1. Complexity Theory Lecture 7
Give the handouts (course web page)
Lecturer: Moni Naor

2. Recap Last week: Non-Uniform Complexity Classes
Polynomial Time Hierarchy
BPP in hierarchy
If NP µ P/Poly the hierarchy collapses
Approxiamte Counting in the hierarchy
This Week:
Randomized Reductions
#P Completeness of Permanent

3. The classes we discussed
EXP
The classes we discussed
PSPACE #P
PSPACE:
Complete problems: TQBF, 2-person games, generalized geography, …
3rd level:
Complete problems: VC dimension for succinct sets
Containment: Approximate #P…
2nd level:
Complete problems: Min DNF, Succinct Set Cover
Containment: Min Circuit, BPP…
1st level:
Complete problems: SAT, UNSAT,…lots…
Containment: factoring, graph isomorphism… PH
Σ3P
Π3P
Δ3P
Σ2P
Π2P
Δ2P NP
coNP
BPP P

4. Approximate Counting is in the Hierarchy
Theorem: For any f 2 #P there is a g 2 Δ3P such that on input x and  g(x, ) outputs a (1+ ) approximation to f(x)
(1-)g(x, ) · f(x) · (1+)g(x, )
Proof: key point – method for proving that sets are of certain size
The set in question:
A = Accept(M(x)) µ {0,1}m
A function f is in #P if there exists a NTM M
running in polynomial time where M(x) has f(x) accepting paths for all x 2 {0,1}n

5. Methods for proving sizes of sets
Let H be family of hash functions where for h 2 H h:{0,1}m  {0,1}ℓ
if h 2 H is 1-1 on A then clearly |A| · 2ℓ
a bit difficult to expect full uniqueness
Relaxed property:
if there are h1, h2, … hℓ 2 H are such that
for all x 2 A there is an 1 · i · l where for all x’ 2 A\{x} we have hi(x)  hi(x’)
then we know |A| < ℓ 2ℓ
Claim: for a pair-wise independent family H, if |A|· 2ℓ-1 , then there exist h1, h2, … hℓ 2 H with the no collisions property. Denote with UH(A,ℓ).
Claim: for A=Accept(M) we have UH(A,ℓ) 2 Σ2P
Can eliminate the quantifier on i by explicitly going over all possibilities!
Conclusion: with oracles calls to UH(A,ℓ) can find smallest ℓ for which UH(A,ℓ) holds
Yields a 2ℓ approximation for f(x)
Can use amplification of f(x) by running M k times (from accepting positions).
New machine M will have fk(x) by accepting paths
No collisions property

6. Pair-wise Independent Hash Functions
A family of hash functions
H={h|h:{0,1}m  {0,1}ℓ }
is pair-wise independent if: for all x ≠ y, for random h 2RH the random variables h(x), h(y) are uniformly distributed and independent
In particular: Pr[h(x)=h(y)]= 1/2ℓ
Already saw the approximate version for distributed equality testing
Pr[h(x)=h(y)]· 
Constructing H:
Construction based on matrix multiplication.
Treat x as a m £ 1 vector.
The function is defined by a random m£ ℓ matrix over GF[2].
H={hA|A 2 GF[2]m£ ℓ and hA(x)=Ax }
Nice property: local computation

7. Proof of No Collisions Claim
For any x 2 A and any x’ 2 A\{x} if h is chosen at random from H
Prh[h(x)=h(x’)]= 1/2ℓ
By the Union bound
Prh[h(x) is unique] ¸ |A| ¢ 1/2ℓ ¸ 1/2
Hitting set: Construct h1, h2, … hℓ one by one, each time covering at least ½ of elements in A that have not been unique so far
Let A’ be the set not covered so far
For any x 2 A’ and any x’ 2 A\{x} for random h 2R H
Prh[h(x) is unique] ¸ |A| ¢ 1/2ℓ ¸ ½.
Note: this is the full A

8. Checking No Collisions Claims
Given h1, h2, … hℓ and A (in the form of Accept(M)) how to check the no collisions claim:
First observation:
8x 2 A 9 i 2 {1,…, ℓ} 8x’ 2 A\{x}: hi(x)  hi(x’)
this is in P3P
Quantification over i‘s is limited: can change order and replace with explicit enumeration
8x,x1,x2,…,xℓ 2A if x {x1,x2,…,xℓ} then h1(x)h1(x1)Ç h2(x)h2(x2) Ç …Ç hℓ(x)hℓ(xℓ)
this is in P1P=CoNP

9. Second Proof of No Collisions Claim
For any x 2 A and any x’ 2 A\{x} if h is chosen at random from H
Prh[h(x)=h(x’)]= 1/2ℓ
By the Union bound
Prh[h(x) is unique] ¸ |A| ¢ 1/2ℓ ¸ ½.
Choose h1, h2, … hℓ at once.
For every x 2 A for each i Pr[hi(x) is unique] ¸ ½.
Probability that at least one i is unique for x is at least 1-1/2ℓ
With probability at least ½ choice is good for all x 2 A.
Conclusions:
There is a choice that is good for all x 2 A.
Approximate Counting is in BPPNP
Choose h1, h2, … hℓ and check the no collisions claim

10. Approximate Counting and Uniform Generation
For self reducible problems:
Exact counting implies exact uniform generation of witnesses/solutions
Choose to set x1=1 with probability proportional to #(1,x2,…xn)/#(x1,x2,…xn)
Approximate counting: almost uniform generation of solutions
Accumulated error is multiplicative (1+p(n))
Theorem: if P=NP then given Boolean circuit C possible to sample almost uniformly at random satisfying assignment to C
Reverse is also true: from uniform sampling to approximate counting

11. From Uniform Generation to Approximate Counting
Reverse is also true: from uniform sampling to approximate counting
For #SAT we know
#(x1,x2,…xn) = #(1,x2,…xn) + #(0,x2,…xn)
To get an  approximation: generate n2/2 assignments
Learn the ration #(1,x2,…xn) and #(0,x2,…xn)
Conclusion: Uniform Generation to Approximate Counting are the same for self reducible problems
Spun a whole industry of Markov Chain/Monte Carlo algorithms
Notable successes:
Approximating the volume of a convex body
Approximating the permanent

12. More on BPP in the Hierarchy Simultaneous Provers
To prove BPP ½ S2P we considered a game between  and  players
 player wants to prove x  L and  player wants to prove x  L
In S2P setting first  makes a move and then  player (based on ’s move)
What if both players have to move simultaneously?
Want strategy where
 player wins whenever x  L even if  player makes move after  player
 player wins whenever x  L even if  player makes move after  player
Homework:
Phrase the above requirements in terms of a polynomial time relation R(x,y,z)
Show that for all L  BPP such a strategy exists
Canetti’s class S_2^P

13. Homework problem discussion: low memory device
A low memory device is given a stream of a1, a2 … an and then b1, b2 … bn where ai, bi 2 {0,1}ℓ and has to determine whether the two multisets are equal.
whether there is a a permutation of the a' s that gives the b's
Solution: fix a finite field F (larger than 2ℓ) and choose random z 2 F
Compute (as you go) i=1n (z-ai) and i=1n (z-bi)
if the two values are not equal – then two streams are not equal
if the two values are equal – then two streams are equal –except with probability at most n/|F|
Can get slightly better probability using -bias probability spaces
Degree n poly in z

14. Application: low memory verification of satisfiability
Suppose that there is a 3-CNF  a 3-CNF and a low memory device wishes to verify that it is satisfiable.
Device has access to : can verify whether the jth clause of  is a given one
Prover: To convince the device that a given assignment satisfies  :
Send first in order of clauses for each clause (x1 Ç x2Ç x3) the value of the assignment on x1, x2, x3
Send in order of variables for each variable xi :
(number of times appears in , value of the assignment on xi )
Verifier:
verify that clauses are correct and assignment satisfies each clause. Problem: how to check consistency between occurrences of xi
translate into two streams of ai and bi
each occurrence of xi in first part to ai =(i, xi )
each occurrence in second part to bi =(i, xi ) (repeat for number of occurrences in )
Check equality of streams

15. Problems with unique solutions
An instance of SAT may have many satisfying assignments
Does the hardness of SAT stem from the multiple assignments
Maybe it is harder to walk towards the right solution if there are several of them
Promise problem: we are told that the number of satisfying assignments is 0 or 1. Can we determine which is the case?
Not responsible if more than a single assignment

16. Randomized reduction to unique solutions Valiant-Vazirani
Theorem: there is a randomized poly-time procedure that given a 3-CNF formula
(x1, x2, …, xn)
outputs a 3-CNF formula ’ such that:
if  is not satisfiable, then ’ is not satisfiable
if  is satisfiable, then with probability at least 1/(8n):
’ has exactly one satisfying assignment
Idea of procedure: project the assignments via hashing and look for the event “only one satisfying assignment hashed to 0k”
k should be log number of satisfying assumptions
want to express it uniquely

17. Expressing uniqueness
Want to use hashing via matrix multiplication
Need to express the statement
(x1, x2, …, xn) (y1, y2, …, yn)=0 over GF[2]
with a unique satisfying assignment
Claim: for each Y=(y1, y2, …, yn) there exists a 3-CNF formula θY on x1, x2, …, xn and additional variables such that:
| θY | = O(n)
θY is satisfiable iff an even number of variables in {xi yi}i=1n are 1
for each setting of the xi variables, this satisfying assignment is unique
Proof: scan from left to right computing the inner product so far
Introduce z0, z1, …, zn and add clause(s) for i=1, …, n
(zi = zi-1 ) if yi=0
(zi = zi-1 © xi ) if yi=1
and (z0 =1) and (zn =1)
Not surprising: can do for any function in P using circuit simulation

18. Prk{0,1,2,…,n-1}[2k-2 ≤ |T| ≤ 2k-1] ≥ 1/n
The Reduction
Procedure:
Choose random k 2R {0,1,2,…,n-1}
for i = 1, 2, …, k
pick random subset Yi 2R {0,1}^n
output ’ = k =   θY1  θY2    θYk
Let T be the set of satisfying assignments for 
Claim: if |T| > 0, then
Prk{0,1,2,…,n-1}[2k-2 ≤ |T| ≤ 2k-1] ≥ 1/n
Good guess

19. Pr[ k has exactly one satisfying assignment] ¸ |T| (½)k+1 ¸ 1/8
Unique Solutions
Denote (Y1 X, Y2 X,  YKX) with h(X)
Claim: if 2k-2 ≤ |T| ≤ 2k-1(good guess), then the probability k has exactly one satisfying assignment is at least 1/8
Fix X  T. Let event Ax be
“h(X)=0 and for all X’  T\{X}: h(X’) ≠ 0”
Pr[h(X)= 0] = (½)k.
For all X’  T\{X}: Pr[h(X’) = 0| h(X) = 0] = (½)k
Pr[8 X’  T\{X} h(X’) ≠ 0| h(X)] ¸ 1-|T|/2k ¸ ½.
Pr[Ak] ¸ (½)k+1
Since events Ax are disjoint from each other
Pr[ k has exactly one satisfying assignment] ¸ |T| (½)k+1 ¸ 1/8
At most 2k-1
At least 2k-2

20. No RP Algorithms for Unique SAT unless…
Since the good guess event is independent of the choice of Y1, Y2, …, Yk we get the probability of success is at least 1/(8n).
Corollary: if there is an RP algorithm for deciding Unique SAT, then NP=RP
Proof:
If there is a an algorithm for deciding Unique SAT then there one for finding the satisfiable assignment if it is unique
Self reducibility
Run the reduction several times and then run the unique sat finding algorithm
Can make no errors if not satisfiable

21. Pr[set with minimum weight is unique in S] ¸ ½.
Alternative Proof
Isolation Lemma
Let (X,S) be a set system
Each s 2 S is a subset of ground set X
|X|=m
Let w: X  {1,…, 2m} be a positive weight function chosen uniformly and independently at random in {1,…, 2m}
Then
Pr[set with minimum weight is unique in S] ¸ ½.

22. Proof of Isolation Lemma
For each element i 2 X and weights for all elements in X\{i}
w1, w2, …, wi-1, wi+1 …, wm
can consider critical value t2Z for w(i):
t may be negative or larger than 2m
If w(i) < t, then minimum weight set must contain i
If w(i) > t, then minimum weight set does not contain i
Pr[w(i)=t] · 1/2m
Claim: If no i 2 X is at critical value (w.r.t. other weight), then minimum is unique
Proof: If two different sets s1 and s2 are the minimum weight sets then there is and i 2 s1 and i s2 (or vice versa).
Pr[some w(i) is at critical value] · m ¢ 1/2m · ½.

23. Applications of the Isolation Lemma
Minimum weight matching: given a graph G=(V,E) with m edges assign random weights to edges from {1,…, 2m} .
With probability at least ½ the minimum weight perfect matching is unique
Let X=E and S={U ½ E| U is a perfect matching} and apply the Isolation Lemma
Homework: use the Isolation Lemma to show another reduction to unique sat
You may find it most useful to consider the max clique problem

24. Reductions between functions
Reduction from function problem A to function problem B
two efficiently computable functions Reduction and ReconstructionIn functions, reducing one problem A to another B involves
A mapping of the instances x of A to instances y of B
A reconstruction process: answering A(x) from the answer to B(y)
Reduction x y A B
Reconstruction
A(x)
B(y)

25. Reductions and #P
A function f is in #P if there exists a NTM M running in polynomial time where M(x) has f(x) accepting paths for all x 2 {0,1}n
problem f is #P-complete if
f is in #P
every problem in #P is in FPf
Parsimonious Reductions:
For counting problems we call a reduction parsimonious if the reconstruction is the identity - preserves the number of solutions
many standard NP-completeness reductions are parsimonious
therefore: if #SAT is #P-complete we get lots of #P-complete problems
par·si·mo·ni·ous     adj.
Excessively sparing or frugal.
More expressive term: witness preserving reductions

26. #SAT Theorem: #SAT is #P-complete. Proof:
#SAT: given a 3-CNF  how many satisfying assignments to  are there?
Theorem: #SAT is #P-complete.
Proof:
clearly in #P: (,x)  R  x satisfies 
take any f  #P defined by relation R  P
Essentially Cook-Levin Reduction, but let’s look inside

27. #P-Completeness of #SAT
…x… …y… C
f(x) =|{y : (x, y)  R}|
Circuit for R
1 iff (x, y)  R
Add new variables z, produce #-CNF  such that
z (x, y, z) = 1  C(x, y) = 1
For (x, y) such that C(x, y) = 1 this z is unique
Can express for each gate using 3-CNF that zi is the correct computation
hardwire x
# satisfying assignments = |{y : (x, y)  R}|
Homework: show that #Clique is #P-Complete

28. Relationship to other classes
To compare to classes of decision problems, usually consider P#P
Easy connections:
NP, coNP  P#P
P#P  PSPACE
Toda’s Theorem: PH  P#P.
So approximate #P  PH  P#P
Big surprise: counting can express iterating quantifiers

29. Counting vs. Deciding
Question: are problem #P complete because they require finding NP witnesses?
or is the counting difficult by itself?
Already saw one counter-example: #DNF
But not really convincing given tight connection to CNF

30. Bipartite Matching
Definition: Let G = (U, V, E) be a bipartite graph with |U| = |V|
a perfect matching in G is a subset M  E that touches every node exactly once
Long history as motivating problem in Complexity Theory
Edmonds, Edmonds-Karp

31. Bipartite Matchings
#MATCHING: given a bipartite graph G = (U, V, E) how many perfect matchings does it have?
Theorem (Valiant): #MATCHING is #P-complete.
But: can find a perfect matching in polynomial time!
No seemingly related problem that is NP-Complete
counting itself must be difficult
The fact that computing #MATCHING is NP-Hard is surprising in itself. The #P completeness is an extra bonus

32. The Permanent The permanent of a matrix A is defined as:
per(A) = ΣπΠiAi, π(i)
Compare to the determinant of a matrix A
det(A) = Σπ sgn(π) ΠiAi, π(i)
# of perfect matchings in a bipartite graph G is exactly the permanent of G’s adjacency matrix AG
a perfect matching defines a permutation that contributes 1 to the sum

33. det(A) = Σπ sgn(π) ΠiAi, π(i)
The Permanent
Valiant’s Theorem says that the permanent of {0,1} matrices is #P-complete
permanent has many nice properties that make it a favorite of complexity theory
Contrast permanent (hard)
per(A) = ΣπΠiAi, π(i)
to determinant (easy):
det(A) = Σπ sgn(π) ΠiAi, π(i)
As a result computing per(A) mod 2 is easy
Important difference between the two:
det(A B) = det(A) det(B) for all matrices
but per(A B) ≠ per(A) per(B) in general
Homework: show

34. Cycle Covers Bipartite graph G = (U, V, E)
Adjacency square matrix AG
Corresponding directed graph with self loops
Cycle cover in a directed graph: collection of node disjoint cycles that covers all the nodes
One-to-one correspondence between matchings and cycle covers
Can think of representation of permutation in two ways
Explicit enumeration
Cycle decomposition
Number of perfect matching = Number of cycle covers in corresponding graph
Weighted graph – the permanent of general matrices
Sum over all cycle covers
For each cycle cover take the product of the weights
Negative weights can zero out positive ones
Cycle cover 0
Green edge: weight -1
Black edge: weight 1
If this gadget appears as a subgraph – zeroes out the whole cycle cover

35. Proof Structure
Show the hardness of the permanent problem for general integer matrices
Build a series of gadgets (weighted graphs)
For a given a 3CNF formula  with n variables and m clauses
map each satisfying assignment into cycle covers of weight 4m of weight
non-satisfying assignments yield cycle covers whose total contribution is 0
Reduce the computation to {0,1} matrices

36. The Clause Gadget All edge weights are all 1
Claim: All cycle covers exclude at least one external edge
Any excluded subset of external edges corresponds to one cycle cover
The inner node will not be connected to any other node
Outer nodes will be connected

37. Literal Gadget … All edge weights are 1 Two possible cycle covers:
Red: the literal is false
Black: the literal is true
The gray nodes are the only ones connected to other parts …
One edge per clause

38. Xor Gadget Edge weight not specified are 1 Claim:
Exactly one of {(u,u’), (v,v’)} is used for non-zero contribution to the cycle covers
If Exactly one of {(u,u’), (v,v’)} is used the contribution is 4
Need to check permanent of 4 submatrices -1 u’ u -1 2 -1 v’ 3 v
u,u’,v,v’ appear elsewhere in the graph. Both (u,u’) and (v,v’) are edges.

39. Embedding the Gadgets For each clause make a clause gadget
Identify each external edge with a literal
Connect each external edge with its literal via an Xor gadget
Connect the two literals of variable via an Xor gadget
Claim: total weight of cycle covers is s 4m where s is the number of satisfying assignments
Any assignment that does comply with the restrictions adds zero

40. Simulating integer weights
For small positive weights
Weight 3
For negative weights consider the permanent mod N=2ℓ +1 where N ¸ 4m s
The -1 becomes N-1 = 2ℓ

41. Simulating negative/large weights
Simulating edge weight 2ℓ … …
ℓ such structures
The integer permanent now allows computing mod N permanent which is equal to 4m ¢ # the satisfying assignments

42. References Unique Sat: Leslie Valiant and Vijay Vazirani, 1985
Isolation Lemma: Ketan Mulmuley, Umesh Vazirani and Vijay Vazirani
Motivation: parallel randomized algorithm for matching based on matrix inversion.
Checking with low memory devices:
Blum and Kannan, Lipton
#P-Completeness of the Permanent: Valiant, 1979.