Complexity Theory Lecture 2 Lecturer: Moni Naor

Recap of last week Computational Complexity Theory: What, Why and How Overview: Turing Machines, Church-Turing Thesis Diagonalization an the Time-Hierarchy (also Space) Communication Complexity –Definition of protocol –Combinatorial Rectangles –Fooling sets an lower bounds –Connection to Time-Space tradeoffs of Turing Machines –Rank lower bound –Covers an non-determinism This week: Non-deterministic and probabilistic communication complexity Space Complexity

Non-deterministic Communication Complexity A non-deterministic function: a:X  {0,1} maps each element to one of the sets {0}, {1} or {0,1}. When evaluating the function the result should be one of the elements in the subset. Let f:X £ Y with range Z. A non-deterministic protocol for verifying that f(x,y)=z is a binary tree where: –Each internal node v is labeled with a non-deterministic function a v :X  {0,1} or b v :Y  {0,1} –Each leaf is either labeled `accept’ or with `failure’ The inputs (x,y) and the non-deterministic choices define a path from the root to a leaf. If f(x,y)=z then there should be at least one path leading to an `accept’ leaf If f(x,y)≠z then there should be no path leading to an `accept’ leaf The cost of a protocol on input (x,y) is the length of the longest path taken on input (x,y) –The cost of a protocol is the maximum path length –The non-deterministic communication complexity of f for verifying z is the cost of the best protocol. Denote it by N z (f) For Boolean f call N 1 (f) the nondeterministic communication complexity and N 0 (f) co- nondeterministic communication complexity

Example Equality : Alice and Bob each hold x,y 2 {0,1} n –want to decide whether x=y or not. N 0 (Equality)= log n What about N 1 (Equality)? Homework : show that for every Boolean function f and z 2 {0,1} N z (f) ¸ log D(f)

Covers and Non-determinism There is a 1-1 correspondence with a z-cover: a collection of not necessary disjoint combinatorial rectangles in f:X x Y where for each (x,y) such that f(x,y)=z there is a rectangle that cover it. Let C z (f) be the size of the smallest z-cover: A non-deterministic protocol for verifying that f(x,y)=z: Alice: Guess a rectangle R intersecting row x, send name to Bob Bob: verify that R intersects column y and report to Alice Accept only if Bob approves –The number of rectangles is the number of leaves Theorem: N z (f) · log C z (f)+1 and C z (f) · 2 N z (f)

Lower bounds A set S µ X x Y is a fooling set for f if there exists a z 2 Z where –For every (x,y) 2 S, f(x,y)=z –For every distinct (x 1,y 1 ), (x 2,y 2 ) 2 S either f(x 1,y 2 )≠z or f(x 2,y 1 )≠z The fooling set argument is still applicable So N 1 (Equality) ¸ n Homework : for x,y 2 {0,1} n let GT(x,y) be 1 iff x>y. Show that N 1 (GT) and N 0 (GT) are  (n)

Deterministic vs. Nondeterministic Communication Complexity There can be a large gap between D(f) and N z (f), but not for both 0 and 1 : Theorem : for any Boolean f:X x Y  {0,1} we have D(f) · N 0 (f) N 1 (f) Proof : Property: if R 0 is a 0-monochromatic rectangle and R 1 is a 1-monochromatic rectangle, then either: R 0 and R 1 do not intersect in columns, or R 0 and R 1 do not intersect in rows Corollary: if R 0 is a 0-monochromatic rectangle and C 1 is a collection of 1-monochromatic rectangles, then either: R 0 intersects in columns less than half of the rectangles in C 1, or R 0 intersects in rows less than half of the rectangles in C 1

The protocol Use the 0 -cover of size C 0 (f) and 1 -cover of size C 1 (f) to construct a protocol Maintain a decreasing list L of potential 0-monochromatic rectangles containing (x,y) Start with the full 0 -cover At each step: If L is empty declare the result to be 1 Alice: search for a 1-rectangle in the 1 -cover that –Contains row x –Intersects in rows at most half the rectangles in L If found report the name ( log C 1 (f) bits) and update L If not found: Bob: search for a 1-rectangle in the 1 -cover that –Contains column y –Intersects in columns at most half the rectangles in L If found report the name and update L If not found declare the result to be 0 Row intersectioncolumn intersection 1 000

Proof Lemma : the protocol is correct If f(x,y)=0 then the 0 -rectangle containing (x,y) is never deleted from L If f(x,y)=1 then the 1 -rectangle containing (x,y) can always be used by either Alice or Bob (from corollary) Lemma : the complexity of the protocol is at most log C 1 (f) log C 0 (f) Each iteration L shrinks by ½ - at most log C 0 (f) rounds Specifying the 1-rectangle log C 1 (f) bits Corollary: if R 0 is a 0-rectangle and C 1 is a collection of 1-rectangles, then either: R 0 intersects in columns less than half of the rectangles in C 1, or R 0 intersects in rows less than half of the rectangles in C 1

The result is tight k- Disjointness : let 1 · k ¿ n/2 let x,y µ {1,…,n} be subsets of size k. let –k-DISJ(x,y)=1 if |x  y| ¸ 1 and –k-DISJ(x,y)=0 otherwise Note |X|=|Y|= ( n k ) N 1 (k-DISJ)=O(log n) N 0 (k-DISJ)=O(k+loglog n) – Using a function h:{1,…,n}  {1,…,2k} which is perfect hash for x [ y Possible to get lower bound of log ( n k ) on D(k-DISJ) using the rank technique

Perfect Hash Functions A family H of functions is (n,k,ℓ) perfect if 8 h 2 H h:{1,…,n}  {1,…, ℓ} 8 S ½ {1,…,n} 9 h 2 H such that h is 1-1 on S A non-deterministic protocol for k- disjointness using an (n,2k,2k) family H of functions Alice : guess h 2 H and send description hoping that h is perfect for x [ y compute h(i) for all i 2 x and send 2k bit vector C x where C x [j]=1 iff 9 i 2 x such that h(i)=j Bob : compute h(i) for all i 2 y and see whether C x and C y intersect Accept only if do not intersect If |x  y| ¸ 1 always reject. If |x  y| ¸ 0 and h is perfect for x [ y – accept Complexity: log|H| + 2k

Existence of Perfect Hash Families: The Probabilistic Method For a fixed S ½ {1,…,n} of size 2k and choose random h:{1,…,n}  {1,…,2k} Pr[h is perfect for S] = k!/(2k) 2k ¼ e -2k Suppose we choose m random h:{1,…,n}  {1,…,2k} Let event A S be `` no h in the collection is perfect for S” Pr[A S ] · (1- e -2k ) m We are interested in showing Pr[ [ S A S ] < 1 This implies that there is a choice of the that is a perfect family of hash function Pr[ [ S A S ] ·  S Pr[A S ] · ( n 2k ) Pr[A S ] The probabilistic method! Union Bound

The parameters Set m= e 2k log ( n 2k ). Then Pr[A S ] · (1- e -2k ) m · (1- e -2k ) e 2k log ( n 2k ) =1/( n 2k ) This means that communication complexity is 2k +log m = 2k +2k +log (k log n) 2 O(k+log log n) Classical constructions: Fredman Komlos Szemeredi More modern one: Pugh

Open Problem: rank lower bound tight? Open Problem : Is there a fixed c>0 such that for all f:X x Y  {0,1} D(f) is O(log c rank(f)) Not true for non-Boolean functions: f(x,y)=  x i y i At least as hard as Inner Product over GF[2]

Probabilistic Communication Complexity Alice an Bob have each, in addition to their inputs, access to random strings of arbitrary length r A and r B (respectively) A probabilistic protocol P over domain X x Y with range Z is a binary tree where –Each internal node v is labeled with either a v (x, r A ) or b v (y, r B ) –Each leaf is labeled with an element z 2 Z Take all probabilities over the choice of r A and r B P computes f with zero error if for all (x,y) Pr[P(x,y)=f(x,y)]=1 P computes f with  error if for all (x,y) Pr[P(x,y)=f(x,y)] ¸ 1-  For Boolean f, P computes f with one-sided  error if for all (x,y) s.t. f(x,y)=0 Pr[P(x,y)=0]=1 and for all (x,y) s.t. f(x,y)=1 Pr[P(x,y)=1] ¸ 1- 

Measuring Probabilistic Communication Complexity For input (x,y) can consider as the cost of protocol P on input (x,y) either worst-case depth average depth over r A and r B Cost of a protocol : maximum cost over all inputs (x,y) The appropriate measure of probabilistic communication complexity: R 0 (f): minimum (over all protocols) of the average cost of a randomized protocol that computes f with zero error. R  (f): minimum (over all protocols) of the worst-case cost of a randomized protocol that computes f with  error. – Makes sense: if 0<  <½  R(f) = R 1/3 (f): R  1 (f): minimum (over all protocols) of the worst-case cost of a randomized protocol that computes f with one-sided  error. – Makes sense: if 0<  <1. R 1 (f) = R ½ 1 (f):

Equality Idea: pick a family of hash functions H={h|h:{0,1} n  {1…m}} such that for all x≠y, for random h 2 R H Pr[(h(x)=h(y)] ·  Protocol: Alice: pick random h 2 R H and send Bob: compare h(x) to h(y) and announce the result This is a one-sided  error protocol with cost log|H|+ log m Constructing H: Fact: over any two polynomials of degree d agree on at most d points Fix prime q such that n 2 · q · 2n 2 map x to a polynomial W x of degree d=n/log q over GF[q] H={h z |z 2 GF[q]} and h z (x)=W x (z)  = d/q= n/q log q · 1/n log n

Relationship between the measures Error reduction (Boolean): for one-sided errors: k repetitions reduces to  k hence R  k 1 (f) · kR  1 (f) for two-sided errors: k repetitions and taking majority reduces the error using Chernoff bounds Derandomization: R(f) = R 1/3 (f) 2  (log D(f))  General Idea: find a small collection of assignments to where the protocol behaves similarly. Problem: to work for all pairs of inputs need to repeat ~n times Instead: jointly evaluate for each leaf ℓ the probability of reaching it, on the given input: P ℓ [x,y] = P ℓ A [x|Bob follows the path] ¢ P ℓ B [y|Alice follows the path] Chernoff: if Pr[x i ] = 1] =1/2 -  then Pr[  i=1 k x i > k/2] · e- 2  2 k Alice computes and sends. Accuracy: log R(f) bits Bob computes

Public coins model What if Alice and Bob have access to a joint source of bits. Possible view: distribution over deterministic protocols Let R  pub (f): be the minimum cost of a public coins protocol computing f correctly with probability at least 1-  for any input (x,y) Example: R  pub (Equality)=(log  Theorem : for any Boolean f: R  (f) is R  pub (f)+O(log n + log 1/  Proof : choose t = 8n/  2 assignments to the public string…

Simulating large sample spaces Want to find among all possible public random strings a small collection of strings on which the protocol behave similarly on all inputs Choose m random strings For input (x,y) event A x,y is more than (  +  ) of the m strings fail the protocol Pr[A x,y ] · e -2  2 t < 2 -2n Pr[ [ x,y A S ] ·  S Pr[A S ] < 2 2n 2 -2n =1 Good 1-  Bad  Collection that should resemble probability of success on ALL inputs

Distributional Complexity Let  be a probability distribution on X x Y and  >0. The ( ,  )- distributional complexity of f (D   (f)) is the cost of the best deterministic protocol that is correct on 1-  of the inputs weighted by  Theorem : for any f: R  pub (f)=max  D   (f)) Protocols of depth d Inputs Is the given protocol correct on the given input Von Neumann’s Minimax Theorem: For all matrices M: max v min q p T M q = min q max p p T M q

Discrepancy and distributional complexity For f:X x Y  {0,1} and rectangle R and be a probability distribution  on X x Y let Disc  (R,f)=|Pr[f(x,y)=1 and (x,y) 2 R] -Pr[f(x,y)=0 and (x,y) 2 R]| Disc  (f)=max R Disc  (R,f)=| Theorem : For f:X x Y  {0,1}, distribution  on X x Y and  D   (f) ¸ log(2  / Disc  (f))

Inner Product Let x,y 2 {0,1} n IP(x,y)=  i=1 n x i y i mod 2 Theorem : R(IP) is  (n). more accurately R 1/2-  pub (IP) ¸ n/2 – log(1/  ) Show that Disc uniform (IP) = 2 -n/2. Therefore D  uniform (IP) ¸ n/2-log(1/  And R 1/2-  pub (IP) ¸ n/2 – log(1/  ) Let H(x,y)=(-1) IP(x,y) Claim: ||H||=√2 n H H T = 2 n I n For rectangle S £ T: Disc uniform (S £ T, IP) = ( 1/2 2n )  x 2 S,y 2 T H(x,y) = ( 1/2 2n ) |1 S ¢ H ¢ 1 T | |1 S ¢ H ¢ 1 T | · || 1 S || ¢ || H || ¢ || 1 T || = √ |S| ¢√ 2 n ¢√ |T| · 2 n/2 ¢ 2 n/2 ¢ 2 n/2 ||A|| =max |v|=1 ||A v|| = max{ | eignevalue of AA T } Cauchy-Schwartz

Summary of Communication Complexity Classes For a `robust’ class f –Closed under composition –Example: polylog communication Nf is not equal Co-Nf Pf= Nf Å Co-Nf BPf is not contained in Nf Å Co-Nf

Summary of techniques and ideas Probabilistic method : to show that a combinatorial object exists: choose a random one and show that it satisfies desired properties with Prob>0 –Constructive vs. non-constructive methods Union bound: define a collection of bad events A 1, A 2, … A n Pr[ [ i A i ] ·  i Pr[A i ] · n Pr[A i ] Simulating large sample spaces by small ones Reversing roles via Minimax Theorem