What does it mean to solve $\mathsf{SVP}_{\gamma}$ in worst-case?

Does it mean that the problem is solvable for any lattice we choose?


2 Answers 2


Solving $\text{SVP}$ in the worst-case means solving a 'hardest' instance of $\text{SVP}$. This isn't what we want for cryptography as we want problems to be hard to solve on average (average-case). Foundational problems like $\text{SIS}$ and $\text{LWE}$ are average-case problems depending on the hardness of worst-case problems like $\text{SVP}$, i.e. if $\text{SVP}$ is hard in the worst-case then $\text{LWE/SIS}$ are hard on average.

I recommend reading 'M. Ajitai. Generating Hard Instances of Lattice Problems' for more information on this.

  • $\begingroup$ can you explain more, why average case is favorable in cryptography?? $\endgroup$
    – Don Freecs
    May 26, 2022 at 13:46
  • 2
    $\begingroup$ @DonFreecs it's just what shows up. When you randomly sample keys, often you are (implicitly) randomly sampling an instance of some underlying problem, so "breaking" things becomes an average-case problem. That being said, precisely what distribution to sample from can sometimes be hard to determine. The worst-case to average-case reduction people are discussing here helps identify the LWE distribution as "the right" one. $\endgroup$
    – Mark
    May 27, 2022 at 7:06

as far as I know, solving a problem in the worst case hardness means solving the problem for any instance given (hence for any distribution of instances) on the other hand, solving a problem in average case means solving this problem for a given distribution ...

"Correct me please, if I am wrong"


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