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On the (M)iNTRU Assumption in the Integer Case

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Provable and Practical Security (ProvSec 2021)

Part of the book series: Lecture Notes in Computer Science ((LNSC,volume 13059))

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In AsiaCrypt 2019, Genise, Gentry, Halevi, Li and Micciancio put forth two novel and intriguing computational hardness hypotheses: The inhomogeneous NTRU (\(\mathsf {iNTRU}\)) assumption and its matrix version \(\mathsf {MiNTRU}\). In this work, we break the integer case of the \(\mathsf {iNTRU}\) assumption through elementary lattice reduction, and we describe how the attack might be generalized to polynomial rings and to the low dimensional \(\mathsf {MiNTRU}\) assumption with small noise.

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  1. 1.

    Hereinafter, we will only use the LLL algorithm for lattice reduction. Better results may be achieved using recent results on the BKZ algorithm (see [14]). However, the LLL algorithm suffices for our elementary analysis.

  2. 2.

    For \(\mathcal {R}=\mathbb {Z}\) and random \(x_0,\dots ,x_{\ell }\), the probability that our first attack cannot be used is only \(\left( 1- \frac{\phi (q)}{q}\right) ^{\ell }\) where \(\phi \) denotes the Euler totient function. In particular, if q is prime our first attack should always work.

  3. 3.

    The only possibility for which this would not be the case takes place when \(g=\gcd (y_0,\dots ,y_{\ell -1},q)>1\) as then \((0,\dots ,0,\frac{q}{g})\) will be the shortest lattice vector, but this scenario is rather improbable.

  4. 4.

    This conclusion also holds in the case of a random tuple.


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Jim Barthel was supported in part by the Luxembourg National Research Fund through grant PRIDE15/10621687/SPsquared. Răzvan Roşie was supported in part by ERC grant CLOUDMAP 787390.

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Correspondence to Jim Barthel .

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A Proof of Lemma 1

A Proof of Lemma 1

We observe that by the properties of the Euclidean and the infinity norm, we have

$$\mathbb {P}\left( \left( \left\| [r\mathbf {y}\mod q^2]\right\| _2\le Bq \right) \cap (r\in S)\right) \le \mathbb {P}\left( \left( \max \big \{ \left| [r\mathbf {y}\mod q^2]\right| \big \}\le Bq\right) \cap (r\in S) \right) $$

where the maximum is taken over all the modulo reduced entries of \(r\mathbf {y}\). The right expression is equal to

$$\mathbb {P}\left( \left( \bigcap _{i=0}^{\ell -1} \underbrace{\big (\left| [ry_iq \mod q^2]\right| \le Bq\big )}_{:=C_i}\right) \cap \underbrace{\big (\left| [r \mod q^2]\right| \le Bq \big )}_{:=C_{\ell }}\cap (r\in S) \right) .$$

Each event \(C_0,\dots ,C_{\ell -1}\) in the probability statement can be written as a union of events \(C_i=\bigcup _{\beta _i=-Bq}^{Bq} ([ry_iq\mod q^2]=\beta _i)\). As this event can only take place whenever \(\beta _i\) is a multiple of q (otherwise, the equality cannot be satisfied), we need only to consider the restricted union of events \(\bigcup _{\beta _i=-B}^{B} ([ry_iq\mod q^2]=\beta _i q)=\bigcup _{\beta _i=-B}^{B} ([ry_i\mod q]=\beta _i)\). Furthermore, \(C_{\ell }=\bigcup _{\beta _{\ell }=-Bq}^{Bq} ([r\mod q^2]=\beta _{\ell })=\bigcup _{\beta _{\ell }=-Bq}^{Bq} (r=\beta _{\ell })\) which is restricted to \(\beta _{\ell }\in S\) by the last condition. Thus, our overall probability is equal to

$$\mathbb {P}\left( \left( \bigcap _{i=0}^{\ell -1} \bigcup _{\beta _i=-B}^{B} ([ry_i\mod q]=\beta _i) \right) \; \cap \; \left( \bigcup _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq} (r=\beta _{\ell }) \right) \right) .$$

Reordering the events gives

$$\mathbb {P}\left( \bigcup _{\beta _0=-B}^{B} \dots \bigcup _{\beta _{\ell -1}=-B}^{B} \bigcup _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq} \left( \bigcap _{i=0}^{\ell -1} ([ry_{i} \mod q]=\beta _i) \cap (r=\beta _{\ell }) \right) \right) .$$

As the events are mutually exclusive, this probability is equal to

$$\sum _{\beta _0=-B}^B \dots \sum _{\beta _{\ell -1}=-B}^B\sum _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq}\mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([ry_{i} \mod q]=\beta _i) \cap (r=\beta _{\ell }) \right) .$$

Using Bayes’ conditional probability rule followed by Euler’s rule of interchanging finite sums, this quantity can be rewritten as:

$$\begin{aligned}&\sum _{\beta _0=-B}^B \dots \sum _{\beta _{\ell -1}=-B}^B\sum _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq}\mathbb {P}\left( r=\beta _{\ell }\right) \mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([ry_{i} \mod q]=\beta _i) \;\Big |\; (r=\beta _{\ell }) \right) \\ =&\sum _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq} \mathbb {P}\left( r=\beta _{\ell } \right) \sum _{\beta _0=-B}^B \dots \sum _{\beta _{\ell -1}=-B}^B\mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([ry_{i} \mod q]=\beta _i) \;\Big |\; (r=\beta _{\ell }) \right) \end{aligned}$$

Naturally \(\mathbb {P}\left( r=\beta _{\ell } \right) =\frac{1}{q^2}\) for any \(\beta _{\ell }\). It remains to investigate the value of the rightmost probability. To do so, we rewrite \(\beta _{\ell }=g_{\ell }\beta '_{\ell }\) where \(g_{\ell }=\gcd (\beta _{\ell },q)\). Then, for fixed \(\beta _0,\dots ,\beta _{\ell -1},\beta _{\ell }\):

$$\begin{aligned}&\mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([ry_{i} \mod q]=\beta _i) \;\Big |\; (r=\beta _{\ell }) \right) \\ =\,&\mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([\beta _{\ell } y_{i} \mod q]=\beta _i) \right) \\ =\,&\mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([g_{\ell }\beta '_{\ell } y_{i} \mod q]=\beta _i) \right) \end{aligned}$$

The events in this probability will only be satisfiable if \(\beta _i\) is a multiple of \(g_{\ell }\), say \(\beta _i=\beta '_ig_{\ell }\). Thus, our cumulative probability rewrites as

$$\begin{aligned}&\sum _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq} \frac{1}{q^2} \sum _{\beta '_0=-\lfloor B/g_{\ell }\rfloor }^{\lfloor B/g_{\ell }\rfloor } \dots \sum _{\beta '_{\ell -1}=-\lfloor B/g_{\ell }\rfloor }^{\lfloor B/g_{\ell }\rfloor } \mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([g_{\ell }\beta '_{\ell } y_{i} \mod q]=\beta '_ig_{\ell }) \right) \\ =&\sum _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq} \frac{1}{q^2} \sum _{\beta '_0=-\lfloor B/g_{\ell }\rfloor }^{\lfloor B/g_{\ell }\rfloor } \dots \sum _{\beta '_{\ell -1}=-\lfloor B/g_{\ell }\rfloor }^{\lfloor B/g_{\ell }\rfloor } \mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([\beta '_{\ell } y_{i} \mod \frac{q}{g_{\ell }}]=\beta '_i) \right) \\ =&\sum _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq} \frac{1}{q^2} \sum _{\beta '_0=-\lfloor B/g_{\ell }\rfloor }^{\lfloor B/g_{\ell }\rfloor } \dots \sum _{\beta '_{\ell -1}=-\lfloor B/g_{\ell }\rfloor }^{\lfloor B/g_{\ell }\rfloor } \mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([ y_{i} \mod \frac{q}{g_{\ell }}]=[\beta '_i\beta '^{-1}_{\ell }\mod \frac{q}{g_{\ell }}]) \right) \end{aligned}$$

where we used the fact that \(g_{\ell }=\gcd (\beta _{\ell },q)\) which implies that \(\beta '_{\ell }\) is invertible modulo \(\frac{q}{g_{\ell }}\). It is now clear that the remaining events are independent as they only depend on \(y_i\). Thus

$$\begin{aligned}&\mathbb {P}\left( \bigcap _{i=0}^{\ell -1} ([ y_{i} \mod \frac{q}{g_{\ell }}] =[\beta '_i\beta '^{-1}_{\ell }\mod \frac{q}{g_{\ell }}]) \right) \\ =&\prod _{i=0}^{\ell -1} \mathbb {P}\left( [ y_{i} \mod \frac{q}{g_{\ell }}]=[\beta '_i\beta '^{-1}_{\ell }\mod \frac{q}{g_{\ell }}] \right) \\ =&\left( \frac{1}{\frac{q}{g_{\ell }}}\right) ^{\ell }\\ =&\left( \frac{g_{\ell }}{q}\right) ^{\ell }. \end{aligned}$$

Thereby, the cumulative probability is given by

$$ \sum _{\underset{\beta _{\ell }\in S}{\beta _{\ell }=-Bq}}^{Bq} \frac{\ell (2\lfloor B/g_{\ell }\rfloor +1)}{q^2} \left( \frac{g_{\ell }}{q}\right) ^{\ell }. $$

   \(\square \)

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Barthel, J., Müller, V., Roşie, R. (2021). On the (M)iNTRU Assumption in the Integer Case. In: Huang, Q., Yu, Y. (eds) Provable and Practical Security. ProvSec 2021. Lecture Notes in Computer Science(), vol 13059. Springer, Cham.

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