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6.3 Probabilistic Turing Transducers

The study of nonprobabilistic computations employed the abstract models of deterministic and

nondeterministic Turing transducers. For the study of probabilistic computations we will usesimilar abstract models, called probabilistic Turing transducers.

Informally, a probabilistic Turing transducer is a Turing transducer that views nondeterminismas randomness. Formally, aprobabilistic Turing transduceris a Turing transducer M = whose computations are defined in the following manner.

A sequence C of the moves of M is said to be a computation if the two conditions below hold.

a. C starts at an initial configuration.b. Whenever C is finite, it ends either at an accepting configuration or a nonaccepting

configuration from which no move is possible.

A computation of M is said to be an accepting computation if it ends at an accepting

configuration. Otherwise, the computation is said to be a nonaccepting, or a rejecting,

computation.

By definition, a probabilistic Turing transducer might have both accepting computations and

nonaccepting computations on a given input.

Each computation of a probabilistic Turing transducer is similar to that of a nondeterministic

Turing transducer, the only exception arising upon reaching a configuration from which more

than one move is possible. In such a case, the choice between the possible moves is maderandomly, with an equal probability of each move occurring.

The function that a probabilistic Turing transducer computes and its error probability are defined

similarly to probabilistic programs.Probabilistic Turing machines are defined similarly to

probabilistic Turing transducers.

A probabilistic Turing machine M is said to accept a language L if

a. On input x from L, M has probability 1 - e(x) > 1/2 for an accepting computation.b. On input x not from L, M has probability 1 - e(x) > 1/2 for a nonaccepting computation.

e(x) is said to be the error probability of M. The error probability is said to be boundedif there

exists a constant < 1/2 such that e(x) for all x. M is said to be a bounded-error probabilisticTuring machine if it has bounded-error probability.

Example 6.3.1 Figure6.3.1
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Figure 6.3.1A segment of a probabilistic Turing machine that generates a random number.

gives the transition diagram of a segment M of a probabilistic Turing machine. On input x, M

finds a random number between 0 and v, with probability 1 - (1/2)k. x is assumed to be a string in

{0, 1}* that starts with 1, and v is assumed to be the natural number represented by x. The binary

representation of the random number is stored in the first auxiliary work tape.

M starts each computation by employing M1 for recording the value of k. Then M repeatedlyemploys M2, M3, and M4 for generating a random string y of length |x|, and checking whether yrepresents an integer no greater than v. M terminates its subcomputation successfully if and only

if it finds such a string y within k tries.

M1 records the value of k in unary in the second auxiliary work tape of M. In the first auxiliary

work tape of M, M2 generates a random string y of length |x| over {0, 1}. M3 checks whether x

represents a number greater than the one y represents. M3 performs the checking by simulating asubtraction of the corresponding numbers. M4 erases the string y that is stored on the first

auxiliary work tape.

The number of executed cycles M2, M3, M4 is controlled by the length k of the string 1 1 that isstored on the second auxiliary work tape. At the end of each cycle the string shrinks by one

symbol.

The probability that M2 will generate a string that represents a number between 0 and v is (v +

1)/2|x|

1/2. The probability that such a string will be generated in k cycles is


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The sum of these probabilities is equal to

The probabilistic Turing machine in the following example is, in essence, a probabilistic

pushdown automaton that accepts a non-context-free language. This automaton can be modified

to make exactly n + 2 moves on each input of length n, whereas each one auxiliary-work-tape,

nonprobabilistic Turing machine seems to require more than n + 2 time to recognize thelanguage.

Example 6.3.2 The one auxiliary-work-tape, probabilistic Turing machine M of Figure6.3.2
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Figure

6.3.2

A one auxiliary-work-tape, probabilistic Turing machine M that accepts the language

L = { w | w is in {a1, b1, . . . , ak, bk}*, and w has the same number of ai's as bi's for

each 1 i k }. ( stands for a sequence of j transition rules that move the head of

the auxiliary work tape, j positions to the right. The sequence does not change thecontent of the tape.)

accepts the language L = { w | w is in {a1, b1, . . . , ak, bk}*, and w has the same number of a i 's as

bi 's for each 1 i k }.

M on any given input w starts its computation by choosing randomly some number r between 1

and 2k. It does so by moving from the initial state q0 to the corresponding state qr. In addition, M

writes Z0 on its auxiliary work tape. Then M moves its auxiliary work-tape head ripositions to

the right for each symbol ai that it reads, and ripositions to the left for each symbol bi that it

reads.

At the end of each computation, the auxiliary work-tape head is located (n(a1) - n(b1))r + (n(a2) -

n(b2))r2

+ + (n(ak) - n(bk))rk

positions to the right of Z0, where n(c) denotes the number of times

the symbol c appears in w. If the head is located on Z0, then the input is accepted. Otherwise, the

input is rejected.

By construction, M accepts each input w from the language L. Alternatively, M might also

accept some strings not in L with probability e(x) = < 1/2, where = (k - 1)/(2k).

The equality = (k - 1)/(2k) holds because if w is not in L then n(ai) - n(bi) 0 for at least one i.In such a case, the equation (n(a1) - n(b1))r + + (n(ak) - n(bk))r

k= 0 can be satisfied by at most k


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- 1 nonzero values of r. However, there are 2k possible assignments for r. As a result the

probability that r will get a value that satisfies the equation is no greater than = (k - 1)/(2k).

The bound on the error probability e(x) can be reduced to any desirable value, by allowing r to

be randomly assigned with a value from {1, . . . , k/ }.

M takes no more than T(n) = (2k)kn + 2 moves on input of length n. M can be modified to make

exactly T(n) = n + 2 moves by recording values modulo (2k)k

in the auxiliary work tape. In sucha case, smaller intermediate values are stored in the finite-state control of M.

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6.3 Probabilistic Turing Transducers

The study of nonprobabilistic computations employed the abstract models of deterministic andnondeterministic Turing transducers. For the study of probabilistic computations we will usesimilar abstract models, called probabilistic Turing transducers.

Informally, a probabilistic Turing transducer is a Turing transducer that views nondeterminismas randomness. Formally, aprobabilistic Turing transduceris a Turing transducer M = whose computations are defined in the following manner.

A sequence C of the moves of M is said to be a computation if the two conditions below hold.

a. C starts at an initial configuration.b. Whenever C is finite, it ends either at an accepting configuration or a nonacceptingconfiguration from which no move is possible.

A computation of M is said to be an accepting computation if it ends at an accepting

configuration. Otherwise, the computation is said to be a nonaccepting, or a rejecting,computation.

By definition, a probabilistic Turing transducer might have both accepting computations andnonaccepting computations on a given input.

Each computation of a probabilistic Turing transducer is similar to that of a nondeterministic

Turing transducer, the only exception arising upon reaching a configuration from which morethan one move is possible. In such a case, the choice between the possible moves is maderandomly, with an equal probability of each move occurring.

The function that a probabilistic Turing transducer computes and its error probability are defined

similarly to probabilistic programs.Probabilistic Turing machines are defined similarly toprobabilistic Turing transducers.
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A probabilistic Turing machine M is said to accept a language L if

a. On input x from L, M has probability 1 - e(x) > 1/2 for an accepting computation.b. On input x not from L, M has probability 1 - e(x) > 1/2 for a nonaccepting computation.

e(x) is said to be the error probability of M. The error probability is said to be boundedif thereexists a constant < 1/2 such that e(x) for all x. M is said to be a bounded-error probabilistic

Turing machine if it has bounded-error probability.

Example 6.3.1 Figure6.3.1

Figure 6.3.1A segment of a probabilistic Turing machine that generates a random number.

gives the transition diagram of a segment M of a probabilistic Turing machine. On input x, M

finds a random number between 0 and v, with probability 1 - (1/2)k. x is assumed to be a string in

{0, 1}* that starts with 1, and v is assumed to be the natural number represented by x. The binaryrepresentation of the random number is stored in the first auxiliary work tape.

M starts each computation by employing M1 for recording the value of k. Then M repeatedlyemploys M2, M3, and M4 for generating a random string y of length |x|, and checking whether y

represents an integer no greater than v. M terminates its subcomputation successfully if and only

if it finds such a string y within k tries.

M1 records the value of k in unary in the second auxiliary work tape of M. In the first auxiliary

work tape of M, M2 generates a random string y of length |x| over {0, 1}. M3 checks whether x


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represents a number greater than the one y represents. M3 performs the checking by simulating a

subtraction of the corresponding numbers. M4 erases the string y that is stored on the first

auxiliary work tape.

The number of executed cycles M2, M3, M4 is controlled by the length k of the string 1 1 that is

stored on the second auxiliary work tape. At the end of each cycle the string shrinks by onesymbol.

The probability that M2 will generate a string that represents a number between 0 and v is (v +

1)/2|x|

1/2. The probability that such a string will be generated in k cycles is

The sum of these probabilities is equal to

The probabilistic Turing machine in the following example is, in essence, a probabilisticpushdown automaton that accepts a non-context-free language. This automaton can be modified

to make exactly n + 2 moves on each input of length n, whereas each one auxiliary-work-tape,

nonprobabilistic Turing machine seems to require more than n + 2 time to recognize thelanguage.

Example 6.3.2 The one auxiliary-work-tape, probabilistic Turing machine M of Figure6.3.2


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Figure

6.3.2

A one auxiliary-work-tape, probabilistic Turing machine M that accepts the language

L = { w | w is in {a1, b1, . . . , ak, bk}*, and w has the same number of ai's as bi's for

each 1 i k }. ( stands for a sequence of j transition rules that move the head of

the auxiliary work tape, j positions to the right. The sequence does not change thecontent of the tape.)

accepts the language L = { w | w is in {a1, b1, . . . , ak, bk}*, and w has the same number of a i 's as

bi 's for each 1 i k }.

M on any given input w starts its computation by choosing randomly some number r between 1

and 2k. It does so by moving from the initial state q0 to the corresponding state qr. In addition, M

writes Z0 on its auxiliary work tape. Then M moves its auxiliary work-tape head ripositions to

the right for each symbol ai that it reads, and ripositions to the left for each symbol bi that it

reads.

At the end of each computation, the auxiliary work-tape head is located (n(a1) - n(b1))r + (n(a2) -

n(b2))r2

+ + (n(ak) - n(bk))rk

positions to the right of Z0, where n(c) denotes the number of times

the symbol c appears in w. If the head is located on Z0, then the input is accepted. Otherwise, the

input is rejected.

By construction, M accepts each input w from the language L. Alternatively, M might also

accept some strings not in L with probability e(x) = < 1/2, where = (k - 1)/(2k).

The equality = (k - 1)/(2k) holds because if w is not in L then n(ai) - n(bi) 0 for at least one i.In such a case, the equation (n(a1) - n(b1))r + + (n(ak) - n(bk))r

k= 0 can be satisfied by at most k


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- 1 nonzero values of r. However, there are 2k possible assignments for r. As a result the

probability that r will get a value that satisfies the equation is no greater than = (k - 1)/(2k).

The bound on the error probability e(x) can be reduced to any desirable value, by allowing r to

be randomly assigned with a value from {1, . . . , k/ }.

M takes no more than T(n) = (2k)kn + 2 moves on input of length n. M can be modified to make

exactly T(n) = n + 2 moves by recording values modulo (2k)k

in the auxiliary work tape. In sucha case, smaller intermediate values are stored in the finite-state control of M.

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