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Automated Theorem ProvingPowerPoint Presentation

Automated Theorem Proving

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Automated Theorem Proving. Lecture 3 Satisfiability modulo theories. Arithmetic programs. In addition, integer-valued variables with affine operations. Formula := A | | A Atom := b | t = 0 | t > 0 | t 0 t Term := c | x | t + t | t – t | ct b SymBoolConst

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Arithmetic programs

- In addition, integer-valued variables with affine operations

- Formula := A | |
A Atom := b | t = 0 | t > 0 | t 0

t Term := c | x | t + t | t – t | ct

b SymBoolConst

x SymIntConst

c {…,-1,0,1,…}

Satisfiability modulo arithmetic

- A formula is a boolean combination of literals
- Each literal is a positive or negative atom
- Each atom is either a boolean variable or a linear constraint over integer variables

x y (a z > 0) (a x > y) y + z x

b x y

c z > 0

d x > y

e y + z x

b (a c) (a d) e

x y (a z > 0) (a x > y) y + z x

b x y

c z > 0

d x > y

e y + z x

b (a c) (a d) e

Arithmetic

Solver

x y (a z > 0) (a x > y) y + z x

b x y

c z > 0

d x > y

e y + z x

b (a c) (a d) e

b = T, e = T

Arithmetic

Solver

Satisfiable

x y (a z > 0) (a x > y) y + z x

b x y

c z > 0

d x > y

e y + z x

b (a c) (a d) e

b = T, e = T

Arithmetic

Solver

a = F

Unsatisfiable

b = T, c = T, e = T

x y (a z > 0) (a x > y) y + z x

b x y

c z > 0

d x > y

e y + z x

b (a c) (a d) e

b = T, e = T

Arithmetic

Solver

a = T

Unsatisfiable

b = T, d = T, e = T

Affine constraints

A collection of m constraints over n variables:

a11 x1 + a12 x2 + … + a1n xn + c1 0

a21 x1 + a22 x2 + … + a2n xn + c2 0

…

am1 x1 + am2 x2 + … + amn xn + cm 0

a1 x1 + a2 x2 + … + an xn + c> 0

a1 x1 + a2 x2 + … + an xn + c-1 0

a1 x1 + a2 x2 + … + an xn + c 0

(-a1)x1 + (-a2)x2 + … + (-an xn) + (-c) 0

a1 x1 + a2 x2 + … + an xn + c= 0

Satisfiability problem for affine constraints

A collection of m constraints over n variables:

a11 x1 + a12 x2 + … + a1n xn + c1 0

a21 x1 + a22 x2 + … + a2n xn + c2 0

…

am1 x1 + am2 x2 + … + amn xn + cm 0

Does there exist an assignment of x1,x2, …,xn over the

integers such that each constraint is satisfied ?

Solving affine constraints

- Integer linear programming
- NP-complete

- Approximate integers by rationals/reals
- Linear programming
- Polynomial time (Khachian 1978, Karmarkar 1984)

- Simplex algorithm (Dantzig 63)
- exponential worst-case time
- polynomial behavior in practice

Affine Constraints

Tableau

x1 x2 …xn

y1 a11 a12 … a1n c1

y2 a21 a22 … a2n c2

…

ym am1 am2 … amn cm

Row variables

Column variables

Read it as:

y1 = a11 x1 + a12 x2 + … + a1n xn + c1

y2 = a21 x1 + a22 x2 + … + a2n xn + c2

…

ym = am1 x1 + am2 x2 + … + amn xn + cm

y1 0

y2 0

…

ym 0

Sample point

x1 x2 …xn

y1 a11 a12 … a1n c1

y2 a21 a22 … a2n c2

…

ym am1 am2 … amn cm

x1 = 0x2 = 0…xn = 0

y1 = c1

y2 = c2

…

ym = cm

- A tableau is feasible if the sample point satisfies
- all sign constraints.
- Otherwise, drop a subset of sign constraints to
- get a feasible tableau.
- For each unsatisfied sign constraint:
- Look for a different point satisfying the constraint
- while preserving existing constraints
- If such a point is found, add the constraint
- Otherwise, declare unsatisfiable
- Declare satisfiable

Pivot operation

Exchange row i and column j:

1. Solve for xj

yi = ai1 x1 + … + aij xj + … + ain xn + ci

xj = (-1/aij) (ai1 x1 + … + (-1)yi + … + ain xn + ci)

2. Substitute in row k i

yk = ak1 x1 + … + akj xj + … + akn xn + ck

yk = (ak1 – akjai1/aij) x1 + … + (akj/aij)yi + … + (akn – akjain/aij) xn + (ck – akjci/aij)

x1 …xj …xn

y1 a11 … a1j … a1n c1

…

yi ai1 … aij … ain ci

…

ym am1 … amj … amn cm

x1 …yi…xn

y1 (a11 – a1jai1/aij)… (a1j/aij) … (a1n – a1jain/aij)(c1 – a1jci/aij)

…

xj (- ai1/aij) … (1/aij) … (- ain/aij)(-ci/aij)

…

ym (am1 – amjai1/aij) … (amj/aij) … (amn – amjain/aij)(cm – amjci/aij)

a 1 -1 1

b 1 1 3

c -1 0 -4

x y

a 1 -1 1

b 1 1 3

c -1 0 -4

Drop sign

constraint

for c

Pivot

a and x

a b

x 1/2 1/2 -2

y -1/2 1/2 -1

c -1/2 -1/2 -2

a y

x 1 1 -1

b 1 2 2

c -1 -1 -3

Pivot

b and y

Manifestly maximized row variable

A row variable is manifestly maximized if every non-zero

entry, other than the entry in the constant column, in its

row is negative and lies in a column owned by a restricted

variable.

m n x

y 1 -1 2 0

l -1 -3 0 -1

- - l is manifestly maximized in the above tableau.
- l is constrained to be at most -1.
- y is not manifestly maximized in the above tableau.

Manifestly unbounded column variable

A column variable is manifestly unbounded if every

negative entry in its column is in a row owned by an

unrestricted variable.

x u

l 1 -1 0

y -1 -1 1

z -1 -2 -1

m 0 1 2

- x is manifestly unbounded in the above tableau.
- x can take arbitrarily large values.
- u is not manifestly unbounded in the above tableau.

Observation

- Given a feasible tableau T and a variable v, there
- is a sequence of pivot operations on T leading to a
- tableau T’ such that either
- v is manifestly maximized in T’, or
- 2. v is manifestly unbounded in T’

Algorithm

- Create initial tableau T with only those sign constraints that are
- satisfied by the sample point of T
- 2. If every row variable satisfies its sign constraint, return satisfiable
- 3. Pick a row k owned by variable y such that the sign constraint is
- not satisfied by the sample point of T
- 4. If y is manifestly maximized in T, return unsatisfiable
- 5. Pick a column j such that akj is positive
- 6. If every restricted row has a non-negative entry in column j,
- perform Pivot(k,j). y becomes manifestly unbounded in T.
- Therefore, add the sign constraint for y. Go to 2.
- 7. (i, j) = ComputePivot(k)
- 8. Perform Pivot(T,i,j)
- 9. If the sample point of T satisfies the sign constraint for y, then
- add the sign constraint for y. Go to 2.
- 9. Go to 4

Observation

- If a row variable y is not manifestly maximized
- either there is a positive entry in some column
- or there is a negative entry in a column owned by an unrestricted variable

Algorithm

- Create initial tableau T with only those sign constraints that are
- satisfied by the sample point of T
- 2. If every row variable satisfies its sign constraint, return satisfiable
- 3. Pick a row k owned by variable y such that the sign constraint is
- not satisfied by the sample point of T
- 4. If y is manifestly maximized in T, return unsatisfiable
- 5’. Pick a column j such that akj is negative and the variable in column j
- is unrestricted.
- 6. If every restricted row has a non-positive entry in column j,
- perform Pivot(k,j). y becomes manifestly unbounded in T.
- Therefore, add the sign constraint for y. Go to 2.
- 7. (i, j) = ComputePivot(k)
- 8. Perform Pivot(T,i,j)
- 9. If the sample point of T satisfies the sign constraint for y, then
- add the sign constraint for y. Go to 2.
- 9. Go to 4

Difference Constraints

Difference constraints

Three different kinds of constraints:

x – y c

x c

-y c

- - very common in program verification
- satisfiability procedure more efficient than
- for general affine constraints
- - satisfiability procedure complete for integers

Vertex x

Constraint x – y c

Edge from y to x

with weight c

Reduction to a graph problemIntroduce a new variable z to denote the value 0

x - z c

x c

z - y c

-y c

- Add a new vertex s.

- Add an edge with weight 0 from s to every other vertex v.

Soundness

If there is a negative cycle in the graph, the set of

constraints is unsatisfiable.

x1 - x2 c1

x2 - x3 c2

…

xn - x1 cn

0 c1 + c2 + … + cn < 0

Bellman-Ford algorithm

d(s) := 0

for each vertex v s:

d(v) :=

for each vertex:

for each edge (u,v):

if d(v) > d(u) + weight(u,v)

d(v) := d(u) + weight(u,v)

for each edge (u,v):

if d(v) > d(u) + weight(u,v)

Graph contains a negative-weight cycle

Completeness

If there is no negative cycle in the graph, then

d(v) - d(u) weight(u,v) for each edge (u,v).

Model: Assign to variable x the value d(x) –d(z).

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