one can be satisfjed. Then we have a new equation over G which can be satisfjed if and only if i are 13 Proof of P OL S YS S AT ( H ) ∈ NPC = ⇒ P OL S YS S AT ( G ) ∈ NPC : Since H is inducible there exists a polynomial p ( x 1 , . . . , x n ) over G such that H = Im ( p ) . Given an equation w 1 · w 2 · · · w s = 1 over H with w i ∈ H ∪ { y 1 , . . . , y m } ∪ { y − 1 1 , . . . , y − 1 m } we can replace every occurrence of y i with p ( x ( i ) 1 , . . . , x ( i ) n ) where x ( i ) with p ( x ( i ) 1 , . . . , x ( i ) new variables over G and every occurrence of y − 1 n ) − 1 .

Proof complexity of inducible subgroups and 14 Proof of P OL S YS S AT ( G / H ) ∈ NPC = ⇒ P OL S YS S AT ( G ) ∈ NPC : Now an equation over G / H looks like ( w 1 · w 2 · · · w s ) H = w 1 H · w 2 H · · · w s H = H with w i ∈ G ∪ { y 1 , . . . , y m } ∪ { y − 1 1 , . . . , y − 1 m } which we can rewrite as w 1 · w 2 · · · w s = p ( x 1 , . . . , x n ) w 1 · w 2 · · · w s · p ( x 1 , . . . , x n ) − 1 = 1 over G for new variables x 1 , . . . , x n .

Commutators Defjnition 15 For two elements a , b ∈ G we write [ a , b ] := aba − 1 b − 1 and call [ a , b ] a commutator. For two subsets A , B ⊆ G we write [ A , B ] := { [ a , b ] = aba − 1 b − 1 : a ∈ A , b ∈ B } and ( A , B ) = ⟨ [ A , B ] ⟩ for the group generated by the commutators [ a , b ] and call ( A , B ) a commutator subgroup. In particular ( G , G ) is the commutator subgroup of G . In fact ( G , G ) is the smallest subgroup of G such that G / ( G , G ) is abelian. Furthermore ( G , G ) = { 1 } if and only if G is abelian.

Commutator subgroup Lemma (Goldmann and Russell 1999, Lemma 5) Hence we can choose the polynomial 16 ( G , G ) ⊆ G is inducible. Reminder: [ a , b ] = aba − 1 b − 1 and ( G , G ) = ⟨{ [ a , b ] : a , b ∈ G }⟩ . Proof: Every element g ∈ ( G , G ) can be written as g = [ a 1 , b 1 ][ a 2 , b 2 ] · · · [ a m , b m ] . Since G is fjnite and [ a , a ] = 1 we have a fjxed m ∈ N such that ( G , G ) = { [ a 1 , b 1 ][ a 2 , b 2 ] · · · [ a m , b m ] : a i , b i ∈ G } . p ( x 1 , y 1 , . . . , x m , y m ) := [ x 1 , y 1 ][ x 2 , y 2 ] · · · [ x m , y m ] . This p induces ( G , G ) , i.e. p ( G 2 m ) = ( G , G ) .