Satisfying Error Conditions 2 - Lecture Slides | CMPS 290, Study notes of Computer Graphics

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• Satisfying Error Conditions
  • Lecture
    • 19 Feb
    • 290G - Lecture

Review: Where Are We?

19 Feb 2004

290G - Lecture 14

Error

Verification

Condition

Program

Specification

Semantics

VC

generation

Invalid

Error

Condition

negate

satisfiable

{a=b}

{f(a)=f(b)}

{b=c})

{f(a)=f(c)}

19 Feb 2004

290G - Lecture 14

{a=b}

{f(a)=f(b)}

{b=c}

{f(a)=f(c)}

Sat?

0

0

0

0

0

0

0

0

1

0

0

0

1

0

0

0

0

1

1

0

0

1

0

0

0

0

1

0

1

0

0

1

1

0

0

0

1

1

1

0

1

0

0

0

1

1

0

0

1

0

1

0

1

0

1

1

0

1

1

0

1

1

0

0

0

1

1

0

1

0

1

1

1

0

1

1

1

1

1

0

a=b

.

f(a)

g

f(b)

.

b

g

c

.

f(a)

g

f(c)

a=b

.

f(a)

g

f(b)

.

b=c

.

f(a)

g

f(c)

a=b

.

f(a)=f(b)

.

b=c

.

f(a)

g

f(c)

{a=b}

{f(a)=f(b)}

{b=c})

{f(a)=f(c)}

19 Feb 2004

290G - Lecture 14

{a=b}

{f(a)=f(b)}

{b=c}

{f(a)=f(c)}

Sat?

0

0

0

0

0

0

0

0

1

0

0

0

1

0

0

0

0

1

1

0

0

1

0

0

0

0

1

0

1

0

0

1

1

0

0

0

1

1

1

0

1

0

0

0

1

1

0

0

1

0

1

0

1

0

1

1

0

1

1

0

1

1

0

0

0

1

1

0

1

0

1

1

1

0

1

1

1

1

1

0

a=b

.

f(a)

g

f(b)

.

b

g

c

.

f(a)

g

f(c)

(a=b

⇒

f(a)=f(b))

{a=b}

⇒

{f(a)=f(b)}

X 0

X 0

Explicated tautology

removes many other

truth assignments

Examples of Theories. Equality.

19 Feb 2004

290G - Lecture 14

Syntax

, f, g, …

Semantics: as expected

E

2

= E

1

E

1

= E

2

E

1

= E

2

E

2

= E

3

E

1

= E

3

E

1

= E

2

f(E

1

) = f(E

2

)

E = E

Example

g(g(g(x)) = x

g(g(g(g(g(x))))) = x

g(x)

x

Satisfiability Algorithm for Equality (1)

19 Feb 2004

290G - Lecture 14

We represent terms as DAGs

Share common subexpressions

E.g. f(f(a, b), b):

f(f(a, b), b)

f

f

a

b

f(a, b) a b

Satisfiability Algorithm for Equality (3)

19 Feb 2004

290G - Lecture 14

t* is representative element for t

For all nodes t = f(t

1

, …, t

n

) and s = f(s

1

, …, s

n

If t

i

= s

i

for all i = 1..n (find)

We add dotted arrow between t* and s*

E.g.

f(a,b) = a

f

f

a

b

Satisfiability Procedure for Equality

19 Feb 2004

290G - Lecture 14

Given F =

i

t

i

= t

i

j

u

j

u

j

Represent all terms in the same DAG

Add dotted edges for t

i

= t

i

Construct the congruence closure of those edges

Check that for all j we have u

j

u

j

Theorem

F is satisfiable

if and only if

for all j

u

j

u

j

Example with Congruence Closure

19 Feb 2004

290G - Lecture 14

13

Consider: g(g(g(x)) = x

g(g(g(g(g(x))))) = x

g(x)

x

g g x

g g g

g g x

g g g

g g g g g x

x

g g g g g

Contra

diction

Presburger Arithmetic

19 Feb 2004

290G - Lecture 14

Syntax:

Semantics: as expected

The most useful in program verification after equality

checking array bounds, etc

Example:

y > 2x + 1

y + x > 1

y < 0

Difference Constraints (cont)

19 Feb 2004

290G - Lecture 14

Hint: Think of a directed graph

with a node for each variable

represent constraint in each graph

and think about graph algorithms

Example

x <= y

y+4 <= w

w-2 <= x

w+1 < =z

What is explanation?

x

y

w 0

z

Difference Constraints

19 Feb 2004

290G - Lecture 14

Theorem

A set of difference constraints is satisfiable

in and only if there is no positive weight cycle in the graph

Can be solved in O(n

2

Algorithm is complete!

Was used successfully in array-bounds checkingelimination and induction variable discovery

How Complete are These Procedures?

19 Feb 2004

290G - Lecture 14

Consider: 3x

2y

3y

2x

x

w

Is it satisfiable?

What is solution?

Is there a solution in Z?

The unsat procedure is sound: unsat Q

unsat Z

But it is incomplete!

Not a problem in practice

Goes away with tricks like:

Transform “ax

b” into “x

b/a

Arithmetic. Discussion

19 Feb 2004

290G - Lecture 14

There are many satisfiability algorithms for Q

even for the general case (e.g. Simplex)

polynomial time

complete

If we add the requirement that solutions are in

Z

then the problem is NP-complete

but

Z

can be handled well with heuristics

No practical satisfiability procedures for (

Q

The satisfiability of (

Z

) is only semi-decidable