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Choosing NTRUEncrypt Parameters. William Whyte NTRU Cryptosystems March 2004. Agenda. Parameter Generation How to pick parameters to obtain a given security level? We present a recipe for parameter generation Will 1363.1use this recipe, or simply the constraints that come out of it?

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Choosing ntruencrypt parameters

Choosing NTRUEncrypt Parameters

William Whyte

NTRU Cryptosystems

March 2004


Agenda
Agenda

  • Parameter Generation

    • How to pick parameters to obtain a given security level?

      • We present a recipe for parameter generation

      • Will 1363.1use this recipe, or simply the constraints that come out of it?

    • Multiple parameter forms

      • Standard form, product form

    • Possible bandwidth savings – NTRU-KEM

  • Key validation

  • But first…


Review ntru parameters
Review: NTRU parameters

  • N, dimension of polynomial ring

    • NTRU works on polynomials of degree N-1

    • Polynomial multiplication is convolution multiplication: terms of degree > N are reduced mod N.

    • For 80-bit security, N = 251.

      • Increases roughly linearly with k for k-bit security

  • q, “big” modulus

    • All coefficients in polynomial are reduced mod q

    • For 80-bit security, q = 239.

      • Increases roughly linearly with k for k-bit security

  • p, “small” modulus

    • Reduce mod p during decryption

    • p = 2, 2+X or 3 for all security levels.

  • Sizes:

    • Public key, ciphertext size = N log2 q = 2004 bits for 80-bit security

    • message size (bits) = N log2 ||p|| = 251 bits for 80-bit security


Review ntruencrypt operations
Review: NTRUEncrypt Operations

  • Key Generation

    • Generate f, g, “small” polynomials in Zq[X]/(XN-1).

    • Public key h = p*f-1*g mod q; private key = (f, fp = f-1 mod p).

  • Encrypt (Raw operation)

    • Encode message as “small” polynomial m.

    • Generate “small” random polynomial r

    • Ciphertext e = r*h + m mod q.

  • Decrypt (Raw operation)

    • Set a = f*e mod q.

      • “mod q” = in range [A, A+q-1].

    • Set m = fp * a mod p.


Review why decryption works
Review: Why Decryption Works

  • a = f * e (mod q) = f * (r*h + m) (mod q) = f * (r*p*g*Fq + m) (mod q) = p*r*g + f*m (mod q) since f*Fq = 1 (mod q)

  • All of the polynomials r, g, f, m are small, so coefficients of p*r*g + f*mwill (usually) all lie within q of each other.

  • If its coefficients are reduced into the right range, the polynomial a(x) is exactly equal to p*r*g + f*m. Then fp * a = p*r*g*fp (mod p) + fp*f*m (mod p) = m (mod p).

  • Current parameter sets for 280 security include means for choosing this range. Choice of range fails on validly encrypted message one time in 2104.

    • “Decryption failures”

    • Attatcker gains information from decryption failures: wants to choose funky r, m.


Review sves 3 encryption
Review: SVES-3 encryption

mLen

00…

b

m

ID

Hash

r

r*h

XOR

Hash

m’

r*h + m’

e


Parameter generation
Parameter Generation

  • Input: k, the desired level of security

  • Process:

    • Choose N

      • Set N to give necessary bandwidth

    • Choose form of f, g, r

      • Ensure combinatorial security

    • Choose q, p

      • Set q to prevent decryption failures

    • Ensure that these parameters give appropriate lattice security

  • There are many different ways of making these choices.

    • These are the proposed ones for X9.98

      • Note: extremely provisional and may change as the analysis proceeds

    • Currently writing up a paper to formalize them


Choose n
Choose N

  • With binary messages, N is the number of bits that can be transported

  • For k bits of security for key transport, want to transport 2k bits of material

    • Prevent birthday-like attacks based on future use of material

  • For SVES-3, want to use at least k bits of random padding

    • Gives security against enumeration attacks if encryption scheme is used to transport low-entropy messages

  • Set N to be the first prime greater than 3k.


Choose form of f g r
Choose form of f, g, r

  • Our choice:

    • Take f = 1+pF

      • Speeds up decryption: f-1 mod p= 1, so we eliminate a convolution

    • Take F, g, r to be binary with df, dg, dr 1s respectively.

      • Number of additions necessary for convolution is df*N.

  • Alternatives:

    • Take f not to be of form 1+pF

      • Slows down decryption but reduces q (see next choice)

    • Take f (or F), g, r to be of the form (e.g.) f = f1 * f2 + f3.

      • “Product form”:

      • Number of additions necessary for convolution is (f1 + f2 + f3)*N.

        • Performance benefit


Binary f g r combinatorial security
Binary F, g, r: Combinatorial Security

  • F, g, r have df, dg, dr 1s respectively

  • Brute force-like search on F, g, r can be speeded up by meet-in-the-middle techniques.

  • Using these techniques, number of binary convolution multiplications needed to break f is

    • Each multiplication requires df.N additions

      • … perhaps divided by 2-8 if we use wordsize cleverly

      • In general, use number of multiplications as security measure

  • Attacker will go for easiest of (f, g), (r, m); pick df = dr = dg.


Df dg dr for different security levels
df, dg, dr for different security levels

  • N and df, using the above criteria:

  • df ~= 0.185 N.


Pick q p
Pick q, p

  • Our choice:

    • Pick p = 2, q to be the first prime greater than p.min(dr, dg) + 1 + p.min(df, N/2)with large order mod N.

      • This gives zero chance of decryption failures

      • Minimum q to do so consistent with choice of p, df.

        • Best lattice security

  • Alternatives:

    • Take p = 2+X or 3, q = first power of 2 greater than p(1).min(dr, dg) + 1 + p(1).min(df, N/2)

      • Taking q to be power of 2 speeds up reductions

      • Larger value of p leads to larger q and worse lattice security

    • Take p = 2, q = largest prime less than first power of 2 greater than p(1).min(dr, dg) + 1 + p(1).min(df, N/2)

      • Speeds up reductions at expense of lattice security

    • Allow a non-zero chance of decryption failures, if it can be determined to be less than 2-k.

      • Reduces q, improves bandwidth and lattice security


Df q for different security levels
df, q for different security levels

  • N, df, q, using the above criteria:


Check lattice strength
Check Lattice Strength

  • We characterize the lattice by two variables:

    • c = (2N) . (2)||f||/s. = 2||f||(pe / q)

      • Length of shortest vector [ (2)||f|| ]…

      • Divided by expected length of shortest vector for lattice of the same determinant [ = (N q/pe) ]…

      • Scaled by (2N) .

    • a = N/q.

  • Experimentally, breaking time is very sensitive to c, somewhat sensitive to a.

  • Experimentally, for fixed c, a, breaking time is exponential in N.

  • For all the parameter sets given in the previous slide, we have a >= 1.25, c >= 2.58.


Lattice strength
Lattice Strength

  • Based on the above experiments:

  • Neglecting zero-forcing; also neglecting fact that the lattices under consideration are stronger than the ones experimented on.


More notes on parameter generation
More Notes on Parameter Generation

  • Note how df affects lattice strength:

    • For these parameters, q ~= 2pdf, ||f|| ~= df, c ~= ||f||/q

      • c is ~independent of df!

      • More precisely: ||f|| >= (df/2). Run expts for c = (pe / p) = 2.066?

      • Rounding q up to next prime reduces min(c) slightly, not much.

  • If we use the number of additions, not multiplications, as measure of combinatorial security, we can reduce df by typically 10-25%

    • Gain decreases as N increases

    • Reducing df reduces q potentially improves bandwidth

  • Using product form (f = f1 * f2 + f3) improves efficiency but increases q

    • Increased bandwidth, but typically only by one bit

  • Using trinary (f, g) gives greater combinatorial security

  • So many appealing choices…


Product form
Product Form

  • Interested in meet-in-the-middle attacks on product form

    • df1 = df2 = df3

  • Standard search, described in Tech Note 4, takes on (f1*f2) and (f3)

    • Could also remove a 1 from f2, add df 1s to f3

  • df =~ 0.032 N for N = 251, 0.028 N for N = 769


Product form speedups keygen lattice consideration
Product Form Speedups, Keygen, Lattice Consideration

  • Speed up typically factor of 2 over standard form

  • Keygen will get slow for larger N -- haven’t done calculations

  • For even larger N, guessing zeroes becomes better than guessing f

  • Increased df doesn’t affect lattice strength (much)

    • c = 2.05 experiments would still be fine

  • Bandwidth increases only for k=160 and 192, by 1 bit per coeff


Ntru kem
NTRU-KEM?

ID

b

Hash1

r

r*h

XOR

Hash2

m’

r*h + m’

e

K

Hash3


Parameters for ntru kem
Parameters for NTRU-KEM

  • Still taking p = 2.

  • Now, only have to transmit about 2k bits, so can save bandwidth

  • For k-bit security:

    • Pick N = 2k

    • Pick df = N/2

    • Increase N until combinatorial security is > 2k.

    • Take df, dr, dg, to be the same

    • Take f=1+pF

    • Set q as before


Parameter sets
Parameter Sets

  • This gives the following (note: c >= 2.05, so we omit it).

  • In some cases, slightly increasing N decreases log2(q); we’ve done this where it helps. Note: q needs rounding up.


Speeding up
Speeding up?

  • The parameters above successfully reduce bandwidth

  • Can we improve speeds?

  • Taking small polynomials to be {-1, 0, 1} improves combinatorial security

    • Taking them to be {-2, -1, 0, 1, 2} would do even better, but…

    • The wider the polynomials are, the wider2 their products are


Trinary polynomials
Trinary polynomials

  • Take p to be 3.

  • f, g, r, m could be trinary

  • Two different forms:

    • Balanced: Equal +1s and –1s

    • Biased: Minimum possible number of –1s

      • Set N/2 1s, N/2 0s

      • If this doesn’t give enough combinatorial security, set some of the 0s to –1s.

      • Once there is adequate combinatorial security, see if we can reduce the number of 1s

      • End with dg+ 1s, dg- -1s

      • Combinatorial security estimated as sqrt ((N pick dg+)(N pick dg-)) / N

        • This needs to be made more precise


Polynomial width
Polynomial width

  • Consider a*b:

    • a has da+ +1s, da- -1s

    • b has db+ +1s, db- -1s

    • Maximum value if all +1s line up, all –1s line up.

    • Minimum value if all +1s line up with –1s.

    • Maximum width is Min(da+, db+) + Min(da-, db-) + Min(da+, db-) + Min(da-, db+)

  • Advantage of having one balanced, one biased is we reduce this width compared to two balanced or two biased.

  • Take f, r to be balanced trinary

    • Gives lowest Hamming weight

  • Take g to be biased trinary

  • Consider m on next slide


Choosing n and encoding m
Choosing N and encoding m

  • Say we choose N to be ~2k

    • Then biased polynomials have very few –1s.

  • Want to transmit k bits of entropy

    • Attacker can meet-in-the-middle on m’

  • Could draw m’ from a space of 3N polynomials

    • But this might be tiresome

    • Open question: exactly how tiresome? Certainly tiresome in that output of Hash2 needs to be encoded as random trinary vector, involving repeated mod 3 divides of big integer

  • Suggestion: Take b, output of Hash2, m’ to be binary

    • Once m’ is generated, flip some terms (only 1s or only 0s) to –1s to obtain combinatorial security

    • If more than (say) 4 need to be flipped, generate another b.


Recipe
Recipe

  • Choose N to be first prime > 2k

  • Choose F, r to be balanced trinary

    • (Actually, choose dr+ = dr- +1 for invertibility)

  • Choose g to be biased trinary

  • Choose f = 1+pF

    • f(1) should not be 0 mod 2

  • Say m’ will have no more than 4 –1s

  • Maximum width is df + dr + dg- + 4

  • Set q = the first power of 2 greater than this width


Parameter sets1
Parameter Sets

  • This gives the following (note: c >= 2.11, so we omit it)

  • df ~= 0.115 N, compared to 0.185 for SVES-3; time = N2.


Notes
Notes:

  • We gain a speedup of a factor of about 2 over standard SVES-3

    • Comparable to (though slightly worse than) speedup from move to product form

    • Could consider product form here too, but there seem to be few advantages

  • Bandwidth is about 0.65 of SVES-3 bandwidth

    • Bandwidth is between 16 k and 18.3 k for security k

    • Goes slightly worse than k, slightly better than k ln k.

  • Lattice strength is a BIG question here

    • Not clear that you can get 80 bits of lattice strength at N=163

    • Equally, not 100% clear that you can’t….

    • Can increase lattice strength by beefing up dg+, but this only goes so far

  • Requires an additional SHA at the end

    • But on fewer SHA compression blocks

    • End up with about 2*0.65 = 1.3 as many SHA calls


Parameter generation summary
Parameter Generation: Summary

  • Outlined a possible parameter generation routine for NTRU

    • Put in k, turn the handle, out come the parameters

    • Parameters can be validated by third parties

  • Specific parameter generation routine may change, but basic method remains the same:

    • Choose N

    • Choose form of f, g, r, m

    • Choose q

    • Check lattice strength; if too low, increase N to next prime and try again.


Open questions
Open Questions

  • How many parameter sets do we want?

    • Optimize speed

    • Optimize bandwidth

    • Optimize for 8-bit processors?

      • Can be done by increasing N  decreasing q < 256 (or 128)

  • Do we ever want to allow decryption failures for k > 80?


Open research questions
Open Research Questions

  • Is it okay to use SHA-160 as the core hash function for k > 80?

    • I think yes, but this needs discussion


Key validation
Key Validation

  • What can go wrong with an NTRUEncrypt public key h? Should be random mod q.

    • Might be all zeroes

      • Reveals message immediately

    • If q is composite and gcd(hi, q) != 1 for all hi, might be possible to recover message from ciphertext by simple modular reduction.

    • If h is too thin, such that r*h will have very few mod q reductions (< 2k effort to guess reduction locations), can recover message from ciphertext by linear algebra.

  • Possible simple key validation procedure:

    • Check that keys are not all the same value

    • Check that sufficient number of hi have gcd(hi, q) = 1

    • Check that width of h > c. q/df, c > 1 some parameter set dependent constant.


Key validation 2
Key Validation (2)

  • More sophisticated:

    • Measure how “random” h looks. For example:

      • Chance that a given mod q value does not occur anywhere in h = (1-1/q)N

      • Find value l such that for random h the probability that l distinct values do not occur anywhere in h is less than 2-k.

        • A different bound may be appropriate

      • Count the number of distinct values that do not occur in h and reject if greater than l.

  • Next draft will contain suggested text.


Issue forward secrecy 1
Issue: Forward Secrecy (1)

  • NTRU key generation is efficient

    • Can generate ephemeral keypairs easily

  • This + next slide propose three ways of getting perfect forward secrecy using this fact

    • Do these actually give forward secrecy?

    • Do they give mutual authentication?

    • Should they be included in the standard?

  • (1) Say Alice has static keypairs (as, As).

    • Bob generates ephemeral keypair (be, Be), sends Alice EAs(Be).

      • This may have to be signed

    • Alice uses Be as the public key for key transport or key agreement

    • Afterwards, Bob disposes of (be, Be).


Issue forward secrecy 2
Issue: Forward Secrecy (2)

  • (2) Say Alice and Bob have static, certified keypairs (as, As), (bs, Bs).

    • Bob generates ephemeral keypair (be, Be), sends Alice EAs(Be).

    • Alice uses both Bs and Be in two runs of a key transport or key agreement mechanism, combines the two transported keys to get a single shared key.

    • Afterwards, Bob disposes of (be, Be).

  • (3) Say Alice and Bob have static, certified keypairs (as, As), (bs, Bs).

    • Bob generates ephemeral keypair (be, Be), sends Alice EAs(Be).

    • Alice generates ephemeral keypair (ae, Ae), sends Bob EBs(Ae).

    • Alice uses Be to transport secret k1 to Bob

    • Bob uses Ae to transport secret k2 to Alice

    • Bob and Alice combine k1, k2 to get shared secret k.

  • Note: need to define encryption carefully above: will probably be symmetric+public-key operation


That s it
That’s it!

  • Questions?


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