COMPLEXITY THEORY

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CSci 5403
COMPLEXITY THEORY
LECTURE X CAUTION MAY BE UNPREDICTABLE
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COMMUNICATION COMPLEXITY
y 2 0,1n
x 2 0,1n
X Y ?
How many bits do Alice and Bob need to send?
Deterministic worst case
X n
Pick random prime r 2 p1,,p2n
r, X mod r
XY mod r
O(log n) bits
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Two cases for x vs y
xy Then Prrxy mod r 1.
x?y imagine xy mod ri, for r1,rm ½ p1,p2n
Claim. m lt n.
Proof. Suppose not. Then ?i ri gt 2n, and by
Chinese Remainder Theorem xy.
Thus Prrxy mod r ? ½.
We can amplify the certainty by iterating k times
If we find r such that x? y mod r, then x?y
Else, say xy. Prr1rkx?y Alice says xy
2-k
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CHECKING LINEAR ALGEBRA
Alice claims that AB C mod p. Can we check?
Best deterministic algorithms O(n2.38) time
Pick random r 2 0,1n. Check A(Br) Cr mod p.
Claim. PrrA(Br)Cr mod p AB ? C mod p ½.
Proof. Let AB ? C. Then D AB-C ? 0.
Fix r1rn-1 and wlog let Di,n ? 0.
Then Pr?jltn Di,jrj rnDi,n mod p ½.
Time for check O(n2).
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RP
Definition. A2RP if there is a polynomial time
NTM such that
1. M accepts A
2. If x 2 A, at least ½ the paths accept
Alternatively x 2 A ) At least half of the paths
of M(x) accept X ? A ) None of the paths
for M(x) accept
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Definition. A probabilistic TM makes all of
its nondeterministic choices by flipping a coin.
RP A exists polytime probabilistic TM M
x 2 A ) PrM(x) accepts ½ x ? A
) PrM(x) accepts 0.
coRP A exists polytime probabilistic TM
M x2 A ) PrM(x) accepts 1 x? A )
PrM(x) accepts lt ½.
BPP
A exists polytime probabilistic TM M
x 2 A ) PrM(x) accepts 2/3 x ? A )
PrM(x) accepts 1/3.
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Definition (alternative)
A 2 RP iff there exists p(n)-time TM M x 2
A ) PrrM(x,r) 1 ½. x ? A ) PrrM(x,r)
0 for uniform r 2 0,1p(n).
Proof.
The bits of r describe a path in the NTM M.
Note. By definition, RP µ NP. More about BPP
and other complexity classes next week.
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PRIMALITY TESTING
Let PRIMES p p is prime.
(nb, not FACTOR (N,k) N has prime factor gt
k)
PRIME-TEST(n) pick random a lt n. if (gcd(a,n)
gt 1) return not prime. if (L(a,n) ? J(a,n))
return not prime. else return probably prime.
Where L(a,n) a(n-1)/2 mod n 1 if a 2 QRn,
-1 o.w. J(a,?i pi) ?i L(a,pi) (computed w/o
p1pn)
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If n 2 PRIMES, then L(a,n) J(a,n).
If n ? PRIMES, a L(a,n) J(a,n) ½.
So PrPRIME-TEST(n)YES n 2 PRIMES
1 PrPRIME-TEST(n)YES n ? PRIMES ½ Thus
Primes 2 coRP µ BPP.
Note AKS 2002 PRIMES 2 P gives O(n6)-time
deterministic algorithm for PRIMES.
The fastest known primality tests are randomized.
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POLYNOMIAL IDENTITY TESTING
Let P(x1,,xm) and Q(x1,,xm) be polynomials
given by arithmetic circuits with ,-, gates
IDENTP (P,Q) 8 X, P(X) Q(X)
Claim. IDENTP P ZEROP (P) 8X P(X)0
Note. There is no known poly-time deterministic
algorithm for identity testing.
ZERO-TEST(P) pick random r1,,rm 2
1,,2n1 if P(r1,,rm) ? 0 return not
zero. else return probably zero.
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Theorem (Schwartz-Zippel). Let P(x1,,xn) ? 0
Have degree d over field F. Then for any
finite S ½ F, PrP(r1,,rn) 0 r1,,rn 2 S
d/S.
Proof. By induction on n.
n1 P(x)?0 has degree d, so at most d roots.
thus PrrP(r)0 d/S.
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ngt1 Write P(x1,,xn) with x1 factored out
Since P?0, for some i, pi ? 0. take max i.
Pick r2,,rn 2R S, then by induction
Prpi(r2,,rn) 0 (d-i)/S.
Let Zi , pi(r2,,rn)0
If pi(r2,,rn) ? 0, then P(x1,r2,,rn) ? 0 and
has degree i, so Pr P(r1,rn) 0 Zi i/S
Thus Pr P(r1,,rn) 0 i/S (d-i)/S
d/S.
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One problem how big could P(r1,,rm) get?
Solution pick random 2m-bit prime q,
check P(r1,,rm) 0 mod q.
Claim. PrP(r1,,rm) 0 mod q not (ln 2)
m2 /2m-1
Proof Number of prime divisors of P(r1,,rm)
m2m. Number of 2m-bit primes 22m/(2m ln 2).
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Define two CNFs to be strongly equivalent if for
every assignment the number of true clauses is
the same for both formulas.
SECNF (?,?) ? and ? are strongly equivalent
Lemma. pOR(x1,,xk) 1 ?j (1-xj) x1 Ç Ç xk
Lemma. Let CNF ? C1 Æ C2 Æ Æ Cm. The
polynomial p?(x1,,xm) ?i pOR(Ci) has
degree at most n and counts the number of true
clauses.
Theorem. SECNF 2 coRP.
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Definition. A 2 ZPP if there exists NTM M so
that 1. x 2 A ) M accepts x 2. x ? A ) M rejects
x 3. The expected running time of M(x) is
poly(x).
Theorem. ZPP RP Å coRP.
Proof. A 2 RP \ coRP means 9 M, M M(x)
accepts ) x 2 A, PrM(x) accepts ½. M(x)
rejects ) x ? A, PrM(x) rejects
½. while(true) run M(x). If it accepts,
ACCEPT. run M(x). If it rejects, REJECT.
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Definition. A 2 ZPP if there exists NTM M so
that 1. x 2 A ) M accepts x 2. x ? A ) M rejects
x 3. The expected running time of M(x) is
poly(x).
Theorem. ZPP RP Å coRP.
Proof. A 2 ZPP means exists M with expected
running time p(n) that is always correct. Run
M(x) for 2p(n) steps. If M accepts,
accept. else reject. Praccept x 2 A ½
Praccept x ? A 0.
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