network profile
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May 11 |
awarded | Yearling |
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Aug 6 |
awarded | Student |
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Aug 6 |
asked | How long does it take to extract a key from a FIPS-140 Level 2 device? |
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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. |
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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. |
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May 16 |
answered | NTRUEncrypt - Choose the initial random polynomial |
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May 13 |
awarded | Necromancer |
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May 11 |
awarded | Teacher |
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May 11 |
awarded | Supporter |
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May 11 |
answered | Does NTRU decrypt correctly now? |
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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. |
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May 11 |
answered | Inverses in Truncated Polynomial Rings |
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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. |
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May 11 |
answered | How can one sign with NTRU? |
