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NTRU: remove a pointless failure check.

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In the key generation step where we invert 3f in the field
Z_q/<x^p-x-1>, I was carefully checking for failure, on the grounds
that even a field does have _one_ non-invertible element, namely zero.
But I forgot that we'd generated f in such a way that it can't
possibly be zero. So that failure check is pointless.

(However, I've retained it in the form of an assertion.)
This commit is contained in:
Simon Tatham 2023-05-28 09:59:41 +01:00
parent 6aca7f1eef
commit 8cf372d4a2
1 changed files with 6 additions and 12 deletions
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18
crypto/ntru.c
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@ -1046,20 +1046,14 @@ NTRUKeyPair *ntru_keygen_attempt(unsigned p, unsigned q, unsigned w)
ntru_scale(f3, f, 3, p, q);
/*
* Try to invert 3*f over Z_q. This should be _almost_ guaranteed
* to succeed, since Z_q/<x^p-x-1> is a field, so the only
* non-invertible value is 0. Even so, there _is_ one, so check
* the return value!
* Invert 3*f over Z_q. This is guaranteed to succeed, since
* Z_q/<x^p-x-1> is a field, so the only non-invertible value is
* 0. And f is nonzero because it came from ntru_gen_short (hence,
* w of its components are nonzero), hence so is 3*f.
*/
uint16_t *f3inv = snewn(p, uint16_t);
if (!ntru_ring_invert(f3inv, f3, p, q)) {
ring_free(f, p);
ring_free(f3, p);
ring_free(f3inv, p);
ring_free(g, p);
ring_free(ginv, p);
return NULL;
}
bool expect_always_success = ntru_ring_invert(f3inv, f3, p, q);
assert(expect_always_success);
/*
* Make the public key, by converting g to a polynomial over q and