In NTRU, we know that $f$ is a ternary polynomial in the ring $$R=\frac{\mathbb{Z}_q[x]}{x^n-1}.$$ Here $f$ has $d+1$ coefficients 1 and $d$ coefficients $-1$ and rest are zero. For computing the public key, it is known that $$h=f^{-1}g$$ where $g$ is also a ternary polynomial with $d$ coefficients 1 and $d$ coefficients $-1$ and rest are zero.

Is there any result that informs about the possible degrees of $f^{-1}$? My main doubt is what if it is small enough?


1 Answer 1


We would typically calculate $f^{-1}$ using the extended Euclidean algorithm, so that it has degree at most $n-1$ and typically has degree exactly $n-1$.

  • $\begingroup$ How it would typically be $n-1$? $\endgroup$
    – PAMG
    Commented Apr 2 at 6:56
  • $\begingroup$ @PAMG Because more than a proportion $(q-1)/q$ of elements of $R^*$ have degree $n-1$. $\endgroup$
    – Daniel S
    Commented Apr 2 at 8:49
  • $\begingroup$ Can you please explain it how? $\endgroup$
    – PAMG
    Commented Apr 2 at 8:54
  • 1
    $\begingroup$ @PAMG balls and bins probability! There are possible $q$ balls one can put into the bin ( the $n-1$ degree monomial) of which one is zero. Zero makes, a degree not equal to $n-1$. Now clear? $\endgroup$
    – kelalaka
    Commented Apr 2 at 10:33
  • $\begingroup$ Yes. Thank you. $\endgroup$
    – PAMG
    Commented Apr 2 at 11:12

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