### Slides

```Probabilistic Public Key Encryption with Equality Test
Duncan S. Wong
Department of Computer Science
City University of Hong Kong
Joint work with Guomin Yang, Chik How Tan and Qiong Huang
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Outline
• What is PKE with Equality Test?
• Is it related to PKE with Keyword Search or Deterministic PKE?
• Applications
• Our construction
• What security level can it achieve?
• Impossibility of achieving IND-ATK (e.g. IND-CPA or IND-CCA1/2)
• Extension: a non-pairing variant
• W-IND-CCA2
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What is PKE with Equality Test (PKE-ET)?
pk1
M1
Enc
C1
M1 =? M2
pk2
M2
Enc
C2
C1
Test
1 iff M1 = M2
C2
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What is PKE with Equality Test (PKE-ET)?
1. Perfect Consistency
For every M in plaintext space PtSp(k),
Pr[ Test(C1, C2) = 1 ] = 1
if (pk1, sk1)  G(1k), (pk2, sk2)  G(1k), C1  E(pk1, M) and C2  E(pk2, M).
2. Soundness
For any PPT A,
Pr[ Test(C1, C2) = 1  M1   M2   M1  M2 ]  (k)
where (C1, C2, sk1, sk2)  A(1k), M1  D(sk1, C1), M2  D(sk2, C2).
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Is PKE-ET related to PKE with Keyword Search?
PKE with Keyword Search (PKES)
• w : keyword
• C = Enc(pk, w)
• TW = Trapdoor(sk, w)
• Test(pk, C, TW) = 1 iff C is an encryption of w under pk
Equality Test
• Test(pk, C1, TW) = 1 & Test(pk, C2, TW) = 1
 Both C1 and C2 are encryptions of the same w.
Limitations
1. A tag TW can only be generated if sk is known.
2. Test: only applicable to ciphertexts generated under the same pk.
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Is PKE-ET related to Deterministic PKE?
Deterministic Public Key Encryption (DPKE)
• S = Enc(pk, M)
• M = Dec(sk, C)
Equality Test
• Given C1 = Enc(pk, M1) & C2 = Enc(pk, M2)
C1 = C2  M1 = M2.
Limitation
1. Only applicable to ciphertexts generated under the same pk.
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Applications of PKE-ET
Outsourced Database, data are stored in encrypted form.
1. Searchable Encryption: anyone is able to search keywords of encrypted
messages even if they are generated under different public keys.
• E.g. building a search engine capable of searching encrypted messages
provided by different vendors
2. Partitioning Encrypted Data: DBMS or the public is able to categorize or
obtain statistical information on messages without any help from the
encrypted message owners.
• E.g. partitioning encrypted files based on file types such as images from
videos
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Our PKE-ET Construction
System Parameters
•
•
•
•
G1, G2: cyclic groups of prime order q
g: generator of G1
Bilinear pairing e: G1 x G1  G2
PtSp: G1\{1}
KeyGen(1k)
• sk = x R Zq*
• pk = y = gx
Enc(pk, m)
1. r R Zq*
2. Ciphertext C := (U, V, W) where U = gr, V = mr, W = H(U, V, yr)  m||r
Dec(sk, C)
1. m||r  WH(U, V, Ux)
2. Verify r  Zq*  m  G1\{1}  U = gr  V = mr
3. If true, return m, else return 
Test(C1, C2)
• Given C1 = (U1, V1, W1) and C2 = (U2, V2, W2), if e(U1, V2) = e(U2, V1), return 1, else return 0.
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What Security Level can our PKE-ET scheme achieve?
(Impossibility of Achieving IND-ATK)
In general, PKE-ET cannot achieve IND-ATK (e.g. IND-CPA or IND-CCA1/2).
IND-ATK:
Reason why PKE-ET cannot achieve IND-ATK: adversary knows the challenge plaintexts
x0 and x1; does not even need to resort its plaintext choosing capability.
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What Security Level can our PKE-ET scheme achieve?
After challenge phase, the adversary knows:
public key: pk
challenge plaintexts: x0 and x1
challenge ciphertext: y
• computes y’ = Enc(pk’, x1)
• returns Test(y, y’)
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What Security Level can our PKE-ET scheme achieve?
It achieves one-way under chosen ciphertext attack (OW-CCA2).
OW-ATK:
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What Security Level can our PKE-ET scheme achieve?
OW-CCA2 security in the random oracle model under the CDH assumption
Proof Idea:
• Game 1: the original scheme
Enc(pk, m) : U = gr, V = mr, W = H(U, V, yr)  m||r
• Game 2: Replace H(U*, V*, yr*) of the challenge ciphertext with a random string
Enc(pk, m*) : U* = gr*, V* = mr*, W* = R*  m||r
• Game 1 and Game 2 are indistinguishable under the CDH assumption.
• The adversary only has a negligible probability to win in Game 2 under
the CDH assumption.
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Extension: a non-pairing variant
• In the PKE-ET, pairing is used in Test only.
• If we remove Test, the scheme is a conventional PKE.
KeyGen(1k)
• sk = x R Zq*
• pk = y = gx
Enc(pk, m)
• r R Zq*
• Compute U = gr, V = mr, W = H(U, V, yr)m||r
• C := (U, V, W)
Dec(sk, C)
• m||r  WH(U, V, Ux)
• Verify r R Zq*  m  G1\{1}  U = gr  V = mr
• If true, return m, else return 
• The PKE can be implemented using a non-bilinear group. So we have
more curves to choose from during implementation.
• Observation: in a non-bilinear group, this PKE achieves a higher security level.
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Extension: a non-pairing variant
• Bad News: still cannot achieve IND-ATK
• A1 chooses x0 = gr0, x1 = gr1 where r0  r1
• challenge stage: b  {0,1}, Enc(pk, xb) = (U = gr, V = xbr, W)
• A2 returns 0 if V = Ur0; otherwise, returns 1.
• Good News: can achieve something stronger than OW-CCA2
W-IND-ATK where the adversary cannot select challenge plaintexts but
the adversary is given the challenge plaintexts.
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W-IND-ATK
• In the random oracle model, the PKE in a non-bilinear group is W-IND-CCA2
secure under the DDH assumption.
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Future Work
• Standard model construction
• Achieving IND-CCA2 for Test-removed version
• Question: is there any application for the property that the same scheme is PKE-ET
on bilinear group while being a PKE on non-bilinear group?
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Q&A
More details can be found in the Proc. of CT-RSA 2010
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