On the State Complexity of Semi-Quantum
Finite Automata
Shenggen Zheng1 (Timothy), Jozef Grusk𝑎1 , Daowen Qi𝑢2
1 Faculty of Informatics, Masaryk University, Czech Republic
2 Department of Computer Science, Sun Yat-sen University, China
arXiv:1307.2499
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1.Introduction
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1.Introduction
 Peter Shor’s algorithm (1994)
A polynomial time algorithm to factor
large numbers on a quantum
computer.
RSA ( public-key encryption)
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1.Introduction
 USA, China, Europe
 Google, D-Wave
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What is quantum computing
 Quantum mechanics + computing
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What is quantum computing
 QM=probability + minus signs
(by Scott Aaronson) Amplitude (complex number)
𝑚𝑒𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡
α
α2
Probability: 𝑝1 + 𝑝2 > 0
Amplitude: 𝛼1 + 𝛼2 = 0
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What is quantum computing
If quantum computers are so great, how
come they have not been built yet?
 They have – and they have proved that 15 = 3 × 5
(with high probability) .
 Perhaps 21 = 3 × 7 will be the next technological
milestone.
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Quantum finite automata
Quantum computer
Quantum finite
automata
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Semi-quantum finite automata
Read only tape head
Classical
Control
C
Unit
Quantum
Memory
C
0 1
1 0
1
0
=
0
1
1/2 1/3 1
1/2
=
1/2 2/3 0
1/2
cos 𝜃
sin 𝜃 1
cos 𝜃
=
− sin 𝜃 cos 𝜃 0
− sin 𝜃
Deterministic transition function: special stochastic matrix
Probabilistic transition function: stochastic matrix
Quantum transition function: unitary matrix
 Two-way finite automata with quantum and classical states
2QCFA (Ambainis, 02’TCS)
 1QCFA ( Zheng, 2011)
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2.State succinctness for promise problem
Promise problem : A = 𝐴𝑦𝑒𝑠, 𝐴𝑛𝑜 , A𝑦𝑒𝑠 ∩ 𝐴𝑛𝑜 = ∅
Language: L = (𝐴𝑦𝑒𝑠, 𝐴𝑛𝑜 ), where 𝐴𝑦𝑒𝑠 ∪ 𝐴𝑛𝑜 = Σ ∗
𝐿
Promise problem: Σ = 0,1, # ,
𝐴𝐸𝑄 (𝑛) = (𝐴𝑦𝑒𝑠 𝑛 , 𝐴𝑛𝑜 (𝑛)), where
𝐴𝑛𝑜
𝐴𝑦𝑒𝑠
𝐴𝑦𝑒𝑠 = {𝑥#𝑦|𝑥 = 𝑦, 𝑥, 𝑦 ∈ {0,1}𝑛 }
𝐴𝑛𝑜 = {𝑥#𝑦|𝑥 ≠ 𝑦, 𝑥, 𝑦 ∈ {0,1}𝑛 , 𝐻 𝑥, 𝑦 = 𝑛/2}
𝐻 𝑥, 𝑦 is the Hamming distance between x and y.
 Deutsch-Jozsa promise problem in query complexity
 In communication complexity (H. Buhrman et al, STOC’98)
Exact quantum communication complexity: 𝑂(log 𝑛)
Exact classical: Ω(𝑛)
 Our result State complexity: 𝑂(𝑛) vs 2Ω(𝑛)
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State succinctness for promise problem
Idea:
α1 + α2 + ⋯
2
=0
…
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State succinctness for promise problem
An exact 1QCFA algorithm: 𝑥 = 𝑥1 ⋯ 𝑥𝑛 , 𝑦 = 𝑦1 ⋯ 𝑦𝑛
𝑥1 ⋯ 𝑥𝑛 # 𝑦1 ⋯ 𝑦𝑛
1
0
⋮
0
𝑈𝑠
1
𝑈𝑓 𝑛
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1
1 1
𝑛 ⋮
1
𝑈1,𝑥1
(−1)
1
1
⋮
𝑛
1
𝑛
𝑥𝑖 +𝑦𝑖
(−1)
𝑖=1
∗
⋮
∗
𝑥1
𝑥
⋯
(−1)𝑥1
1 (−1)𝑥2
⋮
𝑛
(−1)𝑥𝑛
𝑦
⋯
(−1)𝑥1 +𝑦1
1 (−1)𝑥2 +𝑦2
⋮
𝑛
(−1)𝑥𝑛 +𝑦𝑛
If w ∈ 𝐴𝑦𝑒𝑠 , 𝑡ℎ𝑒𝑛 𝑥 = 𝑦,
2
𝑛
1
𝑥
+𝑦
−1 𝑖 𝑖 = 1
𝑛
𝑖=1
If 𝑤 ∈ 𝐴𝑛𝑜 , then H x, y = n/2
2
𝑛
1
−1 𝑥𝑖 +𝑦𝑖 = 0
𝑛
𝑖=1
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State succinctness for promise problem
Theorem 1. The promise problem 𝐴𝐸𝑄 can be solved
exactly by a 1QCFA with 𝑛 quantum basis states and 𝑂 𝑛
classical states, whereas the sizes of the corresponding
1DFA are 2Ω(𝑛) .
 The deterministic communication complexity for 𝐴𝐸𝑄
is at least 0.007n (H. Buhrman et al, STOC’98)
Therefore, the sizes of the corresponding 1DFA are
2Ω(𝑛) (Kushilevitz, Advances in Computers, ‘97).
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3.State succinctness for 𝐿 𝑝
Language 𝐿 𝑝 = 𝑎𝑘𝑝 𝑘 ∈ ℤ+ }.
 1DFA and 1NFA: 𝑝 states
 One-way quantum finite automata (1QFA): 𝑝𝑜𝑙𝑦
1
𝜀
log 𝑝
(Ambainis and Freivalds, FOCS’98)
log 𝑝
1QFA: O( 3 ) (Bertoni et al., TCS ’06)
𝜀
log 2𝑝
1QFA: O(4
) (Ambainis,Nahimovs, TCS ’09)
𝜀
𝛼
𝛼
𝛼
2DFA,2NFA:𝑝1 1 +𝑝2 2 +⋯ + 𝑝𝑠 𝑠 ,where p =
𝛼
𝛼
𝛼
𝑝1 1 𝑝2 2 ⋯ 𝑝𝑠 𝑠
(prime factorization) (Mereghetti, Pighizzini, J.Autom. Lan. Com ‘00)
One-way quantum finite automaton with restart (1QFAR):
1
𝜋
7 states, running time 𝑂( 𝑠𝑖𝑛−2 ( )|𝑤|) (Yakaryilmaz, Cem Say,
DMTCS ‘10)
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𝜀
𝑝
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State succinctness for 𝐿 𝑝
2QCFA, two quantum states, several classical states,
Idea: wheel of fortune, 𝑝 squares,
|𝑞0 =
1
0
,|𝑞1 = , 𝑈𝑝 =
0
1
𝜋
𝑝
𝜋
sin
𝑝
cos
𝜋
𝑝
𝜋
𝑝
𝜋
cos
𝑝
− sin
If 𝑤 ∈ 𝐿 𝑝 , |𝑞 =|𝑞0
If 𝑤 ∉ 𝐿 𝑝 , 𝑤 = 𝑛
|𝑞 = (𝑈𝑝 ) 𝑛 |𝑞0 = cos
𝑛𝜋
|𝑞0
𝑝
+sin
𝑛𝜋
|𝑞1
𝑝
Probability to observe |𝑞1 is 𝑃𝑟 = 𝑠𝑖𝑛2
𝑛𝜋
𝑝
>
4
𝑝2
 Repeat for 𝑂(𝑝2 ) times ?
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State succinctness for 𝐿 𝑝
 Amplification of probabilities in a finite automaton
If 𝑤 ∈ 𝐿: 𝑃𝑎 = 1 and 𝑃𝑟 = 0
If 𝑤 ∉ 𝐿: 𝑃𝑎 = 1 − 𝑃𝑟 and 𝑃𝑟 > 𝑡(𝑝)
If 𝑤 ∈ 𝐿: 𝑃𝑎 = 𝜀 × 𝑡(𝑝) and 𝑃𝑟 = 0
If 𝑤 ∉ 𝐿: 𝑃𝑎 = 𝜀 × 𝑡(𝑝) and 𝑃𝑟 > 𝑡(𝑝)
If 𝑤 ∈ 𝐿: 𝑃𝑎𝑐𝑐 = 1
If 𝑤 ∉ 𝐿: 𝑃𝑟𝑒𝑗 > 1 − 𝜀
1
𝜀
Expected repetitions: ×
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𝑤
1
𝑡 𝑝
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State succinctness for 𝐿 𝑝
Idea: 2QCFA, two quantum states, several classical states,
wheel of fortune, 𝑝 squares,
One iteration:
|𝑞0 =
1
0
,|𝑞1 = , 𝑈𝑝 =
0
1
𝜋
𝑝
𝜋
sin
𝑝
|𝑞 = (𝑈𝑝 ) 𝑛 |𝑞0 = cos
𝜋
𝑝
𝜋
cos
𝑝
cos
If 𝑤 ∈ 𝐿 𝑝 , |𝑞 =|𝑞0
If 𝑤 ∉ 𝐿 𝑝 , 𝑤 = 𝑛
− sin
𝑃𝑎 =
𝑛𝜋
|𝑞0
𝑝
𝜋
𝑝
4𝜀
𝑝2
𝑛𝜋
|𝑞1
𝑝
𝑛𝜋
𝑠𝑖𝑛2
𝑝
+sin
Probability to observe |𝑞1 is 𝑃𝑟 =
>
4
𝑝2
Use 𝑈𝑝,𝜀 to control the expected number of repetitions.
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State succinctness for 𝐿 𝑝
4𝜀
𝑝2
If 𝑤 ∈ 𝐿 𝑝 , |𝑞 = |𝑞0 , 𝑡ℎ𝑒𝑛 𝑃𝑎 =
and 𝑃𝑟 = 0
𝑤
Repeated ad infinitum, the accepting probability
(1 − 𝑃𝑎 )𝑖 1 − 𝑃𝑟 𝑖 𝑃𝑎 = 1
If 𝑤 ∉ 𝐿 𝑝 , 𝑃𝑟 ≥
𝑖≥0
4
and 𝑃𝑎
𝑝2
=
4𝜀
𝑝2
Repeated ad infinitum, the rejecting probability
(1 − 𝑃𝑎 )𝑖 1 − 𝑃𝑟 𝑖 𝑃𝑟 > 1 − 𝜀
𝑖≥0
1
𝜀
 Expected running time 𝑂( 𝑝2 |𝑤|)
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State succinctness for 𝐿 𝑝
 Using a lemma in (Dwork and Stockmeyer, SIAM J. Comput.
’90), we can prove the following theorem.
For any integer 𝑝, any polynomial expected running
time 2PFA recognizing 𝐿 𝑝 with error probability
𝜀 < 1/2 has at least 3 (log 𝑝)/𝑏 states, where b is a
constant.
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4.State succinctness for C m
Language 𝐶 𝑚 = {𝑤|𝑤 ∈ Σ 𝑚 }.
 DFA: 𝑚
 1PFA: 𝑂(log 2 𝑚) (Freivalds, Automatic Control and Computer
Sciences, ‘82)
 1QFA: 𝑂(log 𝑚) (Ambainis and Freivalds, FOCS’98)
 1QFAR: 7 quantum states, expected running time:
1 𝑚
𝑂 2 |𝑤| , exponential of 𝑚
𝜀
(Yakaryilmaz and Cem Say, DMTCS’10)
Our result: 2QCFA, 2 quantum states, several classical
1
states, expected running time: 𝑂 𝑚2 𝑤 4 , polynomial of
𝜖
𝑚 and |𝑤|.
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State succinctness for C m
Idea: |𝑞0 =
1
0
,|𝑞1 = ,
0
1
cos 𝑚 2𝜋 sin 𝑚 2 𝜋
− sin 𝑚 2 𝜋 cos 𝑚 2𝜋
(rotated by the angle − 𝑚 2 𝜋)
𝑈¢ =
𝑈𝜎 = cos 2𝜋 −sin 2 𝜋
sin 2 𝜋 cos 2𝜋
(rotated by the angle 2𝜋)
 𝑛 = |𝑤|,
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𝑛−𝑚
2
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State succinctness for C m
If 𝑤 ∈ 𝐶(𝑚), 𝑞 = 𝑈𝛼
If w ∉ 𝐶 𝑚 , 𝑤 = 𝑛,
𝑞 = 𝑈𝛼

𝑚
𝑚
𝑈¢ 𝑞0 〉 = 𝑞0 .
cos(𝑛 − 𝑚) 2𝜋
𝑈¢ 𝑞0 〉=
− sin(𝑛 − 𝑚) 2 𝜋
Pr =
𝜀
2𝑚2 𝑛 + 1
2
sin(𝑛 − 𝑚) 2 𝜋
|𝑞0 〉
cos(𝑛 − 𝑚) 2𝜋
𝑤
Probability to get |𝑞1 is
sin2 (𝑛
𝑃𝑎 =
1
− 𝑚) 2 𝜋 >
2 𝑚−𝑛
2
+1
(see Ambainis and Watrous, TCS ’02; Zheng et al. TCS ’13)
Use 𝑈𝑚,𝜀 and two random walks to control the expected number of
repetitions.
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State succinctness for C m
If 𝑤 ∈ 𝐶 𝑚 , |𝑞 = |𝑞0 , 𝑡ℎ𝑒𝑛 𝑃𝑎 =
𝜀
2𝑚2 𝑛+1 2
and 𝑃𝑟 = 0
Repeated for ad infinitum, the accepting probability
(1 − 𝑃𝑎 )𝑖 1 − 𝑃𝑟 𝑖 𝑃𝑎 = 1
𝑖≥0
If 𝑤 ∉ 𝐶 𝑚 , 𝑃𝑟 >
1
2 𝑚−𝑛 2 +1
and 𝑃𝑎 <
𝜀
2𝑚2 𝑛+1 2
Repeated for ad infinitum, the rejecting probability
(1 − 𝑃𝑎 )𝑖 1 − 𝑃𝑟 𝑖 𝑃𝑟 > 1 − 𝜀
𝑖≥0
 Expected running time 𝑂
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1 2
𝑚
𝜖
27
𝑤
4
.
23
State succinctness for C m
 Using a lemma in (Dwork and Stockmeyer, SIAMJ. Comput.
’90), we can prove the following theorem.
For any integer 𝑝, any polynomial expected running
time 2PFA recognizing C 𝑚 with error probability
𝜀 < 1/2 has at least 3 (log 𝑚)/𝑏 states, where b is a
constant.
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5.A trade-off property of 1QCFA
 Quantum resources are expensive and hard to deal
with. One can expect to have only very limited
number of qubits in current quantum system.
 𝑝 = 𝑞1 × 𝑞2 and
gcd 𝑞1 , 𝑞2 = 1.
𝐿 𝑝 = 𝐿 𝑞1 ∩ 𝐿(𝑞2 ) ,
 𝐿 𝑞1 1QCFA Α(𝑞1 )
simulated by
 𝐿 𝑞2 1DFA
1QCFA Α(𝑞2 )
 1QCFA is closed under intersection
State complexity: 𝑂 log 𝑞1 + 𝑂(𝑞2 )
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6. Concluding remarks
 We have proved exact 1QCFA state succinctness for a
promise problem.
 We have proved 2QCFA state succinctness for two
languages
 A trade-off property of 1QCFA
 Future work
It has been proved (Klauck, STOC ‘00) that exact 1QCFA have
not state complexity advantage over 1DFA in recognizing a
language. How about exact 2QCFA vs. 2DFA?
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