Finding zero of polynomials of bounded treewidth over finite field

Transcription

1 Finding zero of polynomials of bounded treewidth over finite field Alexander Glikson ID: August 14, 2002 Abstract This paper is submitted as part of a project report for course Advanced Topics in Computer Science, 5 given by prof. J.A. Makowsky at the Technion, Spring This work is based on an article by J.A. Makowsky and K. Meer, Polynomials of bounded tree-width [MM01]. One of the treated cases (finite field) is presented in a slightly simplified and extended form. 1

2 1 Introduction We ve seen during the course that in BSS computational model over the reals the decision problem whether a polynomial of degree 4 has a zero (4-FEAS) is NP-Complete. We are going to show now that in some particular cases this problem could be solved in polynomial (or even linear) time. In this work we are looking at the mentioned decision problem over some finite field, while the polynomials have bounded treewidth - a well-known sparsity condition from graph theory. We are going to prove that in this case there is a linear decision algorithm, although with large constant depending on the field size, the degree and the treewidth. 2 Computational Model The computational model is the one introduced during the course - the Blum-Shub-Smale model (shortly BSS, see also [BCSS98]). It consists of a register machine with registers containing real numbers, which is able to perform basic operations on the registers content, such as arithmetical computations, conditional branches and registers copying. The input is also stored in registers. In this context we are interested in unit cost model, i.e. each register can store any real number, and the complexity is affected only by the number of registers/operations used (every real operation has unit cost). 3 Feasibility of polynomials 3.1 The F EAS problem Definition 1 (FEAS decision problem) Let F be a field, let p( x) be a polynomial in F[ x] in the variables x = (x 1,...,x n ). Problem: Does p have a zero in F n? As mentioned before, here we are interested in the case when the field F is finite, and the polynomials are sparse. 3.2 Reducing F EAS to 4-F EAS Given a polynomial of an arbitrary degree p( x), we can build a new polynomial p ( x,ȳ) with additional variables ȳ = (y 1,...,y m ), such that the 2

3 equation p( x) = 0 has a root if and only if the equation p ( x,ȳ) = 0 has a root. The idea is to replace iteratively every subexpression α β of degree grater than 2 by a new variable y i such that a monomial αβγ will be replaced by y i γ of lower degree, and to demand that in addition an equation α β y i = 0 must hold. In this way we can build a conjunction of equations of degree 2, having a root simultaneously if and only if the original equation had a root. Then, we can build the required polynomial p ( x,ȳ) as follows: p ( x,ȳ) = p 2 0( x,ȳ) + p 2 1( x,ȳ) p 2 m( x,ȳ) = 0 (1) where p 0 is the original polynomial with substituted variables, and p 1...p m are of the form α β y i = 0 (α and β are single variables from {x i, y j }). It is easy to see that p ( x,ȳ) has a zero if and only if p( x) has a zero. The degree of p ( x,ȳ) is at most 4. Note: In similar manner we can deal with system of equations, or even with boolean combination of inequalities (semi-algebraic set), using in addition to the method mentioned above, the following facts: p( x) 0 has a solution iff p( x) y 2 = 0 has a root p( x) > 0 has a solution iff p( x) y 2 1 = 0 has a root p( x) 0 has a solution iff p( x) + y 2 = 0 has a root p( x) < 0 has a solution iff p( x) y = 0 has a root p( x) = 0 has a root or q(ȳ) = 0 has a root iff p( x) q(ȳ) = 0 has a root Example 1 Let Σ be the following system: { x 4 1 x 3 + 2x 2 x 3 x 4 x 2 2 < 0 3x 1 x 2 2 x 1x 3 = 0 (2) We are interested in finding a polynomial p Σ (x 1, x 2, x 3, x 4, ȳ) which has a zero if and only if Σ has a solution. First, we eliminate the inequality: x 4 1x 3 + 2x 2 x 3 x 4 x 2 2 < 0 iff for some y 1 : (x 4 1x 3 + 2x 2 x 3 x 4 x 2 2) y = 0 or: x 4 1x 3 y x 2 x 3 x 4 y 2 1 x 2 2y = 0 3

4 Then, we could reduce the degree of the second equation in (2): 3x 1 y 2 x 1 x 3 = 0 where y 2 = x 2 2 We can summarize the manipulations until now in the following system of equations: x 4 1 x 3y x 2x 3 x 4 y 2 1 x2 2 y = 0 3x 1 y 2 x 1 x 3 = 0 x 2 2 y 2 = 0 Now we can continue the process, and get the following system of equations of degree 2: y 7 y 4 + 2y 4 y 5 y 2 y = 0 3x 1 y 2 x 1 x 3 = 0 x 2 2 y 2 = 0 y 2 1 y 3 = 0 x 3 y 3 y 4 = 0 x 2 x 4 y 5 = 0 x 2 1 y 6 = 0 y 2 6 y 7 = 0 After applying (1), we obtain the following polynomial: p Σ ( x,ȳ) = (y 7 y 4 + 2y 4 y 5 y 2 y 3 + 1) 2 + (3x 1 y 2 x 1 x 3 ) 2 + (x 2 2 y 2 ) (y 2 1 y 3 ) 2 + (x 3 y 3 y 4 ) 2 + (x 2 x 4 y 5 ) 2 + (x 2 1 y 6 ) 2 + (y 2 6 y 7 ) 2 = 0 By construction, it is of degree 4, and it has a zero if and only if the original system has a solution. 4 Polynomials of bounded treewidth 4.1 Polynomial as hypergraph Definition 2 A d-hypergraph G = V, E is a relational structure containing a set of vertices V and a set of hyperedges E V d connecting between 1 and d vertices each. 4

5 Definition 3 We associate with a polynomial p( x) of degree d the d-hypergraph G = V, E such that V = {0, 1,...,n} is the index set of the variables of p( x), x 0 = 1 and E V d is the set of d-tuples of indices (i 1,...,i d ) such that the coefficients c i1,...,i d 0: p( x) = p(x 0, x 1,...,x n ) = (i 1,...,i d ) E c i1,...,i d x i1 x i2... x id A polynomial of degree d with n variables and m monomials will be represented by a d-hypergraph with n + 1 vertices and m hyperedges. The coefficients could be treated as weights of hyperedges, defined by a function C : V d F. Example 2 Given the polynomial p 1 ( x) = x 1 x 3 4 3x2 1 x 5 + 2x 2 x 2 5 x 3x 5, the corresponding 4-hypergraph G 1 = V, E will be: V = {0,1,2,3,4,5}, E = {(1,4,4,4),(0,1,1,5),(0,2,5,5),(0,0,3,5)}. Figure 1: 4-Hypergraph G 1 corresponding to p( x) = x 1 x 3 4 3x2 1 x 5 +2x 2 x 2 5 x 3 x 5 5

6 4.2 Tree decomposition of a hypergraph and treewidth of polynomials Definition 4 A tree decomposition of d-hypergraph G = V, E is a pair (T, X), where T = T, < T is a tree with t < T s iff t is a child of s, and X = {X t t T } is a family of subsets of V, one for each node of T, such that: t TX t = V For each d-hyperedge (i 1,...,i d ) E there is t T such that {i 1,...,i d } X t For each i V the set X(i) = {t T i X t } forms a (connected) subtree of T. The width of a tree decomposition ((T, < T ), {X t t T }) is max t T X t. The treewidth of d-hypergraph G is the minimal width over all tree decompositions of G. Together with definition 3 we can define treewidth of polynomials in a natural way (as the treewidth of the d-hypergraph associated with the given polynomial). Example 3 Considering the 4-hypergraph from example 2, one of the possible tree decompositions of G 1 is shown in figure 3. Therefore, the treewidth of p 1 ( x) is at most 2. It can be shown easily that there is no tree decomposition of G 1 of width 1, since we have monomials with 2 variables. Fortunately, Bodlaender showed that tree decomposition of a hypergraph of treewidth k can be constructed effectively: Theorem 1 (Bodlaender, [Bod97]) There is a linear time algorithm which decides, given a hypergraph G whether it has a tree decomposition of width k, and if yes, constructs one. 4.3 Parse trees of k-boundaried hypergraphs Now we are going to describe an algebra introduced by Abrahamson and Fellows (see [DF99]), that produces terms which are exactly the graphs of tree-width k. 6

7 Definition 5 A hypergraph G is k-boundaried if exactly k of its vertices are labelled by {1,...,k} (i.e. every label appears with one vertex). The labelled vertices are called the boundary δ(g) of G. Now we define operations on such hypergraphs: create: this operation creates a k +1 boundaried hypergraph with no edges (i.e. k + 1 vertices all of which are labelled). join: the join operates on two k + 1 boundaried hypergraphs G 1 = (V 1, E 1 ) with boundary δ(g 1 ) and G 2 = (V 2, E 2 ) with boundary δ(g 2 ). The graph G 1 G 2 is obtained by joining V 1 and V 2 in such a way that those vertices in δ(g 1 ) and δ(g 2 ) having the same label are identified. The vertices in V 1 := V 1 \ δ(g 1 ) and those in V 1 := V 1 \ δ(g 1 ) obtain an own copy. After this identification the hyperedges E 1 and E 2 are unified. change i,j (G): in the k + 1 boundaried hypergraph G labels i and j are interchanged. add i1,i 2,...,i s (G): adds an hyperedge between vertices with labels i 1, i 2,...,i s. Figure 2: Tree decomposition of the 4-Hypergraph G 1 7

8 new i (G): adds a new vertex, labels it i and removes label i from the previously labelled vertex. Let s examine now the hypergraphs that could be obtained using these operations. Definition 6 Let G be a k + 1 boundaried hypergraph. A parse-tree for G is a tree whose vertices are labelled by one of the above operations, such that the following holds: the leafs are labelled by create, the branch nodes are labelled by, the other nodes are labelled by one of the other operations, G is the hypergraph obtained at the root after performing all the operations along the tree bottom-up. The main relation between parse-trees and hypergraphs of bounded treewidth is the following: Theorem 2 (cf. [DF99]) Let G be a hypergraph of treewidth at most k. Starting from a tree decomposition of G we can compute in linear time w.r.t. G a parse-tree of G. Therefore, together with theorem 1, we can construct in linear time a parsetree of a given hypergraph of tree-width k. 4.4 Evaluating MSOL properties on parsable structures, Feferman- Vaught Theorem Assume we want to evaluate (efficiently) some given MSOL-definable property on a given structure. Once we know that it was obtained from some sub-structures using one of the operations introduced in definition 5, we can do it by applying Feferman-Vaught theorem and evaluating the corresponding properties on the sub-structures. We demonstrate this principle on the join operation. Theorem 3 (Feferman-Vaught for join ) Let A and B be k+1 boundaried hypergraphs over τ with universes A and B resp., and boundaries δ(a) A, δ(b) B, Â := A \ δ(a), B := B \ δ(b). Let C := A B be the join of A and B. Let z be an assignment of the variables in a MSOL(τ) formula Φ(x, y, w, X) into the universe C of C such that the following is true: 8

9 - z maps the variables x i of block x to the set  - z maps the variables y i of block y to the set B - z maps the variables w i of block w to the boundary δ(a) = δ(b) (recall that the two boundaries were identified). In particular, z(w i ) is member of the both universes A and B. Denote by z A, z B the assignments with z A (X) = X A, z A (s) = z(s) for s A and z B (X) = X B, z B (s) = z(s) for s B. Then, for every MSOL(τ) formula ψ(x, y, w, X) there are finitely many so called Hintikka formulas h 1,α (x, w, X) and h 2,α (y, w, X) such that for every τ-structure C given as join A B and every assignment z as above we have: (C, z) = ψ(x, y, w, X) α(a, z A ) = h 1,α (x, w, X) (B, z B ) = h 2,α (y, w, X). If ψ itself is a Hintikka formula then there exist uniquely determined Hintikka formulas h 1 and h 2 such that (C, z) = ψ(x, y, w, X) (A, z A ) = h 1 (x, w, X) (B, z B ) = h 2 (y, w, X). In other words, given that C = A B, in order to check whether C satisfies some given MSOL formula, ψ, it is enough to check whether A and B satisfy formulas h 1 and h 2 accordingly. Similar results could be also shown for other operations from definition 5. Therefore, given parse-tree of hypergraph G, we can build Hintikka formulas for leaves and internal nodes of the parse-tree, and their evaluation will give us the unswer whether the whole structure satisfies the original formula. 5 4-F EAS over finite fields for polynomials of bounded treewidth 5.1 Expressing 4-FEAS in logic - CMS Now when we know how to evaluate MSOL properties on graphs of bounded tree-width, we have to express the 4-FEAS problem in some variant of MSOL. Since number of monomials in polynomial is a function of number of variables, we ll need more powerful language than MSOL. We are going to show now that we can express 4-FEAS in MSOL extended with counting abilities - the CMS (Counting Monadic Second Order Logic). In our 9

10 case, the only additional functionality we need is the Card p (X) predicate, which is true if the cardinality of X (modulo F ) is p (see [Cou97] for more details on CMS). The relational structure representing the polynomial will be weighted 4- hypergraph V, E, C (as in definition 3), with V = {0, 1,...,n} representing variables, E V 4 representing monomials, and an hyperedges weight function C : V 4 F over finite field F = {s 1, s 2,...,s m } representing coefficients. The idea is to divide the variable into m subsets U 1,...,U m, defining assignment to variables (i.e., v i U j corresponds to assignment v i := s j ). Next, given assignment to variables, for each possible value s in F we ll calculate how many monomials are equal to s under the given assignment. Then, we are going to examine all the possible combinations of monomial value and the number of monomials having this value, and to check whether between all those combinations some of them correspond to a zero of the polynomial. Formally, we define the following predicates: [ m-part(v, U 1,...,U m ) = U 1,...,U m V i,j {1,...,m} (U i U j = ) i {1,...,m} ] U i = V {(x, num monom eq to(s, k) = Cardk( F y, z, w) {1,...,m} 4 } ) U i1 (x) U i4 (w) U i5 (C(x, y, z, w)) i 1... i 5 = s (i 1,...,i 5 ) {1,...,m} 5 zero-assignment(u 1,...,U m ) = {(t 1,...,t m,k 1,...,k m) {1,...,m} 2m m t i k i =0} i=0 i {1,...,m} and then, the desired feasibility will be expressed as: num monom eq to(t i, k i ) 4-FEAS = m-part(v, U 1,...,U m ) zero-assignment(u 1,...,U m ) Also, the techniques described in previous section (mainly the Feferman- Vaught theorem) are easily extendable to CMS. 10

11 5.2 Finding treewidth of polynomials with reduced degree Question: given polynomial p( x) of treewidth k and degree d, what is the treewidth of p ( x,ȳ), obtained by algorithm described in section 3.2? Idea: since the degree of p is bounded (d), the number of additional variables added by the algorithm is constant (a function of d and k). Given the tree decomposition of p of width k, we could build a new tree decomposition by adding all the additional variables to all the nodes of the tree decomposition. The resulting tree decomposition will be a tree decomposition of p (although, not minimal), and therefore the treewidth of p is still bounded. 5.3 Solving 4-FEAS for finite field and bounded treewidth Now, let s summarize all the steps. Given: Polynomial p( x) of treewidth k over finite field F. Question: Does p( x) has a zero? Step 1 Convert p( x) into p ( x,ȳ) of degree 4 and treewidth k Step 2 Build hypergraph G representing p Step 3 Build tree decomposition of p of width k Step 4 Construct corresponding parse-tree Step 5 Construct CMS formula for 4-FEAS over F Step 6 Construct Hintikka formulas for nodes of parse-tree, top-down starting with the root, according to the corresponding operations Step 7 Evaluate the formulas at the leaves of the parse-tree and conclude about the original formula Since every step could be done in polynomial (or even linear) time, the total time required for the algorithm is also polynomial. 11

12 References [BCSS98] L. Blum, F. Cucker, M. Shub and S. Smale. Complexity and Real Computation. Springer Verlag, [Bod97] H. Bodlaender. Treewidth: Algorithmic techniques and results. In I. Privara, P.Ruzicka, editors, Proc. of the 22nd Symposium Mathematical Foundations of Computer Science, volume 1295 of Lecture Notes in Computer Science, Springer, [Cou97] B. Courcelle. The Expression Of Graph Properties And Graph Transformations In Monadic Second-Order Logic. Handbook of graph grammars and computing by graph transformations, vol.1: Foundations, [DF99] R.G. Downey and M.F. Fellows. Parametrized Complexity. Springer [MM01] J.A. Makowsky and K. Meer, Polynomials of bounded tree width. Available online at admlogic/tr/2001/rev-smalefest.ps 12