Characterizations of symmetric partial Boolean functions with exact quantum query complexity
——Deutsch-Jozsa Problem
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Daowen Qiu School of Data and Computer Science
Sun Yat-sen University, China
Joint work with Shenggen Zheng
arXiv:1603.06505(Qiu and Zheng)
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September 4-8, 2017, NUS, Singapore
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Abstract In this talk, I would like to report a recent work regarding: 1. An optimal exact quantum query algorithm for generalized Deutsch-Jozsa problem 2. The characterization of all symmetric partial Boolean functions with exact quantum 1-query complexity
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Basic background Discover more problems to show that
quantum computing is more powerful than classical computing These problems have the potential of
applications in other areas such as cryptography.
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1. Motivations, Problems, Results 2. Preliminaries 3. Main Results 4. Methods of Proofs 5. Conclusions & Further Problems
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1. Motivations, Problems, Results
(1) Generalized Deutsch-Jozsa (2) Problems with exact quantum 1-query
complexity
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Motivation I
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The Deutsch-Jozsa promise problem [DJ’92]: 𝑥 ∈ 0,1 𝑛, |x| is the Hamming weight of 𝑥,
𝐷𝐷 𝑥 = �01 𝑖𝑖 x = 0 𝑜𝑜 |x| = 𝑛𝑖𝑖 |x| = 𝑛/2
𝑄𝐸 𝐷𝐷 = 1,𝐷 𝐷𝐷 =𝑛2
+ 1
𝐷𝐷𝑛1 = �01 𝑖𝑖|x| ≤ 1 𝑜𝑜|x| ≥ 𝑛 − 1
𝑖𝑖|x| = 𝑛/2 [MJM’15]
𝑄𝐸 𝐷𝐷𝑛1 ≤ 2 [DJ’92] D. Deutsch, R. Jozsa, Rapid solution of problems by quantum computation, In Proceedings of the Royal Society of London, 439A (1992): 553—558. [MJM’15] A. Montanaro, R. Jozsa, G. Mitchison, On exact quantum query complexity, Algorithmica419 (2015) 775--796.
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Problem I
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Motivation II Deutsch-Jozsa problem that is a symmetric partial Boolean function can be solved by DJ algorithm (exact quantum 1-query algorithm). Then how to characterize the other symmetric partial Boolean functions with exact quantum 1-query complexity? Can such functions be solved by DJ algorithm?
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Problem and Result II
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Preliminaries Symmetric partial Boolean functions Classical query complexity Quantum query complexity Multilinear polynomials
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Symmetrical partial Boolean functions
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Representation of symmetric partial Boolean functions
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Isomorphism of symmetric partial Boolean functions
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Classical query complexity
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An exact classical (deterministic) query algorithm to compute a Boolean function 𝑖: {0,1}𝑛 → {0,1} can be described by a decision tree.
If the output of a decision tree is 𝑖(𝑥), for all 𝑥 ∈ {0,1}𝑛, the decision tree is said to "compute" 𝑖. The depthof a tree is the maximum number of queries that can happen before a leaf is reached and a result obtained.
𝐷(𝑖), the deterministic decision tree complexity of 𝑖 is the smallest depth among all deterministic decision trees that compute 𝑖.
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Example
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Deterministic query complexity (how many times we need to query the input bits)
Example: 𝑖 𝑥1, 𝑥2 = 𝑥1 ⊕ 𝑥2
Decision treeThe minimal depth over all decision trees computing 𝑖 is the exact classical query complexity (deterministic query complexity, decision tree complexity) 𝐷(𝑖).
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𝑥1
𝑥2
0 1
0 1
𝑥2
1 01 1
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Quantum query algorithms
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U0 Q Q U1 UT …
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Black box
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𝑄𝑥
𝑄𝑥
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Quantum query complexity
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Multilinear polynomials
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Multilinear polynomials representing symmetric partial Boolean functions
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Methods of Proofs We would like to outline the basic ideas and methods for the proofs of main results.
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𝑫𝑱𝒏𝒌 = �𝟎𝟏 𝒊𝒊 𝒙 ≤ 𝒌 𝒐𝒐 |𝐱| ≥ 𝒏 − 𝒌
𝒊𝒊|𝐱| = 𝒏/𝟐
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𝑄𝐸 𝐷𝐷𝑛𝑘 ≤ 𝑘 + 1
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deg 𝐷𝐷𝑛𝑘 ≥ 2𝑘 + 2
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Theorem:𝑸𝑬(𝒊) = 𝟏 𝒊𝒊 𝒂𝒏𝒂 𝒐𝒏𝒐𝒐 𝒊𝒊 𝒊 can be computed by DJ algorithm
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where 𝑛 − 1 ≥ 𝑘 ≥ ⌊𝑛/2⌋, and 𝑛/2 ≥ 𝑙 ≥ ⌊𝑛/2⌋.
Two Lemmas
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Equivalence transformation
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𝐷𝐷𝑛𝑘 = �01 𝑖𝑖 x ≤ 𝑘 𝑜𝑜 |x| ≥ 𝑛 − 𝑘
𝑖𝑖|x| = 𝑛/2
Theorem. 𝑄𝐸 𝐷𝐷𝑛𝑘 = 𝑘 + 1 and 𝐷 𝐷𝐷𝑛𝑘 = 𝑛/2 + 𝑘 + 1. Theorem. Any symmetric partial Boolean function 𝑖 has 𝑄𝐸(𝑖) = 1 if and only if 𝑖 can be computed by the Deutsch-Jozsa algorithm. Theorem.Any classical deterministic algorithm that solves Simon's problem requires Ω( 𝟐𝐧) queries.
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Conclusions
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Let 𝑖: {0,1}𝑛 → {0,1} be an 𝑛-bit symmetric partial Boolean function with domain of definition 𝐷, and let 0 ≤ 𝑘 < ⌊𝑛
2⌋ . Then, for 2𝑘 + 1 ≤ deg 𝑖 ≤ 2(𝑘 + 1),
how to characterize 𝑖by giving all functions with degrees from 2𝑘 + 1 to 2𝑘 + 2?
For the function 𝐷𝑊𝑛𝑘,𝑙defined as:
can we give optimal exact quantum query algorithms for any 𝑘 and 𝑙?
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Problems
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Thank you for your attention!
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