Headline

GHSA-q3hw-3gm4-w5cr: gnark's Groth16 commitment extension unsound for more than one commitment

Description

The summary is that the proof of knowledge associated to a commitment is crucial to bind the commitment to the actual circuit variables that were supposed to be committed. However, the same σ is used for all proofs of knowledge for the commitments, which allows mixing between them, making it possible to fix the value of all but one commitment before choosing the circuit variable assignments.

In more detail: To simplify notation, let us consider the case of two commitments, each to only a single variable. Let’s say the basis elements for those commitments are K_0 and K_1. Then the proving key will contain K_0 and K_1, and also σ*K_0 and σ*K_1 for the proof of knowledge. The honest prover assigning a to the first circuit variable and b to the second will then produce commitments D_0 = a*K_0 D_1 = b*K_1 Out of the two D’s, a challenge r for the commitment folding will be generated. The folded commitment will then be D_folded = D_0 + r*D_1 = a*K_0 + r*b*K_1 The honest prover will supply a fitting proof of knowledge P = a*(σ*K_0) + r*b*(σ*K_1)

Now the verifier will only use all of this in two ways:

In the check of the Groth16 proof itself, where only the sum D_0 + D_1 is used.
In the proof of knowledge check, where it will be verified that P is indeed σ*(D_0 + r*D_1), with r calculated from D_0 and D_1 as before.

This has the following implications. In the following, a malicious prover’s points will have an apostrophe appended, and we keep D_0 etc. for the legitimate values:

A malicious prover is only forced to provide D'_0 and D'_1 such that the sum is correct. So they can use arbitrary D'_0 as long as they set D'_1 = D_0 + D_1 - D'_0.
After choosing D'_0 and D'_1, the prover can always calculate r. Evaluating σ*(D'_0 + r*D'_1) is then possible as long as both D'_0 and D'_1 are linear combinations of basis elements for which σ times that basis element is known. In particular, this works as long as D'_0 and D'_1 are linear combinations of K_0 and K_1.

The upshot is that a malicious prover can choose arbitrary a’ and b’, and then set D'_0 = a'*K_0 + b'*K_1 D'_1 = (a - a')*K_0 + (b - b')*K_1 Then they calculate r for this, and set P = (a' + r*(a-a'))*(σ*K_0) + (b' + r*(b-b'))*(σ*K_1) This will then be accepted as a valid proof. Yet the first commitment point can be chosen completely independently of a and b, so in particular the malicious prover can use a constant for this, so that they will know the in-circuit challenge that will be added to the public inputs before they have to choose the witness assignments. For most use cases of such challenges (for proving things with Fiat-Shamir, random linear combinations etc.) this causes a critical soundness problem.

The problem generalizes to more than two commitments and commitments to more than one circuit variable each; one can freely choose all but one commitment as arbitrary linear combinations of the basis elements for all commitments, and then must choose the one remaining commitment in such a way that the sum is correct.

The root cause of the issue is that the σ used for the proofs of knowledge is the same, allowing to mix between the basis elements, as one has σ times them available for all of them. So the fix is to have a separate σ for each commitment. So in our example above, the proving key would have the basis elements K_0 and K_1, and for the proofs of knowledge now σ_0*K_0 and σ_1*K_1. Folding the commitments would not be possible in the same way now, so the verifier will have to do more pairings. The prover could still provide a folded proof of knowledge however. With D_0 = a*K_0 D_1 = b*K_1 the proof of knowledge would be P = a*(σ_0*K_0) + r*b*(σ_1*K_1) For later, let us use notation for the unfolded proofs of knowledge P_0 = a*(σ_0*K_0) P_1 = b*(σ_1*K_1) so that P = P_0 + r*P_1

The verifying key would need G and σ_0*G and σ_1*G. To check the two unfolded proofs of knowledge would be the checks e(P_0, G) = e(D_0, σ_0*G) e(P_1, G) = e(D_1, σ_1*G) As r is a challenge derived from D_0 and D_1, we may instead check e(P_0, G) + r*e(P_1, G) = e(D_0, σ_0*G) + r*e(D_1, σ_1*G) The left hand side is e(P_0, G) + r*e(P_1, G) = e(P_0 + r*P_1, G) = e(P, G) So the prover can just provide P and then the verifier checks e(P, G) = e(D_0, σ_0*G) + r*e(D_1, σ_1*G) Unfortunately, the right hand side can’t be folded as before, as there isn’t a side of the pairing that is kept constant between the pairings as before. So the verifier will need to have a pairing for each commitment on the right hand side.

Impact

It is a soundness issue - in case of multiple commitments used inside the circuit the prover is able to choose all but the last commitment. As we use the commitments for optimized non-native multiplication, lookup checks etc. as random challenges, then it could impact the soundness of the whole circuit.

However, using multiple commitments has been discouraged due to the additional cost to the verifier and it has not been supported in the recursive in-circuit Groth16 verifier and Solidity verifier. So we expect the impact of the issue be very small - only for the users who have implemented the native Groth16 verifier or are using it with multiple commitments. We do not have information of such users.

Patches

The issue has been patched in e7c66b000454f4d2a4ae48c005c34154d4cfc2a2

Workarounds

The recommendation has been to use only a single commitment and then derive in-circuit commitments as needed using std/multicommit package.

References

See the correspondence above.

2 months ago

ghsa

Open in Source

#git

Description

In more detail:
To simplify notation, let us consider the case of two commitments, each to only a single variable. Let’s say the basis elements for those commitments are K_0 and K_1. Then the proving key will contain K_0 and K_1, and also σK_0 and σK_1 for the proof of knowledge. The honest prover assigning a to the first circuit variable and b to the second will then produce commitments
D_0 = aK_0
D_1 = bK_1
Out of the two D’s, a challenge r for the commitment folding will be generated. The folded commitment will then be
D_folded = D_0 + rD_1 = aK_0 + rbK_1
The honest prover will supply a fitting proof of knowledge
P = a(σK_0) + rb(σ*K_1)

Now the verifier will only use all of this in two ways:

In the check of the Groth16 proof itself, where only the sum D_0 + D_1 is used.
In the proof of knowledge check, where it will be verified that P is indeed σ(D_0 + rD_1), with r calculated from D_0 and D_1 as before.

This has the following implications. In the following, a malicious prover’s points will have an apostrophe appended, and we keep D_0 etc. for the legitimate values:

A malicious prover is only forced to provide D’_0 and D’_1 such that the sum is correct. So they can use arbitrary D’_0 as long as they set D’_1 = D_0 + D_1 - D’_0.
After choosing D’_0 and D’_1, the prover can always calculate r. Evaluating σ(D’_0 + rD’_1) is then possible as long as both D’_0 and D’_1 are linear combinations of basis elements for which σ times that basis element is known. In particular, this works as long as D’_0 and D’_1 are linear combinations of K_0 and K_1.

The upshot is that a malicious prover can choose arbitrary a’ and b’, and then set
D’_0 = a’K_0 + b’K_1
D’_1 = (a - a’)K_0 + (b - b’)K_1
Then they calculate r for this, and set
P = (a’ + r(a-a’))(σK_0) + (b’ + r(b-b’))(σK_1)
This will then be accepted as a valid proof. Yet the first commitment point can be chosen completely independently of a and b, so in particular the malicious prover can use a constant for this, so that they will know the in-circuit challenge that will be added to the public inputs before they have to choose the witness assignments. For most use cases of such challenges (for proving things with Fiat-Shamir, random linear combinations etc.) this causes a critical soundness problem.

The root cause of the issue is that the σ used for the proofs of knowledge is the same, allowing to mix between the basis elements, as one has σ times them available for all of them.
So the fix is to have a separate σ for each commitment. So in our example above, the proving key would have the basis elements K_0 and K_1, and for the proofs of knowledge now σ_0K_0 and σ_1K_1. Folding the commitments would not be possible in the same way now, so the verifier will have to do more pairings. The prover could still provide a folded proof of knowledge however. With
D_0 = aK_0
D_1 = bK_1
the proof of knowledge would be
P = a(σ_0K_0) + rb(σ_1K_1)
For later, let us use notation for the unfolded proofs of knowledge
P_0 = a(σ_0K_0)
P_1 = b(σ_1K_1)
so that
P = P_0 + rP_1

The verifying key would need G and σ_0G and σ_1G. To check the two unfolded proofs of knowledge would be the checks
e(P_0, G) = e(D_0, σ_0G)
e(P_1, G) = e(D_1, σ_1G)
As r is a challenge derived from D_0 and D_1, we may instead check
e(P_0, G) + re(P_1, G) = e(D_0, σ_0G) + re(D_1, σ_1G)
The left hand side is
e(P_0, G) + re(P_1, G) = e(P_0 + rP_1, G) = e(P, G)
So the prover can just provide P and then the verifier checks
e(P, G) = e(D_0, σ_0G) + re(D_1, σ_1*G)
Unfortunately, the right hand side can’t be folded as before, as there isn’t a side of the pairing that is kept constant between the pairings as before. So the verifier will need to have a pairing for each commitment on the right hand side.