Weak space over complex numbers Pushkar Joglekar Raghavendra Rao B - PowerPoint PPT Presentation

Weak space over complex numbers Pushkar Joglekar Raghavendra Rao B V Siddhartha Sivakumar September 12, 2017 IIT Madras, Chennai 1

Computation over Real and Complex Numbers Motivation: To provide theoretical foundation for characterizing intrinsic nature of numberical computations over real/complex numbers. 2

The Blum-Shub-Smale Model A BSS machine M over F ( F ∈ { R , C } ) can be viewed as a Turing machine where a complex (real) number stored in each cell. • M has a finite number of constants α 1 , . . . , α k ∈ F , called the parameters of M . • M can perform + , × , − and ÷ operations with full precision where the operands are either the contents of the cells or the parameters. • M can compare content of any cell with 0 (= 0 test) and branch based on the result of the comparision. • Input is an element from F ∗ = � n ≥ 0 F n . • Output is 0 or 1. 3

THe Blum-Shub-Smale Model Definition P F is the set of all subsets of F ∗ that can be computed by polynomial time bounded BSS machines. NP F is the set of all subsets of F ∗ that can be computed by polynomial time bounded non-deterministic BSS machines 4

Time vs Space • Theory of time complexity is well established (notion of completeness, relativized computation, parallel complexity etc,.) • Separations of time complexity classes is known (NC R � = P R ). • Defining a feasible notion for space is challenging and widely open. 5

Notions of space for the BSS model Known Measures of space: • Unit cost space: Content of the cell (a value from R or C ) counted as one unit of space. • Weak space: a more careful cost model for the contents each cell. 6

How to count space? Unit cost model : Count as one cell per non-blank cell of the BSS machine. 7

How to count space? Unit cost model : Count as one cell per non-blank cell of the BSS machine. Features : • Counts the maximum number of cells used during any point of the computation. 7

How to count space? Unit cost model : Count as one cell per non-blank cell of the BSS machine. Features : • Counts the maximum number of cells used during any point of the computation. • Represents width of the algebraic circuit computing the same language. 7

How to count space? Unit cost model : Count as one cell per non-blank cell of the BSS machine. Features : • Counts the maximum number of cells used during any point of the computation. • Represents width of the algebraic circuit computing the same language. Limitations : • Does not measure the size of individual cells. • Number of configurations is infinite, hence no feasible comparison with time complexity. 7

Unit cost : too strong • Each cell in the tape can hold a value from R . • A step can perform any arithmetic operations or compare two real numbers. Theorem (Michaux ’89) Any set L ⊆ R ∗ that is computable using the BSS model can also be computed by a BSS machine that uses only a constant number of cells. Constant space is enough for any computation!! 8

Weak space Introduced by Naurois ’06. M be a BSS machine with parameters α 1 , . . . , α k . • On a given input x ∈ F n , at any stage of computation, every cell c of M represents a rational function f c = p c ( x 1 , . . . , x n , α 1 , . . . , α k ) / q c ( x 1 , . . . , x n , α 1 , . . . , α k ). • For the term c γ x γ 1 1 · · · x γ n n α γ n +1 · · · α γ n + k , space 1 k ≈ log c γ + � n + k i =1 log γ i . • For the polynomial p = � γ c γ X γ n + k � � space ( p ) = log c γ + log γ i . γ, c γ � =0 i =1 9

Weak space • For a rational function f c = p c ( x 1 , . . . , x n , α 1 , . . . , α k ) / q c ( x 1 , . . . , x n , α 1 , . . . , α k ), space ( f c ) = space ( p c ) + ( q c ) . • for a configuration Γ with non-empty cells c 1 , . . . c m , m � space (Γ) = space ( c j ) . j =1 • Space of M on x is the max of all configuratons. • Gives a reasonable definition of LOGSPACE W and PSPACE W . 10

Weak space :Properties Theorem (Naurois 06) • LOGSPACE W ⊆ P W ∩ NC 2 R . • PSPACE W ⊆ P R 11

Weak space: properties Conjecture (Nauraois 06) 1. NC 1 R �⊆ LOGSPACE W . 2. LOGSPACE W ⊆ NC 1 ⇒ DLOG ⊆ NC 1 . R = 12

Our Results - 1 Theorem (1) NC 1 C �⊆ PSPACE W , i.e., there is a set L ∈ NC 1 C but L / ∈ PSPACE W . 13

Our Results - 2 For a complexity class C ⊆ F ∗ , let BP( C ) = C ∩ { 0 , 1 } ∗ . We prove: Theorem (2) BP(LOGSPACE W ) ⊆ DLOG . 14

Proof sketch Theorem (1) NC 1 C �⊆ PSPACE W , i.e., there is a set L ∈ NC 1 C but L / ∈ PSPACE W . Sketch. • Let Sym n , n / 2 ( x 1 , . . . , x n ) = � � i ∈ S x i . S ⊂ [ n ] , | S | = n / 2 • Let L n = { ( x 1 , . . . , x n ) | = Sym n , n / 2 ( x 1 , . . . , x n ) = 0 } , and L = � n ≥ 0 L n . • L ∈ NC 1 C . We show that L / ∈ PSPACE W . 15

Proof sketch Lemma Let L ∈ Space( s ( n )) , then for every n > 0 , there exist t ≥ 1 and polynomials f i , j , 1 ≤ i ≤ t, 1 ≤ j ≤ m i , g i , j and 1 ≤ i ≤ t, 1 ≤ j ≤ m i in Z [ x 1 , . . . , x n ] such that: 1. space ( f i , j ) ≤ s ( n ) , for every 1 ≤ i ≤ t 1 , 1 ≤ j ≤ m i ; and 2. space ( g i , j ) ≤ s ( n ) , for every 1 ≤ i ≤ t 2 , 1 ≤ j ≤ m i ; and 3. L ∩ F n = � t � m i j =1 [ f i , j = 0] ∩ � j = 1 m i [ g i , j � = 0] . i =1 16

Proof Sketch • Suppose L ∈ space ( n c ) for some c > 0. • Let f i , j , g i , j , 1 ≤ i ≤ t , 1 ≤ j ≤ m i as given in the lemma, and • L n = � t � m i j =1 [ f i , j = 0] ∩ � j = 1 m i [ g i , j � = 0]. i =1 • Let V i = � m i j =1 [ f i , j = 0], W i = � m i j =1 [ g i , j � = 0] and T i = V i ∩ W i . i =1 � T i , where � • Then, L n = ∪ t i =1 T i . Then L n = ∪ t T i is the Zariski closure. • Since L n is irreducible, L n = � T i for some i , i.e., L n ⊆ [ f i , j = 0] for some j , therefore Sym n , n / 2 | f i , j . 17

Proof sketch Lemma Any polynomial divisible by Sym n , n / 2 has 2 Ω( n ) monomials. • A contradiction to Sym n , n / 2 | f i , j . ∈ Space ( n c ) for any c ≥ 0. • Conlusion: L / 18

Thank You !! 19