Generating Sharper and Simpler Nonlinear Interpolants for Program - PowerPoint PPT Presentation

Generating Sharper and Simpler Nonlinear Interpolants for Program Verification Takamasa Okudono 1 , Yuki Nishida 2 , Kensuke Kojima 2 , Kohei Suenaga 2 , Kengo Kido 1 and Ichiro Hasuo 3 1 University of Tokyo, Japan 2 Kyoto University, Japan 3 National Institute of Informatics, Japan APLAS 2017, Suzhou, China. November 28 th 2017 Takamasa Okudono (University of Tokyo) 1

Interpolant is effective Purpose of This Work at program verification Disjointness of the regions • Automatic generation of polynomial interpolants. Def. [interpolant] • 𝐵, 𝐶: Formulas satisfying ⊨ ¬(𝐵 ∧ 𝐶) . Essential to • Formula 𝐽 is an interpolant of 𝐵 and 𝐶 if: separate the regions 1. ⊨ 𝐵 → 𝐽 2. ⊨ ¬(𝐶 ∧ 𝐽) 3. Variables in 𝐽 appear in both of 𝐵, 𝐶 For polynomial interpolants, atomic propositions are: (Poly.) ≧ 0, (Poly.)>0, (Poly.)=0 2 Takamasa Okudono (University of Tokyo)

Existing Work • [Dai+, CAV’13]: generation of polynomial interpolants with numerical optimization • Challenge 1: Unable to generate any interpolants in “ touching ” cases • Challenge 2: Incorrect and complex due to numerical errors Touching 54.1800𝑦 + 108.3601𝑧 ≥ 0 𝑦 + 2𝑧 ≥ 0 Takamasa Okudono (University of Tokyo) 3

Our Contribution • [Dai+, CAV’13]: generation of polynomial interpolants with numerical optimization • Challenge 1: Unable to generate any interpolants in “ touching ” cases • Challenge 2: Incorrect and complex due to numerical errors Touching 54.1800𝑦 + 108.3601𝑧 ≥ 0 Solved! 𝑦 + 2𝑧 ≥ 0 Solved! (Contribution 1) (Contribution 2) Takamasa Okudono (University of Tokyo) 4

Challenge 1 in [Dai+]: Sharpness • If two regions of 𝐵, 𝐶 are “touching”, [Dai+, CAV’13] always fails at generating their interpolant. • 𝐵 = (𝑧 − 𝑦 > 0 ∧ 𝑧 + 𝑦 > 0) Touching • 𝐶 = (−𝑧 ≥ 0) Takamasa Okudono (University of Tokyo) 5

Challenge 1 in [Dai+]: Sharpness • If two regions of 𝐵, 𝐶 are “touching”, [Dai+, CAV’13] always fails at generating their interpolant. • 𝐵 = (𝑧 − 𝑦 > 0 ∧ 𝑧 + 𝑦 > 0) Touching • 𝐶 = −𝑧 ≥ 0 • 𝐽 = 𝑧 > 0 • There is an interpolant, but [Dai, CAV’13] cannot find it! Takamasa Okudono (University of Tokyo) 6

Challenge 1 in [Dai+]: Sharpness • If two regions of 𝐵, 𝐶 are “touching”, [Dai+, CAV’13] always fails at generating their interpolant. Takamasa Okudono (University of Tokyo) 7

Challenge 1: Flow of [Dai+] [Parrilo , Mathematical Programming’03] (2) Polynomial (1) Formulas (3) SDP ↦ ↦ Optimization Use SDP Solver 𝐵, 𝐶 Problem Problem ↦ (5) Numerical (4) Numerical (6) Interpolant ↦ ↦ Solution Solution 𝐽 of (2) of (3) Takamasa Okudono (University of Tokyo) 8

Contribution 1: Method for Sharpness [Parrilo , Mathematical Programming’03] Method for Sharpness (2) Polynomial (1) Formulas (3) SDP ↦ ↦ Optimization Use SDP Solver 𝐵, 𝐶 Problem Problem ↦ (5) Numerical (4) Numerical (6) Interpolant ↦ ↦ Solution Solution 𝐽 of (2) of (3) Takamasa Okudono (University of Tokyo) 9

Contribution 1: Example Method for Sharpness (2) Polynomial (1) Formulas ↦ Optimization 𝐵, 𝐶 Problem 𝐵 = y − x > 0, y + x > 0 , 𝐶 = (−𝑧 ≥ 0) Takamasa Okudono (University of Tokyo) 10

Contribution 1: Example [Dai+, CAV’13] • 𝐵 = (y − x > 0, y + x > 0), 𝐶 = (−𝑧 ≥ 0) • [Dai+, CAV’13] • Find polynomials 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 ∈ ℝ[𝑌] s.t. 1 𝑧 + 𝑦 + 𝑧 − 𝑦 2 𝑧 + 𝑦 2 • 𝐽 ≔ 2 + 𝜏 1 + 𝜏 2 𝑧 − 𝑦 + 𝜏 3 𝑧 + 𝑦 + 𝜏 4 𝑧 − 𝑦 • 𝐽 ′ ≔ 1 2 + 𝜏 5 (−𝑧) • 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 are sums of squares • 𝐽 + 𝐽 ′ = 0 𝜏 is a sum of squares • ( 𝐽 contains only 𝑧 ) ⟺ • Then 𝐽 > 0 is an interpolant 2 + ⋯ + 𝜒 𝑜 2 ∃𝜒 1 , … , 𝜒 𝑜 ∈ ℝ 𝑌 ; 𝜏 = 𝜒 1 Takamasa Okudono (University of Tokyo) 11

Contribution 1: Example [Dai+, CAV’13] • 𝐵 = (y − x > 0, y + x > 0), 𝐶 = (−𝑧 ≥ 0) • [Dai+, CAV’13] • Find polynomials 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 ∈ ℝ[𝑌] s.t. 1 𝑧 + 𝑦 + 𝑧 − 𝑦 2 𝑧 + 𝑦 2 • 𝐽 ≔ 2 + 𝜏 1 + 𝜏 2 𝑧 − 𝑦 + 𝜏 3 𝑧 + 𝑦 + 𝜏 4 𝑧 − 𝑦 • 𝐽 ′ ≔ 1 2 + 𝜏 5 (−𝑧) • 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 are sums of squares • 𝐽 + 𝐽 ′ = 0 • ( 𝐽 contains only 𝑧 ) • Then 𝐽 > 0 is an interpolant Infeasible and unable to generate any interpolants! Takamasa Okudono (University of Tokyo) 12

Contribution 1: Example [Dai+, CAV’13] • 𝐵 = (y − x > 0, y + x > 0), 𝐶 = (−𝑧 ≥ 0) • [Dai+, CAV’13] • Find 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 ∈ ℝ[𝑌] s.t. 1 𝑧 + 𝑦 + 𝑧 − 𝑦 2 𝑧 + 𝑦 2 • 𝐽 ≔ 2 + 𝜏 1 + 𝜏 2 𝑧 − 𝑦 + 𝜏 3 𝑧 + 𝑦 + 𝜏 4 𝑧 − 𝑦 • 𝐽 ′ ≔ 1 2 + 𝜏 5 (−𝑧) ∵ Assume the feasibility. • 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 are sums of squares 0 = 𝐽 + 𝐽 ′ 0, 0 • 𝐽 + 𝐽 ′ = 0 = 1 + 𝜏 1 0, 0 > 0. • ( 𝐽 contains only 𝑧 ) Contradiction. □ • Then 𝐽 > 0 is an interpolant Infeasible and unable to generate any interpolants! Takamasa Okudono (University of Tokyo) 13

Contribution 1: Example [Dai+, CAV’13] • 𝐵 = (y − x > 0, y + x > 0), 𝐶 = (−𝑧 ≥ 0) • Our method for sharpness • Find polynomials 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 ∈ ℝ[𝑌] and 𝑠 1 , 𝑠 2 , 𝑠 3 ∈ ℝ ≥0 s.t. • 𝐽 ≔ 𝜏 1 + 𝜏 2 𝑧 − 𝑦 + 𝜏 3 𝑧 + 𝑦 + 𝜏 4 𝑧 − 𝑦 𝑧 + 𝑦 + 𝑠 1 + 𝑠 2 𝑧 − 𝑦 + 𝑠 3 𝑧 + 𝑦 • 𝐽 ′ ≔ 𝜏 5 −𝑧 • 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 are sums of squares • 𝑠 1 + 𝑠 2 + 𝑠 3 > 0 • 𝐽 + 𝐽 ′ = 0 • 𝐽 contains only 𝑧 • Then 𝐽 > 0 is an interpolant Takamasa Okudono (University of Tokyo) 14

Contribution 1: Example [Dai+, CAV’13] 𝑠 3 = 0 𝑠 2 = 1 𝑠 1 = 0 • 𝐵 = (y − x > 0, y + x > 0), 𝐶 = (−𝑧 ≥ 0) 𝜏 3 = 1 𝜏 4 = 0 𝜏 2 = 0 • Our method for sharpness 𝜏 1 = 0 • Find polynomials 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 ∈ ℝ[𝑌] and 𝑠 1 , 𝑠 2 , 𝑠 3 ∈ ℝ ≥0 s.t. • 𝐽 ≔ 𝜏 1 + 𝜏 2 𝑧 − 𝑦 + 𝜏 3 𝑧 + 𝑦 + 𝜏 4 𝑧 − 𝑦 𝑠 5 = 2 𝑧 + 𝑦 + 𝑠 1 + 𝑠 2 𝑧 − 𝑦 + 𝑠 3 𝑧 + 𝑦 • 𝐽 ′ ≔ 𝜏 5 −𝑧 • 𝜏 1 , 𝜏 2 , 𝜏 3 , 𝜏 4 , 𝜏 5 are sums of squares • 𝑠 1 + 𝑠 2 + 𝑠 3 > 0 𝐽 = 2𝑧 • 𝐽 + 𝐽 ′ = 0 • 𝐽 contains only 𝑧 • Then 𝐽 > 0 is an interpolant Feasible and able to generate an interpolant! Takamasa Okudono (University of Tokyo) 15

Contribution 1: Completeness A and B s.t. ⊨ ¬(A∧ B) Method for Sharpness “touching” cases [Dai+] Takamasa Okudono (University of Tokyo) 16

Challenge 2: Numerical Error in [Dai+] (2) Polynomial (1) Formulas (3) SDP ↦ ↦ Optimization Use SDP Solver 𝐵, 𝐶 Problem Problem ↦ (5) Numerical (4) Numerical (6) Interpolant ↦ ↦ Solution Solution 𝐽 of (2) of (3) Numerical Error Takamasa Okudono (University of Tokyo) 17

Challenge 2: Numerical Error in [Dai+] (2) Polynomial (1) Formulas (3) SDP ↦ ↦ Optimization Use SDP Solver 𝐵, 𝐶 Problem Problem ↦ (5) Numerical (4) Numerical (6) Interpolant ↦ ↦ Solution Solution 𝐽 of (2) of (3) Numerical Numerical Error Error Takamasa Okudono (University of Tokyo) 18

Challenge 2: Numerical Error in [Dai+] (2) Polynomial (1) Formulas (3) SDP ↦ ↦ Optimization Use SDP Solver 𝐵, 𝐶 Problem Problem Less ↦ simple (5) Numerical (4) Numerical (6) Interpolant ↦ ↦ Maybe Solution Solution 𝐽 spurious of (2) of (3) Numerical Numerical Numerical Error Error Error Numerical error spreads Takamasa Okudono (University of Tokyo) 19

Challenge 2: Numerical Error in [Dai+] (2) Polynomial (1) Formulas (3) SDP ↦ ↦ Optimization Use SDP Solver 𝐵, 𝐶 Problem Problem Less ↦ simple The same problem occurs in our method for sharpness (Contribution 1) (5) Numerical (4) Numerical (6) Interpolant ↦ ↦ Maybe Solution Solution 𝐽 spurious of (2) of (3) Numerical Numerical Numerical Error Error Error Numerical error spreads Takamasa Okudono (University of Tokyo) 20

Challenge 2: Example • Example: 𝐵 = 𝑦 = 0 ∧ 𝑧 = 0 , 𝐶 = (𝑦 + 2𝑧 < 0) • Spurious interpolant 𝐽 = 54.1800𝑦 + 108.3601𝑧 ≥ 0 • 𝑦, 𝑧 = (−108.3601, 54.1800) satisfies both 𝐶 and 𝐽 Def. [interpolant] • 𝐵, 𝐶: Formulas satisfying ⊨ (𝐵 ∧ 𝐶) . • Formula 𝐽 is an interpolant of 𝐵 and 𝐶 if: 1. ⊨ 𝐵 → 𝐽 2. ⊨ ¬(𝐶 ∧ 𝐽) 3. Variables in 𝐽 appears in both of 𝐵, 𝐶 Takamasa Okudono (University of Tokyo) 21

Contribution 2: Observation Spurious Interpolant: 𝐽 = 54.1800𝑦 + 108.3601𝑧 ≥ 0 ≈× 2 Simplified Interpolant: 𝐽 = 𝑦 + 2𝑧 ≥ 0 Correct and simple interpolant of 𝐵 = 𝑦 = 0 ∧ 𝑧 = 0 , 𝐶 = (𝑦 + 2𝑧 < 0) Takamasa Okudono (University of Tokyo) 22