✬ ✩ Synthesizing from Components: Building from Blocks Ashish Tiwari SRI International 333 Ravenswood Ave Menlo Park, CA 94025 Joint work with Sumit Gulwani (MSR), Vijay Anand Korthikanti (UIUC), Susmit Jha (UC Berkeley), Sanjit Seshia (UC Berkeley), Thomas Sturm (Munich), Ankur Taly (Stanford), Ramarathnam Venkatesan (MSR) ✫ ✪ MSR, Redmond Component-based Synthesis: 1
✬ ✩ Component-Based Synthesis .... Problem: How to wire the components to synthesize a desired system ? ✫ ✪ MSR, Redmond Component-based Synthesis: 2
✬ ✩ Concrete Examples Desired System F spec Components f i ’s sort an array comparators compute x + y modulo arithmetic ops 2 find rightmost one bitwise ops, arithmetic ops compute x 243 multiplication accept ω -regular language Buchi automata safe hybrid system multiple operating modes geometry construction ruler-compass steps deobfuscated code parts of obfuscated code verification proof verification inference rules Question: ∃ C : ∀ x : F spec ( x ) = C ( f 1 , f 2 , . . . )( x ) ✫ ✪ MSR, Redmond Component-based Synthesis: 3
✬ ✩ Synthesis Problem Classes “This is difficult” “This is ill posed” “This is too general to be solvable” ∃ C : ∀ x : F spec ( x ) = C ( f 1 , f 2 , . . . )( x ) Parameters that define the synthesis problem: • composition operator C • class of specifications F spec • class of component specifications f i Fixing the synthesis problem: ✫ ✪ fix these parameters, fix representation of F spec , f i MSR, Redmond Component-based Synthesis: 4
✬ ✩ Bounded Synthesis The synthesis problem is still hard We make it feasible by replacing the unbounded quantifier, ∃ C , by a bounded quantifier ∃ C : ∀ x : F spec ( x ) = C ( f 1 , f 2 , . . . )( x ) ⇓ ∃ c : ∀ x : F spec ( x ) = c ( f 1 , f 2 , f 3 )( x ) , c in some finite set This bounded synthesis problem is solved by deciding the ∃∀ formula ✫ ✪ MSR, Redmond Component-based Synthesis: 5
✬ ✩ Straight-Line Program Synthesis composition operator function composition components primitive functions system complex function Bounded synthesis version: • fix length of program • fix upper bound on number of each component ∃ P : ∀ x : F spec ( x ) = P ( x ) , P a straight-line program composing f i ’s ⇓ ∃ π : ∀ x : F spec ( x ) = f π (1) ( f π (2) ( f π (3) ( x ))) ✫ ✪ MSR, Redmond Component-based Synthesis: 6
✬ ✩ Example: Straight-Line Program Synthesis Specification: Evaluate polynomial a ∗ h 2 + b ∗ h + c Budget: two multiplication and two addition operators Finite search space Synthesized Program: 1. o 1 := a ∗ h ; 2. o 2 := o 1 + b ; 3. o 3 := o 2 ∗ h ; 4. return o 3 + c ; Correctness: ( a ∗ h + b ) ∗ h + c = a ∗ h 2 + b ∗ h + c ✫ ✪ MSR, Redmond Component-based Synthesis: 7
✬ ✩ Example: Straight-Line Program Synthesis Specification: Turn-off rightmost contiguous 1 bits Example: 010101100 �→ 010100000 Budget: two addition and at most four bitwise Boolean operators Finite search space: Also need some constants Synthesized Program: 1. o 1 := x + ( − 1) ; 2. o 2 := o 1 | x ; 3. o 3 := o 2 + 1 ; 4. return o 3 & x ; Correctness on sample input: ✫ ✪ 010101100 �→ 010101011 �→ 010101111 �→ 010110000 �→ 010100000 MSR, Redmond Component-based Synthesis: 8
✬ ✩ Loop-free Program Synthesis composition operator function composition components primitive functions, if-then-else system complex function Bounded synthesis version: • fix length of program • fix upper bound on number of each component including if-then-else ∃ P : ∀ x : F spec ( x ) = P ( x ) , P a straight-line program composing f i ’s ⇓ ∃ π : ∀ x : F spec ( x ) = f π ( ǫ ) ( f π (1) ( f π (11) ( x 1 ) , f π (12) ( x 2 , x 1 ))) ✫ ✪ MSR, Redmond Component-based Synthesis: 9
✬ ✩ Example: Loop-free Program Synthesis Specification: Obfuscated code Example: We are given if (h(x)) if (x*(x+1)% 2 == 1) y := f(x) else y := g(x) else y := f(g(x)) Components Budget: f, g, h, if-then-else Synthesized Program: o := g(x); if (h(x)) y := o; else y := f(o); Correctness: Equivalence of two loop-free programs ✫ ✪ MSR, Redmond Component-based Synthesis: 10
✬ ✩ Loop-free Program Synthesis ∃ π : ∀ x : F spec ( x ) = f π ( ǫ ) ( f π (1) ( f π (11) ( x 1 ) , f π (12) ( x 2 , x 1 ))) Enumerate all possible programs and check Enumerate all permutations π and check Checking if a synthesized program is the desired program is a verification problem Bounded Synthesis := iteratively perform verification But we can learn from failures ... ✫ ✪ MSR, Redmond Component-based Synthesis: 11
✬ ✩ ∃∀ φ Solvers Bounded Synthesis �→ ∃∀ solving How to solve ∃ u : ∀ x : φ formulas? A1 Counter-example guided iterative solver A2 Distinguishing input solver • Applies even when φ not fully known A3 Numerical solver ✫ ✪ MSR, Redmond Component-based Synthesis: 12
✬ ✩ A1: Solving ∃∀ φ Counter-example guided iterative procedure for solving ∃ � u : ∀ � x : φ ( � x ) u, � 1. Guess � u 0 for � u 2. (Verification) Check if ∀ � x : φ ( � u 0 , � x ) 3. If true, then return � u 0 4. Get counterexample � x 0 , add it to X 5. (Finite Synthesis) Find new � u 0 such that � ∃ � u 0 : φ ( � x 0 ) u 0 , � � x 0 ∈ X 6. Go to Step 2 ✫ ✪ MSR, Redmond Component-based Synthesis: 13
✬ ✩ A1: Counter-example Guided Iterative ∃∀ Solving Needs a backend quantifier-free solver That can return counterexamples We use an SMT solver The structure of φ , and additional knowledge about what φ encodes, is used optimize the above procedure to expedite convergence Related Work: Sketch, Aha Reference: Synthesis of loop-free programs, PLDI 2011 ✫ ✪ MSR, Redmond Component-based Synthesis: 14
✬ ✩ A2: Distinguishing Input Solver Solving ∃ � u : ∀ � x : φ ( � u, � x ) 1. X := some finite set of choices for � x 2. Find two programs that work for X , but differ on some � x 0 � ∃ � x 0 : ( ( φ ( � x ) ∧ φ ( � x ))) ∧ ( φ ( � x 0 ) �⇔ φ ( � x 0 )) u 1 , � u 2 , � u 1 , � u 2 , � u 1 , � u 2 , � � x ∈ X 3. If satisfiable, we add � x 0 to X and go to (2) 4. If unsatisfiable, then find one program that works for X � ∃ � u 1 : φ ( � x ) u 1 , � � x ∈ X 5. If satisfiable, return � u 1 ✫ ✪ 6. Otherwise, return “not synthesizable” MSR, Redmond Component-based Synthesis: 15
✬ ✩ A2: Properties of the A2 Solver The second algorithm for solving ∃ � u : ∀ � x : φ ( � x ) u, � • Does not need the full specification of the desired program • We only need the knowledge of the specification on the set X • Does not perform the verification step An interative implementation of A2: 1. Tool asks user for the expected output on input � x 0 2. Tool synthesizes internally two programs that work correctly for X := { � x 0 } , but differ on input � x 1 3. Tool asks user for the expected output on input � x 1 4. Add � x 1 to X and repeat ✫ ✪ MSR, Redmond Component-based Synthesis: 16
✬ ✩ A3: Nonsymbolic ∃∀ Solver A third algorithm for solving ∃ � u : ∀ � x : φ ( � x ) u, � 1. Find finite set X of input-output pairs of the specification 2. Synthesize program that works for finite set X 3. Verify the synthesized program on randomly sampled inputs We solved Step (2) using an SMT solver previously We can avoid the SMT solver and instead 1. hierarchical program synthesis: first synthesize high-level components 2. enumerate composition of high-level components guided by goal ✫ ✪ MSR, Redmond Component-based Synthesis: 17
✬ ✩ Example: Synthesis Without Symbolic Reasoning Specification: Construct a triangle, given its base, a base angle and sum of the other two sides. Components: Ruler compass constructions Formal specification: Given points p 1 , p 2 and numbers a, r , find point p := r > length ( p 1 , p 2 ) φ pre := Angle ( p, p 1 , p 2 ) = a ∧ length ( p, p 1 ) + length ( p, p 2 ) = r φ post Construction: L1 := ConstructLineGivenAngleLine(L,a); C1 := ConstructCircleGivenPointLength(p1,r); (p3,p4) := LineCircleIntersection(L1,C1); L2 := PerpendicularBisector2Points(p2,p3); ✫ ✪ p5 := LineLineIntersection(L1,L2); MSR, Redmond Component-based Synthesis: 18
✬ ✩ Example: Geometry Construction Synthesis Step 1 find concrete input-output pair consistent with specification = Line ( � 81 . 62 , 99 . 62 � , � 99 . 62 , 83 . 62 � ) L = 88 . 07 r = 0 . 81 radians a Compute output for this input: p := � 131 . 72 , 103 . 59 � Step 2 Start enumerating partial programs built using an extended library Step 3 Evaluate if intermediate objects generated by the partial program are good and try other choices in Step (2) otherwise ✫ ✪ MSR, Redmond Component-based Synthesis: 19
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