Software Synthesis using Automated Reasoning

Software Synthesis using Automated Reasoning

SVARM 2011Saarbrücken, Germany

Ruzica Piskac

1

SOFTWARE SYNTHESIS

2

specification(formula)

automated reasoning(theorem prover) code

Software Synthesis

val bigSet = ....

val (setA, setB) = choose((a: Set, b: Set) ) => ( a.size == b.size && a union b == bigSet && a intersect b == empty))

val bigSet = ....


Codeval n = bigSet.size/2val setA = take(n, bigSet)val setB = bigSet −− setA

Codeval n = bigSet.size/2val setA = take(n, bigSet)val setB = bigSet −− setA

3

Software Synthesis

val bigSet = ....


val bigSet = ....


Codeassert (bigSet.size % 2 == 0)val n = bigSet.size/2val setA = take(n, bigSet)val setB = bigSet −− setA

Codeassert (bigSet.size % 2 == 0)val n = bigSet.size/2val setA = take(n, bigSet)val setB = bigSet −− setA

4

Software Synthesis• Software synthesis = a technique for automatically

generating code given a specification• Why?

– ease software development– increase programmer productivity– fewer bugs

• Challenges– synthesis is often a computationally hard task– new algorithms are needed

5

Software Synthesis – Then and Now• Studied for a long time

– Church Synthesis Problem– Zohar Manna, Richard J. Waldinger: “A Deductive

Approach to Program Synthesis” [1980]

• Recent increased interests– due to increasing computational power and

improvements in automated reasoning– Sumit Gulwani (MSR): search-based techniques– Ras Bodik (Berkley): Programming by Sketching

6

COMPLETE FUNCTIONAL SYNTHESIS

JOINT WORK WITH VIKTOR KUNCAK, MIKAEL MAYER AND PHILIPPE SUTER[PLDI 2010, CAV 2010, CACM 2011/2]

7

Synthesis as programming language construct

… val x = readInteger() + 4

val y1 =

println(“y1 = “ + y1)…

… val x = readInteger() + 4

val y1 =

println(“y1 = “ + y1)…

choose(y 5*x + 7*y == 31)⇒

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“choose” Construct

• specification is part of the Scala language• two types of arguments: input and output

• a call of the form

• corresponds to constructively solving the quantifier elimination problem

where a is a parameter

val x1= choose(x F( x, a ))⇒val x1= choose(x F( x, a ))⇒

),(. axFx

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Complete Functional Synthesis

• Note: pre(a) is the “best” possible

Definition (Synthesis Procedure)A synthesis procedure takes as input formula F(x, a) and outputs:

1. a precondition formula pre(a)2. list of terms Ψ

such that the following two implications are valid:

Definition (Synthesis Procedure)A synthesis procedure takes as input formula F(x, a) and outputs:

1. a precondition formula pre(a)2. list of terms Ψ

such that the following two implications are valid:

]:[)(

)(),(.

xFapre

apreaxFx

10

From Decision Procedure to Synthesis Procedure

• based on quantifier elimination / model generating decision procedures

• implemented for logic of linear integer (rational, real) arithmetic, for Boolan Algebra with Presburger Arithmetic (BAPA)

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Complete Functional Synthesis

• complete = the synthesis procedure is guaranteed to find code that satisfies the given specification

• functional = computes a function that satisfies a given input / output relation

• Important features:– code produced this way is correct by

construction – no need for further verification– a user does not provide hints on the

structure of the generated code

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Synthesis for Linear Integer Arithmetic – Example / Overview

choose((h: Int, m: Int, s: Int) (⇒ h * 3600 + m * 60 + s == totalSeconds && h ≥ 0 && m ≥ 0 && m < 60 && s ≥ 0 && s < 60 ))


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Returned code:

assert (totalSeconds ≥ 0) val h = totalSeconds div 3600val temp = totalSeconds + (-3600) * hval m = temp div 60val s = totalSeconds + (-3600) * h + (-60)* m

Synthesis Procedure - Overview

• process every equality: take an equality Ei, compute a parametric description of the solution set and insert those values in the rest of formula

– for n output variables, we need n-1 fresh new variables

– number of output variables decreased for 1– based on Extended Euclidean Algorithm– compute preconditions

14

Synthesis Procedure - Overview

• at the end there are only inequalities – similar procedure as in [Pugh 1992]

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}|0

0

60

1

0

3600

0

1

{ , Z

dstotalSecons

m

h

60603600

06036006000

dstotalSecon

dstotalSecon

Parametric Solution of EquationTheoremFor an equation with S we denote the set of solutions.

• Let SH be a set of solutions of the homogeneous equality: SH = { y | }

SH is an “almost linear” set, i.e. can be represented as a linear

combination of vectors: SH = λ1s1 + ... λn-1sn-1

• Let w be any solution of the original equation

• S = w + λ1s1 + ... λn-1sn-1 + preconditions: gcd(i)| C

TheoremFor an equation with S we denote the set of solutions.






01

Cxn

i

ii

01

n

i

iiy

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Solution of a Homogenous EquationTheoremFor an equation with SH we denote the set of solutions.

where values Kij are computed as follows:• if i < j, Kij = 0 (the matrix K is lower triangular)• if i =j

• for remaining Kij values, find any solution of the equation

TheoremFor an equation with SH we denote the set of solutions.



01

n

i

iiy

}|{

)1(

)1(1

1

1

11

1 Z

K

K

K

K

S i

nn

n

n

n

H

))gcd((

))gcd(( 1

jkk

jkkjjK

01

n

ji

ijijjj zK 17

Finding any Solution (n variables)

• Inductive approach– 1x1 + 2x2 +... + nxn = C

1x1 + gcd(2,...,n )[λ2x2 +... + λnxn] = C

1x1 + xF = C

• find values for x1 (w1) and xF (wF) and then solve inductively: λ2x2 +... + λnxn = wF

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Finding any Solution (2 variables)

• based on Extended Euclidean Algorithm (EEA)– for every two integers n and m finds numbers p and q such

that n*p + m*q = gcd(n, m)

• problem: 1x1 + 2x2 = C

• solution:– apply EEA to compute p and q such that 1p + 2q = gcd(1, 2)

– solution: x1 = p*C/ gcd(1, 2)

x2 = q*C/ gcd(1, 2)

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Linear Integer Arithemtic: Example

val (x1, y1) = choose(x: Int, y: Int => 2*y − b =< 3*x + a && 2*x − a =< 4*y + b)val (x1, y1) = choose(x: Int, y: Int => 2*y − b =< 3*x + a && 2*x − a =< 4*y + b)

val kFound = false for k = 0 to 5 do { val v1 = 3 * a + 3 * b − k if (v1 mod 6 == 0) { val alpha = ((k − 5 * a − 5 * b)/8).ceiling val l = (v1 / 6) + 2 * alpha val y = alpha val kFound = true break } } if (kFound) val x = ((4 * y + a + b)/2).floor else throw new Exception(”No solution exists”)


Precondition: k. 0 ≤ k ≤ 5 6|3a + 3b − k∃ ∧Precondition: k. 0 ≤ k ≤ 5 6|3a + 3b − k∃ ∧ 20

Handling of Inequalities

• Solve for one by one variable:– separate inequalities depending on polarity of x:

• Ai ≤ αix

• βjx ≤ Bj

– define values a = maxi A⌈ i/αi and b = min⌉ j ⌊Bj/ βj⌋

• if b is defined, return x = b else return x = a• further continue with the conjunction of all

formulas A⌈ i/αi ≤ ⌉ ⌊Bj/ βj⌋

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Algorithm for more than one Variable

⌈(2y − b − a)/3 ≤ (4y + a + b)/2⌉ ⌊ ⌋ ⇔ ⌈(2y − b − a) 2/6 ≤ (4y + a + b) 3/6∗ ⌉ ⌊ ∗ ⌋ ⇔ (4y − 2b − 2a)/6 ≤

[(12y + 3a + 3b) − (12y + 3a + 3b) mod 6]/6

⇔ (12y + 3a + 3b) mod 6 ≤ 8y + 5a + 5b ⇔ 12y + 3a + 3b = 6 l + k k ≤ 8y + 5a + 5b∗ ∧

Consider the formula 2y − b ≤ 3x + a 2x − a ≤ 4y + b∧Consider the formula 2y − b ≤ 3x + a 2x − a ≤ 4y + b∧

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• 12y + 3a + 3b = 6 l + k k ≤ 8y + 5a + 5b∗ ∧• upon applying the equality, we obtain

– preconditions: 6|3a + 3b − k– solutions: l = 2λ + (3a + 3b − k)/6 and y = λ

• substituting those values in the inequality results in k − 5a − 5b ≤ 8λ

• final solution: λ = (⌈ k − 5a − 5b)/8⌉


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Synthesis for Sets

val (setA, setB) = choose((a: Set[O], b: Set[O]) ) => (−maxDiff <= a.size − b.size && a.size − b.size <= maxDiff && a union b == bigSet && a intersect b == empty))


Precondition⌈|bigSet| − maxDiff/2 ≤ |bigSet| + maxDiff/2⌉ ⌊ ⌋Precondition⌈|bigSet| − maxDiff/2 ≤ |bigSet| + maxDiff/2⌉ ⌊ ⌋

Codeval kA = ((bigSet.size + maxDiff)/2).floorval setA = take(kA, bigSet)val setB = bigSet −− setA

Codeval kA = ((bigSet.size + maxDiff)/2).floorval setA = take(kA, bigSet)val setB = bigSet −− setA

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INTERACTIVE SYNTHESIS OF CODE SNIPPETS

JOINT WORK WITH TIHOMIR GVERO AND VIKTOR KUNCAK[UNDER SUBMISSION, CAV2011]

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isynth - Interactive Synthesis of Code Snippets

def fopen(name:String):File = { ... }def fread(f:File, p:Int):Data = { ... }var currentPos : Int = 0var fname : String= null...def getData():Data =

def fopen(name:String):File = { ... }def fread(f:File, p:Int):Data = { ... }var currentPos : Int = 0var fname : String= null...def getData():Data =

Returned value:fread(fopen(fname),currentPos)Returned value:fread(fopen(fname),currentPos)

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Interactive Synthesis of Code Snippets

Complete functional synthesis– for specific theories (linear arith, sets, multisets)– based on generating models

isynth:– works for any operations in API– based on finding proofs– code completion using automated reasoning– specification is given through type constraints and

test cases27

isynth – Polymorphic Types

def map[A,B](f:A => B, l:List[A]): List[B] = { ... }def stringConcat(lst : List[String]): String = { ... }...def printInts(intList:List[Int], prn: Int => String): String =

def map[A,B](f:A => B, l:List[A]): List[B] = { ... }def stringConcat(lst : List[String]): String = { ... }...def printInts(intList:List[Int], prn: Int => String): String =

Returned value:stringConcat(map[Int, String](prn, intList))Returned value:stringConcat(map[Int, String](prn, intList))

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isynth - Interactive Synthesis of Code Snippets

• supports method combinations, type polymorphism, user preferences

• based on first-order resolution – combines forward and backward reasoning

• ranking of returned solutions is obtained through a system of weights

• http://lara.epfl.ch/w/isynth

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http://lara.epfl.ch/w/isynth

System of weights in isynth

• Symbol weights

• Term weights– Recalculate weight of terms with user

preferred symbols

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User preferred

Local symbols

Method and field symbolsAPI symbolsArrow

Low High

isynth - Evaluation

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Program # Loaded Declaratio

ns

# Methods in Synthesized

Snippets

Time [s]

FileReader 5 4 0.001

Map 3 3 < 0.001

FileManager 7 3 0.001

Calendar 29 3 0.001

FileWriter 442 6 0.72

SwingBorder 161 2 0.02

TcpService 112 3 0.62

Related Work / Reading MaterialComfusy project:

– M. Mayer, R. Piskac, P. Suter: “Complete Functional Synthesis”, PLDI 2010– V. Kuncak, M. Mayer, R. Piskac, P. Suter: “Comfusy: A Tool for Complete Functional

Synthesis”, CAV 2010– A. Solar-Lezama, L. Tancau, R. Bodík, S. A. Seshia, V. A. Saraswat: “Combinatorial sketching

for finite programs”. ASPLOS 2006– S. Jha, S. Gulwani, S. A. Seshia, A. Tiwari: “Oracle-guided component-based program

synthesis”. ICSE (1) 2010– S. Srivastava, S. Gulwani, J.S. Foster: From program verification to program synthesis. POPL

2010– S. Gulwani: Dimensions in program synthesis. PPDP 2010

isynth project:– T. Gvero, R.Piskac, V. Kuncak. “Interactive Synthesis of Code Snippets”, CAV 2011– The Agda Project [http://wiki.portal.chalmers.se/agda/]– D. Mandelin, L. Xu, R. Bodık, D. Kimelman: “Jungloid mining: helping to navigate the API

jungle”, PLDI '05

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http://wiki.portal.chalmers.se/agda/

Conclusions

Software Synthesis– method to obtain correct software from the given

specification– Complete Functional Synthesis: extending

decision procedures into synthesis algorithms– Interactive Synthesis of Code Snippets

finding a term of a given type

Contributions to/from automated reasoning– decision procedures– methods to combine them

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SYNTHESIS FOR LINEAR INTEGER ARITHMETIC

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Quantifier Elimination for Linear Integer Arithmetic

• Problem of great interest:– [Presburger, 1929], [Cooper, 1972]– [Pugh, 1992], – [Weispfenning, 1997]– [Nipkow 2008] – verified in Isabelle

• Our algorithm for integers:– Efficient handling equalities– Handling of inequalities as in [Pugh 1992]– Computes witness terms , builds a program from them

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36

Returned code:

val h = totalSeconds div 3600val temp = totalSeconds + ((-3600) * h)val m = min(temp div 60, 59)val s = totalSeconds + ((-3600) * h) + (-60 * m)

Synthesis Procedure for Linear Integer Arithmetic

• Works on disjunctive normal form• Preprocessing:– standard elimination of negations and divisibility

constraints:

• From now on: we only consider formulas that are a conjunction of equalities and inequalities

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122121

.|

EEEEEE

ckEkEc

Overview of a Synthesis Procedure

1. process every equality: take an equality Ei, compute a parametric description of the solution set and insert those values in the rest of formula

• for n output variables, we need n-1 fresh new variables

• number of output variables decreased for 1• compute preconditions

2. at the end there are only inequalities – same procedure as in [Pugh 1992]

38


1.Corresponding quantifier elimination problem: x y . 5x + 7y = a x ≤ y∃ ∃ ∧2.give parametric description for all solutions of 5x + 7y = a:

x = -7z + 3ay = 5z - 2a

3.Rewrite inequation x ≤ y in terms of z: 5a ≤ 12z → z ≥ ceil(5a/12)

• a

choose((x, y) 5 * x + 7 * y == a && x ≤ y))⇒choose((x, y) 5 * x + 7 * y == a && x ≤ y))⇒

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• Obtain synthesized program:

choose((x, y) 5 * x + 7 * y == a && x ≤ y))⇒choose((x, y) 5 * x + 7 * y == a && x ≤ y))⇒

val z = ceil(5*a/12)val x = -7*z + 3*aval y = 5*z + -2*a

val z = ceil(5*a/12)val x = -7*z + 3*aval y = 5*z + -2*a

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Parametric Solution of EquationTheoremFor an equation with S we denote the set of solutions.






TheoremFor an equation with S we denote the set of solutions.






01

Cxn

i

ii

01

n

i

iiy

41

Solution of a Homogenous EquationTheoremFor an equation with SH we denote the set of solutions.



TheoremFor an equation with SH we denote the set of solutions.



01

n

i

iiy

}|{

)1(

)1(1

1

1

11

1 Z

K

K

K

K

S i

nn

n

n

n

H

))gcd((

))gcd(( 1

jkk

jkkjjK

01

n

ji

ijijjj zK 42

Finding any Solution (n variables)

• Inductive approach– 1x1 + 2x2 +... + nxn = C

1x1 + gcd(2,...,n )[λ2x2 +... + λnxn] = C

1x1 + xF = C

• find values for x1 (w1) and xF (wF) and then solve inductively: λ2x2 +... + λnxn = wF

43

Finding any Solution (2 variables)

• based on Extended Euclidean Algorithm (EEA)– for every two integers n and m finds numbers p and q such

that n*p + m*q = gcd(n, m)

• problem: 1x1 + 2x2 = C

• solution:– apply EEA to compute p and q such that 1p + 2q = gcd(1, 2)

– solution: x1 = p*C/ gcd(1, 2)

x2 = q*C/ gcd(1, 2)

44

Handling of Inequalities

• Solve for one by one variable:– separate inequalities depending on polarity of x:

• Ai ≤ αix

• βjx ≤ Bj

– define values a = maxi A⌈ i/αi and b = min⌉ j ⌊Bj/ βj⌋

• if b is defined, return x = b else return x = a• further continue with the conjunction of all

formulas A⌈ i/αi ≤ ⌉ ⌊Bj/ βj⌋

45


⌈(2y − b − a)/3 ≤ (4y + a + b)/2⌉ ⌊ ⌋ ⇔ ⌈(2y − b − a) 2/6 ≤ (4y + a + b) 3/6∗ ⌉ ⌊ ∗ ⌋ ⇔ (4y − 2b − 2a)/6 ≤

[(12y + 3a + 3b) − (12y + 3a + 3b) mod 6]/6

⇔ (12y + 3a + 3b) mod 6 ≤ 8y + 5a + 5b ⇔ 12y + 3a + 3b = 6 l + k k ≤ 8y + 5a + 5b∗ ∧


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• 12y + 3a + 3b = 6 l + k k ≤ 8y + 5a + 5b∗ ∧• upon applying the equality, we obtain

– preconditions: 6|3a + 3b − k– solutions: l = 2λ + (3a + 3b − k)/6 and y = λ

• substituting those values in the inequality results in k − 5a − 5b ≤ 8λ

• final solution: λ = (⌈ k − 5a − 5b)/8⌉


47

Linear Integer Arithemtic: Example

val (x1, y1) = choose(x: Int, y: Int => 2*y − b =< 3*x + a && 2*x − a =< 4*y + b)val (x1, y1) = choose(x: Int, y: Int => 2*y − b =< 3*x + a && 2*x − a =< 4*y + b)



Precondition: k. 0 ≤ k ≤ 5 6|3a + 3b − k∃ ∧Precondition: k. 0 ≤ k ≤ 5 6|3a + 3b − k∃ ∧ 48

Synthesis for Sets



Precondition⌈|bigSet| − maxDiff/2 ≤ |bigSet| + maxDiff/2⌉ ⌊ ⌋Precondition⌈|bigSet| − maxDiff/2 ≤ |bigSet| + maxDiff/2⌉ ⌊ ⌋

Codeval kA = ((bigSet.size + maxDiff)/2).floorval kB = bigSet.size − kAval setA = take(kA, bigSet)val setB = take(kB, bigSet −− setA)

Codeval kA = ((bigSet.size + maxDiff)/2).floorval kB = bigSet.size − kAval setA = take(kA, bigSet)val setB = take(kB, bigSet −− setA)

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Generic Techniques

• Multiple output variables - computed one by one

• Disjunctions – computed one by one, combined using if ... else ... expressions (NP-hardness)

output variables: y1, ..., yn

assert pren+1(x)val y1 = Ψ1(x)val y2 = Ψ2 (x, y1)....val yn = Ψn (x , y1, ..., yn-1)

output variables: y1, ..., yn

assert pren+1(x)val y1 = Ψ1(x)val y2 = Ψ2 (x, y1)....val yn = Ψn (x , y1, ..., yn-1)

pre1(x, y1, ..., yn-1)

pren(x,y1)

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Software Synthesis using Automated Reasoning

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