Re: [sfc] [Last-Call] Secdir last call review of draft-ietf-sfc-proof-of-transit-08

[Christian Huitema <[e-mail]>]

On 9/23/2021 12:31 PM, Frank Brockners (fbrockne) wrote:
> Hi Christian,
>
> Thanks a lot for your detailed review. Please see inline.
>
>> -----Original Message-----
>> From: Christian Huitema via Datatracker <[e-mail]>
>> Sent: Monday, 20 September 2021 05:48
>> To: [e-mail]
>> Cc: [e-mail]; [e-mail]; [e-mail]
>> Subject: Secdir last call review of draft-ietf-sfc-proof-of-transit-08
>>
>> Reviewer: Christian Huitema
>> Review result: Serious Issues
>>
>> I have reviewed this document as part of the security directorate's  ongoing
>> effort to review all IETF documents being processed by the  IESG.  These
>> comments were written primarily for the benefit of the security area directors.
>> Document editors and WG chairs should treat these comments just like any
>> other last call comments.
>>
>> This document proposes a security mechanism to prove that traffic transited
>> through all specified nodes in a path. The mechanism works by adding a short
>> option to each packet for which transit shall be verified. The option consists of a
>> random number set by the originator of the packet, and a sum field to which
>> each transit node adds a value depending on public parameters, on the random
>> number and on secrets held by the node. The destination has access to all the
>> secrets held by the nodes on the path, and can verify whether or not the final
>> sum corresponds to the sum of expected values. The proposed size of the
>> random number and the sum field is 64 bits.
>>
>> In the paragraph above, I described the mechanism without mentioning the
>> algorithm used to compute these 64 bit numbers. The 64 bit size is obviously a
>> concern: for cryptographic applications, 64 bits is not a large number, and that
>> might be a weakness whatever the proposed algorithm. The actual algorithm
>> appears to be a bespoke derivation of Shamir's Secret Sharing algorithm (SSS). In
>> other word, it is a case of "inventing your own crypto".
> ...FB: SSS is a well know algorithm and draft-ietf-sfc-proof-of-transit does not modify it.
> All draft-ietf-sfc-proof-of-transit does is to operationalize the SSS algorithm for the proof of transit use case.
>
> Also note that the draft does not require the use of 64 bit numbers.
> Nor does draft require a minimum time between changing the secrets.
> What particular attack are you concerned about where 64 bit numbers are a concern?
>
>> SSS relies on the representation of polynomials as a sum of Lagrange Basis
>> Polynomials. Each of the participating nodes holds a share of the secret
>> represented by a point on the polynomial curve. A polynomial of degree K on the
>> field of integers modulo a prime number N can only be revealed if at list K+1
>> participants reveal the value of their point. The safety of the algorithm relies on
>> the size of the number N and on the fact that the secret shall be revealed only
>> once. But the algorithm does not use SSS directly, so it deserves its own security
>> analysis instead of relying simply on Shamir's work.
>>
>> The proposed algorithm uses two polynomials of degree K for a path containing
>> K+1 nodes, on a field defined by a prime number N of 64 bits. One of the
>> polynomial, POLY-1, is secret, and only fully known by the verifying node.
>> The other, POLY-2 is public, with the constant coefficient set at a random value
>> RND for each packet.
>>
>> For each packet, the goal is compute the value of POLY-1 plus POLY-2 at the
>> point 0 -- that is, the constant coefficient of POLY-3 = POLY-1 + POLY-2.
>>
>> Without going in too much details, one can observe that the constant
>> coefficient of POLY-3 is equal to the sum of the constant coefficients of POLY-1
>> and POLY-2, and that the constant coefficient of POLY-2 is the value RND
>> present in each packet. In the example given in section 3.3.2, the numbers are
>> computed modulo 53, the constant coefficient of POLY-1 is 10, and the value
>> RND is 45. The final sum  CML is indeed
>> 10 + 45 = 2 mod 53.
>>
>> To me, this appears as a serious weakness in the algorithm. If an adversary can
>> observe the value RND and CML for a first packet, it can retrieve the constant
>> coefficient of POLY-1, and thus can predict the value of CML for any other
>> packet. That does not seem very secure.
> ...FB: There seems to be a bit of confusion or misreading of how the method works. In the above statement you seem to assume that the verifier would not be part of the proof-chain, so that the final CML value would be somehow exposed to an external entity along with RND. This is not the case. The verifier is the last node (k+1) in the proof-chain.
>
> At concept level, the method reconstructs the polynomial hop by hop, picking up a point on the curve at every hop. Only final node in the proof-chain, which is also the verifier, acts on the information of all the k+1 points and as such is able to reconstruct the polynomial.
>
> In section 3.2.1, the draft explicitly states that the verifier *is* part of the proof-chain: "Each of the k+1 nodes (including verifier) are assigned a point on the polynomial i.e., shares of the SECRET." The fact that the verifier, i.e., the last node in the proof-chain ("k+1"),  can retrieve the secret, is desired and intentional, because the verifier needs to compare the result of the iterative construction of the secret with the secret value it received from the controller. This is how the system is designed, and the calculation of (10+45) mod 53 = 2 is part of the verification.

OK. That's slightly less bad. But it is still very bad crypto, because 
you are effectively doing a linear combination.

You are evaluating POLY-3 = POLY-1 + POLY-2

POLY-2 can be written as POLY-2 = RND + POLY-2-NC, in which POLY2-NC 
only contains the non constant terms -- that is, POLY-2-NC(0) = 0

Then for any point X, we get POLY-3(X) = POLY-1(X) + POLY2-NC(X) + RND
For a given value Xj of X, this means we can express : POLY-3(Xj) = Vj + RND
In which Vj is a constant term = POLY-1(Xj) + POLY2-NC(Xj)

Each node will increment the cumul by the value LPCj * POLY-3(Xj) = LPCj 
* (Vj + RND)

Suppose that an adversary can observe the value of CML before and after 
being incremented by node Xj. Suppose that it could do that twice. Then 
it has the values:

CML1-before-j = C1b
CML1-after-j = C1a
D1 = C1a - C1b = LPCj * (Vj + RND1)

CML1-before-j = C2b
CML1-after-j = C2a
D2 = C2a - C2b = LPCj * (Vj + RND2)

D2-D1 = LPCj*(RND2-RND1)

LPCj = (RND2-RND1)/(D2-D1)
Vj = D2/LPCj - RND2

The inverse of numbers modulo a prime P is easily computed -- see 
Fermat's little theorem.

Once the input and output of a node have been observed twice, it becomes 
easy to update the cumulative sum CML while bypassing these nodes.

The scheme described in the draft is definitely not equivalent to SSS. 
It boils down to linear combinations of coefficients, and it is not secure.

-- Christian Huitema