Quantum Algorithms for the Resiliency of Vectorial Boolean Functions

Abstract

Recently, Cui and Guo (Quantum Inform. Process. 18, 182, 2019) gave some quantum testing algorithms of multi-output Boolean functions, one of them was for resiliency testing. However, through the algorithm, we cannot get the probability distribution of a Boolean function. Improving the algorithm there, we present quantum algorithms for this problem. Under the condition of non-resiliency, we obtain not only the conclusion that the function F(x) is statistically dependent of which t variables but also the distance of the probability distribution from the uniform distribution.

Appendix: Proof of Lemma 1

For the convenience of description, some notations are introduced first. Let $\eta =(a_{i_{1}}, a_{i_{2}}, \cdots a_{i_{t}})$, I = I_t = i₁⋯i_t, $\alpha =\alpha _{i_{1}}{\cdots } \alpha _{i_{t}}$, $\text {Pr}[F(x)=y | x_{i_{1}}{\cdots } x_{i_{t}}=\alpha ]=P_{I,\alpha }^{y}$, $\text {Pr}[F(x)=y | x_{i_{1}}\cdots x_{i_{t}}=\alpha ]-\frac {1}{2^{m}}=\varepsilon _{I,\alpha }^{y}$.

In order to make the proof more clearly, let’s look at some propositions.

Proposition 1

If $b\in {F_{2}^{m}}\setminus \{0\}$, then

$$ S(a_{I},b) =\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha}\cdot \sum\limits_{y\in {F_{2}^{m}}}(-1)^{b\cdot y}\varepsilon_{I,\alpha}^{y}. $$

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Proof

If b≠ 0, let $x^{\prime }\in \{x_{1},\cdots , x_{n}\}-\{x_{i_{1}},\cdots , x_{i_{t}}\}$, then

$$ \begin{array}{@{}rcl@{}} S(a_{I},b)&=&\frac{1}{2^{n}}\sum\limits_{x\in {F^{n}_{2}}}(-1)^{a_{I}\cdot x+b\cdot F(x)}\\ &=&\frac{1}{2^{n}}\sum\limits_{x_{i_{1}}{\cdots} x_{i_{t}}}(-1)^ {a_{i_{1}}x_{i_{1}}+{\cdots} +a_{i_{t}}x_{i_{t}}} \sum\limits_{x^{\prime}}(-1)^{b\cdot F(x)}\\ &=&\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha}\cdot (\text{Pr}\left[b\cdot F(x)=0 | x_{i_{1}}{\cdots} x_{i_{t}}=\alpha\right]\\ &&-\text{Pr}\left[b\cdot F(x)=1 | x_{i_{1}}{\cdots} x_{i_{t}}=\alpha\right])\\ &=&\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha}\cdot \left( \sum\limits_{y\in \{b\cdot y=0\}}P_{I,\alpha}^{y}- \sum\limits_{y\in \{b\cdot y=1\}}P_{I,\alpha}^{y}\right)\\ &=&\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha}\cdot \left( \sum\limits_{y\in \{b\cdot y=0\}}\varepsilon_{I,\alpha}^{y}- \sum\limits_{y\in \{b\cdot y=1\}}\varepsilon_{I,\alpha}^{y}\right)\\ &=&\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha}\cdot \sum\limits_{y\in {F_{2}^{m}}}(-1)^{b\cdot y}\varepsilon_{I,\alpha}^{y}.\\ \end{array} $$

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Similarly, we have

Proposition 2

If b = 0, and a_I≠ 0, then

$$ S(a_{I},b) =\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha}\cdot \sum\limits_{y\in {F_{2}^{m}}}(-1)^{b\cdot y}\varepsilon_{I,\alpha}^{y}. $$

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Proposition 3

If b = 0, and a_I = 0, then

$$ S(a_{I},b) =\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha}\cdot \sum\limits_{y\in {F_{2}^{m}}}(-1)^{b\cdot y}\varepsilon_{I,\alpha}^{y}+1. $$

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On the other hand, by the definition of Walsh transform, we have

$$ S(a_{I},0) =\frac{1}{2^{t}}\sum\limits_{x_{i_{1}}{\cdots} x_{i_{t}}}(-1)^ {a_{i_{1}}x_{i_{1}}+{\cdots} +a_{i_{t}}x_{i_{t}}}. $$

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Accordingly, S(0, 0) = 1, S(a_I, 0) = 0 for a_I≠ 0.

1.1 Proof of Lemma 1

Proof

Considering that for any fixed α, β, y, v, we have ${\sum }_{b\in {F_{2}^{m}}}(-1)^{b\cdot (y+v)} \varepsilon _{I,\alpha }^{y}\varepsilon _{I,\beta }^{v}=0$ unless y = v; ${\sum }_{\eta \in {F_{2}^{t}}}(-1)^{\eta \cdot (\alpha +\beta )} \varepsilon _{I,\alpha }^{y}\varepsilon _{I,\beta }^{y}=0$ unless α = β, then

$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{b\in {F_{2}^{m}}\setminus\{0\}}\sum\limits_{{a}\in W_{I}}[S(a,b)]^{2}\\ &=&\sum\limits_{b\in {F_{2}^{m}}}\sum\limits_{{a}\in W_{I}} \left[\frac{1}{2^{t}}\sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha} \sum\limits_{y\in {F_{2}^{m}}}(-1)^{b\cdot y}\varepsilon_{I,\alpha}^{y}\right]^{2}\\ &=&2^{-2t}\sum\limits_{b\in {F_{2}^{m}}}\sum\limits_{\eta\in {F_{2}^{t}}} \left[\left( \sum\limits_{\alpha\in {F_{2}^{t}}}(-1)^{\eta\cdot\alpha} \sum\limits_{y\in {F_{2}^{m}}}(-1)^{b\cdot y}\varepsilon_{I,\alpha}^{y}\right) \left( \sum\limits_{\beta\in {F_{2}^{t}}}(-1)^{\eta\cdot\beta} \sum\limits_{v\in {F_{2}^{m}}}(-1)^{b\cdot v}\varepsilon_{I,\beta}^{v}\right)\right]\\ &=&2^{-2t}\sum\limits_{\alpha\in {F_{2}^{t}}}\sum\limits_{\beta\in {F_{2}^{t}}} \sum\limits_{\eta\in {F_{2}^{t}}}(-1)^{\eta\cdot(\alpha+\beta)} \sum\limits_{y\in {F_{2}^{m}}}\sum\limits_{v\in {F_{2}^{m}}}\left( \sum\limits_{b\in {F_{2}^{m}}}(-1)^{b\cdot (y+v)} \varepsilon_{I,\alpha}^{y}\varepsilon_{I,\beta}^{v}\right)\\ &=&2^{-2t}\sum\limits_{\alpha\in {F_{2}^{t}}}\sum\limits_{\beta\in {F_{2}^{t}}} \sum\limits_{\eta\in {F_{2}^{t}}}(-1)^{\eta\cdot(\alpha+\beta)} \sum\limits_{y\in {F_{2}^{m}}}2^{m}\varepsilon_{I,\alpha}^{y}\varepsilon_{I,\beta}^{y}\\ &=&2^{m-2t}\sum\limits_{y\in {F_{2}^{m}}}\sum\limits_{\alpha\in {F_{2}^{t}}}\sum\limits_{\beta\in {F_{2}^{t}}} \left( \sum\limits_{\eta\in {F_{2}^{t}}}(-1)^{\eta\cdot(\alpha+\beta)} \varepsilon_{I,\alpha}^{y}\varepsilon_{I,\beta}^{y}\right)\\ &=&2^{m-t}\sum\limits_{y\in {F_{2}^{m}}}\sum\limits_{\alpha\in {F_{2}^{t}}} (\varepsilon_{I,\alpha}^{y})^{2}. \end{array} $$

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