Download Introduction to the Theory of Computation Chapter 10.2

Introduction to the Theory of Computation
Chapter 10.2
Takagi Lab. M1
Sho Nishimura
Outline
• Probabilistic algorithms,
probabilistic Turing machines,
and complexity class BPP
• Example of probabilistic algorithms :
primality test
• Example of probabilistic algorithms :
equivalence of read-once branching programs
Probabilistic algorithms
• A probabilistic algorithm:
designed to use the outcome of a random
process
– Some problems seem to be more easily solvable
by probabilistic algorithms
than by deterministic algorithms
A Probabilistic Turing Machine
• Def 10.3: a probabilistic Turing machine 𝑀
– A type of nondeterministic TM in which each
nondeterministic step is called a coin-flip step and
has TWO legal next moves.
• The probability of branch 𝑏:Pr 𝑏 = 2−𝑘
(𝑘: the number of coin-flip steps)
– The probability that 𝑀 accepts the input 𝑤:
Pr 𝑀 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑤 =
Pr[𝑏]
𝑏 𝑖𝑠
𝑎𝑛 𝑎𝑐𝑐𝑒𝑝𝑡𝑖𝑛𝑔 𝑏𝑟𝑎𝑛𝑐ℎ
• Of course, the probability that 𝑀 rejects the input 𝑤
is the complementary event of this
A Probabilistic Turing Machine
A Probabilistic Turing Machine
• accept/reject: error probability 𝜖 0 ≤ 𝜖 <
• “𝑀 recognizes language 𝐴 with error
probability 𝜖”:
1. 𝑤 ∈ 𝐴 → Pr 𝑀 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑤 ≥ 1 − 𝜖
2. 𝑤 ∉ 𝐴 → Pr 𝑀 𝑟𝑒𝑗𝑒𝑐𝑡𝑠 𝑤 ≥ 1 − 𝜖
1
2
Class BPP
• Def 10.4: class BPP
– BPP is the class of languages that are recognized
by probabilistic polynomial Turing machines with
1
an error probability of .
3
Actually, any constant error probability(0 < 𝜖 <
would yield an equivalent definition.
1
)
2
Lemma 10.5:Amplification Lemma
• Lemma 10.5:Amplification Lemma
– 𝜖: a fixed constant strictly between 0 and
1
2
– For any polynomial 𝑝𝑜𝑙𝑦 𝑛 , a probabilistic
polynomial time TM 𝑀1 that operates with error
probability 𝜖 has an equivalent probabilistic
polynomial time TM 𝑀2 that operates with an
error probability of 2−𝑝𝑜𝑙𝑦 𝑛 .
Lemma 10.5:Amplification Lemma
• How to prove Lemma 10.5?
– 𝑀2 simulates 𝑀1 by running 𝑀1 polynomial times
and taking the majority vote of outcomes.
• Error rate decreases exponentially.
Lemma 10.5:Amplification Lemma
(Formal proof)
• 𝑀1 (error prob.:𝜖
1
< ), a polynomial
2
−𝑝𝑜𝑙𝑦 𝑛
→ 𝑀2 (error prob.:2
𝑝𝑜𝑙𝑦(𝑛)
)
– 𝑀2 =“ On input 𝑥,
1. Calculate 𝑘 (describe later)
2. Run 2𝑘 independent simulations of 𝑀1 on input 𝑥
3. If most accept, then accept; otherwise reject.
Lemma 10.5:Amplification Lemma
(Formal proof)
• How to calculate 𝑘(boundary)?
– 𝑆: any sequence of results 𝑀2 might obtain in
Stage 2
• 𝑝𝑆 : probability 𝑆 is obtained
• 𝑐: correct results, 𝑤: wrong results(𝑐 + 𝑤 = 2𝑘)
• 𝑆 is a bad sequence if 𝑐 ≤ 𝑤
– 𝜖𝑥 : the probability 𝑀1 is wrong on input 𝑥
– →If 𝑆 is a bad sequence, then
𝑝𝑠 ≤ 𝜖𝑥 𝑤 1 − 𝜖𝑥 𝑐
Lemma 10.5:Amplification Lemma
(Formal proof)
• How to calculate 𝑘(boundary)?
– If 𝑆 is a bad sequence, then
𝑝𝑠 ≤ 𝜖𝑥 𝑤 1 − 𝜖𝑥
≤ 𝜖𝑤 1 − 𝜖 𝑐
𝑐
1
2
• 𝜖𝑥 ≤ 𝜖 < and 𝑐 ≤ 𝑤
therefore 𝜖𝑥 1 − 𝜖𝑥 ≤ 𝜖(1 − 𝜖)
– Also,
𝑝𝑠 ≤ 𝜖 𝑤 1 − 𝜖
• 𝑘 ≤ 𝑤 and 𝜖 < 1 − 𝜖
𝑐
≤ 𝜖𝑘 1 − 𝜖
𝑘
Lemma 10.5:Amplification Lemma
(Formal proof)
• How to calculate 𝑘(boundary)?
– The number of bad sequences are at most 22𝑘
– Therefore,
Pr 𝑀2 𝑜𝑢𝑡𝑝𝑢𝑡𝑠 𝑖𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑙𝑦 𝑜𝑛 𝑖𝑛𝑝𝑢𝑡 𝑥
=
𝑝𝑠
𝑏𝑎𝑑 𝑆
2𝑘
≤2
𝑘
⋅𝜖 1−𝜖
𝑘
= 4𝜖 1 − 𝜖
𝑘
• This upper bound decreases exponentially in 𝑘
– Because 4𝜖 1 − 𝜖 < 1
– Specific value that allows us to bound the error probability
by 2−𝑡 :
𝑡
𝑘≥
log 2 (4𝜖(1 − 𝜖))
(QED)
Primality
• A prime number ↔ A composite
– Testing whether a integer is prime or not
• A polynomial time algorithm is now known
(“AKS primality test”) but difficult
• A much simpler probabilistic polynomial time algorithm
for primality testing
• A number= ∏factors
– No probabilistic polynomial algorithm for finding
factors is known to exist
• The algorithm is based on
“Fermat’s little theorem”.(Theorem 10.6)
Primality
• Notations
– “𝑥 and 𝑦 are equivalent modulo 𝑝”:
𝑥 ≡ 𝑦(mod 𝑝)
• 𝑥 mod 𝑝: the smallest nonnegative 𝑦 such that
𝑥 ≡ 𝑦(mod 𝑝)
– Every number is equivalent modulo 𝑝 to
some member of the set 𝑍𝑝+ = {0, … , 𝑝 − 1}
• 𝑍𝑝 = {1, … , 𝑝 − 1}
• If 𝑥 ≡ 𝑝 − 1(mod 𝑝), then we may refer 𝑥 as
𝑥 ≡ −1(mod 𝑝)
Theorem 10.6:Fermat’s Little Theorem
• Theorem 10.6:Fermat’s little theorem
– If 𝑝: prime, and 𝑎 ∈ 𝑍𝑝+
– Then 𝑎𝑝−1 ≡ 1(mod 𝑝)
• Example
– 𝑝 = 7, 𝑎 = 2 → 27−1 = 64 ≡ 1(mod 7)
– 𝑝 = 6, 𝑎 = 2 → 26−1 = 32 ≡ 2(mod 6)
Fermat’s Little Theorem, pseudoprimes,
and the Carmichael numbers
• A Fermat test
– If 𝑎𝑝−1 ≡ 1(mod 𝑝), then we say
𝑝 passes the Fermat test at 𝑎
• Some composites pass this test.
– Pseudoprimes
– The Carmichael numbers
Correct definition of
Pseudoprime and the Carmichael numbers
• Pseudoprime
– a composite number that passes a test or
sequence of tests that fail for most composite
numbers
– c.f. Fermat pseudoprime to a base 𝑎(𝑝𝑠𝑝(𝑎))
• A composite 𝑛 such that 𝑎𝑛−1 ≡ 1(mod 𝑛)
• The Carmichael number
– An odd composite 𝑛 which satisfies Fermat’s little
theorem for every choice of 𝑎 which is relatively
prime to 𝑛
Reference: Wolfram Mathworld
http://mathworld.wolfram.com/
Algorithm “PSEUDOPRIME”
• Go through all Fermat tests?
→would require exponential time
• How to reduce the number of tests?
– “If a number is not pseudoprime, it fails at least
half of all tests.”(c.f. Problem 10.16)
• Probabilistic algorithm containing a parameter
𝑘 that determines the error probability
Algorithm “PSEUDOPRIME”
• PSEUDOPRIME = “On input 𝑝:
1. Select 𝑎1 , … , 𝑎𝑘 randomly in 𝑍𝑝+
𝑝−1
2. Compute 𝑎𝑖 mod 𝑝 for each 𝑖
3. If all computed values are 1, then accept;
otherwise, reject”
• Works properly?
– 𝑝 is prime→passes all tests and accept
– 𝑝 is not pseudoprime
→it passes at most half of all tests
1
• Probability it passes each test≤
2
• Probability it passes all random-selected tests≤ 2−𝑘
• Time complexity?: polynomial
Algorithm “PSEUDOPRIME”
• Eg. Check primality of 𝑝 = 17,18
(𝑘 = 4)
– 𝑝 = 17
1. 𝑎1 = 2, 𝑎2 = 3, 𝑎3 = 5, 𝑎4 = 7
2. 𝑎116 ≡ 1, 𝑎216 ≡ 1, 𝑎316 ≡ 1, 𝑎416 ≡ 1(mod 17)
3. Accept
– 𝑝 = 18
1. 𝑎1 = 2, 𝑎2 = 3, 𝑎3 = 5, 𝑎4 = 7
2. 𝑎117 ≡ 14, 𝑎217 ≡ 9, 𝑎317 ≡ 11, 𝑎417 ≡ 13(mod 18)
3. Reject
Algorithm “PSEUDOPRIME”
• Eg. Check primality of
𝑝 = 2821 = 7 × 13 × 31(𝑘 = 3)
1. 𝑎1 = 2, 𝑎2 = 3, 𝑎3 = 5
2. 22820 ≡ 1, 32820 ≡ 1, 52820 ≡ 1(mod 2821)
3. Accept……?
Actually, 2821 is one of the Carmichael numbers.
72820 ≡ 2016, 132820 ≡ 2171,
312820 ≡ 1457(mod 2821)
Improve “PSEUDOPRIME”
• How to deal with the Carmichael numbers?
– Number 1 has exactly two square roots
±1 modulo any prime 𝑝
– For composites, 1 has 4 or more square roots
• Eg.±1, ±8 are the square roots of 1, modulo 21
• If 𝑝 passes the Fermat test at 𝑎, we can obtain
a square root of 1 from 𝑎
– 𝑎𝑝−1 mod 𝑝 = 1
𝑝−1
2
mod 𝑝
Algorithm “PRIME”
• PRIME = “ On input 𝑝,
1. If 𝑝 is even, accept if 𝑝 = 2; otherwise, reject
2. Select 𝑎1 , … , 𝑎𝑘 randomly in 𝑍𝑝+
3. For each 𝑖 from 1 to 𝑘:
𝑝−1
4. Compute 𝑎𝑖
mod 𝑝 and reject if different from 1
5. Let 𝑝 − 1 = 𝑠𝑡 where 𝑠 is odd and 𝑡 = 2ℎ
𝑠⋅20 𝑠⋅21
𝑠⋅2ℎ
𝑎𝑖 , 𝑎𝑖 , … , 𝑎𝑖
6. Compute the sequence
modulo 𝑝
7. If some of the sequence is not 1, find the last element
that is not 1 and reject if it is not -1
8. All tests passed, so accept”
Algorithm “PRIME”
• Example(𝑝 = 2741, 𝑘 = 1)
– 𝑎1 = 2 is selected, 2741 passes Fermat test at 2:
22740 ≡ 1(mod 2741)
– 2741 − 1 = 685 × 22 (𝑠 = 685, 𝑡 = 4, ℎ = 2)
– The sequence:
2685 ≡ 656,
1
685×2
2
≡ 2740 ≡ −1,
2
685×2
2
≡ 1(mod 2741)
1
• The last element that is not 1 is “2685×2 modulo 2741”,
which is −1.
– Therefore accept
Algorithm “PRIME”
• Example(𝑝 = 2821, 𝑘 = 1)
– 𝑎1 = 2 is selected, 2821 passes Fermat test at 2.
– 2821 − 1 = 705 × 22 (𝑠 = 705, 𝑡 = 4, ℎ = 2)
– The sequence:
2705 ≡ 2605,
705×21
2
≡ 1520,
2
705×2
2
≡ 1(mod 2821)
– Therefore reject
Algorithm “PRIME”
• Works correctly?
– If 𝑝 is even, obviously it works
– Theorem 10.7:
If 𝑝 is an odd prime number,
Pr 𝑃𝑅𝐼𝑀𝐸 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑝 = 1
– Theorem 10.8:
If 𝑝 is an odd composite number,
Pr 𝑃𝑅𝐼𝑀𝐸 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑝 ≤ 2−𝑘
• 𝑎𝑖 is a (compositeness) witness
– if the algorithm rejects at either stage 4 or 7, using
𝑎𝑖
Theorem 10.7(proof)
• If 𝑎 were a stage 4 witness, 𝑝 is composite
(Fermat’s little theorem)
• If 𝑎 were a stage 7 witness?
∃𝑏 ∈ 𝑍𝑝+ 𝑠. 𝑡. 𝑏 ≢ 1 mod 𝑝 𝑎𝑛𝑑 𝑏 2 ≡ 1(mod 𝑝)
Therefore 𝑏 2 − 1 ≡ 0(mod 𝑝)
Factor this: (𝑏 + 1)(𝑏 − 1) ≡ 0(mod 𝑝)
Which implies: 𝑏 + 1 𝑏 − 1 = 𝑐𝑝
(𝑐: some positive integer)
– 𝑏 − 1, 𝑏 + 1: strictly between 0 and 𝑝(𝑏 ≢ 1 mod 𝑝 )
– A multiple of prime can’t be expressed as a product of
numbers that are smaller than it is.
– Therefore 𝑝 is composite.
–
–
–
–
The Chinese Remainder Theorem
• The Chinese remainder theorem
– If 𝑝 and 𝑞 are relatively prime,
a one-to-one correspondence between
𝑍𝑝𝑞 and 𝑍𝑝 × 𝑍𝑞 exists
• Each number 𝑟 ∈ 𝑍𝑝𝑞 corresponds to a pair 𝑎, 𝑏
where 𝑎 ∈ 𝑍𝑝 , 𝑏 ∈ 𝑍𝑞 such that
𝑟 ≡ 𝑎 mod 𝑝 , 𝑟 ≡ 𝑏 mod 𝑞
Theorem 10.8
• Theorem 10.8
– If 𝑝 is an odd composite number,
Pr 𝑃𝑅𝐼𝑀𝐸 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑝 ≤ 2−𝑘
Theorem 10.8(proof)
• We show:
if 𝑝 is an odd composite number and 𝑎 is
selected randomly in 𝑍𝑃+ ,
1
Pr 𝑎 𝑖𝑠 𝑎 𝑤𝑖𝑡𝑛𝑒𝑠𝑠 ≥
2
– At least as many witnesses as non-witnesses exist
in 𝑍𝑝+
• By finding a unique witness for each non-witness.
• If above is proved,
Pr 𝑃𝑅𝐼𝑀𝐸 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑝 ≤ 2−𝑘
is obvious.
Theorem 10.8(proof)
• In every non-witness, the sequence(stage 6):
all 1s
/ contains -1 at some position followed by 1s
– Among all non-witnesses of the second kind, find
a non-witness ℎ for which the -1 appears in the
largest position 𝑗 in the sequence.
(position for first element: 0)
•
𝑗
𝑠⋅2
ℎ
≡ −1(mod 𝑝)
Theorem 10.8(proof)
• 𝑝:composite→
𝑝: the power of a prime
/ 𝑝 = 𝑞𝑟 (𝑞, 𝑟: relatively prime)
• According to the Chinese remainder theorem:
some number 𝑡 ∈ 𝑍𝑝 whereby
𝑡 ≡ ℎ mod 𝑞 , 𝑡 ≡ 1(mod 𝑟)
• Therefore
𝑠⋅2𝑗
𝑠⋅2𝑗
𝑡
≡ −1 mod 𝑞 , 𝑡
≡ 1(mod 𝑟)
• Hence 𝑡 is a witness.
–𝑡
𝑠⋅2𝑗
≢ ±1 mod 𝑝 , 𝑡
𝑠⋅2𝑗+1
≡ 1(mod 𝑝)
Theorem 10.8(proof)
• One witness 𝑡 → more witnesses?
– 𝑑𝑡 mod 𝑝 is a unique witness for each nonwitness 𝑑
• Really a witness?: 𝑑 is a non-witness, therefore
𝑗
𝑗+1
𝑠⋅2
𝑠⋅2
𝑑
≡ ±1, 𝑑
≡ 1 mod 𝑝
𝑗
𝑗+1
𝑠⋅2
𝑠⋅2
(𝑑𝑡) ≢ ±1, (𝑑𝑡)
≡ 1 mod 𝑝
• Unique?: 𝑑1 , 𝑑2 are distinct non-witnesses,
𝑑1 𝑡 mod 𝑝 ≠ 𝑑2 𝑡 mod 𝑝
– 𝑡 𝑠⋅2
𝑗+1
≡ 𝑡 ⋅ 𝑡 𝑠⋅2
𝑗+1 −1
≡ 1(mod 𝑝)
if 𝑡𝑑1 ≡ 𝑡𝑑2 (mod 𝑝), then
𝑗+1 −1
𝑗+1 −1
𝑠⋅2
𝑠⋅2
𝑑1 = 𝑡𝑑1 ⋅ 𝑡
mod 𝑝 = 𝑡𝑑2 ⋅ 𝑡
mod 𝑝 = 𝑑2
Theorem 10.8(proof)
• 𝑝:composite→
𝑝: the power of a prime
/ 𝑝 = 𝑞𝑟 (𝑞, 𝑟: relatively prime)
– 𝑝 = 𝑞 𝑒 where 𝑞 is prime and 𝑒 > 1
– Let 𝑡 = 1 + 𝑞 𝑒−1 , and expanding 𝑡 𝑝 using the
binomial theorem,
𝑡 𝑝 = 1 + 𝑞 𝑒−1 𝑝 = 1 + 𝑝𝑞 𝑒−1
+𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑠 𝑜𝑓 ℎ𝑖𝑔ℎ𝑒𝑟 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 𝑞 𝑒−1
– This is equivalent to 1 mod 𝑝
– Therefore 𝑡 is a witness.
(QED)
PRIMES ∈ BPP
• PRIMES={𝑛| 𝑛 is a prime number in binary}
• Theorem 10.9: 𝑃𝑅𝐼𝑀𝐸𝑆 ∈BPP
One-sided error
• The algorithm “PRIME” has one-sided error
– When the algorithm rejects
→ the input is 100% composite.
– When the algorithm accepts
→ there may be incorrect answer
• One-sided error is a common feature in
probabilistic algorithms
– Special complexity class “RP” is designated
Class RP
• Definition 10.10: Class RP
– RP is the class of languages that are recognized by
probabilistic polynomial time Turing machine
where inputs in the language are accepted with a
1
probability of at least and inputs not in the
2
language are rejected with a probability of 1.
• Ex. COMPOSITES ∈ RP
Branching Programs
• A branching program
– Model of computation used in complexity theory
and in certain practical areas(ex. CAD).
– Testing whether two branching programs are
equivalent
• Generally it is coNP-complete
• A certain restriction on the class of branching programs
→a probabilistic polynomial time algorithm for testing
equivalence
Branching Programs
• Definition 10.11: A branching program
– A branching program is a directed acyclicgraph
where all nodes are labeled by variables, except
for two output nodes labeled 0 or 1. The nodes
that are labeled by variables are called query
nodes. Every query node has two outgoing edges,
one labeled 0 and the other labeled 1. Both
output nodes have no outgoing edges. One of the
nodes in a branching program is designated the
start node.
Read-once Branching Programs
• A read-once branching program
– A branching program that can query each variable
at most one time on every directed path from the
start node to an output node
Read-once Branching Programs
Theorem 10.13
• 𝐸𝑄𝑅𝑂𝐵𝑃 ={ 𝐵1 , 𝐵2 |𝐵1 and 𝐵2 are equivalent
read-once branching programs}
• Theorem 10.13
– 𝐸𝑄𝑅𝑂𝐵𝑃 ∈ BPP
Theorem 10.13
• Proof idea
– “Assign random values to the variables 𝑥1 , … , 𝑥𝑚 ,
and accept if 𝐵1 and 𝐵2 agree on the
assignment“?
• If 𝐵1 and 𝐵2 disagree only on a single assignment out of
the 2𝑚 possible assignments?
→this doesn’t work.
– “Randomly select a non-Boolean assignment to
the variables and evaluate 𝐵1 and 𝐵2 in a suitably
defined manner”
Theorem 10.3(proof)
• Assign polynomials over 𝑥1 , … , 𝑥𝑚 to the
nodes/edges of a read-once branching
program 𝐵
– Start node: assign the constant function 1
– A node labeled 𝑥 has been assigned polynomial 𝑝
→outgoing 1-edge: assign the polynomial 𝑥𝑝,
outgoing 0-edge: assign the polynomial 1 − 𝑥 𝑝
– A node: assign the sum of the polynomials of the
incoming edges
– 𝐵 itself: assign
the polynomial of the output node 1
Theorem 10.13(proof)
𝑝=1
𝑝=
1 − 𝑥1
𝑝=
1 − 𝑥1
(1 − 𝑥2 )
𝑝 = 𝑥1
𝑝 = 𝑥1
(1 − 𝑥2 )
𝑝 = 1 − 𝑥1 1 − 𝑥2 𝑥3
+𝑥1 1 − 𝑥2 𝑥3 + 𝑥1 𝑥2
Theorem 10.13(proof)
• ℱ: a finite field with at least 3𝑚 elements
• Algorithm:
𝐷 =“On input 𝐵1 , 𝐵2 :
1. Select elements 𝑎1 , … , 𝑎𝑚 randomly from ℱ
2. Evaluate the assigned polynomials 𝑝1 , 𝑝2 at 𝑎1
through 𝑎𝑚
3. If 𝑝1 𝑎1 , … , 𝑎𝑚 = 𝑝2 𝑎1 , … , 𝑎𝑚 then accept;
otherwise reject
• This runs in polynomial time
• Error probability: at
1
most
3
Theorem 10.13(proof)
• Relation between a read-once branching
program 𝐵 and the assigned polynomial 𝑝?
– For any Boolean assignments to 𝐵’s variables, all
polynomials assigned to its nodes evaluate to
either 1 or 0
• The polynomials that evaluate to 1 are those on the
computation path for the assignment
– Since 𝐵 is read-once, the polynomial is written as
𝑝 = ∑ ∏𝑦𝑖
(𝑦𝑖 = 𝑥𝑖 , 1 − 𝑥𝑖 , 1)
• 𝑦𝑖 = 1: when a path doesn’t contain variable 𝑥𝑖
Theorem 10.13(proof)
• split “𝑦𝑖 = 1” in the polynomial 𝑝:
𝑦𝑖 = 1 = 𝑥𝑖 + (1 − 𝑥𝑖 )
• The end result is an equivalent polynomial 𝑞
that contains a product term for each
assignment on which 𝐵 evaluates to 1
Theorem 10.13(proof)
• 𝐵1 and 𝐵2 are equivalent:
this algorithm always accepts
– They evaluate to 1 on exactly the same
assignments, so 𝑞1 and 𝑞2 are equal.
Therefore 𝑝1 and 𝑝2 are always the same
• 𝐵1 and 𝐵2 are inequivalent:
this algorithm rejects with a probability of at
2
least
3
– Shown by Lemma 10.15
• We need Lemma 10.14 to prove Lemma 10.15
(QED)
Lemma 10.14
• Lemma 10.14
– For every 𝑑 ≥ 0, a degree-𝑑 polynomial 𝑝 on a
single variable 𝑥 either has at most 𝑑 roots, or is
everywhere equal to 0.
Lemma 10.14(proof)
• Induction on 𝑑
– Basis:𝑑 = 0
• A polynomial of degree 0 is constant. If that constant is
not 0, the polynomial has no roots.
– Induction step: Assume true for 𝑑 − 1 and prove
true for 𝑑
• If 𝑝: non-zero polynomial of degree 𝑑 with a root at 𝑎
→the polynomial (𝑥 − 𝑎) divides 𝑝 evenly.
• Then the polynomial 𝑝/(𝑥 − 𝑎) is an non-zero
polynomial of degree 𝑑 − 1, and it has at most 𝑑 − 1
roots by virtue of the induction hypothesis.
(QED)
Lemma 10.15
• Lemma 10.15
– Let ℱ be a finite field with 𝑓 elements and let 𝑝 be
a nonzero polynomial on the variables 𝑥1 through
𝑥𝑚 , where each variable has degree at most 𝑑.
If 𝑎1 through 𝑎𝑚 are selected randomly in ℱ, then
𝑚𝑑
Pr 𝑝 𝑎1 , … , 𝑎𝑚 = 0 ≤
𝑓
Lemma 10.15(proof)
• Induction on 𝑚
– Basis: 𝑚 = 1
By Lemma 10.14, 𝑝 has at most 𝑑 roots, so the
probability that 𝑎1 is one of them is at most 𝑑/𝑓
Lemma 10.15(proof)
• Induction on 𝑚
– Induction step: true for 𝑚 − 1 and proof for 𝑚
• 𝑥1 : one of 𝑝’s variables, then
𝑑
𝑝=
𝑥1
𝑖
× 𝑝𝑖 (𝑥2 , … , 𝑥𝑚 )
𝑖=0
• If 𝑝 𝑎1 , … , 𝑎𝑚 = 0, two possible cases.
1.
2.
All 𝑝𝑖 evaluate to 0
Some 𝑝𝑖 doesn’t evaluate to 0 and
𝑎1 is the root of the single variable polynomial obtained by
evaluating 𝑝0 through 𝑝𝑑 on 𝑎2 through 𝑎𝑚
Lemma 10.15(proof)
• Induction on 𝑚
– Bound the probability of the first case
• 𝑝 is non-zero, so one of 𝑝𝑗 must not be non-zero
• The probability of the first case
≤ 𝑝𝑗 evaluates to 0
• By induction hypothesis,
𝑝𝑗 evaluates to 0 ≤ 𝑚 − 1 𝑑/𝑓
– Bound the probability of the second case
• On the assignment 𝑎2 , … , 𝑎𝑚 , 𝑝 reduces to a non-zero
polynomial in the single variable 𝑥1 .
• The basis of induction shows the probability is at most 𝑑/𝑓
– Therefore the probability that 𝑎1 , … , 𝑎𝑚 is a root of
the polynomial is at most 𝑚𝑑/𝑓
(QED)
Pseudorandom generators
• Here, we assume these algorithms use true
randomness
– True randomness is difficult (or impossible) to obtain
• → pseudorandom generators
– Deterministic algorithms whose output appears random
• Probabilistic algorithms may work equally with
pseudorandom generators
– Proving this is difficult
• Sometimes they may not work well with certain
pseudorandom generators
• Sophisticated pseudorandom generators
– Under the assumption of a one-way function
Summary
• A probabilistic Turing machine
– A type of nondeterministic TM with coin-flip steps
• class BPP
– The complexity class associated with polynomial
probabilistic TMs
• Examples of probabilistic algorithms
– Primality testing
– Checking equivalence of read-once branching
programs
• One-sided error and class RP
Thank you for listening!