stp: a decision procedure for bit-vectors and arrays david l. dill stanford university

STP: A Decision Procedure for

Bit-vectors and Arrays

David L. DillStanford University

Software analysis tools present unique challenges for decision

procedures● Theories must match programming

language semantics– Operations are on bit-vectors, not integers– Arrays (for modelling memories)

● Must handle very large inputs with – Many array reads – Deeply nested array writes – Many linear equations – Many variables

● Decision procedure is called many times.

What went before● Series of decision procedures: SVC, CVC, CVCL● All of these had combinations of first-order

theories– Equality– Uninterpreted functions and predicates– Boolean connectives– Linear arithmetic over real numbers (and integers,

in the case of CVCL)– But not quantifiers.

● CVCL was in use in EXE (or it’s predecessor – Dawson Engler research group)

Combination of theories● The core strategy of SVC, CVC, CVCL was

based on dynamically breaking down formulas into conjunctions of “atomic formulas”– Atomic formulas have no Boolean connectives

(correspond to propositional variables).– Recursively assert/deny alpha (deny = assert

negation of)– Simplify after assertion– When simplified formula is conjunction of

literals, use special decision procedures.

SAT vs. CVCL

● CVC/CVCL used CHAFF-like SAT solver to choose splitting variables

● … but puts lots of slow stuff in the inner loop!

STP● CVCL was already used by Engler’s group,

but was unfixably slow.– There existed many examples generated by

Engler– Made correctness/performance testing easy.

● Vijay Ganesh and I decide to try a different approach, inspired by UCLID (Seshia & Bryant).– Put SAT at the bottom, with unmodified inner

loop.– Preprocess formula for higher-level reasoning

(bit-vectors and arrays).

STP● Decides satisfiability of formulas over

– Bit-vector Terms● Constants● +, -, *, (signed) div, (signed) mod● Concatenation, Extraction● Left/Right Shift, Sign-extend, bitwise-Booleans

– Array Terms● Read(Array, index)● Write(Array, index, val)● But no array equality

– Predicates: =, signed & unsigned comparisons● If satisfiable, produces a model.

Comparison with Saturn● STP is a separate “component” (can be

stand-alone, or used through an API).– Programming language or other tool is separate

● General input language (can define 23-bit bit-vector types if you want).

● Signed/unsigned encoded in operators, not in data types.

● No “points to”, heap, etc.● Implements signed/unsigned multiply,

divide, remainder (but no floating point).

Some projects using STP

● STP has evolved (maintained by someone I don’t know in Australia)

● Several projects have used it.– EXE : Bug Finder by Dawson Engler, Cristian

Cadar and others at Stanford– Klee : Cadar, Dunbar, Engler– MINESWEEPER: Bug Finder by Dawn Song

and her group at CMU– …

Main Ideas of STP

● Eager translation to CNF with word-level pre-processing – Theories not in “inner loop” of SAT solver,

unlike Nelson-Oppen approaches (e.g. CVCL).

● Abstraction-Refinement for arrays.– Laziness to counterbalance eagerness

● Solve linear formulas mod 2n in P-time

Bitvector theory

● Data type: BV(n), where n is constant.● Almost all machine bitvector operations

– Change length (sign extended and not).– Concatenate bitvectors, extract bits from

BVs.– Signed and unsigned arithmetic: +, -, *, /,

%, <,>, etc.– AND, OR, NOT, XOR, etc.

Array theory● Array type: BV(n) -> B(m)● read(A, i) – value of A[i]● write(A, i, v) – copy of A updated at

index i with value v.● No destructive modification – write

returns a new array, which is updated old array.

● Sometimes used to represent heap storage.

Array theory

Identities:read(write(A, i, v), j) = ite(i = j, v, read(A, j))

STP has a limited theoryNo comparison of whole arrays, e.g.

write(A, i, v) = write(A, j, w)This makes things easier (see

http://sprout.stanford.edu/PAPERS/LICS-SBDL-2001.pdf

if you don’t like “easier”).

Implementation

● DAG representation of expressions– Same subexpression structure = same

pointer.– Maintained by hashing.– No destructive operations on DAGs

(modification requires new nodes).● Makes substitution, equality check very

efficient.● Often log size of tree expression

representation.

DAGs

● All recursive traversals must be “memoized”– Want traversal to be linear in size of DAG,

not tree.– First thing to think about when functions

don’t finish: “Maybe I messed up memoization.”

● Updating nodes near the root less expensive than updating nodes near leaves.

STP ArchitectureInput Formula

Refinement Loop

Substitutions

Simplifications

Linear Solving

BitBlast

CNF Coversion

SAT Solver

Array Abstraction

Result

Substitution is Important

● Inputs often have many simple equations (in EXE, this is how constant arrays are defined):– x = 4– A[4] = e

● Early pass to substitute these– Allows constant evaluation– Enables other optimizations– Reduces non-constant indices in array reads.

Word-level simplifications● Many simple local rewrites:

– Bitwise Boolean identities (e.g. a AND !a = 00000, a XOR a = 11111, (a + b)[0:3] = a[0:3]+b[0:3], etc.

– Generally, avoid distributive laws because they cause blow-up.

– Be careful about “destroying sharing” in DAG.● Flatten trees of associative operators● Sort operands of commutative operation

– Tweak ordering for easy simplification– Expressions numbered in order of creation (children <

parents). Use this order to sort, but:– Put constants first (F AND a AND …), (1 + 3 + x +)– Arrange for x, !x to be adjacent (also x, -x)

Array Reads

Read(A,i0) = t0Read(A,i1) = t1...Read(A,in) = tn

v0 = t0v1 = t1...vn = tn

(i1=i0) => v1=v0(i2=i0) => v2=v0(i2=i1) => v2=v1…

• Problem : O(n2) axioms added, n is number of read indices• Lethal, if n is large: n = 10000, # of axioms: ~ 100 million • Blowup seems hard to avoid (e.g. UCLID).• This is “aliasing” from another perspective.

Replace array readswith fresh variables

and axioms

Abstraction/Refinement in STP

● Pose problem as a conjunction of formulas– E.g., instantiated array read axioms

● Abstraction: solve for a proper subset of the formulas– E.g., omit array read axioms

● Early exit if:– Unsatisfiable– Satisfiable and model actually satisfies unabstracted

formula.● Otherwise, add some omitted formulas to the

abstracted formula and solve again.– If at least one of these formulas is false in the model, that

model will not be regenerated.

Abstraction-Refinement Algorithm for Array Reads

Read(A,0)=0Read(A,i)=1

Input

v0 = 0vi = 1

After Abstraction

SATSolver

i = 0v0 = 0vi = 1

Check Inputon Assignment

Counterexample

Refinement Step:Add Axiom

(i=0) => vi = 0Rerun SATSolver

i=1vi=1 False:

Read(A,0)=0Read(A,0)=1

Experience with Read Abstraction-Refinement

Works well, even in satisfiable cases– Satisfier often finds a model that minimizes

aliasing– Few axioms need to be added during

refinement– Typical number of refinement loops : < 3

The Problemwith Array Writes

● Standard transformation read(write(A, i, v), j) = ite(i=j, v, read(A, i)) <- “if-then-else” causes term blow-up● Many different read expressions share

write sub-terms.● O(n*m) blow-up in expr DAG

– n is write term nesting levels– m is number of read indices

The Problem with Array Writes

R(W(W(A,i0,v0),i1,v1),j) =R(W(W(A,i0,v0),i1,v1),k)

If (i1=j) v1 elsif (i0=j) v0else R(A,j)

v0

If (i1=k) v1 elsif(i0=k) v0

else R(A,j)=

=

R

j

R

k

W

i0A

W

v1i1

=

ite=

i1 k

ite=

i1 jv1 v1

ite=

i0 jv0

ite=

i0 kv0

RA j

Write transformation

Replace read(write(A, i, v), j)with a fresh variable (e.g., v0)and “axiom” v0 = ite(i=j, v, read(A, j))

Abstraction omits axiom.

Abstraction-Refinement Algorithm

for Array WritesR(W(A,i,v),j)= 0R(W(A,i,v),k)=1i = j /= kv /= 0

After Abstraction v1=0v2=1

i = j /=kv/=0

SATSolver

v1=0, v2=1i = j =0, k=1,

v = 1

Check modelon original

formula

Refinement Step

Add Axiom to SATv1=ite(i=j,v,R(A,j))

UNSAT

False:R(W(A,0,1),0)=0R(W(A,0,1),1)=10 = 0 /= 11 /= 0

Experimental Results:Array Writes

Example (Size in unique nodes)

Result Write Abstraction (sec)

NO Write Abstraction

(sec)Grep0084 (69K)

Sat 109 506

Grep0106 (69K)

Sat 270 TimeOut

Grep0117 (70K)

Sat 218 TimeOut

610dd9dc (15k)

Sat 188 101

Testcase20 (1.2M)

Sat 67 Memory Out

Examples courtesy Dawn Song (CMU) and David Molnar (Berkeley)

Algorithm for SolvingLinear Bit-vector Equations

● Inspired by Barrett et al., DAC 1998● Basic Idea in STP

– Solve for a variable and substitute it away– If cannot eliminate a whole variable,

eliminate as many bits as possible.● Previous Work

– Mostly variants of Gaussian Elimination– Solve-and-substitute is more convenient in

a general decision procedure.

Algorithm for Solving Linear Bit-vector Equations

(3 bits)3x + 4y + 2z = 02x + 2y + 2 = 04y + 2x + 2z = 0

Solve for x infirst eqn:

3-1 mod 8 = 3,

x = 4y + 2z(3 bits)

2y + 4z + 2 = 04y + 6z = 0

Substitute x

Algorithm for SolvingLinear Bit-vector Equations

(3 bits)2y + 4z + 2 = 04y + 6z = 0

All Coeffs EvenNo Inverse

Divide by 2Ignore high-order

bits

(2 bits)y[1:0] + 2z[1:0] + 1 = 02y[1:0] + 3z[1:0] = 0

Algorithm for SolvingLinear Bit-vector Equations

(2 bits)y[1:0] + 2z[1:0] + 1 = 02y[1:0] + 3z[1:0] = 0

Solve for y[1:0]

(2 bits)y[1:0]=2z + 3

Substitute y[1:0]

(2 bits)3z[1:0] + 2 = 0

Algorithm for SolvingLinear Bit-vector Equations

(2 bits)3z[1:0] + 2 = 0 Solve for z[1:0]

(2 bits)z[1:0]=2

Solution (3 bits):z[1:0] = 2y[1:0] = 2z[1:0] + 3 = 3y = y’ @ 2z = z’ @ 3x = 4y + 2z

Experimental Results: Solver for Linear Equations

Name (Node Size)

Result All On

(sec)

Arr-On,Linear OFF

(sec)

Arr OFF,Linear On(sec)

All OFF

(sec)

Test15 (0.9M)

Sat 66 192 64 Memory Out

Test16(0.9M)

Sat 67 233 66 MemoryOut

Thumb1(3.2M)

Sat 115 111 113 MemoryOut

Thumb2(4.3M)

Sat 1920 MO 1920 MemoryOut

Thumb3(2.7M)

Sat 840 MO 840 MemoryOut

Equivalence checking of block cipher implementations● Problem: Prove correctness of block

ciphers (e.g., AES).– Constant number of loop iterations– No interesting heap usage

● Approach:– Given two implementations, AES1 and AES2– Turn them into big expressions by unrolling

loops, etc.– Prove that AES1(x) ≠ AES2(x) is

unsatisfiable.

How can this possibly work?

● Round 1

● Round 2

● Round 3

● Round 4

● Round 1

● Round 2

● Round 3

● Round 4

Many block ciphers consist of a fixed sequence of “rounds”.

Implementations of rounds in two algorithms may vary, but bits “between” rounds are equal.

So, we only have to prove individual rounds are equivalent.

Equivalence checking

inputs

Equal for many testinputs.

Only try to prove Equivalence when nodes pass this test.

● ●

Equivalence checking

Prove equal usingSTP

● ●

Equivalence checking

inputs

Replace by everywhere in DAG.

This makes higher-level expressions more similar.

●

Project ideas

● Use abstraction/refinement to improve efficiency of verification with multiplication/division and other ugly operators.

● Hack Klee (open source, based on LLVM, uses STP).

● Extend to floating point.● Formal verification of crypto functions in

“C” (get Eric Smith thesis from me).