- 77 Views
- Uploaded on
- Presentation posted in: General

Randomized Computation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Randomized Computation

Roni Parshani 025529199

Orly Margalit037616638

Eran Mantzur 028015329

Avi Mintz017629262

Denotation:

- L is Language
- M is probabilistic polynomial time turning machine
Definition:

L RPif M such that

- x L Prob[ M(x) = 1 ]
- x L Prob[ M(x) = 0 ] = 1

The disadvantage of RP (coRP) is when the Input doesnt belong to language (does belong to the language) the machine needs to return a correct answer at all times.

Definition:

- x L L(x) = 1
- x L L(x) = 0

- Proof:
- Given: L RP
- Aim : L NP
LRP

xL M such that more than 50% of y give M(x,y) = 1 y : M(x,y) = 1

xL y M(x,y) = 0

- LNP

Definition:

L coRPif M such that

- x L Prob[ M(x) = 1 ] = 1
- x L Prob[ M(x) = 0 ]
An alternative way to define coRP is

coRP = { : L RP }

- Proof:
- Give: L coRP
- Aim : L co-NP
LcoRP RP NP Lco-NP

P(.) is a polynomial

Definition:

L RP1 if M, p(.) such that

- x L Prob[ M(x,r) = 1 ]
- x L Prob[ M(x,r) = 0 ] = 1

P(.) is a polynomial

Definition:

L RP2 if M, p(.) such that

- x L Prob[ M(x,r) = 1 ] 1 2-p(|x|)
- x L Prob[ M(x,r) = 0 ] = 1

- Aim: RP1 = RP2
RP2 RP1

we can always select a big enough x such that

< 1 2-p(|x|)

RP1 RP2

L RP1 M, p(.) such that

xL Prob[ M(x,r) = 1 ]

we run M(x,r) t(|x|) times:

- If in any of the runs M(x,r) = 1 output is 1
- If in all of the runs M(x,r) = 0 output is 0

Select t(|x|)

Therefore if xL output is 0

If xL the probability of outputting 0 is only if M(x,r) = 0 all t(|x|) times

- ( Prob[M(x,r) = 0] )t(|x|) (1-)t(|x|)
- [1-] 2-p(|x|)

So the probability of outputting 1 is larger than 1- 2- p(|x|)

- L RP2
Conclusion:

- RP1 RP RP2 RP1
Therefore RP1 = RP = RP2

- RP1 RP RP2 RP1

Definition:

L BPP if M such that

- x L Prob[ M(x) = 1 ]
- x L Prob[ M(x) = 1 ] <
In other words:

- x : Prob[ M(x) = L(x) ]

coBPP = { : L BPP }

= { : M : Prob[ M(x) = L(x) ] }

= { : : Prob[ (x) = (x) ] }

= BPP

= 1 M(.)

(M(.) exists iff (.) exists)

Previously we defined stricter and weaker definition for RP, in a similar way we will define for BPP.

Denotation:

- p(.) positive polynomial
- f polynomial time computable function
Definition:

L BPP1 if M, p(.), f such that

- x L Prob[ M(x) = 1 ] f(|x|) +
- x L Prob[ M(x) = 1 ] < f(|x|) -

Proof:

Aim: BPP BPP1

f(|x|) = andp(|x|) = 6

This gives the original definition of BPP.

Proof:

Aim: BPP1 BPP

L BPP1 M such that

xL Prob [ M(x) = 1] f(|x|) +

xL Prob [ M(x) = 1] < f(|x|)

we want to know with Prob >

if 0 p f(|x|) 1/p(|x|)

or iff(|x|) + 1/p(|x|) p 1

Define: M runs M(x) n times, and each M(x) returns

If > f(|x|)M returns YES, else NO

Calculation of n

We run n independent Bernoulli variables with p and

Prob < 2 =

Choose : and

Result: M decides L(M) with Prob >

Denotation:

- p(.) positive polynomial
Definition:

L BPP2 if M, p(.) such that

- x : Prob[ M(x) = L(x) ] 1-2-p(|x|)

Proof:

Aim: BPP BPP2

p(|x|) =

This gives the original definition of BPP.

Proof:

Aim: BPP BPP2

L BPP M : x Prob[ M(x) = L(x) ]

Define: M runs M(x) n times, and each M(x) returns

If > M returns YES, else NO

We know : Exp[M(x)] > xL

Exp[M(x)] < x L

Chernoffs Equation :

Let {X1 , X2 , , Xn} be a set of independent Bernoulli variables with the same expectations p,and : 0< p(p-1)

Then

Prob

From Chernoffs equation :

Prob[|M(x) Exp[M(x)]| ]

But if |M(x) Exp[M(x)]|

then M returns a correct answer

Prob[M(x)= L(x) ]

polynomial P(x) we choose n such that

Prob[M(x) = L(x) ]

L BPP2

Proof:

L RP if M such that

- x L Prob[ M(x) = 1 ]
- x L Prob[ M(x) = 0 ] = 1
We previously proved BPP = BPP1

If we place in BPP1 formula with

f(.) and p(.)4

this gives the original definition of RP.

Proof:

L P M such that M(x) = L(x)

- x : Prob[ M(x) = L(x) ] =1
- L BPP

Definition:

L PSPACEif M such that M(x) = L(x)

and p such that M uses p(|x|) space.

(No time restriction)

Definition:

L PP if M such that

- x L Prob[ M(x) = 1 ] >
- x L Prob[ M(x) = 1 ]
In other words

- x : Prob[ M(x) = L(x) ] >

Definition: (reminder)

L PP if M such that

- x : Prob[ M(x) = L(x) ]
Proof:

L PP M, p(.) such that

x: Prob[ M(x,r) = L(x) ] >

and M is polynomial time.

Aim: L PSPACE

- Run M on every single r.
- Count the number of received 1 and 0.
- The correct answer is the greater result.

- By the definition of PP, every L PPthis algorithm will always be correct.
- M(x,r) is polynomial in space
- New algorithm is polynomial in space
- L PSPACE

Claim: PP = PP1

If we have a machine that satisfies PP it also satisfies PP1

(Since PP is stricter then PP1 and demands grater then 1/2 and PP demands only, equal or grater to ) so clearly

Let M be a language in PP1

Motivation

The trick is to build a machine that will shift the answer of M towards the NO direction with a very small probability that is smaller than the smallest probability difference that M could have. So if M is biased towards YES our shift will not harm the direction of the shift. But if there is no bias(or bias towards NO) our shift will give us a bias towards the no answer.

Proof:

Let M be defined as:

M chooses one of two moves.

- With probability return NO
- With probability invoke M

If :

If :

Suppose that is decided by a non deterministic

machine M with a running time that is bounded by the polynomial p(x).

The following machine M then will decide L according to the following definition:

M uses its random coin tosses as a witness to M with only one toss that it does not pass to M. This toss is used to choose its move. One of the two possible moves gets it to the ordinary computation of M with the same input(and the witness is the random input).

The other choice gets it to a computation that always accepts.

Consider string x.

If M doesn't have an accepting computation then the probability that M will answer 1 is exactly 1/2.

On the other hand, if M has at least one accepting computation the probability that M will answer correctly is greater then 1/2.

So we get that:

Meaning and by the

previous claim (PP = PP1) we get that .

We define a probabilistic turning machine which is allowed to reply I Dont Know which will be symbolized by .

Definition:

L ZPP if M such that

- x : Prob[ M(x) = ]
- x : Prob[ M(x) = L(x) or M(x) = ] = 1

Take . Let M be a ZPP machine.

We will build a machine M that decides L according to the definition of RP.

If

then by returning 0 when

we will always answer correctly because in this case

If

the probability of getting the right answer with M is greater then 1/2 since M returns a definite answer with probability greater then 1/2 and Ms definite answers are always correct.

In the same way it can be seen that by defining M(x) as:

we get that

If

then we will get a YES answer from

and hence from M with probability greater then 1/2.

If

then we will get a NO answer from

and hence from M with probability greater

then 1/2.

Definition:

RSPACE (s)=L RPsuch that MRPuses at most s(|x|) space and exp( s(|x|) ) time.

BadRSPACE (s) = RSPACE (s) without time restriction.

Claim: badRSPACE = NSPACE

badRSPACE NSPACE

Let L badRSPACE.

If x L

that means there is at least one witness and the non deterministic machine of NSPACE will choose it.

If x L

that means there are no witnesses at all therefore the non deterministic machine of NSPACE also will not find a solution.

NSPACE badRSPACE

L NSPACE. M is the Non - deterministic Turing machine which decides L in space S(|x|).

If x L

there exists r of length exp(S(|x|), so that M(x,r) = 1,

where r is the non-deterministic guess used by M. Therefore the probability of selecting r so that

M(x,r) = 1 is at least

So if we repeatedly invoke M(x,.) on random rs we can expect that after tries we will see an accepting computation.

So what we want our machine M to do is run M on x and a newly randomly selected r (of length exp(S(|x|))) for about

times and accept iff M accepts in one of these tries.

Problem:

In order to count to

we need a counter that uses space of exp(S(|x|)),

and we only have S(|x|).

Solution:

We will use a randomized counter that will use only S(|x|) space.

We flip k = coins.

if all are heads then stop else go on. The expected num of tries .

But the real counter only needs to count to k and therefore only needs space of

.