On formal complexity measures Pavel Pudl ak Mathematical - PowerPoint PPT Presentation

On formal complexity measures Pavel Pudl´ ak Mathematical Institute, Academy of Sciences, Prague and Charles University, Prague Workshop on Randomness and Enumeration November 24-28, 2008 La Suiza Andina, Curacaut´ ın, Chile 1

Overview 1st lecture: boolean formulas, NC 1 formal complexity measures examples convexity rank as a formal complexity measure 2nd lecture n log n lower bound on monotone formulas linear upper bound for general formulas 2

Literature (main): M. Karchmer, E. Kushilevitz, N. Nisan, Fractional covers and communication complexity, 1995 A.A. Razborov, Applications of Matrix Methods to the Theory of Lower Bounds in Computational Complexity 1990 A.A. Razborov, On submodular complexity measures, 1992 P. Hrubeˇ s, S. Jukna, A. Kulikov, P. Pudl´ ak and P. Savick´ y, On convex complexity measures (in preparation) 3

NC 1 = “the class of problems with fast parallel algorithms” = computable by log-depth circuits = computable by polynomial size formulas NC 1 ⊆ P Problem NC 1 = P ? , NC 1 = NP ? etc. 4

NC 1 = “the class of problems with fast parallel algorithms” = computable by log-depth circuits = computable by polynomial size formulas NC 1 ⊆ P Problem NC 1 = P ? , NC 1 = NP ? etc. Problem Find explicitly defined boolean functions that cannot be computed by polynomial size formulas. By counting argument: there exist functions that can only be computed by exponential size formulas. 4

formula size complexity basis of connectives ∨ , ∧ , ¬ variables x 1 , . . . , x n size of a formula = number of occurrences of variables Example. ( x 1 ∧ ¬ x 2 ) ∨ ( ¬ x 1 ∧ x 2 ) has size 4. Definition Let f : { 0 , 1 } n → { 0 , 1 } . The formula size complexity of f is L ( f ) := the minimal s such that f can be computed by a formula of size s Example. Let f ( x 1 , x 2 ) := x 1 ⊕ x 2 , then L ( f ) = 4. To prove NC 1 � = P it suffices to prove super polynomial lower bounds on the formula size complexity of a function in P . 5

functions, rectangles, measures Let f : { 0 , 1 } n → { 0 , 1 } . The associated rectangle of f is S f := f − 1 (0) × f − 1 (1) , M ⊆ S f is a monochromatic rectangle, if for some ǫ ∈ { 0 , 1 } and 1 ≤ i ≤ n , for all ( x , y ) ∈ M , x i = ǫ and y i = 1 − ǫ There are 2 n maximal monochromatic rectangles M i ,ǫ = { ( x , y ) ∈ S f ; x i = ǫ and y i = 1 − ǫ } . 6

Let R be the set of all subrectangles of S f . A rectangle function is a real-valued function µ : R → R . A rectangle measure is a rectangle function µ : R → R satisfying the following condition: (i) Normalization : µ ( M ) ≤ 1 for every monochromatic rectangle M ∈ R (ii) Subadditivity : µ ( R ) ≤ µ ( R 1 ) + µ ( R 2 ), for every rectangle R ∈ R and for every its partition into disjoint union of rectangles R 1 , R 2 ∈ R . 7

Let R be the set of all subrectangles of S f . A rectangle function is a real-valued function µ : R → R . A rectangle measure is a rectangle function µ : R → R satisfying the following condition: (i) Normalization : µ ( M ) ≤ 1 for every monochromatic rectangle M ∈ R (ii) Subadditivity : µ ( R ) ≤ µ ( R 1 ) + µ ( R 2 ), for every rectangle R ∈ R and for every its partition into disjoint union of rectangles R 1 , R 2 ∈ R . Theorem If µ is a rectangle measure, then µ ( S f ) ≤ L ( f ) 7

Proof: Let φ be a formula computing f . To every subformula ψ of φ , assign a rectangle A × B such that x ∈ A → ψ ( x ) = 0 and y ∈ B → ψ ( y ) = 1 (1) as follows. assign S f to φ ; 1 if ψ ↔ A × B and ψ = ψ 1 ∨ ψ 2 , then split vertically into A × B 1 and A × B 2 2 while preserving (1); if ψ ↔ A × B and ψ = ψ 1 ∧ ψ 2 , then split horizontally into A 1 × B and 3 A 2 × B while preserving (1); Then we have the rectangles assigned to x i and ¬ x i are monochromatic; if ψ ↔ R , then µ ( R ) ≤ the size of ψ (proof by induction on the size of ψ ) Hence µ ( S f ) ≤ size of φ , whence µ ( S f ) ≤ L ( f ). QED 8

Let λ ( A × B ) := minimal size of ψ satisfying (1). Then λ is a measure; 1 λ ( S f ) = L ( f ); 2 λ ( R ) = max µ µ ( R ). 3 Thus in principle we can prove exponential lower bounds. So to prove NC 1 � = P we need only to find a suitable measure! 9

why are we not able to prove exponential lower bound? simple measures give only small bounds 1 measures that could prove exponential lower bounds are difficult to estimate 2 10

Example 1 [Khrapchenko/Paterson] Let f ( x 1 , . . . , x n ) = x 1 ⊕ · · · ⊕ x n . Define w ( R ) = |{ ( x , y ) ∈ R ; d ( x , y ) = 1 }| and µ ( R ) := w ( R ) 2 . | R | Then µ is a measure and µ ( S f ) = n 2 . Whence L ( f ) ≥ n 2 . Proof. Normalization: Suppose A × B is monochromatic, w.l.o.g. | A | ≤ | B | . Then for every x ∈ A , there exists at most one y with d ( x , y ) = 1. Hence w ( A × B ) ≤ | A | . | A |·| B | ≤ | A | 2 | A | 2 Whence µ ( A × B ) ≤ | A | 2 = 1 . ≤ w 2 r 1 + w 2 Subadditivity follows from the inequality ( w 1 + w 2 ) 2 1 r 2 , for r 1 , r 2 > 0. 2 r 1 + r 2 w ( S f ) = n 2 n − 1 , | S f | = 2 n − 1 · 2 n − 1 . Remark. L ( x 1 ⊕ · · · ⊕ x n ) ≤ 9 8 n 2 ; if n is a power of 2, then ≤ n 2 . 11

Example 2 [Rychkov] Let n be a power of 2, let H be the words of the Hamming code of length n . Let g ( x ) = 1 iff x ∈ H . Define w ( R ) = |{ ( x , y ) ∈ R ; d ( x , y ) = 2 }| and µ ( R ) := w ( R ) 2 n | R | . Then µ is a measure and µ ( S g ) = Ω( n 2 ). Whence L ( g ) = Ω( n 2 ). This is not known to be optimal! We only know L ( g ) = O ( n 2 log n ). 12

(iii) Additivity : µ ( R ) = µ ( R 1 ) + µ ( R 2 ), for every rectangle R ∈ R and for every its partition into disjoint union of rectangles R 1 , R 2 ∈ R . µ is additive iff ∃ α : S f → R such that ∀ R ∈ R � µ ( R ) = α ( x , y ) . ( x , y ) ∈ R Think of α as a matrix. Lemma Let F : R → R be a convex function, let w be an additive measure. Then µ , defined by � w ( R ) � µ ( R ) = | R | · F , | R | is subadditive. If F ( x ) = x r , for some r ≥ 1, and µ is normalized, we call µ a polynomial measure. Khrapchenko measure is a polynomial measure with F ( x ) = x 2 and α ( x , y ) = 1 if d ( x , y ) = 1, and = 0 otherwise. 13

Theorem Let µ be a polynomial measure defined by µ ( R ) = w ( R ) r | R | r − 1 , where w is a nonnegative additive measure. Then for 1 ≤ r ≤ 2 , µ ( S f ) ≤ (2 n ) r ; 1 for r > 2 , µ ( S f ) = O ( n ) . 2 Thus always µ ( S f ) = O ( n 2 ) . Proof of 1. r − 1 r . From normalization we get w ( M ) ≤ | M | Since w is nonnegative and additive and S f is the union of the 2 n maximal monochromatic rectangles, we have r − 1 � r . w ( S f ) ≤ w ( M i ,ǫ ) ≤ 2 n | S f | w ( S f ) Hence ≤ 2 n . Raising it to the r we get the theorem. r − 1 | S f | r 14

Remark. Using an additive measure with some negative values one can obtain an n 2 lower bound. Example. [Karchmer, Kushilevitz, Nisan] Let f be the parity function. Define � 1 / 2 n − 1 if d ( x , y ) = 1 α ( x , y ) = − 1 / 2 n (2 n − 1) otherwise. The corresponding additive measure gives the lower bound n 2 . 15

Remark. Using an additive measure with some negative values one can obtain an n 2 lower bound. Example. [Karchmer, Kushilevitz, Nisan] Let f be the parity function. Define � 1 / 2 n − 1 if d ( x , y ) = 1 α ( x , y ) = − 1 / 2 n (2 n − 1) otherwise. The corresponding additive measure gives the lower bound n 2 . What if the measure is defined using F ( x ) = x 2 log x (for x ≥ 1)? 15

strong subadditivity and convexity (iv) Strong subadditivity : if µ ( R ) ≤ � m i =1 µ ( R i ), for every rectangle R and every its partition into disjoint union of rectangles R i ⊆ R . The partition number of a rectangle R is defined by κ ( R ) = min { m ; R can be decomposed into m disjoint monochromatic rectangles } . κ is a strongly subadditive measure; 1 µ ≤ κ for every strongly subadditive measure µ ; 2 for almost all f , κ ( S f ) ≥ 2 Ω( √ n ) . 3 Problem. What is max f κ ( S f )? 16

A fractional partition of R is R 1 , . . . , R m ∈ R and r 1 , . . . , r m ∈ (0 , 1], such that m � χ R = r i χ R i i =1 where χ R , χ R i denote the characteristic functions of the rectangles. Definition A measure µ is convex, if for every R ∈ R and every fractional partition R 1 , . . . , R m ∈ R , r 1 , . . . , r m ∈ (0 , 1] of R m � µ ( R ) ≤ r i µ ( R i ) . i =1 17

A fractional partition of R is R 1 , . . . , R m ∈ R and r 1 , . . . , r m ∈ (0 , 1], such that m � χ R = r i χ R i i =1 where χ R , χ R i denote the characteristic functions of the rectangles. Definition A measure µ is convex, if for every R ∈ R and every fractional partition R 1 , . . . , R m ∈ R , r 1 , . . . , r m ∈ (0 , 1] of R m � µ ( R ) ≤ r i µ ( R i ) . i =1 Theorem It is not possible to prove more than quadratic lower bounds using convex measures. 8 n 2 for all R ∈ Rect. More precisely, if µ is a convex measure, then µ ( R ) ≤ 9 17