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Let matrix <math>\mathbf{A} \in \mathcal{R}_q^{n \times m}</math> be chosen uniformly at random and <math>k</math> hint vectors <math>\mathbf{s}_i</math> from <math>D_{\Lambda_q^\perp(\mathbf{A}), s}</math> with <math>\lVert \mathbf{s}_i \rVert</math>. Given <math>\mathbf{A}</math> and <math>\{ \mathbf{s}_i \}_{i \in [k]}</math>, an adversary is asked to find a new short non-zero vector <math>\mathbf{s}^* \in \mathcal{R}^m</math> satisfying <math>\mathcal{A} \cdot \mathbf{s}^* = \mathbf{0} \bmod q \land \lVert \mathbf{s}^* \rVert \leq \beta \land \mathbf{s}^* \notin \mathcal{K}-\text{span}\left( \{ \mathbf{s}_i \}_{i \in [k]} \right).</math>
Let matrix <math>\mathbf{A} \in \mathcal{R}_q^{n \times m}</math> be chosen uniformly at random and <math>k</math> hint vectors <math>\mathbf{s}_i</math> from <math>D_{\Lambda_q^\perp(\mathbf{A}), s}</math> with <math>\lVert \mathbf{s}_i \rVert</math>. Given <math>\mathbf{A}</math> and <math>\{ \mathbf{s}_i \}_{i \in [k]}</math>, an adversary is asked to find a new short non-zero vector <math>\mathbf{s}^* \in \mathcal{R}^m</math> satisfying <math>\mathcal{A} \cdot \mathbf{s}^* = \mathbf{0} \bmod q \land \lVert \mathbf{s}^* \rVert \leq \beta \land \mathbf{s}^* \notin \mathcal{K}-\text{span}\left( \{ \mathbf{s}_i \}_{i \in [k]} \right).</math>


The provided definition is the module-variant, which was defined by Albrecht et al.<ref>Albrecht, Martin R., et al. [https://eprint.iacr.org/2022/941.pdf Lattice-based SNARKs: publicly verifiable, preprocessing, and recursively composable]. ''Annual International Cryptology Conference''. Cham: Springer Nature Switzerland, 2022.</ref> The original version can be recovered by setting <math>\mathcal{R} = \mathbb{Z}</math> and <math>\mathcal{K} = \mathbb{Q}</math>.
The provided definition is the module-variant, which was defined by Albrecht et al.<ref>Albrecht, M.R., Cini, V., Lai, R.W., Malavolta, G. and Thyagarajan, S.A. [https://eprint.iacr.org/2022/941.pdf Lattice-based SNARKs: publicly verifiable, preprocessing, and recursively composable]. ''Annual International Cryptology Conference''. Cham: Springer Nature Switzerland, 2022.</ref> The original version can be recovered by setting <math>\mathcal{R} = \mathbb{Z}</math> and <math>\mathcal{K} = \mathbb{Q}</math>.


Intuitively, k-SIS asks for a SIS solution that is not a linear combination of the provided hints.  
Intuitively, k-SIS asks for a SIS solution that is not a linear combination of the provided hints.  
The condition <math>\mathbf{s}^* \notin \mathcal{K}-\text{span}\left( \{ \mathbf{s}_i \}_{i \in [k]} \right) </math> can be dropped when <math>k < m^{1/4} </math> as then the probability that there is an additional short vector in the <math>k </math>-dimensional sublattice spanned by <math>\{ \mathbf{s}_i \}_{i \in [k]}</math> is negligible.<ref name=":0" />


== Hardness of k-SIS ==
== Hardness of k-SIS ==
k-SIS (over <math>\mathbb{Z}</math>) is at least as hard as SIS. Boneh and Freeman<ref name=":0" /> proved this result for constant <math>k \in \mathcal{O}(1)</math> and Ling et al.<ref>Ling, S., Phan, D.H., Stehlé, D. and Steinfeld, R. [https://www.iacr.org/archive/crypto2014/86160136/86160136.pdf Hardness of k-LWE and applications in traitor tracing]. ''Algorithmica'' 79.4 (2017): 1318-1352.</ref> improved this result to <math>k \in \mathcal{O}(m)</math>.
k-SIS<sub>n,m,q,β,s</sub> (over <math>\mathbb{Z}</math>) is at least as hard as SIS<sub>n,m-k,q,β'</sub>. Boneh and Freeman<ref name=":0" /> proved this result for constant <math>k \in \mathcal{O}(1)</math> and Ling et al.<ref>Ling, S., Phan, D.H., Stehlé, D. and Steinfeld, R. [https://www.iacr.org/archive/crypto2014/86160136/86160136.pdf Hardness of k-LWE and applications in traitor tracing]. ''Algorithmica'' 79.4 (2017): 1318-1352.</ref> improved this result to <math>k \in \mathcal{O}(m)</math>.
 
The initial proof<ref name=":0" /> relies on the following observation. Let <math>\mathbf{A} \in \mathbb{Z}_q^{n \times m-1}, \mathbf{e} \leftarrow D_{\mathbb{Z}^{m-1}, s}</math>, and <math>e_m</math> a short <math>\mathbb{Z}_q</math>-invertible entry. Define <math>\mathbf{A}' = \begin{bmatrix} \mathbf{A} &-\mathbf{A} \cdot \mathbf{e} \cdot e_m^{-1} \end{bmatrix}</math> and <math>\mathbf{e}' = \begin{bmatrix} \mathbf{e} \\ e_m \end{bmatrix}</math>. Then, <math>\mathbf{A}' \cdot \mathbf{e}' = \mathbf{A} \cdot \mathbf{e} -\mathbf{A} \cdot \mathbf{e} \cdot e_m^{-1} \cdot e_m = \mathbf{0}.</math> In this way, the proof embeds a hint for each added column to the challenge matrix. Embedding multiple hints and recovering a SIS solution requires several technical details, which we omit here.


TODO - Provide intuition of reduction
No proof was provided for the module variant.


== Constructions based on k-SIS ==
== Constructions based on k-SIS ==


* Linearly homomorphic signatures<ref name=":0" />
* Linearly homomorphic signatures<ref name=":0" />
* k-time GPV signatures in the standard model<ref name=":0" />


== Related Assumptions ==
== Related Assumptions ==

Revision as of 10:12, 26 July 2025

The k-SIS assumption was introduced in 2011 by Boneh and Freeman[1]. The assumption hands out k hints additionally to the SIS challenge matrix restricting the solution space by any linear combination of these hints.

Formal Definition

k-SISn,m,d,q,β,s

Let matrix 𝐀qn×m be chosen uniformly at random and k hint vectors 𝐬i from DΛq(𝐀),s with 𝐬i. Given 𝐀 and {𝐬i}i[k], an adversary is asked to find a new short non-zero vector 𝐬*m satisfying 𝒜𝐬*=𝟎modq𝐬*β𝐬*𝒦span({𝐬i}i[k]).

The provided definition is the module-variant, which was defined by Albrecht et al.[2] The original version can be recovered by setting = and 𝒦=.

Intuitively, k-SIS asks for a SIS solution that is not a linear combination of the provided hints.

The condition 𝐬*𝒦span({𝐬i}i[k]) can be dropped when k<m1/4 as then the probability that there is an additional short vector in the k-dimensional sublattice spanned by {𝐬i}i[k] is negligible.[1]

Hardness of k-SIS

k-SISn,m,q,β,s (over ) is at least as hard as SISn,m-k,q,β'. Boneh and Freeman[1] proved this result for constant k𝒪(1) and Ling et al.[3] improved this result to k𝒪(m).

The initial proof[1] relies on the following observation. Let 𝐀qn×m1,𝐞Dm1,s, and em a short q-invertible entry. Define 𝐀=[𝐀𝐀𝐞em1] and 𝐞=[𝐞em]. Then, 𝐀𝐞=𝐀𝐞𝐀𝐞em1em=𝟎. In this way, the proof embeds a hint for each added column to the challenge matrix. Embedding multiple hints and recovering a SIS solution requires several technical details, which we omit here.

No proof was provided for the module variant.

Constructions based on k-SIS

  • Linearly homomorphic signatures[1]
  • k-time GPV signatures in the standard model[1]

Related Assumptions

  • k-LWE

References

  1. 1.0 1.1 1.2 1.3 1.4 1.5 Boneh, Dan, and David Mandell Freeman. Linearly homomorphic signatures over binary fields and new tools for lattice-based signatures. International Workshop on Public Key Cryptography. Berlin, Heidelberg: Springer Berlin Heidelberg, 2011.
  2. Albrecht, M.R., Cini, V., Lai, R.W., Malavolta, G. and Thyagarajan, S.A. Lattice-based SNARKs: publicly verifiable, preprocessing, and recursively composable. Annual International Cryptology Conference. Cham: Springer Nature Switzerland, 2022.
  3. Ling, S., Phan, D.H., Stehlé, D. and Steinfeld, R. Hardness of k-LWE and applications in traitor tracing. Algorithmica 79.4 (2017): 1318-1352.