William Whyte
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# 8 Comments

 Feb 1 comment how to define the Truncated Polynomial? See for example purplemath.com/modules/polydiv.htm Dec 21 comment Secure, patent-free alternative to NTRU You're right, the statement in PATENTS.md is not consistent with the GPL. That wasn't the intent, it is meant to be fully GPL-compliant. We're changing that now. Jan 7 comment Implementations of Ntru TLS BTW, NTRU is now available under the GPL and other FOSS licenses: github.com/NTRUOpenSourceProject/ntru-crypto. The open-source version has not yet been fully integrated into CyaSSL though. Nov 28 comment Is the “secure-as-worst-case” version of NTRU patented? The patents are now available under GPL. Go use them! May 19 comment NTRUEncrypt - Choose the initial random polynomial If p = 3 then you're using trinary, so let's say that r, g and m have dr, dg, dm +1s and -1s respectively and f has df +1s and (df-1) -1s. To avoid decryption failures altogether you need p*2dr*2dg + (2df-1)*2dm < 128, so df, dm, dr, dg all need to be around sqrt(128) or, basically, 11. If you're taking them much larger than this you'll see significant numbers of decryption failures. You want to be able to take df etc to be around N/3. If you take q to be the power of 2 above (N^2)/9 you will certainly have no decryption failures. Lower values of q give you some risk. May 17 comment NTRUEncrypt - Choose the initial random polynomial For binary, df is the number of 1s; for trinary with "flat" f, you have df 1s and df-1 -1s (or you could define df so that there are df+1 1s and df -1s, it doesn't matter much); for trinary with form 1+pF, you have df +1s and df -1s. May 11 comment Does NTRU decrypt correctly now? Decryption isn't probabilistic: running the decryption algorithm multiple times always gives the same result. (Paulo Ebermann asks the right question here). However, it may be inconsistent with encryption, which is a different thing. May 11 comment How can one sign with NTRU? Encryption is probabilistic, but decryption isn't, and the question was about decryption as a means of signing, so I don't think this is the right way of thinking about it.