Linear projections of the Vandermonde polynomial

Theoretical Computer Science 795 (2019) 165–182

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Theoretical Computer Science www.elsevier.com/locate/tcs

Linear projections of the Vandermonde polynomial C. Ramya ∗ , B.V. Raghavendra Rao Department of Computer Science and Engineering, IIT Madras, Chennai, India

a r t i c l e

i n f o

a b s t r a c t An n-variate Vandermonde polynomial is the determinant of the n × n matrix where the ith column is the vector (1, xi , x2i , . . . , xni −1 ) T . Vandermonde polynomials play a crucial role in the theory of alternating polynomials and are useful in Lagrangian polynomial interpolation which arise in the theory of error correcting codes. In this work we study structural and computational aspects of linear projections of Vandermonde polynomials. First, we consider the problem of testing if a given polynomial is equivalent to the Vandermonde polynomial. We obtain a deterministic polynomial time algorithm to test if f is linearly equivalent to the Vandermonde polynomial when f is given as product of linear factors. In the case when f is given as a black-box our algorithm runs in randomized polynomial time. Exploring the structure of projections of Vandermonde polynomials further, we describe the group of symmetries of a Vandermonde polynomial and obtain a basis for the associated Lie algebra. Finally, we study arithmetic circuits built over projections of Vandermonde polynomials. We show universality property for some of the models and obtain a lower bounds against sum of projections of Vandermonde determinant. © 2019 Elsevier B.V. All rights reserved.

Article history: Received 11 October 2017 Received in revised form 14 August 2018 Accepted 19 June 2019 Available online 2 July 2019 Communicated by V.Th. Paschos Keywords: Algebraic complexity theory Vandermonde polynomials Polynomial equivalence problem

1. Introduction The n × n symbolic Vandermonde matrix is given by

⎡

1 ⎢ x1 ⎢

⎢ x2 ⎢ 1 ⎢ V = ⎢ .. ⎢ . ⎢ . ⎢ . ⎣ .

xn1−1

1 x2 x22

.. . .. .

xn2−1

1

1

··· ··· .. . .. . ···

··· ··· .. . .. . ···

⎤

1 xn ⎥ ⎥

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

xn2 ⎥

.. . .. .

xnn−1

(1)

where x1 , . . . , xn are variables. The determinant of the symbolic Vandermonde matrix is a homogeneous polynomial of n degree 2 given by VDn (x1 , . . . , xn )  det( V ) = i < j (xi − x j ) and is known as the n-variate Vandermonde polynomial. An alternating polynomial is one that changes sign when any two variables of {x1 , . . . , xn } are swapped. Vandermonde polynomi-

*

Corresponding author. E-mail addresses: [email protected] (C. Ramya), [email protected] (B.V. Raghavendra Rao).

https://doi.org/10.1016/j.tcs.2019.06.010 0304-3975/© 2019 Elsevier B.V. All rights reserved.

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als are central to the theory of alternating polynomials. In fact, any alternating polynomial is divisible by the Vandermonde polynomial [1,2]. Further, Vandermonde matrix and Vandermonde polynomial are useful in Lagrangian interpolation of polynomials which arise often in the theory of error correcting codes. Linear projections are the most natural form of reductions in Algebraic Complexity Theory developed by Valiant [3]. Comparison between classes of polynomials in Valiant’s theory depends on the types of linear projections. (See [4] for a detailed exposition.) Taking a geometric view on linear projections of polynomials, Mulmuley and Shohoni [5] proposed the study of geometry of orbits of polynomials under the action of G L (n, F), i.e., the group of n × n non-singular matrices over F . This led to the development of Geometric Complexity Theory, whose primary objective is to classify families of polynomials based on the geometric and representation theoretic structure of their G L (n, F) orbits and the topological closures of these orbits. In this article, we investigate computational and structural aspects of linear projections of the family VD = (VDn )n≥0 of Vandermonde polynomials over the fields of real and complex numbers. To begin with, we consider the polynomial equivalence problem. Recall that, in the polynomial equivalence problem (POLY-EQUIV) given a pair of polynomials f , g ∈ F[ X ] where X = {x1 , . . . , xn } we ask if f is equivalent to g under a non-singular linear change of variables. That is, Problem : POLY-EQUIV Input : f , g ∈ F[ X ] where X = {x1 , . . . , xn } Output : An n × n matrix A ∈ G L (n, F) such that f ( X ) = g ( A · X ) if one exists. Else output ‘No such equivalence exists’. POLY-EQUIV is one of the fundamental computational problems concerning polynomials and received significant attention in the literature. The Polynomial Equivalence Problem was extensively studied in [6]. It is closely related to several other questions in algebraic complexity theory. For instance, problems such as determinantal complexity of permanent and depth three arithmetic formula lower bounds can be posed as instances of the POLY-PROJ problem, a generalization of the POLY-EQUIV problem where we do not require the matrix A to be invertible. That, is POLY-PROJ is POLY-EQUIV with A ∈ G L (n, F) replaced by A ∈ F n×n , b ∈ F n such that f ( X ) = g ( A X + b). (See section 1.1 in [6] for details.) POLY-EQUIV can be solved in PSPACE over reals [7] as well as over any algebraically closed field [8], and is in NP ∩ co-AM [9] over finite fields. When the coefficients of the input polynomial are integral and we require the matrix A in the definition of POLY-EQUIV to be in G L (n, C), the POLY-EQUIV problem reduces to Hilbert Nullstellensatz (see [10] and references therein) which is known to be in the class AM assuming Generalized Riemann Hypothesis due to [11]. However, it is not known to be decidable over the field of rational numbers [8]. Saxena [8] showed that POLY-EQUIV is at least as hard as the graph isomorphism problem even in the case of degree three forms. Given the lack of progress on the general problem, there have been significant focus on special cases over the recent years. Kayal [12] showed that testing if a polynomial f given as a black-box is equivalent to the elementary symmetric polynomial, or to the power symmetric polynomial can be done in randomized polynomial time. Further, in [6], Kayal obtained randomized polynomial time algorithms for POLY-EQUIV when one of the polynomials is either the determinant or permanent. In this work, we consider the special case of testing equivalence to Vandermonde polynomial:

Problem : VD-EQUIV Input : f ∈ F[x1 , . . . , xn ] Output : Homogeneous linearly independent linear functions L 1 , L 2 , . . . , L n such that f = VD( L 1 , L 2 , . . . , L n ) if they exist, else ‘No such equivalence exists.’. Remark 1. Although Vandermonde polynomial is a special form of determinant, randomized polynomial time algorithm to test equivalence to determinant polynomial due to [6] does not directly give an algorithm for VD-EQUIV. Further, VD-EQUIV is in striking similarity to the well known reconstruction problem for polytopes: Given the edges of a simple polytope P , reconstruct the vertices of P . The latter is a well studied problem in combinatorial geometry and has efficient algorithms [13]. However the reconstruction problems for polytopes is a promise problem and the algorithms for the same are not adaptable to testing equivalence to Vandermonde polynomials. We show that VD-EQUIV can be solved in deterministic polynomial time when f is given as a product of linear factors (Theorem 1). Theorem 1. There is a deterministic polynomial time algorithm for VD-EQUIV when the input polynomial f is given as a product of linear forms. Combining the above theorem with Kaltofen’s factorization algorithm, [14], we get a randomized polynomial time algorithm for VD-EQUIV when the input polynomial f is given as a black-box. Given a randomized polynomial time algorithm it is natural to ask if VD-EQUIV can be derandomized. We show that derandomizing the above algorithm will derandomize polynomial identity testing problem.

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Corollary 1. PIT is polynomial time equivalent to VD-EQUIV when the polynomial f is given as a black-box. For an n-variate polynomial f ∈ F[x1 , . . . , xn ], the group of symmetry G f of f is the set of non-singular matrices that fix the polynomial f . The group of symmetry of a polynomial and the associated Lie algebra have significant importance in geometric complexity theory. Kayal [6] used the structure of Lie algebras of permanent and determinant in his algorithms for special cases of POLYEQUIV. Motivated by this, Grochow [15] studied the problem of testing conjugacy of matrix Lie algebras. More recently Kayal et al. [16], gave an efficient randomized reconstruction algorithm for full rank arithmetic branching programs. Their algorithm involves a complete characterization of the group of symmetries of the iterated matrix multiplication polynomial (denoted by IMMn,d - the product of d n × n symbolic matrices). Further, in [16] the equivalence test for iterated matrix multiplication is obtained by proving a characterization of IMMn,d by its group of symmetries. Thus, obtaining explicit description of the group of symmetry and the associated Lie algebra is an interesting task. In the same vein as in [6], we describe the group of symmetries of the Vandermonde polynomial, give a basis for the Associated Lie algebra. Theorem 2. Let GVD denote the group of symmetries of Vandermonde polynomial and let gVD denote the Lie algebra associated with GVD . Then, group on n elements. (i) GVD = {( I + ( v ⊗ 1)) · P | P ∈ A n , v ∈ F n }, where A n is the alternating n (ii) Let f ∈ F[x1 , . . . , xn ] be a homogeneous polynomial of degree 2 . If G f = GVD then f (x1 , . . . , xn ) = α · VD(x1 , . . . , xn ) for some α ∈ F. (iii) gVD = { v ⊗ 1 | v ∈ F n }. Projections of a family of polynomials can be viewed as a computational model, where resource would be any measure of complexity on the polynomials in the family. For example, determinantal complexity of an n-variate polynomial f (denoted by dc ( f )) is the smallest m such that f can be expressed as a linear projection of the m × m determinant polynomial. Note that proving exponential size lower bound for permn against arithmetic circuits is equivalent to showing that dc ( permn ) is exponential in n. However, the best known lower bound on dc ( permn ) is n2 /2 due to Mignon and Ressayre [17]. In [18], Shpilka studied linear projections of the elementary symmetric polynomials as a computational model, i.e., polynomials that can be written as a symmetric function in a number of linear polynomials. The symmetric complexity of a polynomial f is defined as the smallest  number m such

that f = Symm,d (1 , . . . , m ) for some affine linear forms 1 , . . . , m and d ≥ 1, where Symm,d ( y 1 , . . . , ym ) = S ⊆{1,...,m} i ∈ S y i , the elementary symmetric polynomial of degree d. Shpilka [18] obtained linear lower bounds on the symmetric complexity of the determinant polynomial. In a similar spirit, in this article, we explore linear projections of Vandermonde polynomials as a computational model. Though every polynomial cannot be written as a linear projection of the Vandermonde polynomial, we observe that every polynomial f ∈ F[x1 , . . . , xn ] can be written as sum of linear projections of Vandermonde polynomials. Further, we prove closure properties (or lack of) and a few lower bounds. Organization of the rest of the paper is as follows. In Section 2 we give the preliminary definitions. In Section 3, we describe the algorithm to test equivalence to Vandermonde polynomials. In Section 4, we describe the group of symmetries of the Vandermonde polynomial and obtain a basis for its lie algebra. In Section 5, we discuss closure properties and lower bounds for projections of Vandermonde polynomials as a computational model. 2. Preliminaries Throughout the paper, unless otherwise stated, F ∈ {C, R}. We briefly review different types of projections of polynomials that are useful for the article. For more details, see [4]. Let [n] = {1, 2, . . . , n}. For a set P of polynomials, let var( P ) denote the set of variables that appear in at least one of the polynomials in P . Definition 1 (Projections). Let f , g ∈ F[x1 , x2 , . . . , xn ]. We say that f is projection reducible to g denoted by f ≤ g, if there are linear functions 1 , . . . , n ∈ F[x1 , . . . , xn ] such that f = g (1 , . . . , n ). Further, we say

• f ≤proj g if 1 , . . . , n ∈ F ∪ {x1 , . . . , xn }. • f ≤homo g if 1 , . . . , n are linear forms. • f ≤aff g if 1 , . . . , n are affine linear forms. For instance, let g = x1 x2 + x3 and f = x21 + 3x1 + 1 then f = g (x1 + 1, x1 + 1, x1 ). Thus f ≤aff g. Based on the types of projections, we consider the following classes of polynomials that are projections of the Vandermonde polynomial.

VD = {VD(x1 , x2 , . . . , xn ) | n ≥ 1}; and

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VDproj = {VD(ρ1 , ρ2 , . . . , ρn ) | ρi ∈ ( X ∪ F), ∀i ∈ [n]}; and VDhomo = {VD(1 , 2 , . . . , n ) | i ∈ F[x1 , x2 , . . . , xn ], deg(i ) = 1, i (0) = 0 ∀i ∈ [n]}; VDaff = {VD(1 , 2 , . . . , n ) | i ∈ F[x1 , x2 , . . . , xn ], deg(i ) ≤ 1 ∀i ∈ [n]}; The group of symmetry of a polynomial is one of the fundamental objects associated with a polynomial: Definition 2. Let f ∈ F[x1 , x2 , . . . , xn ]. The group of symmetries of f (denoted by G f ) is defined as:

G f = { A | A ∈ GL(n, F), f ( A · X ) = f ( X )}. i.e., the group of invertible n × n matrices A such that f ( A · X ) = f ( X ). The Lie algebra gf of a polynomial f is the tangent space of G f at the identity matrix and is defined as: Definition 3 ([6]). Let f ∈ F[x1 , x2 , . . . , xn ]. Let matrices A ∈ F n×n such that

 be a formal variable with  2 = 0. Then g f is defined to be the set of all

f ((1n +  A ) · X ) = f ( X ), where 1n is the n × n identity matrix. It can be noted that g f is non-trivial only when G f is a continuous group. For a random polynomial, both G f as well as

g f are trivial [6].

For a polynomial f ∈ F[x1 , . . . , xn ], k ≥ 0, let ∂ =k ( f ) denote the F -linear span of the set of all partial derivatives of f of order k, i.e.,

∂k f ( f )  F -Span i 1 , . . . , ik ∈ [n] . ∂ xi 1 . . . ∂ xi k

∂

=k

3. Testing equivalence to Vandermonde polynomials In this section, we obtain an efficient algorithm for VD-EQUIV . Our algorithm runs in deterministic polynomial time when the input polynomial f is given as a product of its linear factors, and runs in randomized polynomial time when f is given as a black-box. Let f , g ∈ F[x1 , . . . , xn ]. Throughout this section we use f ≡lin g to denote the fact that there exists linearly independent linear forms L 1 , . . . , L n such that f = g ( L 1 , . . . , L n ). In the language of invariant theory, f ≡lin g if and only if g is in the G L (n, F) orbit of f . We begin with the case when f is given by its linear factors. 3.1. Testing equivalence when f is given as product of linear forms When the polynomial is given as product of linear forms, we show VD-EQUIV ∈ P. Algorithm 1 is the main procedure and uses Algorithm 2 and Algorithm 3 as subroutines. Approach Suppose f = 1 · 2 · · ·  p , where 1 , . . . ,  p are linear forms. Checking if f ≡lin VD is equivalent to testing if there are linearly independent linear forms L 1 , . . . , L n such that 1 · · ·  p = i < j ( L i − L j ). A first approach would be to solve the linear equations i = L j i − L ki 1 ≤ i ≤ p treating the coefficients of the variables in L j as unknowns. Clearly, if f ≡lin VD then there is a bijection π : {1, . . . , p } → {(i , j ) | 1 ≤ i < j ≤ n} such that the system of linear equations i = L j i − L ki , where π (i ) = ( j i , ki ) has a solution. However, it is not guaranteed that there will be a solution for an arbitrary bijection π . It is not clear if there is an efficient way of constructing π when f ≡lin VD. An alternate approach arises from the observation that to test if f ≡lin VD, it is enough to get hold of indices i 1 , . . . , in−1 such that

 i 1 = L 1 − L 2 ,  i 2 = L 1 − L 3 , . . . ,  i n −1 = L 1 − L n ,

(2)

for some linear forms L 1 , . . . , L n . Thus it boils down to getting the indices i 1 , . . . , in−1 that satisfies (2). A first guess would be to get a set of n − 1 linearly independent linear forms from 1 , 2 , . . . ,  p , say i 1 , . . . , in−1 . However, these may not satisfy (2), for an arbitrary linearly independent set i 1 , . . . , in−1 . While it is not clear if there is an efficient way of selecting the required indices, looking at a combinatorial structure based on i 1 , . . . , in−1 leads us to a simple iterative algorithm for testing if f ≡lin VD. An overview of the algorithm is given below.

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Outline of the algorithm 1. Suppose f ≡lin VD. Let P m = { L i − L m | i < m} ∪ { L m − L i | m < i } for m ∈ [n]. Then we have {1 , . . . ,  p } ⊆ { ±  | ,  ∈ P m }. Therefore, to detect if f ≡lin VD, it suffices to check if for some m ∈ [n], the differences of the linear forms in the P m contains the set {1 , . . . ,  p }. We do this by visualizing the set P m as labels of vertices of the complete graph G m on n − 1 vertices V m = {1, . . . , n} \ {m}. For i ∈ V m label(i ) = L i − L m for i < m and label(i ) = L m − L i otherwise. For i = j , label(i , j ) = L i − L j . (See Definition 4 for a formal definition of G m .) Now, checking if f ≡lin VD reduces to obtaining the vertex labels of G m for some 1 ≤ m ≤ n. However, since the linear forms L 1 , . . . , L n are not known, we will have to construct the graph G m implicitly for some 1 ≤ m ≤ n. Our crucial observation is that if f ≡lin VD, any set R of n − 1 linearly independent factors of f induces a spanning forest of G m for every 1 ≤ m ≤ n (Observation 2). Further, each connected component in the spanning forest induced by R will have at least one vertex label in R (Lemma 2). Once we have a vertex each from each of the connected components in the spanning forest induced by R in some G m , the vertex labels of G m can be retrieved from a simple traversal of the forest (procedure UNCOVER-VERTICES given in Algorithm 2). The main steps in the algorithm can be summarized as follows: (a) Choose a set R of n − 1 many linearly independent forms from S = {1 , . . . ,  p }. (b) Identify the set of vertex labels in R for some graph G m . This step is done implicitly by a generic procedure DETECT-VERTICES given in Algorithm 3 which can be used to identify set of vertex labels from R that label vertices in some G m . The procedure is independent of the index m, requires a linear form r and returns two sets of linear forms in R that will be candidate labels of vertices in G m and labels of edges incident on r in G m , if r is a label of some vertex in G m for some 1 ≤ m ≤ n. In fact, the procedure DETECT-VERTICES will return the correct value only if the parameter r is a label of a vertex in G m for some 1 ≤ m ≤ n. Thus we need to run the procedure DETECT-VERTICES with every possible r ∈ R as a parameter, and discard the wrong choices by a simple check on the sets returned (Lines 10 and 12 in VD-EQUIV). (c) Once DETECT-VERTICES returns a set of vertices, we can use the procedure UNCOVER-VERTICES which then will traverse the spanning forest induced by R in G m for some 1 ≤ m ≤ n, and obtain the set of all vertex labels in G m and hence L 1 , . . . , L n . Then, all we need to ensure is that the vertex labels uncovered by UNCOVER-VERTICES indeed generate the linear forms in the set S. The correctness of the approach is crucially dependent on the following facts • The outcome of the algorithm is independent of the choice of linear forms in R. • The set of linear forms in R that label vertices in some G m can be detected by a simple test (DETECT-VERTICES). 2. When f ≡lin VD then for no set of linear forms L 1 , . . . , L n the equation {1 , . . . ,  p } = { L i − L j |i < j } and the algorithm will always output ‘No such equivalence exists’. Algorithm 1 gives a detailed algorithm implementing the ideas described above. It uses two subroutines described in Algorithms 3 and 2.

Algorithm 1: VD-EQUIV.

1 2 3 4 5 6 7 8 9 10

Input : S = {1 , 2 . . .  p } such that f = 1 · 2 · · ·  p Output : L 1 , L 2 , · · · , L n such that f = VD( L 1 , L 2 , . . . , L n ) if f ≡lin VD. n Else ‘No such equivalence exists’ if p = 2 for any n < p or dim(span{1 , 2 , . . . ,  p }) = n − 1 then return ‘No such equivalence exists’ end Choose R = {r1 , r2 , . . . , rn−1 } linearly independent linear forms in S. for r ∈ R do W ←∅ ( K 1 , K 2 ) ← Detect-Vertices( R , r ). If Error Continue with next choice of r in line 5 for r ∈ R \ {r } do ( T 1 , T 2 ) ← Detect-Vertices( R , r ). If Error Continue with next choice of r if K 1 = T 1 or K 1 = T 2 then

11 12

N 0 = K 1 and go to line 16 ; else if K 2 = T 1 or K 2 = T 2 then

13 14 15

else

N 0 = K 2 and go to line 16 ; Continue with next choice of r in line 8 ; W ← Uncover-vertices( R , S , N 0 ). If Error then Continue with next choice of r in line 8 if | W | = n − 1 and S = W ∪ ({ −  | ,  ∈ W } ∩ S ) then Let L be any linear form such that W ∪ { L } is linearly independent. Output { L } ∪ { L −  |  ∈ W }

16 17 18 19 20

end

21 end 22 Output ‘No such equivalence exists’.

 here f ≡lin VD(x1 , . . . , xn ) ;

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Algorithm 2: UNCOVER-VERTICES(R , S , N 0 ). : R = {r1 , . . . , rn−1 } linearly independent of linear forms. S = {1 , . . . ,  p } such that f = 1 · · ·  p . N 0 ⊆ F[x1 , . . . , xn ] set of linear forms. Output : A set W of linear forms that are labels of all vertices of the graph G m , provided N 0 is the set of vertex labels of G m contained in the set R, for some 1 ≤ m ≤ n. 1 W = N0. 2 for i = 1 to n − 1 do 3 Compute N i = { −  ,  −  ,  +  |  ∈ N i −1 ,  ∈ R \ W } ∩ S Input

4 5

if N i is linearly dependent set then return Error;

6

else if { −  ,  − ,  +  | ,  ∈ W ∪ N i } ⊆ S then

7 8 9

else

W = W ∪ Ni ; return Error;

10 end 11 return W

Algorithm 3: DETECT-VERTICES(R , h). : R = {r1 , . . . , rn−1 } linearly independent set of linear forms. h∈R Output : ( S 1 , S 2 ) ⊆ F[x1 , . . . , xn ] sets of linear forms with the following property. If h is a label of a vertex in G m , for some 1 ≤ m ≤ n, then, one of S 1 or S 2 is the set of all vertex labels in G m from the set R and the other set consists of set of all edge labels incident on h in G m . D h ← ∅. for  ∈ R \ {h} do Compute D h = D h ∪ { + h,  − h, h − } ∩ S  D h is set of differences if  + h ∈ S then sign( + h) = 1 if h −  ∈ S then sign(h − ) = 0 if  − h ∈ S then sign( − h) = −1 Input

1 2 3 4 5 6

7 end 8 Compute greedily sets A h1 , A h2 , . . . , A th ⊆ D h such that each A hi is maximal with respect to the property that ∀,  ∈ A hi one of  −  ,  −  or

 +  ∈ S.

9 if t > 2 then 10 return Error; 11 if t = 1 then 12 Set A h2 = ∅; 13 for i = 1 to 2 do 14 forms( A hi ) = ∅

if {h} ∪ A hi is a diff-closed set and A hi = ∅ then

15

forms( A hi ) ← {h} ∪ forms( A hi ) ;

16

for g ∈ A hi do  = |sign( g )| · ( g − sign( g ) · h) + (1 − |sign( g )|) · (h − g ).

17 18

forms( A hi ) ← {} ∪ forms( A hi )

19

end

20 21 end

22 return (forms( A h1 ), forms( A h2 ))

We now prove the correctness and complexity of the above algorithm. 3.2. Correctness of Algorithm 1 As a first step, we observe that lines (1)-(3) of Algorithm 1 are correct: Observation 1. If f = 1 · · · · ·  p ≡lin VD then p =

n 2

and dim(span{1 , 2 , . . . ,  p }) = n − 1.

n

Proof. Clearly if f ≡lin VD we have p = 2 . For the second part if f = VD( L 1 , . . . , L n ), for linearly independent linear forms L 1 , . . . , L n then {1 , . . . ,  p } = { L i − L j | i < j }. Therefore dim(span{1 , 2 , . . . ,  p }) = dim(span{ L i − L j | i < j }) = n − 1. 2 We now prove that whenever f ≡lin VD, Algorithm 1 outputs correctly. The proof follows the sequence of observations stated in the outline of the algorithm. Our proof of correctness depends crucially on considering the linear forms 1 , . . . ,  p as labels of vertices and edges in a complete graph. We begin with the definition of the graph.

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Definition 4. Suppose f ≡lin VD, then there exists n linearly independent linear forms L 1 , . . . , L n such that



f = 1 ·  2 · · ·  p =

(L i − L j )

(3)

1≤i < j ≤n

S = {1 , 2 , . . . ,  p } = { L i − L j | i < j }

(4)

For any m ∈ [n], let G m denote the complete undirected graph on vertices { v 1 , v 2 , . . . , v n } \ { v m }. For every vertex v j ∈ V (G m ), j = m, define



label( v j ) =

Lm − L j

if m < j

L j − Lm

if j < m

(5)

Similarly for every edge e = ( v j , v k ) ∈ E (G m ) in the graph G m , let label(e ) be defined as



label(e ) =

L j − Lk

if j < k

Lk − L j

if k < j





(6)



Note that {1 , . . . ,  p } = v ∈ V (G m ) {label( v )} ∪ e∈ V (G m ) {label( v )}. Fix any linearly independent R = {r1 , r2 , . . . , rn−1 } ⊆ S. For every j ∈ [n − 1], the linear form r j in R corresponds to either the label of a vertex or of an edge in the graph G m for any m ∈ [n]. For m ∈ [n], let Q m,1 be the linear forms in R that correspond to the labels of vertices in the graph G m and Q m,2 be the linear forms in R that correspond to labels of edges in G m . More formally, let

Q m,1 

Q m,2 



label( v ) ∩ R = {label(u 1 ), . . . , label(uk )}

v ∈ V (G m )



for u 1 , u 2 , . . . , uk ∈ V (G m )

label(e ) ∩ R = {label(e 1 ), . . . , label(en−k−1 )}

for e 1 , . . . , en−k−1 ∈ E (G m ).

e∈ E (G m )

Let G m [r1 , . . . , rn−1 ] denote the sub-graph {u 1 , . . . , uk } ∪ {e 1 , . . . , en−k−1 } in G m , i.e., consisting of edges with labels in Q m,2 and vertices incident on them and vertices with labels in Q m,1 . We begin with the observation that linearly independent set of labels of G m cannot contain a cycle: Observation 2. For any set R = {r1 , . . . , rn−1 } ⊂ S of linearly independent linear forms and 1 ≤ m ≤ n, the graph G m [r1 , . . . , rn−1 ] is acyclic. For Observation 3 and Lemmas (1)-(5) we assume f ≡lin VD and hence the existence of the graph G m as in Definition 4. The following simple property of edge labels in G m is useful: Lemma 1. Suppose L i = xi , ∀i ∈ [n]. Then, for any set of edges e 1 , . . . et in G m , every non-zero linear form in F -Span{label(e 1 ), . . . , label(et )} is dependent on at least two variables. Proof. Clearly, ∀ j ∈ [n], x j ∈ / F -Span{xi − xi | i < i }. Since L i = xi , every edge label in G m is of the form xi − xi , i < i . Therefore, for any j ∈ [n] we have x j ∈ / F -Span{label(e 1 ), . . . , label(et )} for any subset of edges e 1 , . . . , et . Hence every non-zero linear form in F -Span{label(e 1 ), . . . , label(et )} is dependent on at least two variables. 2 Using the above, we now show that every connected component of G m [r1 , . . . , rn−1 ] has a unique vertex label in R: Lemma 2. Let f ≡lin VD. For linearly independent linear forms R = {r1 , . . . , rn−1 } ⊆ S and for any m ∈ [n], every connected component C in G m [r1 , r2 , . . . , rn−1 ] has a unique vertex w C ∈ V (C )1 such that label( w C ) ∈ R. Proof. First, we argue that every connected component in G m [r1 , . . . , rn−1 ] has at most one vertex whose label is in R. Suppose C is a connected component in G m [r1 , r2 , . . . , rn−1 ] with at least two vertices v α , v β ∈ V (C ) such that label( v α ), label( v β ) ∈ R. Assume without loss of generality that α < β . Consider the path P¯ = ( v α , e c1 , e c2 , . . . , e c| P¯ |−1 , v β ) between v α and v β in the connected component C , where e c1 , . . . , e c| P¯ | −1 are edges. From the definition of G m , there are constants γ1 , . . . γ| P¯ |−1 ∈ {−1, 1} such that

1

V (C ) refers to the set of vertices in the component C of G m .

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γ1 label(ec1 ) + γ2 label(ec2 ) + · · · + γ| P¯ |−1 label(ec| P¯ |−1 ) = label( v α ) − label( v β ).

(7)

Equation (7) is a contradiction since {label( v α ), label(e c1 ), . . . , label(e c| P¯ |−1 ), label( v β )} ⊆ R and hence linearly independent. Now, suppose there exists a connected component C such that for every vertex w ∈ V (C ), label( w ) ∈ / R. Let v be any vertex in C in G m [r1 , . . . , rn−1 ]. We first argue for the case when L i = xi for all 1 ≤ i ≤ n. It is enough to argue that {label( v )} ∪ {r1 , . . . , rn−1 } is a linearly independent set, a contradiction since dim(F -Span( S )) = n − 1. Now, suppose that {label( v )} ∪ {r1 , . . . , rn−1 } is linearly dependent, i.e., label( v ) = δ1 r1 + · · · + δn−1 rn−1 for some δ1 , . . . , δn−1 ∈ F . Since v ∈ V (G m ), label( v ) = ±(xm − x j ) for some j = m. Now,

±(xm − x j ) = δ1 r1 + · · · + δn−1 rn−1 Let V

(8)

label( E (C )). Clearly, the linear forms in V and label( E (C )) are variable disjoint. Further, xm

var( V ) and

=R\ ∈ x j ∈ var(label( E (C ))). Therefore, (8) implies that x j ∈ F -Span(label( E (C ))), a contradiction to Lemma 1. When L i = xi for some 1 ≤ i ≤ n, we note that there is a non-singular matrix A such that A · L i = xi ∀i. Now repeating the above argument for A · L i instead of L i completes the proof. 2 Though, G m [r1 , . . . , rn−1 ] has at least one vertex for each connected component, there is no definite guarantee on the number of connected components. In the following, we show that, there is always at least one 1 ≤ m ≤ n such that G m [r1 , . . . , rn−1 ] at least two connected components: Lemma 3. For any linearly independent set of linear forms R = {r1 , r2 , . . . , rn−1 } ⊆ S, there exists an m ∈ [n] such that 

     label(u ) ∩ R  ≥ 2.  u ∈ V ( G m ) 

Proof. Assume for the sake of  contradiction that there is a choice of linearly independent linear forms R = {r1 , . . . , rn−1 } ⊆ S such that for every m ∈ [n], | u ∈ V (G m ) label(u ) ∩ R | < 2. Then, for any pair of linear forms  = L i − L j ,  = L i − L j in the set R, we have {i , j } ∩ {i , j } = ∅. Suppose not, {i , j } ∩ {i , j } = ∅. Without loss of generality, let i = i . Then  = L i − L j and

 = L i − L j . Observe that by  the definition of graphs G 1 , . . . , G m in Definition 4, linear forms  and  are labels of vertices in graph G i . In this case, | u ∈ V (G i ) label(u ) ∩ R | > 1, a contradiction to our assumption. Let R = {(i , j ) | L i − L j ∈ R }. As { L 1 , . . . , Ln } are linearly independent linear forms, we have | R | = | R |. For any pair of linear forms  = L i − L j ,  = L i − L j in R, we must have {i , j } ∩ {i , j } = ∅. It follows that | R | ≤ n/2 which is a contradiction for n > 2 as | R | = n − 1. Note that n n ≥ 2 is without loss of generality as p = 2 ≥ 1. 2 We need the following  assumption for the remainder of the section:



Let m be such that 

τ ,  ∈ [n − 1].

u ∈ V (G m ) label(u ) ∩



R  ≥ 2 as guaranteed by Lemma 3. Let {rτ , r } ⊆



u ∈ V (G m ) label(u ) ∩

R for some

We need a notation for subset of linear forms in S whose pairwise differences lie in S: Definition 5. A set T ⊆ S of linear forms is diff-closed if for every ,  ∈ T either  −  ∈ S or  +  ∈ S or  −  ∈ S. A diff-closed set T ⊂ D ⊆ S is said to diff-maximal with respect to set D if no subset of T in D is diff-closed. That is, for any d ∈ D \ T , T ∪ {d} is not diff-closed. The following is an easy characterization of diff-closed sets in S: Observation 3. Let T ⊆ S be a linearly independent set. T is diff-closed if and only if there is an 1 ≤ i ≤ n and I ⊆ [n] \ {i } such that T = {±( L i − L j ) | j ∈ I }. Proof. Clearly, if T = {±( L i − L j ) | j ∈ I } for some 1 ≤ i ≤ n and I ⊆ [n] \ {i }, then T is diff-closed. For the converse, suppose T is a diff-closed set and for any 1 ≤ i ≤ n and I ⊆ [n] \{i }, T = {±( L i − L j ) | j ∈ I }. Let H T be the graph with V ( H T ) = {1, . . . , n} and E ( H T ) = {(i , j ) | ±( L i − L j ) ∈ T }. It suffices to show that H T = K 1,α for some α ≤ n which is equivalent to showing that H T is a star graph. We argue that H T is a connected acyclic graph and that H T has diameter at most 2.

• H T is acyclic. If H T contains a cycle C , then the linear forms labeling the edges of C form a linearly dependent set, a contradiction to T being linearly independent.

• H T is a connected graph. If H T has at least two connected components C 1 and C 2 , there exists edges (i , j ) ∈ E (C 1 ) and (k, t ) ∈ E (C 2 ). Then,  = ±( L i − L j ) ∈ T ,  = ±( L i − L j ) ∈ T such that {i , j } ∩ {i , j } = ∅. This implies that none of  +  ∈ S or  −  ∈ S or  −  ∈ S is true, a contradiction to T being diff-closed set. • H T has diameter at most 2. Suppose not, let p be a path in H T of length at least 3. Then there exists vertex disjoint edges (i , j ) and (k, t ) in p. Then, by an argument similar to the one above we get linear forms ,  ∈ T such that none of  +  ∈ S or  −  ∈ S or  −  ∈ S is true, a contradiction to T being diff-closed set.

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173

From (1) and (2), H T = K 1,α for some α ≤ n. K 1,α being a bipartite graph, let {i } be the vertex in the left partition and I be the set of vertices in the right partition with | I | ≤ α . Then, T = {±( L i − L j ) | j ∈ I } as required. 2 The following lemma obtains the structure of diff-closed sets inside D h for h ∈ {rτ , r }. Lemma 4. For h ∈ {rτ , r }, the set D h computed in line (3) of Algorithm 3 can be partitioned into two unique diff-closed sets A h1 and A h2 such that A h1 ⊆



u ∈ V (G m ) label(u ),

A h2 ⊆



e ∈ E (G m ) label(e ).

Then forms( A h1 ) ⊆ Q m,2 , forms( A h2 ) ⊆ Q m,1 .

Proof. Let h ∈ {r , rτ }. Recall that {r , rτ } ∈ V (G m ), h = ±( L m − L t ) for some t ∈ [n], t = m. Then D h = { L i − L j | i < j , {i , j } ∩ {m, t } = ∅}. Let

A h1 = { ∈ D h |  = ±( L m − L q ), q ∈ [n], q = m}; A h2 = { ∈ D h |  = ±( L q − L t ), t ∈ [n], q = t }. Then D h = A h1 ∪ A h2 is a partition. By Observation 3, it follows that A h1 and A h2 are diff-closed sets. Further from computation of sets A h1 and A h2 in line (8) of Algorithm 3, A h1 and A h2 are diff-maximal with respect to set D h . Hence the partition of D h

into A h1 and A h2 is unique. By the construction of forms( A h1 ) and forms( A h2 ) in line (13) − (19) of Algorithm 3, forms( A h1 ) =





e ∈ E (G m ) label(e )



∩ R and

forms( A h2 )

=



u ∈ V (G m ) label(u )

∩ R. This completes the proof. 2

Using Lemmas 1-4 we can now argue that irrespective of the choice of linearly independent linear forms made by Algorithm 1, it successfully recovers the labels of vertices in G m for some m ≥ 1, provided f ≡lin VD. Lemma 5. For any choice of linearly independent linear forms R in line (4) of Algorithm 1, there is a choice for r and r in the for loops at lines (5) and (8) respectively, such that (i) N 0 = {label(u 1 ), label(u 2 ), . . . , label(uk )}; and (ii) Uncover-vertices( R , S , r , N 0 ) returns W = V (G m ) in line (10) of Algorithm 2 where m is such that | Proof. (i) Let r = r and r = rτ in lines (5) and (8) of Algorithm 3. Observe that {rτ , r } ⊆

 u ∈ V (G m )

 u ∈ V (G m ) 2

label(u ) ∩ R | ≥ 2.

label(u ) ∩ R. For ease

of notation we use A i , A τi instead of A ri  , A ri τ for i ∈ [2]. Let K 1 = forms( A 1 ), K 2 = forms( A  ) and T 1 = forms( A 1τ ), T 2 = forms( A 2τ ). By Lemma 4, one of K 1 or K 2 is in Q m,1 and the other is in Q m,2 . Similarly one of T 1 or T 2 is in Q m,1 and the other is in Q m,2 . Consider the case when K 1 ⊆ Q m,1 and T 1 ⊆ Q m,1 . Then by Lemma 4, we have



K 1 = forms( A 1 ) =

label( v ) ∩ R

v ∈ V (G m )



T 1 = forms( A 1τ ) =

label( v ) ∩ R

v ∈ V (G m )

Therefore, K 1 = T 1 in line (10) of Algorithm 1 and

N0 = K1 =



label( v ) ∩ R = Q m,1 = {label(u 1 ), label(u 2 ), . . . , label(uk )},

v ∈ V (G m )

where the last equality follows from definition of Q m,1 . The cases when K 1 ⊆ Q m,1 & T 2 ⊆ Q m,1 , K 2 ⊆ Q m,1 & T 1 ⊆ Q m,1 and K 2 ⊆ Q m,1 & T 2 ⊆ Q m,1 are analogous and checks are made in lines (10) and (12) of Algorithm 1. (ii) Let C be a connected component in G m [r1 , . . . , rn−1 ]. By Claim 2 let w be the unique vertex such that label( w ) ∈ R. We argue by induction on j that if label( w ) ∈ N 0 then for every vertex v ∈ V (C ) with dist ( w , v ) ≤ j, label( v ) ∈ N j . Base case is when j = 0 and follows from Claim 5(i). For the induction step, let u ∈ V (C ) be such that (u , v ) ∈ E (G m [r1 , . . . , rn−1 ]) and dist ( w , u ) ≤ j − 1. By the induction hypothesis, we have label(u ) ∈ N j −1 . Since (u , v ) ∈ E (G m [r1 , . . . , rn−1 ]) we have that label(u , v ) ∈ R and label(u , v ) = r , r as r = rτ and r = r are labels of a vertex in G m [r1 , . . . , rn−1 ] By line 3 of Algorithm 1, label(u , v ) + label(u ) ∈ N j as required. Either u < v or v < u. In any case label( v ) ∈ N j . 2 Finally, we conclude the correctness of Algorithm 1: Theorem 3. Suppose f ≡lin VD then for any choice of R = {r1 , r2 , . . . , rn−1 } from {1 , · · · ,  p } in line (4) Algorithm 1 outputs f is equivalent to VD(x1 , . . . , xn ) in line (19).

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Proof. Suppose f = 1 · 2 · · ·  p and f ≡lin VD. Let L 1 , . . . , L n be linear forms as in the discussion preceding  Lemma 2. Let R   { label ( v )} ∩ R  ≥ 2 as guaranteed v ∈ V (G m )

be the set of linear forms chosen in line (4) of Algorithm 1. Let m be such that  in Lemma 3. Suppose {rτ , r } ⊆



v ∈ V (G m ) {label( v )} ∩ R. Consider the for loops in lines (4) and (8) of Algorithm 1 with {r , r } = {rτ , r }. By Lemma 5 (i), N 0 is the label of vertices and hence W = V (G m ) in line (15) of Algorithm 1 by Lemma 5 (ii). Therefore S = W ∪ { −  | ,  ∈ S } ∩ S as required. 2

Next, we observe that Algorithm 1 is correct whenever it outputs “ f is equivalent to VD”: Lemma 6. If Algorithm 1 outputs f is equivalent to VD(x1 , . . . , xn ) in line (19), then f ≡lin VD. Proof. If Algorithm 1 outputs f is equivalent to VD(x1 , . . . , xn ) in line (19), then there exists a set W of size n − 1 and S = W ∪ ({ −  | ,  ∈ W } ∩ S ) in line (17) of Algorithm 1. Suppose W = { L 1 , . . . , L n −1 }. Then for any linear form L such that W ∪ { L } is linearly independent, f = ±VD( L , L − L 1 , . . . , L − L n −1 ) and hence f ≡lin VD. 2 Time complexity of Algorithm 1 The construction of linearly independent set R can be done in time O (n6 ). Overall, at most n2 calls are made to the procedure DETECT-VERTICES which in turn requires at most O (n2 ) arithmetic and set operations. The procedure UNCOVER-VERTICES can be implemented with at most O (n2 ) algebraic and set operations. Thus the overall time complexity of the algorithm is O (n6 ). 3.3. Testing equivalence when f is given as black-box In this section, we note that combining the non-black box algorithm given in the Section 3.1 with Kaltofen’s polynomial factorization algorithm [14], we get: Theorem 4. There is a randomized polynomial time algorithm for VD-EQUIV when f is given as a black-box. Proof. The result immediately follows from Algorithm 4 and Theorem 1.

Algorithm 4: VD-EQUIV

1 2 3 4 5 6 7 8 9 10

− 2.

Input : f ∈ F[x1 , x2 , . . . , xn ] as a black-box Output : ‘ f is equivalent to VD(x1 , . . . , xn )’ if f ≡lin VD. Else ‘No such equivalence exists’ Run Kaltofen’s factorization Algorithm [14] if f is irreducible then Output ‘No such equivalence exists’ end else Let B 1 , B 2 , . . . , B p be black-boxes to the irreducible factors of f obtained from Kaltofen’s Algorithm. n if p = 2 for any n < p then return Error ; Interpolate black-boxes B 1 , . . . , B p to get linear forms 1 , . . . ,  p and verify if B i ≡ i for i ∈ [ p ] in randomized polynomial time. Run Algorithm 1 with 1 · 2 · · ·  p as input.

11 end

2 Grochow in [15] posed the question of whether PIT is necessary and sufficient to test equivalence to the determinant polynomial. Here, we show that given black-box access to a polynomial f testing equivalence to Vandermonde polynomial is equivalent to PIT. Corollary 2. PIT is polynomial time equivalent to VD-EQUIV in the black-box setting. Proof. Since polynomial factorization is polynomial time equivalent to PIT in the black-box setting [19], by Theorem 1 we have, VD-EQUIV ≤ P PIT. For the converse direction, let f ∈ F[x1 , . . . , xn ] be a polynomial of degree d. Given black-box access to f we construct black-box to a polynomial g such that f ≡ 0 if and only if g ≡lin VD. Consider the polynomial n

+1

n

g = x12 f + VD(x1 , x2 , . . . , xn ). If f ≡ 0 then clearly g = VD(x1 , x2 , . . . , xn ). If f ≡ 0 then deg( g ) > 2 and hence g is not equivalent to VD. Observe that given black-box access to f we can construct in polynomial time a black-box for g. 2

C. Ramya, B.V. Raghavendra Rao / Theoretical Computer Science 795 (2019) 165–182

Fig. 1. The map

175

σ on vertex i in G.

4. Group of symmetries and Lie algebra of the Vandermonde polynomial In this section we characterize the group of symmetries and Lie algebra of the Vandermonde polynomial. In the following theorem by P ∈ A n we mean the n × n permutation matrix corresponding to an even permutation in the alternating group A n . Theorem 5. Let VD denote the determinant of the symbolic n × n Vandermonde matrix. Then, GVD = {( I + ( v ⊗ 1)) · P | P ∈ A n , v ∈ F n }, where An is the alternating group on n elements. Proof. We first argue the forward direction. Let A = B + ( v ⊗ 1) where B ∈ A n and v = ( v 1 , v 2 , . . . , v n ) ∈ F n . We show that A ∈ GVD :  Let σ be the permutation defined by the permutation matrix B. Then the transformation defined by A is

n A · xi = xσ (i ) + i =1 v i xi . Now it is easy to observe that i < j (xi − x j ) = i < j (( A · xi ) − ( A · x j )). Therefore A ∈ GVD . For the converse direction, consider A ∈ GVD . We need to show that A = B + ( v ⊗ 1) where B ∈ A n and v = ( v 1 , v 2 , . . . , v n ) ∈ F n . A defines a linear transformation on the set of variables {x1 , x2 , . . . , xn } and let i = A · xi . We have

i < j ( xi − x j ) = i < j (i −  j ). By unique factorization of polynomials, we have that there exists a bijection σ : {(i , j ) | i < j } → {(i , j ) | i < j } such that σ (i , j ) = (i , j ) iff i −  j = xi − x j . We now show that the σ is induced by a permutation π ∈ S n : Claim 5.1. Let σ be as defined above. Then there exists a permutation π of {1, . . . , n} such that σ (i , j ) = (π (i ), π ( j )). Proof of Claim 5.1. Let G be a complete graph on n vertices such that edge (i , j ) is labelled by (i −  j ) for i < j. Let H be the complete graph on n vertices with the edge (i , j ) labelled by (xi − x j ) for i < j. Now σ can be viewed as a bijection from E (G ) to E ( H ). It is enough to argue that for any 1 ≤ i ≤ n,

σ ({(1, i ), (2, i ), . . . , (i − 1, i ), (i , i + 1), . . . , (i , n)}) = {(1, ki ), (2, ki ), . . . , (ki − 1, ki ), (ki , ki + 1), . . . , (ki , n)}

(9)

for some unique ki ∈ [n]. Then π : i → ki is the required permutation. For the sake of contradiction, suppose that (9) is not satisfied for some i ∈ [n]. Then there are distinct j , k, m ∈ [n] such that the edges {(i , j ), (i , k), (i , m)} in G under σ map to edges in {(α , β), (γ , δ), (η, κ )} in H where the edges (α , β), (γ , δ) and (η, κ ) do not form a star in H . Note that α , β, γ , δ, η, κ need not be distinct. Various possibilities for the vertices α , β, γ , δ, η, κ and the corresponding vertex-edge incidences in H are depicted in the Fig. 1. Observe that in the figure the edges are labelled with a ± sign to denote that based on whether i < j or j < i one of + or − is chosen. Recall that we have,

∀i < j |var(i −  j )| = 2.

(10)

We denote by P the edges {(i , j ), (i , k), (i , m)} in G. Consider the following two cases: Case 1: P in G maps to one of (a), (b) or (c ) in H under σ (see Fig. 1). In each of the possibilities, it can be seen that there exist linear forms  and 

in {i ,  j , k , m } such that |var(±( − 

))| = 4 which is a contradiction to Equation (10).

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Case 2: P in G maps to (d) in H under σ (see Fig. 1). Without loss of generality suppose σ (i , j ) = (α , β), σ (i , k) = (α , γ ) and σ (i , m) = (β, γ ). Recall that σ (i , j ) = (i , j ) if and only if i −  j = xi − x j . Then we get  j − k = xβ − xγ by the definition of σ . Therefore, we have σ ( j , k) = (β, γ ) = σ (i , m) which is a contradiction since σ is a bijection. Therefore, for all 1 ≤ i ≤ n, Equation (9) is satisfied and there exists a permutation

π such that σ (i , j ) = (π (i ), π ( j )).



Let P π be the permutation matrix corresponding to the permutation π obtained from the claim above. To complete the proof, we need to show that A = P π + ( v ⊗ 1) for v ∈ F n . Let 1 = a11 x1 + a12 x2 + · · · + a1n xn , 2 = a21 x1 + a22 x2 + · · · + a2n xn , . . ., n = an1 x1 + an2 x2 + · · · + ann xn Now suppose π is the identity permutation, i.e., σ (i , j ) = (i , j ) for all i < j, therefore 1 − 2 = x1 − x2 , 1 − 2 = x1 − x2 , . . . , 1 − n = x1 − xn . Now, we have the following system of linear equations

a11 − a21 = 1, a12 − a22 = −1, a13 − a23 = 0, a14 − a24 = 0, . . . , a1n − a2n = 0 a11 − a31 = 1, a12 − a32 = 0, a13 − a33 = −1, a14 − a34 = 0, . . . , a1n − a3n = 0

.. . a11 − an1 = 1, a12 − an2 = 0, a13 − an3 = 0, a14 − an4 = 0, . . . , a1n − ann = −1. From the equations above, it follows that when π is the identity permutation, A − I = v ⊗ 1 for some v ∈ F n where 1 is the −1 A = I + v ⊗ 1 for some vector with all entries as 1. When π is not identity, it follows from the above arguments that P π v ∈ F n . Since VD(( I + v ⊗ 1) X ) = VD( X ), we conclude that π ∈ A n . 2 Now, we show that if any homogeneous polynomial of degree f should be a scalar multiple of VD(x1 , . . . , xn ). Vandermonde polynomial is characterized by GVD .

n 2

has its group of symmetries identical to that of GVD , then

Lemma 7. Let f ∈ F[x1 , . . . , xn ] be a homogeneous polynomial of degree some α ∈ F .

n 2

. If G f = GVD then f (x1 , . . . , xn ) = α · VD(x1 , . . . , xn ) for

Proof. Let f ∈ F[x1 , . . . , xn ]. Since G f = GVD = {( I + ( v ⊗ 1)) · P | P ∈ A n , v ∈ F n }, G f ∩ S n = A n . Hence f is an alternating polynomial. By the fundamental theorem of alternating polynomials [20,1], there n exists a symmetric polynomial g ∈ F[x1 , . . . , xn ] such that f (x1 , . . . , xn ) = g (x1 , . . . , xn ) · VD(x1 , . . . , xn ). Since deg( f ) = 2 = deg(VD(x1 , . . . , xn )), g = α for some α ∈ F . 2 Using the description of GVD above, we now describe the Lie algebra of GVD . Lemma 8. We have gVD = { v ⊗ 1 | v ∈ F n }. Proof. We have

A ∈ gVD ⇐⇒



(xi − x j +  ( A (xi ) − A (x j ))) =

i> j



(xi − x j ) By Definition 3

i> j

⇐⇒ A (xi ) = A (x j ) ∀i = j ⇐⇒ A = v ⊗ 1 for some v ∈ F n . 2 We also note that gVD is solvable: Lemma 9. gVD is a 2-step solvable Lie algebra. Proof. Let L (0) = gVD = { v ⊗ 1 | v ∈ F n }. Then,

L ( 1 ) = [ L ( 0) , L [0] ]

= {[u , v ] | u , v ∈ L (0) } = {[a ⊗ 1, b ⊗ 1] | a, b ∈ F n } = {[wt(a) · b − wt(b) · a] ⊗ 1 | a ⊗ 1, b ⊗ 1 ∈ L (0) }

C. Ramya, B.V. Raghavendra Rao / Theoretical Computer Science 795 (2019) 165–182

where for any v = ( v 1 , . . . , v n ) ∈ F n , wt( v ) = wt(b)a) = 0. Then,

n

i =1

177

v i . Before computing L (2) , we note that for a, b ∈ F n , wt(wt(a)b −

L (2 ) = [ L (1 ) , L (1 ) ]

= {[b · wt(a) − a · wt(b)] ⊗ 1 | a ⊗ 1, b ⊗ 1 ∈ L (1) } =0 Thus, gVD is a 2-step solvable Lie algebra.

2

Further, it may be noted that gVD has a commutative sub-algebra of co-dimension 1: Lemma 10. Let C = {a ⊗ 1 | wt(a) = 0}. Then C is a commutative sub-algebra of gVD of dimension n − 1. Proof. If wt(a) = wt(b) = 0, then [a ⊗ 1, b ⊗ 1] = 0, therefore, C is a commutative sub-algebra of gVD . Since the space defined by wt(x) = 0 in F n has co-dimension one, we conclude that C is a commutative sub-algebra of gVD with dimension n − 1. 2 Remark 2. The facts in Lemmas 9 and 10 were pointed to us by an anonymous referee. 5. Expressive power of projections of Vandermonde polynomials In this section we study polynomials that can be represented as projections of Vandermonde polynomials. Recall the definitions of the classes VD, VDproj , VDhomo and VDaff from Section 2. For any arithmetic model of computation, universality and closure under addition and multiplication are among the most fundamental and necessary properties to be investigated. Here, we study these properties for projections of the Vandermonde polynomial and their sums. Most of the proofs follow from elementary arguments. By definition, VD, VDproj , VDhomo ⊆ VDaff . Also, any polynomial with at least one irreducible non-linear factor cannot be written as a projection of VD. As expected, we observe that there are products of linear forms that cannot be written as a projection of VD. Lemma 11. (x1 − y 1 )(x2 − y 2 ) ∈ / VDaff . Proof. Suppose f = (x1 − y 1 )(x2 − y 2 ) ∈ VDaff

, then there are affine forms 1 , . . . , n such that (x1 − y 1 )(x2 − y 2 ) = 1≤i < j ≤n (i −  j ) are non constant polynomials. Without loss of generality, 1≤i < j ≤n (i −  j ). Clearly, only two factors of let i −  j = x1 − y 1 and i −  j = x2 − y 2 . Then, we must have i − i ,  j −  j , i −  j and  j − i as constant polynomials, as they are factors of VD(1 , . . . , n ) and hence i −  j = i − i − ( j − i ) is a constant, which is a contradiction. 2 Corollary 3. Let F = { f m }m≥1 , f m = (x1 − y 1 ) · · · (xm − ym ) be a polynomial family on 2m-variables. Then, for any f m ∈ F , f m ∈ / VDaff . Proof. Suppose f = (x1 − y 1 ) · · · (xm − ym ) ∈ VDaff , then f |x¯ =¯a with a¯ = (x3 = 1, x4 = 1, . . . , xm = 1, y 3 = 0, y 4 = 0, . . . , ym = 0) is also computable in VDaff . This implies f |x¯ =¯a = (x1 − y 1 )(x2 − y 2 ) is in VDaff , a contradiction to Lemma 11. 2 Lemma 12. The classes VD, VDproj , VDhom and VDaff are not closed under addition and multiplication. Proof. Since sum of any two variable disjoint polynomials is irreducible, it is clear that VD, VDproj , VDhom and VDaff are not closed under addition. For multiplication, take f 1 = x1 − y 1 , and f 1 = x2 − y 2 . By Lemma 11, f 1 f 2 ∈ / VDaff and hence f1 f2 ∈ / VD ∪ VDproj ∪ VDhom . Since f 1 , f 2 ∈ VD ∩ VDproj ∩ VDhom ∩ VDaff , we have that VD, VDproj , VDhom and VDaff are not closed under multiplication. 2 It can also be seen that the classes of polynomials VD, VDproj , VDhom and VDaff are properly separated from each other: Lemma 13. (1) VDproj  VDaff and VDhomo  VDaff . (2) VDproj ⊂ VDhomo and VDhomo ⊂ VDproj .

 Proof.

• VDproj  VDaff : Let f = (x1 − y 1 ) + (x2 − y 2 ). Then f = det

seen that (x1 − y 1 ) + (x2 − y 2 ) ∈ / VDproj .

1

1

y 2 − x2

x1 − y 1

 . By comparing factors it can be

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 • VDhomo  VDaff : Let g = x1 + x2 − 2. Then g = det

1

1



. Suppose g ∈ VDhomo , then there exists an n × n 1 − x2 x1 − 1

Vandermonde matrix M such that g ≤homo det( M ). In other words, x1 + x2 − 2 = i < j i , j ∈[n] ( j − i ), where i ’s are linear forms which is impossible since x1 + x2 − 2 is non-homogeneous. • VDhomo  VDproj : Let h = (x1 − 1)(x1 − 2). Observe that h ∈ VDproj . However, since VDhomo consists only of polynomials with linear factors, h ∈ / VDhomo .   1 1 • VDproj  VDhomo : Let f = (x1 − y 1 ) + (x2 − y 2 ). For M = , we have det( M ) ∈ VDhomo and f = det( M ). y 2 − x2 x1 − y 1 It can be seen that (x1 − y 1 ) + (x2 − y 2 ) ∈ / VDproj . 2 In this section, we consider polynomials that can be expressed as sum of projections of Vandermonde polynomials. Definition 6. For a class C of polynomials, let  · C be defined as



·C =

f

f = ( f n )n≥0 where ∀n ≥ 0 ∃ g 1 , g 2 , . . . , g s ∈ C , α1 , . . . , αs ∈ F such that f =

α1 g1 + α2 g2 + · · · + αs g s , s = n O (1)

.

Lemma 14. x1 · x2 ∈ /  · VD. Proof. Suppose there exists g 1 , g 2 , . . . , g s ∈ VD. Note that for every i, either deg( g i ) ≤ 1 or deg( g i ) ≥ 3. Since deg( g ) = 2, it is impossible that x1 x2 = g 1 + · · · + g s for any s ≥ 0. 2 Lemma 15. The class  · VD is closed under addition but not under multiplication. (i) If f 1 , f 2 ∈  · VD then f 1 + f 2 ∈  · VD. (ii) There exists f 1 , f 2 ∈  · VD such that f 1 · f 2 ∈ /  · VD. Proof. (i) Closure under addition follows by definition. (ii) Let f 1 = x1 − y 1 and f 2 = x2 − y 2 , clearly f 1 , f 2 ∈  · VD. Since for any g ∈ VD, deg( g ) = 2, one can conclude that f1 f2 ∈ /  · VD. 2 We now consider polynomials in the class VDproj . Any univariate polynomial f of degree d can be computed by depth-2 circuits of size poly(d). However there are univariate polynomials not in VDaff which is a subclass of depth 2 circuits (consider any univariate polynomial irreducible over F ). Here, we show that the class of all univariate polynomials can be computed efficiently by circuits in VDproj . Lemma 16. Let f = a0 + a1 x + a2 x2 + · · · + ad xd be a univariate polynomial of degree d. Then there are g i ∈ VDproj , 1 ≤ i ≤ s ≤ O (d2 ) for some αi ∈ F such that f = g 1 + · · · + g s . Proof. Consider the (d + 1) × (d + 1) Vandermonde matrix M 0 ,

⎡

1 ⎢ x

1

β1 ⎢ 2 2 ⎢ x β 1 ⎢ ⎢ . .. ⎢ . . M0 = ⎢ . ⎢ . .. ⎢ . . ⎢ . ⎢ d −1 d −1 ⎣x β1 xd β1d

1

1

··· ··· .. . .. . ··· ···

··· ··· .. . .. . ··· ···

1

⎤

βd−1 ⎥ ⎥ βd2−1 ⎥ ⎥ .. ⎥ . ⎥ ⎥ .. ⎥ ⎥ . ⎥ −1 ⎥ ⎦ βdd− 1 d βd−1

Let g 0 = det( M 0 ) = γ00 + γ01 x + γ02 x2 +· · ·+ γ0,d−1 xd−1 + γ0d xd where γ00 , . . . , γ0d ∈ F and γ0d = 0. Note that g 0 ∈ VDproj . a γ a γ a γ a Setting α0 = γ d we get α0 g 0 = ad xd + d γ0,d−1 xd−1 + · · · + dγ 01 x + dγ 00 . Now, let M 1 be the d × d Vandermonde matrix, 0d

0d

0d

0d

C. Ramya, B.V. Raghavendra Rao / Theoretical Computer Science 795 (2019) 165–182

⎡

1 ⎢ x

1

⎢ ⎢ x2 ⎢ ⎢ M 1 = ⎢ .. ⎢ . ⎢ . ⎢ . ⎣ .

xd−1

1

1

β1 · · · β12 · · · .. .. . . .. .. . . β1d−1 · · ·

··· ··· .. . .. . ···

1

179

⎤

βd−1 ⎥ ⎥ βd2−1 ⎥ ⎥ .. ⎥ ⎥ . ⎥ .. ⎥ ⎥ . ⎦ −1 βdd− 1

Then

g 1 = det( M 1 ) = γ10 + γ11 x + γ12 x2 + · · · + γ1,d−1 xd−1 , where ad γ01

γ10 , . . . , γ1d ∈ F . Observe that xd is not a monomial in α1 g1 . Set α1 = ad γ00

ad γ0,d−1 ad−1 γ1,d−1 − γ0d γ1,d−1 . Then

α1 g1 = ad−1 xd−1 + · · · +

γ0d x + γ0d . Extending this approach: Let M i be a (d − (i − 1)) × (d − (i − 1)) Vandermonde matrix,

⎡

1 ⎢ x

⎢ ⎢ ⎢ ⎢ Mi = ⎢ ⎢ ⎢ ⎢ ⎣

1

2

x

.. . .. . xd−i

1

1

β1 · · · β12 · · · .. .. . . .. .. . . β1d−i · · ·

··· ··· .. . .. . ···

Now observe that by setting

αi =

1

⎤

βd−i ⎥ ⎥ βd2−i ⎥ ⎥ .. ⎥ ⎥ . ⎥ .. ⎥ ⎥ . ⎦ d −i βd−i a

d −i γi,d−i − (α0 γ0,d−i + α1 γ1,d−i + · · · + αi −1 γi −1,d−i ) we ensure that

d

contain any term of the form x for d − i ≤ p ≤ d − 1. Thus summands. Then, using O (d2 ) summands f can be obtained. 2 p

k=0

i

j =0

α j g j does not

αk gk = ad x . Hence to compute ad x we require d d

d

Recall that the n-variate power symmetric polynomial of degree d is defined as Pown,d = xd1 + xd2 + · · · + xnd . From the arguments in Lemma 16, it follows that Pown,d can be expressed by polynomial size circuits in  · VDproj . Corollary 4. There are polynomials f i ∈ VDproj 1 ≤ i ≤ nd such that Pown,d =

s

i =1

αi f i .

Now, to argue that VDhomo and VDaff are universal, we need the following: Lemma 17 ([21]). Over any infinite field containing the set of integers, there exists 2d linear forms L 1 , . . . , L 2d such that d  i =1

d

xi =

2 

L di

i =1

Combining Corollary 4 with Lemma 17 we establish the universality of  · VDhomo and  · VDaff . Lemma 18. The classes  · VDhomo and  · VDaff are universal. Also, in the following, we note that  · VDaff is more powerful than depth three  ∧  circuits: Lemma 19. poly − size  ∧   poly−size  · VDaff . Proof. Let f ∈  ∧ . Then f =

n k

s  i =1

di . Then for any k ≥ 1, dim ∂ =k ( f ) ≤ s. Now, for any 1 ≤ k ≤ n/2 we have dim ∂ =k (VD) ≥

. Therefore, if f = VD we have s = 2(n) by setting k = n/2. Hence poly − size  ∧   poly−size  · VDaff .

2

5.1. A lower bound against  · VDproj Observe that  · VDproj is a subclass of non-homogeneous depth three circuits of bottom fan-in 2, i.e., [2] . It is known that S ym2n,n can be computed by non-homogeneous [2] circuits of size O (n2 ). We show that any  · VDproj computing S ymn,n/2 requires a top fan-in of 2(n) and hence  · VDproj P [2] where the subscript P denotes containment up to polynomial size. The lower bound is obtained by a variant of the evaluation dimension as a complexity measure for polynomials.

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Definition 7 (Restricted evaluation dimension). Let f ∈ F[x1 , . . . , xn ] and S = {i 1 , . . . , ik } ⊆ [n]. Let a¯ = (ai 1 , ai 2 , . . . , aik ) ∈ {0, 1, ∗}k and f | S =¯a be the polynomial obtained by substituting for all i j ∈ S,

xi j =

⎧ ⎪ ⎨1 ⎪ ⎩

if ai j = 1

0

if ai j = 0

xi j

if ai j = ∗

def

Let f | S = { f | S =¯a | a¯ ∈ {0, 1, ∗}k }. The restricted evaluation dimension of f is defined as: def

REDS ( f ) = dim(F -span( f | S )) It is not hard to see that the measure REDS is sub-additive: Lemma 20. For any f , g ∈ F[x1 , x2 , . . . , xn ], REDS ( f + g ) ≤ REDS ( f ) + REDS ( g ). In the following, we show that Vandermonde polynomials and their projections have low restricted evaluation dimension: Lemma 21. Let f = VD( y 1 , . . . , yn ) such that for every i ∈ [n], y i ∈ {x1 , . . . , xn } ∪ F . Then for any S ⊂ {1 . . . , n} with | S | = k, we have REDS ( f ) ≤ (k + 1)2 . Proof. Without loss of generality suppose S = { j 1 , j 2 , . . . , jk } ⊆ [n] and | S | = k. Let T = {x j 1 , x j 2 , . . . , x jk } ∩ var( f ) = {xi 1 , xi 2 , . . . , xim }. Observe that m ≤ k. For a vector v ∈ {0, 1, ∗}n and b ∈ {0, 1}, let #b ( v ) denote the number of occurrences of b in the vector v. Then, for any a¯ = (a j 1 , a j 2 , . . . , a jk ) ∈ {0, 1, ∗}k ,

• If #0 ((ai 1 , ai 2 , . . . , aim )) ≥ 2 or #1 ((ai 1 , ai 2 , . . . , aim )) ≥ 2 then f | S =¯a = 0. • If #0 ((ai 1 , ai 2 , . . . , aim )) = #1 ((ai 1 , ai 2 , . . . , aim )) = 1. Let T 1 be the set of polynomials obtained from such evaluations of m k f d . The number of such assignments is at most 2 · 2 ≤ 2 · 2 ≤ k2 and hence | T 1 | ≤ k2 . • If #0 ({ai 1 , ai 2 , . . . , aim }) = 1 or #1 ({ai 1 , ai 2 , . . . , aim }) = 1. Let T 2 denote the set of polynomials obtained from such eval m m k uations. Since number of such assignments is 2 m−1 ≤ 2 1 ≤ 2 1 ≤ 2k, we have | T 2 | ≤ 2k. • If, #0 ({ai 1 , ai 2 , . . . , aim }) = #1 ({ai 1 , ai 2 , . . . , aim }) = 0, in this case, the polynomial f does not change under these evaluations. From the above case analysis, we have F -span( f | S =a ) = F -span( T 1 ∪ T 2 ∪ { f }). Therefore REDS ( f ) ≤ k2 + 2k + 1 ≤ (k + 1)2 . 2 Lemma 22. Let S ymn,k be the elementary symmetric polynomial in n variables of degree k. Then for any S ⊂ {1, . . . , n}, | S | = k, we have REDS ( S ymn,k ) ≥ 2k − 1. Proof. Let S ymn,k be the elementary symmetric polynomial in n variables of degree k. For T ⊆ S , T = ∅, define a¯ T =

(a1 , . . . , ak ) ∈ {1, ∗}k as:



ai =

∗ if xi ∈ T 1 if xi ∈ S \ T

Note that it is enough to prove:

dim({ S ymn,k | S =¯a T | T ⊆ S , T = ∅}) ≥ 2k − 1

(11)

Since { S ymn,k | S =¯a T | T ⊆ S , T = ∅} ⊆ {F -span( S ymn,k | S =¯a ) | a¯ = (a1 , . . . , ak ) ∈ {1, ∗}k }, by Equation (11) we have

REDS ( S ymn,k ) ≥ dim({ S ymn,k | S =¯a T | T ⊆ S , T = ∅}) = 2k − 1. To prove (11), note that for any distinct T 1 , T 2 ⊆ S, we have S ymn,k | S =¯a T and S ymn,k | S =¯a T have distinct leading monomials 1 2 with respect to the lex ordering since they have distinct supports. Since the number of distinct leading monomials in a space of polynomials is a lower bound on its dimension, this concludes the proof of (11). 2 Theorem 6. If

s  i =1

αi f i = S ymn,n/2 where f i ∈ VDproj then s = 2(n) .

C. Ramya, B.V. Raghavendra Rao / Theoretical Computer Science 795 (2019) 165–182

Proof. The proof is a straightforward application of sub-additivity of REDS combined with Lemmas 22 and 21.

181

2

Observation 4. An exponential lower bound for  · VDhomo and a linear lower bound for  · VDaffine follow directly from lower bounds for  circuits. 5.2. Connection to  circuits In this subsection, we show the connection of classes built over Vandermonde polynomials to the well-studied depth three  circuits. Lemma 23. [d]  ⊆ 2 ∧ VDaff . Proof. Let C be a  circuit computing a polynomial f ∈ F[x1 , . . . , xn ] of degree d. Let m1 , . . . , mk be the multiplication gates in C . Let mi be a multiplication gate in C computing the product of linear forms say i 1 · · · id . Consider the linear forms

L 0 = 0 , L 1 = i 1 , L 2 = i 2 , . . . , L d =  i d Then



VD( L 0 , L 1 , . . . , L d ) = i 1 i 2 · · · id

(L i − L j )

1
VD( L 0 , L 1 , . . . , L d )

i 1 i 2 · · · i d =

1
(12)

− L j)

For any g , h ∈ F[x1 , . . . , xn ] with h(0, . . . , 0) = 1 and h| g, we have

g h

=

g 1 − (1 − h)

=

d 

g (1 − h) j =

j =0

d 

c j gh j

(13)

j =0

where d = deg( g /h) and c j are integers. Note that after translating the input variables {x1 , . . . , xn } and multiplying by suitable constants from F , we can get h(0, . . . , 0) = 1. Also, if h ∈ VDaff such that h(0, . . . , 0) = 1, then after translation of variables {x1 , . . . , xn } and multiplication by constants from F , we have h ∈ VDaff and h(0, . . . , 0) = 1. Hence  ⊆ 2 ∧d VDaff . 2 The above observation is interesting as powering gates seem to be easier to handle as opposed to product gates in the context of proving lower bounds. 6. Discussion and open questions We initiated the study of projections of the Vandermonde polynomial and sum of such projections. Due to the geometric nature of Vandermonde polynomials, we believe that study of sum of projections of Vandermonde polynomials would lead to useful measures of complexity for polynomials. We conclude with the following open questions:

• A polynomial f ∈ F[x1 , . . . , xn ] is said to be characterized by its symmetries if for any polynomial g ∈ F[x1 , . . . , xn ] with deg( f ) = deg( g ) such that G g contains G f , we have g is a scalar multiple of f (Definition 1.2.5.3, [22]). Theorem 2(ii) shows a weaker version of this characterization when g = VD(x1 , . . . , xn ) and G f = GVD . In fact, if we want to show that Vandermonde polynomial is characterized by it symmetries, the current proof of Lemma 7 breaks down as we can only conclude that G f ∩ S n contains A n . From this, it is unclear if f is an alternating polynomial or a symmetric polynomial. In this context, we pose the following problem: n – Let f be a homogeneous polynomial on n variables of degree 2 . If G f contains GVD , then show that f = α · VD(x1 , . . . , xn ) with α ∈ F . • Show that  · VDproj is universal, i.e., for any multivariate polynomial f , there are f i ∈ VDproj 1 ≤ i ≤ s such that f =  f , for some s > 0. i i • Obtain exponential lower bounds against  · VDaff for an explicit polynomial. Declaration of Competing Interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

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