By Newman, Moskowitz, Chang, Brahmadesam

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R` afols 1. The challenger speciﬁes a universe of attributes P of size m and gives it to the attacker A. 2. A selects a subset S ⊂ P of s attributes and a threshold t such that 1 ≤ t ≤ s. 3. Setup(1λ , P) and gives params to A. 4. Ext(params, B, msk) as the answer. 5. A outputs two messages M0 , M1 of the same length. 6. Enc(params, S, t, Mb ) and gives C to A. 7. Step 4 is repeated. 8. A outputs a bit b. The advantage of such an adversary A in breaking the IND-sCPA security of the ABE scheme is deﬁned as AdvIND-sCPA A,ABE (λ) = |2 Pr[b = b ] − 1| .

The main trick in the proof will be to use the input of the aMSE-DDH problem to compute evaluations of some polynomials in γ “in the exponent”. → Let I(− x 2m+t−1−s , κ, α, γ, ω, T ) be the input of the algorithm B. First, B speciﬁes a universe of attributes, P = {at1 , . . , atm }. Next, the adversary A chooses a set S ⊂ P of cardinal s that he wants to attack, and a threshold t such that 1 ≤ t ≤ s. Without loss of generality, we assume S = {atm−s+1 , . . , atm } ⊂ P. From now on, we will denote by AS the subset A ∩ S, for any subset of attributes A.

For any adversary A against the IND-sCPA security of our attribute-based encryption scheme, for a universe of m attributes P, and a challenge pair (S, t) with s = |S|, there exists a solver B of the ( ˜, m, ˜ t˜)aMSE-DDH problem, for ˜ = m − s, m ˜ = m + t − 1 and t˜ = t + 1, such that (λ) ≥ AdvaMSE-DDH B 1 · AdvIND-sCPA (λ). A 2 Proof. We are going to construct an algorithm B that uses the adversary A as a black-box and that solves the (m − s, m + t − 1, t + 1)-augmented multi-sequence of exponents decisional Diﬃe-Hellman problem.