50.530: Software Engineering Sun Jun SUTD

Week 10: Invariant Generation

Problem {pre}while B do program{post} if there exists an invariant inv such that the following are satisfied: (1) pre => inv (2) {inv && B} program {inv} (3) inv && !B => post and the loop terminates. How do we find inv so as to complete the proof?

Big View inv pre pre => inv

Big View B !B inv post pre inv && !B => post

Big View B !B inv post pre one iteration {inv && B}program{inv}

Static/Dynamic Analysis Static analysis: infer (loop) invariants based on source code without executing the program (treating programs a mathematical formula) Dynamic analysis: infer (loop) invariants based on testing results. It’s about learning something about the invariants and making guesses!

Exercise 1 x = 0.1; y = 0; while (x < 2) { k = 4 – x*x; y = sqrt(4-k); x += 0.001; } if (y < 0) { error(); Show that the error is not occurring.

Ernst et al. IEEE Transactions on Software Engineering 2001 Dynamically Discovering Likely Program Invariants to Support Program Evolution

The Approach Seem familiar?

Instrumentation Instrument at the beginning/end of each method and the start of loops. Daikon only supports two forms of data: scalar numbers (including characters and Booleans) and sequence of scalars; Convert other values into one of these forms.

Example: Instrumentation public int sumUp (int[] B, int N) { int i = 0; int s = 0; while (i != N) { i = i+1; s = s +B[i] } return s; public int sumUp (int[] B, int N) { //add code to output values int i = 0; int s = 0; while ( i != N) { i = i+1; s = s +B[i]; } return s;

Example: Testing 100 randomly-generated arrays of length 7 to 13, in which each element was a random number in the range of -100 to 100. The following s is the learned pre-condition.

Example: Testing 100 randomly-generated arrays of length 7 to 13, in which each element was a random number in the range of -100 to 100. The following s is the learned post-condition.

Example: Testing 100 randomly-generated arrays of length 7 to 13, in which each element was a random number in the range of -100 to 100. The following loop invariants are learned .

Discussion What invariants should we infer?

What Invariants to Infer? Invariants over any variables Constant value, e.g., x = a; Uninitialized, e.g., x = uninit; Invariants over a single numeric variable Range limit, e.g., x >= a, x <= b, a <= x <= b Nonzero, e.g., x != 0 Modulus, e.g., x mod b = a Non-modulus, e.g., x mod b != a

What Invariants to Infer? Invariants over two numeric variables Linear relationship, e.g., y = ax+b Ordering comparison: x < y, x <= y, x > y, x >= y, x = y, x != y Functions, e.g., y = fn(x) or x = fn(y) where fn is one of Python’s built-in unary functions like absolute values, negation, etc. Invariants over x+y: any invariant from the list of invariants over a single numeric variable, such as (x+y) mod b = a Invariants over x-y: as for x+y;

What Invariants to Infer? Invariants over three numeric variables Linear relationship, e.g., z = ax+by+c Functions, e.g., z = fn(x, y) or x = fn(y) where fn is one of Python’s built-in binary functions like min, max, GCD, and, or, etc. How about four variables and more?

What Invariants to Infer? Invariants over a single sequence variable Range: minimum and maximum sequence values, ordered lexicographically; for instance, this can indicate the range of string or array values Element ordering: whether the elements of each sequence are non-decreasing, non-increasing, or equal Invariants over all the sequence elements (treated as a single large collection)

What Invariants to Infer? Invariants over two sequence variables Linear relationship: y = ax + b, element-wise Comparison: x < y, x <= y, x > y, x >= y, x = y, x != y, perform lexicographically Subsequence relationship: x is a subsequence of y or vice versa Reversal: x is the reverse of y Invariants over a sequence and a numeric variable Membership: i in s

What Invariants to Infer? Derived variables Derived from any sequence s Length: size(s) Extremal elements: s[0], s[1], s[size(s)-1], s[size(s)-2] Derived from any numeric sequence s sum: sum(s) Minimum elements: min(s) Maximum elements: max(s) Derived from any sequence s and any numeric variable i Element at the index: s[i], s[i-1] Subsequences: s[0..i], s[0..i-1] Derived from function invocations: number of calls so far

Algorithm Collect samples at a program point (through instrumentation and testing) For all variables, test every potential invariant (defined above) Remove an invariant if it is violated by a sample.

Exercise 2 int inc(int *x, int y) { *x += y; return *x; } Given the program and the collected data, what are the invariants?

Filtering Invariants Too many potentially invariants could discourage programmers from looking through them. A better test suite could help. Daikon filters invariants by computing an invariant confidence: assume a random input, what is the chance of the invariant would appear?

Invariant Confidence: Example A range for numeric ranges like x in [32..126] are reported only if the limits appear to be non-coincidental: if several values near the extremes all appear about as often as would be expected (assuming uniform distribution).

Invariant Confidence: Example Suppose the reported value for variable x fall in a range of size r that includes 0 Suppose that x != 0 holds for all test cases The probability of x != 0 is: (1-1/r)^n where n is the number of samples If the probability is less than a user-defined confidence threshold, then x != 0 is reported.

Scalability Daikon’s invariant detection time is Potentially cubic in the number of variables in scope at a program point (not the total number of variables in the program) Linear in the number of samples (the number of times a program point is executed) Linear in the number of instrumented program points.

Case Study: Invariant Stability Warming: One program!

Case Study: Invariant Stability Conclusion: Stable?

More Invariants, Better Programs? Experiment setup 424 student programs from a single assignment for CSE 142 at University of Washington The quality of the programs is measured by their scores. Invariant detection was performed over 200 executions of each program, resulting in 3 to 28 invariants per program. Conclusion: No co-relation

Discussion For invariant generation, shall we use random test case generation or systematic test case generation? How do we measure the usefulness of the generated invariants? How do we test whether a generated invariant is really a loop invariant? How do we identify the useful templates for invariants? Can we discover disjunctive invariants?

Unbounded Symbolic execution for program Verification Jaffar et al. RV’11 Unbounded Symbolic execution for program Verification

Motivation Symbolic execution doesn’t handle loops well: path explosion Loop invariants are essential to handle loops. Idea: learn loop invariant through symbolic execution

Iterative Deepening Step 1: execute path L0,1,4,5 symbolically x = 0 && //from L0 x >= n && //from L1 x < 0 //from L4 Interpolant at L4: x >= 0 L0 x = 0; L1 while (x < n) { L2 x++; L3 } L4 if (x < 0) { L5 error(); L6 }

Iterative Deepening Step 2: check if x >= 0 is a loop invariant by checking whether the following is satisfiable. x >= 0 && x < n && x1 = x+1 && x1 < 0 No! Thus x >= 0 is a loop invariant. Complete the proof with Hoare logic rules. L0 x = 0; L1 while (x < n) { L2 x++; L3 } L4 if (x < 0) { L5 error(); L6 }

Another Look Initially, L0 L0 x = 0; L1 while (x < n) { L2 x++; L4 if (x < 0) { L5 error(); L6 } L1 x<n x>=n L2 L4 x<0 L3 error

Another Look With the loop invariant, L0 L0 x = 0; L1 while (x < n) { L2 x++; L3 } L4 if (x < 0) { L5 error(); L6 } L1 x>=0 x<n x>=n L2 L4 x<0 L3 error This serves as a proof that error is not reachable. Finding a loop invariant is to find this label at this a loop head!

Iterative Deepening L0 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L1 new=old new!=old L2 L6 lock=0 L3 error L4 L5 Is error happening? What label shall we generate at L1?

Iterative Deepening Step 1: execute path L0,1,6,7 symbolically lock=0&&new=old+1&& //from L0 new==old && //from L1 lock==0 //from L6 Interpolant at L6: lock!=0 Is lock!=0 an invariant during the loop? L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error();

Iterative Deepening Step 1: execute path L0,1,6,7 symbolically lock=0&&new=old+1&& //from L0 new=old && //from L1 lock==0 //from L6 What is the interpolant at L1? That is, A is lock=0&&new=old+1 B is new=old&&lock=0 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error();

Ideal Case The interpolant at L1 is new!=old || lock != 0 Exercise 3: Is this a loop invariant strong enough to prove that error is not possible? L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); Recall existing techniques only return conjunctive interpolants. The interpolant at L1 thus may be either new!=old or lock!=0, neither of which is a loop invariant.

Iterative Deepening Step 2: execute path L0,1,2,3,5,1,6,7 symbolically lock=0&&new=old+1&& //from L0 new!=old && //from L1 lock1=1&old1=new && //from L2 new=old1&& //from L1 lock1==0 //from L6 Interpolant at L1? L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error();

Iterative Deepening Step 2: execute path L0,1,2,3,4,5,1,6,7 symbolically lock=0&&new=old+1&& //from L0 new!=old && //from L1 lock1=1&old1=new && //from L2 lock2=0&new1=new+1 && //from L2 new=old1&& //from L1 lock1==0 //from L6 Interpolant at L1? L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); It doesn’t help to execute more iterations

Alternative Approach L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L0 L1 lock=0&&new=old+1 L2 L3 Assume there is a label Inv at L1 which is a loop invariant; The following is true. lock=0&&new=old+1 => Inv lock=1&&old=new => Inv L5 L1’ lock=1&&old=new

Alternative Approach L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L0 L1 lock=0&&new=old+1 L2 lock=0&&new=old+1 Ideally, we let Inv be (lock=0&&new=old+1) || (lock=1&&old=new) || (lock=0&&new=old+1) Exercise: check if Inv is indeed a loop invariant. L3 L4 L5 L1’ L5 L1’ lock=1&&old=new

Invariant Validation L0 L1 (lock=0&&new=old+1) || (lock=1&&old=new) error L4 Since it is a loop invariant, we can label L1 now. Is it strong enough? L5

An Ideal Algorithm Identify paths which end at the loop head for the first time. Test if the disjunction of the path conditions is a loop invariant strong enough for the proof If positive, terminate Otherwise, identify paths which end at the loop head for the second time. …

Discussion First time: i = 0; Second time: i = 1; int i = 0; Third time: i = 2; … How do we make the jump to i <= 1000? int i = 0; while (i < 1000) { i++; }

Another Look at Daikon L0 {(lock=0,old=*, new=*+1), (lock=1,old=*+1, new=*+1), …} L1 new=old Pre-defined abstraction new!=old L2 L6 lock=0 lock=0 new=old new=old+1 L3 error L4 L5 Can Daikon find the right invariant in this case?

New Approach: USE Step 1: execute symbolically L0 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L1 L6 L7

New Approach: USE Step 2: Compute interpolant L0 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L1 L6 lock!=0 L7

New Approach: USE Step 3: Label loop head L0 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L1 {lock=0, new=old+1} L6 lock!=0 L7

New Approach: USE Step 4: abstract loop head labels based on the new condition. The loop head L1 is visited with a different path with a new condition. Abstract the labels on L1 so that it is implied by the new condition. L0 L1 lock=0&&new=old+1 L2 L3 L5 L1’ lock=1&&old=new

New Approach: USE Step 4: abstract loop head labels based on the new condition. Remove labels at L1 until the conjunction of the remaining labels is implied by the new condition L0 L1 lock=0&&new=old+1 true L2 L3 L5 Do we need to continue from L1’ given now it is stronger than an ancestor L1? L1’ lock=1&&old=new

New Approach: USE Step 5: execute symbolically Since lock=0&&new=old+1 (at L1’) implies true (at L1). We stop. L0 L1 true L2 L3 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L4 L5 L1’ lock=0&&new=old+1

USE: First Abstraction L0 L1 true Is this abstraction safe or not? It is safe iff error is not reachable if it is not reachable based on this abstraction. new=old new!=old L2 L6 lock=0 L3 error Is error reachable or not based on this abstraction? L4 L5

USE: Checking Run DFS/BFS algorithm on this graph shows that error is reachable. L0 -> L1 -> L6 -> error A counterexample based on the abstraction might not be a real counterexample! L0 L1 true new=old new!=old L2 L6 lock=0 L3 error L4 L5

USE: Spuriousness Checking Run DFS/BFS algorithm on this graph shows that error is reachable. L0 -> L1 -> L6 -> error Symbolically execute the above path and conclude that it is spurious. L0 L1 true new=old new!=old L2 L6 lock=0 L3 error L4 L5 Why it is spurious?

USE: Refinement The path L0,L1,L6,error is spurious. One (or more) loop head in this path must be too abstract. Find an interpolant at the loop head (L1) L0 L1 true new=old new!=old L2 L6 lock=0 L3 error L4 L5 lock=0&&new=old+1&& new=old && lock=0 Assume the interpolant found at L1 is: new!=old

USE: Refinement The path L0,L1,L6,error is spurious. One (or more) loop head in this path must be too abstract. Find an interpolant at the loop head (L1) L0 new!=old new=old L6 lock=0 error

USE: Re-explore Since the label at L1 has changed, we need to re-explore. This time, we can’t remove the label at L1. We continue instead. L0 L1 new!=old L2 L3 L5 L1’ lock=1&&old=new

USE: Re-explore Continue with L6, symbolic execution proves that it is not possible. L0 L1 new!=old L2 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L3 L5 L1’ lock=1&&old=new L6

USE: Re-explore Continue with L6, symbolic execution proves that it is not possible. L0 L1 new!=old L2 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L3 L5 L1’ lock=1&&old=new We can’t go further since lock==1. L6

USE: Re-explore Backtrack to L1’ and continue with L2, symbolic execution shows it is not feasible. L0 L1 new!=old L2 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L3 L5 L1’ lock=1&&old=new L2’ We can’t go further since old==new.

USE: Re-Explore Backtrack to L3, continue with L4,L5,L1. We can stop at L1’ because lock=0&&new=old+1 implies new!=old. L0 L1 new!=old L2 L0 lock=0;new=old+1 L1 while (new!=old) { L2 lock=1;old=new; L3 if (*) { L4 lock=0;new++;} L5 }; L6 if (lock==0) L7 error(); L3 L4 L5 L1’ lock=0&&new=old+1

Recap: the USE Approach L0 L1 new!=old new!=old L2 L3 L5 L4 L1 L5 L6 L2 L1 subsumed by new!=old

Recap: the USE Approach This approach acknowledges the difficulty in finding (disjunctive) loop invariants and compensates it with a combination of state space exploring and abstraction-refinement.

Case Study Iterative Deepening New Approach

Exercise 4 The path L0,L1,L6,error is spurious. One (or more) loop head in this path must be too abstract. Find an interpolant at the loop head (L1) L0 L1 new=old new!=old L2 L6 lock=0 L3 error L4 L5 lock=0&&new=old+1&& new=old && lock=0 What if the interpolant at L1 is: new=old+1?