Is NTRU still hard if $G$ is set to 1?

I'm looking at the description of NTRUEncrypt given on page 21 of http://archive.dimacs.rutgers.edu/Workshops/Post-Quantum/Slides/Silverman.pdf and using its notation. So in NTRU there are always two private parameters $$F$$ and $$G$$. However, decryption only requires knowledge of $$F$$. I'm confused about what the purpose of $$G$$ is because it doesn't come into the decryption and I don't see why removing $$G$$ hurts the security of NTRU - more specifically, is NTRU still hard if $$G$$ is always set to 1 (i.e., $$(1,0,0,...,0)\in \{-1,0,1\}^N$$)?

If $$G$$ is set to $$1$$, then the adversary can easily decrypt the ciphertext because in this case $$h = pf^{-1} \mod q, p$$ is coprime with $$q$$, then inverting $$p \mod q$$ is possible and after that he calculates $$f \mod q$$ from $$h$$, then he calculates $$f^{-1} \mod p$$ from $$f$$
• Thanks, I have a follow-up, in the case where $g$ is no longer set to $1$ so that $h=pf^{-1}g \mod q$, would being able to factor $h$ break NTRU? May 28 at 16:14