© 2011 Carnegie Mellon University Binary Decision Diagrams Part 2 15-414 Bug Catching: Automated...

© 2011 Carnegie Mellon University

Binary Decision Diagrams Part 2

15-414 Bug Catching: Automated Program Verification and Testing

Sagar ChakiSeptember 14, 2011

2

Binary Decision Diagrams – Part 2Sagar Chaki, Sep 14, 2011


BDDs Recap

Typically mean Reduced Ordered Binary Decision Diagrams (ROBDDs)• Can be viewed as reduced forms of Ordered Binary Decision Trees

• Obtained by eliminating duplicate nodes and redundant nodes• Often substantially smaller than the OBDT

Canonical representation of Boolean formulas• Unlike other normal forms like CNF and DNF

Size of BDD depends critically on variable ordering

In practice, BDDs are built up from their components• Via (efficient) Boolean operations• Dynamic variable ordering used to manage BDD size

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Running Example: Comparator

Comparator

a1 a2 b1 b2

(A) f = 1 , a1 = b1 Æ a2 = b2

(B) f = 1 , a1 = b1 Ç a2 = b2

4



Conjunctive Normal Form

(: a1 Ç b1 ) Æ (: b1 Ç a1) Æ (: a2 Ç b2 ) Æ (: b2 Ç a2)

(: b1 Ç a1 ) Æ (: a1 Ç b1) Æ (: a2 Ç b2 ) Æ (: b2 Ç a2)

(b1 Ç a1 ) Æ (: a1 Ç : b1) Æ (: a2 Ç b2 ) Æ (: b2 Ç a2)

(: a1 Ç b1 ) Æ (: b1 Ç a1) Æ (: a2 Ç : b2 ) Æ (b2 Ç a2)

5



Truth TableRow# a1 b1 a2 b2 f

0 0 0 0 0 0

1 0 0 0 1 1

2 0 0 1 0 0

3 0 0 1 1 1

4 0 1 0 0 0

5 0 1 0 1 1

6 0 1 1 0 0

7 0 1 1 1 0

8 1 0 0 0 0

9 1 0 0 1 1

10 1 0 1 0 1

11 1 0 1 1 0

12 1 1 0 0 1

13 1 1 0 1 0

14 1 1 1 0 0

15 1 1 1 1 0

6



Representing a Truth Table using a Graph

a1

b1 b1

a2 a2

b1 b2 b2 b2

a2 b2

b2 b2 a2 a2

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Representing a Truth Table using a Graph

a1

b1 b1

a2 a2

b2 b2

1 0 1 1

b2 b2

1 0 0 0

a2 a2

b2 b2

1 0 0 0

b2 b2

0 0 0 0

8



OBDT to ROBDD

a1

b1 b1

a2

b2 b2

10

a2

b2 b2

A B C D

Which pairs are isomorphic?

(1) {A,B} and {C,D}

(2) {A,C} and {B,D}

(3) {A,D} and {B,C}

9



OBDT to ROBDD

a1

b1 b1

b2

10

a2

b2

Is this a ROBDD?

(1) YES

(2) NO

10



OBDT to ROBDD

a1

b1 b1

b2

10

a2

b2

What function does X represent?

(1) a2 = b2 (2) a2 = (: b2)

(3) a2 ) b2 (4) a2 © b2

(5) :(a2 © b2)

(6) (a2 Æ b2) Ç (: a2 Æ : b2)

X1,5,6

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ROBDD (a.k.a. BDD) Summary

If BDD(f1) and BDD(f2) are isomorphic then:

1. f1 = f2

2. f1 and f2 have the same variables

3. BDD(f1) and BDD(f2) have the same variable ordering

If BDD(f) is the leaf node “1” then f is: 4. Satisfiable5. Unsatisfiable6. Valid

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ROBDD and variable ordering

a1

a2 a2

b1 b1

b2

1 0

b1 b1

b2 b2

Is this a ROBDD?

(1) YES

(2) NO

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a1

a2 a2

b1 b1

b2

1 0

b1 b1

b2

Is this a ROBDD?

(1) YES

(2) NO

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There exists a function whose BDD grows polynomially in the number of variables for some ordering and exponentially for others?• TRUE

There exists a function whose BDD grows exponentially for all variable orderings?• TRUE

There exists a function whose BDD grows linearly for all variable orderings?• TRUE

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BDD Operations

True : BDD(TRUE)

False: BDD(FALSE)

Var : v BDD(v)

Not : BDD(f) BDD(:f)

And : BDD(f1) £ BDD(f2) BDD(f1 Æ f2)

Or : BDD(f1) £ BDD(f2) BDD(f1 Ç f2)

Exist : BDD(f) £ v BDD(9 v. f)

16



Basic BDD Operations

True False

Var(v)

1 0

10

v

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BDD Operations: Not

1 00 1

10

v

O(1) O(1)

18



BDD Operations: Not

1 00 1

01

v

Swap “0” and “1”

O(1) O(1)

O(n)

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BDD Operations: And

vWhat formula does this

represent?

What formula does this

represent?

Suppose this is the BDD for f

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BDD Operations: And

vfv=0


fv=1

fv=0 and fv=1 are known as the co-factors of f w.r.t. v

f = (X Æ fv=0) Ç (Y Æ fv=1)

21



BDD Operations: And

vfv=0


fv=1

fv=0 and fv=1 are known as the co-factors of f w.r.t. v

f = (: v Æ fv=0) Ç (v Æ fv=1)

22



BDD Operations: And (Simple Cases)

And (f, ) = 0 0

And (f, ) = 1 f

And ( ,f ) = 1 f

And ( ,f ) = 0 0

23



BDD Operations: And (Complex Case)

v1

f1 g1

v2

f2 g2

(: v1 Æ f1) Ç (v1 Æ g1) (: v

2 Æ f2) Ç (v2 Æ g2)

ÆÆ

24



BDD Operations: And (Complex Case 1)

v1

f1 g1

v1

f2 g2

(: v1 Æ f1) Ç (v1 Æ g1) (: v

1 Æ f2) Ç (v1 Æ g2)

ÆÆ

v1 = v2

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(: v1 Æ f1) Ç (v1 Æ g1) (: v

1 Æ f2) Ç (v1 Æ g2)Æ

v1 = v2

(: v1 Æ X) Ç (v1 Æ Y)

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(: v1 Æ f1) Ç (v1 Æ g1) (: v

1 Æ f2) Ç (v1 Æ g2)Æ

v1 = v2

(: v1 Æ (f1 Æ f2)) Ç (v1 Æ (g1 Æ g2))

Compute recursivelyCompute recursively

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(: v1 Æ f1) Ç (v1 Æ g1) (: v

1 Æ f2) Ç (v1 Æ g2)Æ

v1 = v2

(: v1 Æ (f1 Æ f2)) Ç (v1 Æ (g1 Æ g2))

v1

f1 Æ f2 g1 Æ g2

What if f1 Æ f2 = g1 Æ g2 ?

Return f1 Æ f2

28




v1

f1 g1

v2

f2 g2

(: v1 Æ f1) Ç (v1 Æ g1) (: v

2 Æ f2) Ç (v2 Æ g2)

ÆÆ

v1 < v2

v1 appears before v2 in the variable

ordering

d2

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(: v1 Æ f1) Ç (v1 Æ g1) d2Æ

v1 < v2

(: v1 Æ (f1 Æ d2)) Ç (v1 Æ (g1 Æ d2))

v1

f1 Æ d2 g1 Æ d2

What if f1 Æ d2 = g1 Æ d2 ?

Return f1 Æ d2

O(n1 £ n2)

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BDD Operations: Or

Or(d1,d2)

=

Not ( And ( Not(d1), Not(d2) ) )

O(n1 £ n2)

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BDD Operations: Exist

Exist(“0”,v) = ?

32




Exist(“0”,v) = “0”

Exist(“1”,v) = ?

33




Exist(“0”,v) = “0”

Exist(“1”,v) = “1”

Exist((: v Æ f) Ç (v Æ g) , v) = ?

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Exist(“0”,v) = “0”

Exist(“1”,v) = “1”

Exist((: v Æ f) Ç (v Æ g) , v) = Or(f,g)

Exist((: v’ Æ f) Ç (v’ Æ g) , v) = ?

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Exist(“0”,v) = “0”

Exist(“1”,v) = “1”

Exist((: v Æ f) Ç (v Æ g) , v) = Or(f,g)

Exist((: v’ Æ f) Ç (v’ Æ g) , v) =

(: v’ Æ Exist(f,v)) Ç (v’ Æ Exist(g,v))

O(n2)

But f is SAT iff 9 V. f is not “0”. So why doesn’t this imply P = NP?

Because the BDD size changes!

36



BDD Applications

SAT is great if you are interested to know if a solution exists

BDDs are great if you are interested in the set of all solutions• How many solutions are there?• How do you do this on a BDD?

Or if your problem involves computing a fixed point• Set of nodes reachable from a given node in a graph

37



BDD Application: Counting Sudoku Solutions

1

3 2

1 2 3 4

1

2

3

4

How many ways can you solve this puzzle?

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0 3 1 2

2 1 0 3

3 0 2 1

1 2 3 0

1 2 3 4

1

2

3

4

How many ways can you solve this puzzle? At least 2.

39




0 3 1 2

1 2 0 3

3 0 2 1

2 1 3 0

1 2 3 4

1

2

3

4

How many ways can you solve this puzzle? At least 2.

40




1

3 2

1 2 3 4

1

2

3

4

a11,b11

a44,b44

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1

3 2

1 2 3 4

1

2

3

4

: a13 Æ b13

a33 Æ : b33

a31 Æ b31

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1

3 2

1 2 3 4

1

2

3

4

Distinct Elements

a11 © a12 Ç b11 © b12

Æ

a11 © a13 Ç b11 © b13

Æ

a11 © a14 Ç b11 © b14

Æ

a12 © a13 Ç b12 © b13

Æ

a12 © a14 Ç b12 © b14

Æ

a13 © a14 Ç b13 © b14

Repeat for each row, column and sub-square

Construct BDD

Count number of solutions

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Graph Reachability

0

1

2

3

4

5

6

7

Which nodes are reachable from “7”?

{2,3,5,6,7}

But what if the graph has trillions of nodes?

44



Graph Reachability

0

1

2

3

4

5

6

7

Use three Boolean variables (a,b,c) to encode each node?

: a Æ : b Æ : ca Æ b Æ c

45



Graph Reachability

0

1

2

3

4

5

6

7



a Æ : b Æ : c

46



Graph Reachability

0

1

2

3

4

5

6

7



a Æ : b Æ : c

a Æ : b Æ c

47



Graph Reachability

0

1

2

3

4

5

6

7

Key Idea 1: Every Boolean formula represents a set of nodes!

a Æ b Æ : c = ?

The nodes whose encodings satisfy the formula.

48



Graph Reachability

0

1

2

3

4

5

6

7


a Æ b Æ : c = {6}

49



Graph Reachability

0

1

2

3

4

5

6

7


a Æ b = ?

50



Graph Reachability

0

1

2

3

4

5

6

7


a Æ b = {6,7}

53



Graph Reachability

0

1

2

3

4

5

6

7

• Key Idea 2: Edges can also be represented by Boolean formulas

• An edge is just a pair of nodes

• Introduce three new variables: a’, b’, c’

• Formula © represents all pairs of nodes (n,n’) that satisfy © when n is encoded using (a,b,c) and n’ is encoded using (a’,b’,c’)

54



Graph Reachability

0

1

2

3

4

5

6

7

: a Æ : b Æ : c Æ : a’ Æ : b’ Æ c’

Key Idea 2: Edges can also be represented by Boolean formulas

55



Graph Reachability

0

1

2

3

4

5

6

7

a Æ : b Æ c Æ : a’ Æ b’ Æ : c’


56



Graph Reachability

0

1

2

3

4

5

6

7

a Æ : b Æ c Æ : a’ Æ b’ Æ : c’

Ç

: a Æ : b Æ : c Æ : a’ Æ : b’ Æ c’


a Æ : b Æ c Æ : a’ Æ b’ Æ : c’

Ç

: a Æ : b Æ : c Æ : a’ Æ : b’ Æ c’

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Graph Reachability

0

1

2

3

4

5

6

7

Key Idea 3: Given the BDD for a set of nodes S, and the BDD for the set of all edges R, the BDD for all the nodes that are adjacent to S can be computed using the BDD operations

Image(S,R) =

(9 a,b,c . (S Æ R)) [ a \ a’, b \ b’, c \ c’]

Variable renaming : replace a’ with a

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Graph Reachability Algorithm

S = BDD for initial set of nodes;R = BDD for all the edges of the graph;

while (true) { I = Image(S,R); //compute adjacent nodes to S if (And(Not(S),I) == False) //no new nodes found

break; S = Or(S,I); //add newly discovered nodes to result}

return S;

Symbolic Model Checking. Has been done for graphs with 1020 nodes.

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Questions?

Sagar ChakiSenior Member of Technical StaffRTSS ProgramTelephone: +1 412-268-1436Email: [email protected]

U.S. MailSoftware Engineering InstituteCustomer Relations4500 Fifth AvenuePittsburgh, PA 15213-2612USA

Webwww.sei.cmu.edu/staff/chaki

Customer RelationsEmail: [email protected]: +1 412-268-5800SEI Phone: +1 412-268-5800SEI Fax: +1 412-268-6257

mailto:[email protected]

© 2011 Carnegie Mellon University Binary Decision Diagrams Part 2 15-414 Bug Catching: Automated...

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