I'm trying to implement a naive version of CKKS in Python. It was great until I start implementing the modulus.

For this kind of schemes, the modulus $q$ is in the range $(-q/2,q/2]$. How does this work?

In CKKS paper (I think BGV and others do the same) use something like this (a toy example): $c = m (mod$ $q)$. Where $c$ and $m$ are polynomials, so the coefficients of m are mod q. So c and m are congruent mod q.

So for each coefficient $n$ (if the range of mod is the typical $[0,1,...q-1]$), a way is to solve:

$n = m*q + R$

Where $m$ is the greatest integer that makes $n<m*q$ and $R$ is the remainder that solves the equation being this the same reminder that for c, so n is the result for the coefficient.

But if the modulus range is other like I say, how is this done? The intuition says me that is like wrapping the number in a ''clock'' that starts in -q/2+1 and goes up to q/2 before starting again. But I don't know how to apply this intuition for negative numbers that are smaller than -q/2.

For example which is the answer of -771 (mod 1280)


1 Answer 1


If you first have some modulus function $x\bmod q$ that always returns a value in $[0, q)$, it is straightforward to switch this to your desired modulus function, for example one can define

$$x\bmod^{\pm} q := (x\bmod q) - \lfloor q/2\rfloor.$$

This is to say that to get a modulus function that returns a value in $[c, c+q)$ rather than $[0,q)$, it suffices to (manually) shift the result of calling the standard modulus function.

  • $\begingroup$ This was my first attempt and didn't work.... So I have a bug in other please! At least I know that the mod was not the problem. $\endgroup$
    – mmazz
    Nov 19, 2022 at 13:13
  • $\begingroup$ For some programming languages there is a difference between "modulus" and "remainder", usually with how they treat negative inputs. I would try to look into the documentation of your programming language for this, as it is likely the issue. $\endgroup$
    – Mark Schultz-Wu
    Nov 19, 2022 at 20:37
  • $\begingroup$ Thats true. I think that Python %, implementes modulus, so thats its correct. Tell me if what I think its correct. The idea of this congruence in the new interval its that $c\equiv x\pmod q$, its to find witch $c \in (-q/2, q/2]$ has the same modulus that $m \bmod q$? For what I understand $q/2\bmod q=q/2$, but if at that I subtract $q/2$ will give me a wrong answer (for what I understand). What I'm missing? $\endgroup$
    – mmazz
    Nov 21, 2022 at 15:04
  • $\begingroup$ If your issue is one of the rounding boundary, i.e. $(-q/2, q/2]$ vs $[-q/2, q/2)$, you could simply add/subtract 1 from the result as necessary. $\endgroup$
    – Mark Schultz-Wu
    Nov 21, 2022 at 16:14
  • 1
    $\begingroup$ @mmazz: one way to implement $c \in (-q/2, q/2]$ (given a % operation that gives a result in $[0, q)$ is $c = ((x + (q/2-1)) \% q) - (q/2-1)$ $\endgroup$
    – poncho
    Dec 19, 2022 at 19:13

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