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# Model Checking of Systems Employing Commutative Functions PowerPoint PPT Presentation

This talk is about how you can find lots of bugs in real code by making compilers aggressively system specific. Model Checking of Systems Employing Commutative Functions. A.Prasad Sistla, Min Zhou, Xiaodong Wang presented by Min Zhou University of Illinois at Chicago. Outline.

Model Checking of Systems Employing Commutative Functions

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#### Presentation Transcript

This talk is about how you can find lots of bugs in real code by making compilers aggressively system specific

## Model Checking of Systems Employing Commutative Functions

A.Prasad Sistla, Min Zhou, Xiaodong Wang

presented by Min Zhou

University of Illinois at Chicago

### Outline

• Transition Diagram(TD) and Symbolic State Graph

• Predicate Template and bisimulation ~0

• Extended Predicate Template and bisimulation ~k

• Experiment Results

• Conclusion

We only consider such TDs who only have assignments:

x:=c

c is a constant

x:=y

y is another variable

x:=φ(x)

φ is a unary function;

each such φ in a TD is commutative with each other:

φ1 φ2 = φ2φ1

### Transition Diagram(TD)

a≤y y++

1

a:=x,x++

0

b:=x,x++

2

b≤y x:=0,y:=0

Variables:a,b,x,y

### Symbolic State Graph

• Sym_Reach(G, u) = (S0,R0, L0)

s.lc

s.val

s.exp

location

• variables  expressions

• s.exp(x): the composition of functions that were applied to x since last time a constant was assigned

• variables  values

• s.val(x): the latest constant assigned to variable x

### How to construct Symbolic State Graph

TD

x:=c

x:=φ(x)

x:=y

s1.exp(x)=φ(s0.exp(x))

s1.exp(x)=x

s1.exp(x)=s0.exp(y)[x/y]

q1

q1

q1

s1

s1

s1

s0

q0

q0

s0

s0

q0

s1.val(x)=s0.val(x)

s1.val(x)=c

s1.val(x)=s0.val(y)

### Symbolic States

• act_state(s) = (s.lc, h) where h(x) = s.exp(x){s.val(x)}

Symbolic

States

actual

states

a  y y++

1

1,(0,0,0,0)

(a,b,x+1,y)

s1

0,(0,0,0,0)

(a,b,x,y)

a:=x,x++

0

b:=x,x++

s0

s2

2

0,(0,0,0,0)

(a,b,x+1,y+1)

b  y x:=0,y:=0

s3

2,(0,0,0,0)

(a,b,x+1,y)

val

TD:

lc

exp

### Our Goal

• Define a bisimulation relation over symbolic states

• For every location q, define a predicate template ptemplates(q)

• s ~0 t require they are equivalent w.r.t ptemplates(s.lc)

p

f

### Predicate Template

var(p) X*

predicate, derived from guards and correctness formula

x:=y

x:=c

p(x)

p(x)

q1

q1

q

q

qi

qi

### What should be in ptemplates(q)

• (AP,fid) U (guard(q), fid)Є ptemplates(q)

x:= φ1(x)

x:=φ2(x)

(P, fid) Є ptemplates(q)

p(x)

q1

q

qi

x:=φ(x)

(P,f(x) = y) Є ptemplates(q)

（P, f(x)=) Є ptemplates(q)

p0: x  y

p1: a  y

p2: b  y

Formula: (x  y)

Ptemplates(1)={

(p0, fid),

(p1, fid),

(p1, a  x),

(p2, b  x)}

a  y y++

1

a:=x,x++

0

b:=x,x++

2

b  y x:=0,y:=0

### Example

Instantiate predicate templates in states:

(p(xi), xi yi) [s] = p [(s.exp (yi) /xi ) { xi/ yi }], where yi  

Eg:

Define ~0as follows: for any two states s and t, s ~0 t iff

s.lc = t.lc, s.val = t.val

(p, f) Є ptemplates(s.lc), (p, f)[s] (p, f)[t]

### Bisimulation ~0

• p: x1 < c

• s.exp(x1): x1+1

• s.exp(x2): x2+2

• (p,x1  x2 ) [s] =

• (s.exp (x2) < c){ x1/x2 } =

• (x1+2 < c)

• an implicit universal quantifier over the free variables

Proof idea:

assume s0 ~0t0 (p,fid) Є ptemplates(q1)

In this case, (p,fid) Є ptemplates(q0)

so we have (p,fid)[s0] (p,fid)[t0]

x:=φ(x)

s1.exp(x)=φ(s0.exp(x))

q1

q0

s1.val(x)=s0.val(x)

s1

t1

s0

t0

### Theorem 1 ~0is a bi-simulation on the symbolic state graph Sym_Reach(G, u).

Now We show

(p,fid)[s1] (p,fid)[t1]

(p,fid)[s0] (p,fid)[t0] 

x P[s0.exp(x)]  P[t0.exp(x)] 

x P[s0.exp(φ (x))]  P[t0.exp(φ (x))]

x P[φ (s0.exp(x))]  P[φ (t0.exp(x))] 

x P[s1.exp(x)]  P[t1.exp(x)]

x:=φ(x)

s1.exp(x)=φ(s0.exp(x))

q1

q0

s1.val(x)=s0.val(x)

s1

t1

s0

t0

By commutation

P(x)

P(x)

P(x)

si

ti

qi

s

q

t

### Extension of Bisimulation ~0

• If (p,f) Є ptemplates(q), we require (p,f)[s]  (p,f) [t] even in above case.

• Not necessary. Only need when this path is feasible for these two states

TD:

X

X

P(x)

qi

q

P(x)

ti

t

ti-k

### Bisimulation ~k

• Only in this case, we require (p,f)[s]  (p,f) [t]

P(x)

si

s

si-k

feasible, length = k

### ~k

• In ~k , we require a conditional equivalence

• Lemma ~k+1  ~k,

• but ~k+1 need more computation

q2

x2=0 

x1++

x1 20 

x1++, x2++

q0

q1

q4

x1=0 

x2 ++

q3

x2  20 

### Example of a TD for which ~0  ~1

any two states of the form

(q1,(0,0), (x1 + c 0, x2 + c’ 0))

are bisimular w.r.t ~1

taken from T.Bultan 1999

### Experiment Results

taken from T.Bultan 1999

### Conclusion and future work

• Defined a non decreasing chain of bisimulation

• Can be used in a class of infinite systems

• ~k can be checked on-the-fly

• Need investigate how to combine with static analysis