William Whyte
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366
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 Dec12 answered Reduced message expansion in NTRU Sep24 awarded Autobiographer Apr9 answered RSA: Letting $p$ and $q$ have different bit-size Jan7 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. Nov28 comment Is the “secure-as-worst-case” version of NTRU patented? The patents are now available under GPL. Go use them! Sep10 answered Is the “secure-as-worst-case” version of NTRU patented? Jul24 answered Secret sharing scheme with possibility to change the secret May11 awarded Yearling Aug6 awarded Student Aug6 asked How long does it take to extract a key from a FIPS-140 Level 2 device? May19 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. May17 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. May16 answered NTRUEncrypt - Choose the initial random polynomial May13 awarded Necromancer May11 awarded Teacher May11 awarded Supporter May11 answered Does NTRU decrypt correctly now? May11 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. May11 answered Inverses in Truncated Polynomial Rings May11 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.