# Size of Quantum Finite State Transducers Ruben Agadzanyan, Rusins Freivalds.

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Size of Quantum Finite State Transducers Ruben Agadzanyan, Rusins Freivalds

Outline Introduction Previous results When deterministic transducers are possible Quantum vs. probabilistic transducers

Introduction Probabilistic transducer definition Computing relations Quantum transducer definition

Introduction Transducer definition Finite state transducer (fst) is a tuple T = (Q, Σ 1, Σ 2, V, f, q 0, Q acc, Q rej ), V : Σ 1 x Q → Q a  Σ 1 :

Introduction Transducer definition R  Σ 1 * x Σ 2 * R = {(0 m 1 m,2 m ) : m ≥ 0} Σ 1 = {0,1} Σ 2 = {2} Input: #0 m 1 m \$ Output: 2 m Transducer may accept or reject input

Introduction Transducer types Deterministic (dfst) Probabilistic (pfst) Quantum (qfst)

Introduction Computing relations R  Σ 1 * x Σ 2 * R = {(0 m 1 m,2 m ) : m ≥ 0} For α > 1/2 we say that T computes the relation R with probability α if for all v, whenever (v, w)  R, then T (w|v) ≥ α, and whenever (v, w)  R, then T (w|v)  1 - α 01 α

Introduction Computing relations R  Σ 1 * x Σ 2 * R = {(0 m 1 m,2 m ) : m ≥ 0} For 0 0 such that for all v, whenever (v, w)  R, then T (w|v) ≥ α + ε, but whenever (v, w)  R, then T (w|v)  α - ε. 01 α ε

Introduction Computing relations R  Σ 1 * x Σ 2 * R = {(0 m 1 m,2 m ) : m ≥ 0} We say that T computes the relation R with probability bounded away from ½ if there exists ε > 0 such that for all v, whenever (v, w)  R, then T (w|v) ≥ ½ + ε, but whenever (v, w)  R, then T (w|v)  ½ - ε. 01 ½ ε

Outline Introduction Previous results When deterministic transducers are possible Quantum vs. probabilistic transducers

Previous results Probabilistic transducers are more powerful than the deterministic ones (can compute more relations) Computing relations with quantum and deterministic transducers Computing a relation with probability 2/3

Previous results pfst and qfst more powerful than dfst? For arbitrary ε > 0 the relation R 1 = {(0 m 1 m,2 m ) : m ≥ 0} can be computed by a pfst with probability 1 – ε. can be computed by a qfst with probability 1 – ε. cannot be computed by a dfst.

Previous results other useful relation The relation R 2 = {(w2w, w) : w  {0, 1}* } can be computed by a pfst and qfst with probability 2/3.

Outline Introduction Previous results When deterministic transducers are possible Quantum vs. probabilistic transducers

When deterministic transducers are possible Comparing sizes of probabilistic and deterministic transducers Not a big difference for relation R(0 m 1 m,2 m ) Exponential size difference for relation R(w2w,w), probability of correct answer: 2/3 Relation with exponential size difference and probability: 1-ε

When deterministic fst are possible fst for R k = {(0 m 1 m,2 m ) : 0  m  k} For arbitrary ε > 0 and for arbitrary k the relation R k = {(0 m 1 m,2 m ) : 0  m  k} Can be computed by pfst of size 2k + const with probability 1 – ε For arbitrary dfst computing R k the number of the states is not less than k

When deterministic fst are possible fst for R k’ = {(w2w,w) :  m  k, w  {0, 1} m } The relation R k’ = {(w2w,w) :  m  k, w  {0, 1} m } Can be computed by pfst of size 2k + const with probability 2/3 (can’t be improved) For arbitrary dfst computing R k’ the number of the states is not less than a k where a is a cardinality of the alphabet for w.

When deterministic fst are possible improving probability For arbitrary ε > 0 and k the relation R k’’ = {( code(w) 2 code(w),w) :   m  k, w  {0, 1} m } Can be computed by pfst of size 2k + const with probability 1 - ε For arbitrary dfst computing R k’’ the number of the states is not less than a k where a is a cardinality of the alphabet for w

Outline Introduction Previous results When deterministic transducers are possible Quantum vs. probabilistic transducers

Exponential size difference for relation R(0 m 1 n 2 k,3 m ) Relation which can be computed with an isolated cutpoint, but not with a probability bouded away from 1/2

Quantum vs. probabilistic fst exponential difference in size The relation R s’’ = {(0 m 1 n 2 k,3 m ) : n  k & (m = k V m = n) & m  s & n  s & k  s} Can be computed by qfst of size const with probability 4/7 – ε, ε > 0 For arbitrary pfst computing R s’’ with probability bounded away from ½ the number of the states is not less than a k where a is a cardinality of the alphabet for w

Quantum vs. probabilistic fst qfst with probability bounded away from 1/2? The relation R s’’’ = {(0 m 1 n a,4 k ) : m  s & n  s & (a = 2 → k = m ) & (a = 3 → k = n )} Can be computed by pfst and by qfst of size s + const with an isolated cutpoint, but not with a probability bounded away from ½

Conclusion Comparing transducers by size: probabilistic smaller than deterministic quantum smaller than probabilistic and deterministic

Thank you!