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ECE 474a/575aSusan Lysecky

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ECE 474A/57AComputer-Aided Logic Design

Lecture 11Binary Decision Diagrams (BDDs)


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Boolean Logic Functions Representations

Function can be represented in different waysTruth table, equation, K-map, circuit, etcSome representations not unique (not canonical)

a

a

b

F

F

Circuit 1

Circuit 2

a

0

0

1

1

b

0

1

0

1

F

1

1

0

0

The function F

Truth table

Equation 2: F(a,b) = a

Equation 1: F(a,b) = ab+ ab

1 1

0 1

0 0

0

1

F ba


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Why BDDs An Efficient Representation

Synthesis, optimization, verification, and testing algorithms/tools manipulatelarge Boolean functions

Important to have efficient way to represent these functionsBinary Decision Diagrams (BDDs) have emerged as a popular choice for representingthese functions

BDDsGraph representation similar to a binary tree (i.e. decision trees from previouslectures)

Able to efficiently represent large functionsSome representations are canonical (unique)

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Mux Representation of Boolean Functions

MUX circuit to implement logicfunction S

S(x1, x2, x3)

x3

x2

x1

10MUX

01

10MUX

1

10MUX

01

10MUX

x3x2x1 S

000 1100 0010 1110 1001 1101 1011 0111 0

S(0, 0, 0) = 1

X3 (0)

x2 (0)

X1 (0)

10MUX

01

10MUX

1

10MUX

01

10MUX

S(0, 0, 1) = 0

X3 (1)

x2 (0)

X1 (0)

10MUX

01

10MUX

1

10MUX

01

10MUX

S(0, 1, 0) = 1

X3 (0)

x2 (1)

X1 (0)

10MUX

01

10MUX

1

10MUX

01

10MUX

S(1, 1, 1) = 0

X3 (1)

x2 (1)

X1 (1)

10MUX

01

10MUX

1

10MUX

01

10MUX


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Mux Representation of Boolean FunctionsRelation to BDDs

Corresponding BDD to implement function SOne-to-one correspondence to the MUX gates inthe flipped circuit

x3x2x1 S

000 1100 0010 1110 1001 1101 1011 0111 0

S(x1, x2, x3)

x3

x2

x1

10MUX

01

10MUX

1

10MUX

01

10MUX

Same circuit, justflipped

Corresponding BDD

1 0

x1

x2x2

x3

S(x1, x2, x3)

S(x1, x2, x3)

x3

x2

x1

10MUX

01

10MUX

1

10MUX

01

10MUX


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Binary Decision Diagram (BDD)Example 1

How does it work?Line with bubble represent value = 0Lines without bubble represent value = 1

x3x2x1 S

000 1100 0010 1110 1001 1101 1011 0111 0

1 0

x1

x2x2

x3

S(x1, x2, x3)

1 0

x1

x2x2

x3

S(0, 0, 0)

1 0

x1

x2x2

x3

S(1, 1, 1)

1 0

x1

x2x2

x3

S(0, 1, 0)

1 0

x1

x2x2

x3

S(0, 0, 1)

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Binary Decision Diagram (BDD)Edge Notations

Several ways to represent value = 1 and value = 0Bubble vs. Non-bubble lineDashed vs. Solid line

T (then) vs. E (else) labelsWe will adopt T vs. E labels consistent with most of the book (Hatchel) examples

1 0

x1

x2x2

x3

S(x1, x2, x3)

1 0

x1

x2x2

x3

S(x1, x2, x3)

1 0

x1

x2x2

x3

S(x1, x2, x3)

T

E

TE

T

E

T

E


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Binary Decision Diagram (BDD)Example 2

Lets consider another functionf(a,b,c,d) = abc + bd + cd

1 0

a

bb

dT

E

T TE

E

T

E

f

cc

T T

EE

1 0

a

bb

dT

E

T TE

E

T

E

f

cc

T T

EE

What is the value of f(1,0,1,0)?

Notice that if a=1 and b=0, thefunction does not depend on a

value for c.

1 0

a

bb

dT

E

T TE

E

T

E

f

cc

T T

EE

What is the value of f(1,1,0,1)?


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Ordered Binary Decision Diagram (OBDD)What is a OBDD?

Ordered binary decision diagrams ensure the variables appear inthe same order along all paths from the root to the leaves

0 1

a

cc

T

T TE

E

E

f

bT

E

Ordering : a c b Not ordered

1 0

a

cb

T

T

T

E

E

E

f

bc

T

TE

E

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Ordered Binary Decision Diagram (OBDD)Different Ordering Lead to Different Complexity Example1

Variable ordering important, may result in a more complex (or simple) BDD All three BDDs below represent the same functionThird ordering (b c a d) optimal because there is exactly one node foreach variable

1 0

b

c

dT

E

T

E

T

E

f

a

T

E

Order : b c a d

1 0

a

bb

dT

E

T TE

E

T

E

f

cc

T T

EE

Order : a b c d

1 0

a

dd

cT

E

T TE

E

T

E

f

bb

T TEE

Order : a d b c

cT

E


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Ordered Binary Decision Diagram (OBDD)Different Ordering Lead to Different Complexity Example 2

Consider F = ab + cd + ef, again both BDDs represent same function Variable order has a large impact on resulting BDD, first variable ordering (a b c d e f) yields a much simpler BDD

Order : a b c d e f Order : a c e b d f

0 1

a

b

e

dT

E

TT

E

E

T

E

f

f

c

T

T

E

E

0

a

e

c

f

bb

c

e e e

bb

d d

f

1


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BDDs for Basic Logic Functions

1 0

aTE

f

NOTa F

0 11 0

ANDba F

00 010 001 011 1

1 0

a

b

E

T

T

E

f

ORba F

00 010 101 111 1

1 0

a

b

E

ET

T

f

XORba F

00 010 101 111 0

1 0

a

bb

E

E E

T

T

T

f

NANDba F

00 110 101 111 0

f

01

a

b

E

T

T

E

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Formal Definition of BDDs

A BDD is a direct acyclic graph (DAG) representing amultiple-output switching function F

Nodes are partitioned into three subsets

Function nodeRepresents the function symbol (f)Indegree = 0Outdegree = 1

Internal nodeRepresents variable in function (a, b, c, d)Indegree 1Outdegree = 2

Terminal nodeRepresents a value (1 or 0)Indegree 1Outdegree = 0

1 0

b

c

dT

E

T

E

T

E

f

a

T

E

Function node

Internalnodes

Terminal nodes


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A BDD definition (cont)Three types of edges

Incoming edge

From the function node and defines functionT edge

From an internal node and represents whenthe corresponding variable is 1

E edge

From an internal node and represents whenthe corresponding variable is 0

1 0

b

c

dT

E

T

E

T

E

f

a

T

E

Incoming edge

E edge

T edge


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A BDD definition (cont)The function f represented by a BDD isdefined as follows

The function of the terminal node is aconstant value (1 or 0)The function of a T edge is the function ofthe head nodeThe function of a E edge is the complementof the function of the nodeThe function of a node v is given by vf T +vf E, when f T is the function of the T edge andf E is the function of the E edgeThe function of the function node is thefunction of its outgoing edge

1 0

b

c

dT

E

T

E

T

E

f

a

T

E

Function of the functionnode is the function ofits outgoing edge

Function of thisT edge is a

Function of this terminal node is 1

Function of this terminalnode is 0

Function of thisE edge is a

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Building BDDs - Exercise 1

Build a BDD for f = abc+ abc + abc+ abc Use the variable ordering a b c

f

Compute cofactors of f a and f a with respect to b(second variable in ordering)

f a = bc+ bc

f a = bc+ bc

(f a)b = f ab = c

aET

f

partial expansionwith respect to b

(f a)b = f ab = c

bET

f ab f ab

(f a)b = f ab = c

(f a)b = f ab = c

bET

f ab f ab

aET

f a f a


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Build a BDD for f = abc+ abc + abc+ abc Use the variable ordering a b c

f

Compute cofactors of f ab , f ab, f ab, f ab with respectto c (third variable in ordering)

f ab = c

f ab = c

f ab

= c

f ab = c

expansion withrespect to c, finalBDD

f abc = 0

f abc = 1a

ET

f

bET

bET

cET

1 0

f abc = 1f abc = 0 c

ET

1 0f abc = 0

f abc = 1

cET

0 1

f abc = 0

f abc = 1

cET

0 1

aET

bET

f ab f ab

bET

f ab f ab


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f = abc + abc + abc + abc

aET

f

bET

bET

cET

cET

cET

cET

1 0 1 0 0 1 0 1

Does it work?

f b00110011

c01010101

a00001111

10100101

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Build a BDD for f = abc+ bd + cdUse the variable ordering b c d a

f

Compute cofactors of f with respectto b (first variable in ordering)

f = abc + bd+ cd

f b = d + cd

bET

f

f b f b

partial expansionwith respect to b

f b = ac + cd


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Compute cofactors of f b and f b with respect to c(second variable in ordering)

f b = ac + cd

f b = d + cd

bET

f

f b f b

bET

f

f bc

cE

T

f bc = f bc = f bc

equivalentcofactors, we cancreate a singlenode (reduced)

f bc = d

f bc = d

f bc

cET

f bcf bc

cET

f bc

bET

f

f bc = a

f bc = d

partial expansionwith respect to c


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Compute cofactors of f bc, f bc, and f bc withrespect to d (third variable in ordering)

f bc = f bc = f bc = d

bET

f

f bc

cET

f bc = f bc = f bc

f bcd = f bcd = f bcd = 1

f bcd = f bcd = f bcd = 0

bET

f

f bc

cET

dET

1 0

partial expansionwith respect to d

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Compute cofactors of f bc with respect to a (fourthvariable in ordering)

bET

f

f bc

c ET

f bc = a

dET

1 0

bET

f

cE

T

dET

1 0

f bca = 1

f bca = 0

aET

expansion withrespect to a, finalBDD


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Build a BDD for f = abc + bd + cd

Does it work?

bET

f

cET

dET

1 0

aET

f b00001111

c00110011

d01010101

a00000000

0000111

1

0011001

1

0101010

1

1111111

1

010101000101011

1


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BDD to Boolean Function Exercise 1

Can we go from a BDD to a Boolean function ?Sum all paths from function node to terminal nodes

bET

f

cET

dET

1 0

aET

F = bca + bcd+ bd

= aF a + aF a= a(1) + a(0)= a + 0= a

= cF c + cF c= c(a) + c(d)= ca + cd

= bF b + bF b= b(ca + cd) + b(d)

= bca+ bcd + bd

= d

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BDD to Boolean Function Exercise 2

Another Example

a ET

f

bE

T

0 1

cET

F = abc + ab+ a

= c

= bF b + bF b= b(c) + b(1)= bc+ b

= aF a + aF a= a(bc + b) + a(1)= abc + ab+ a


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Reducing BDDs

aET

f

bET

bET

cET

1 0

cET

1 0

cET

0 1

cET

0 1

isomorphic subgraphs

aET

f

bET

bET

cET

1 0

cET

1 0

cET

0 1

aET

f

bET

cET

1 0

cET

1 0

cET

0 1

When building BDDs, result not always reduced (Example 1 - slide 19)We have isomorphic subgraphs, potential for reduction

Transform a non-reduced BDD into a reduced BDD by iterativel y applyingIdentify isomorphic subgraphsRemove redundant nodes

A ordered reduced BDD (ROBDD) is unique so we have a canonical form

F b = F b(redundant node)


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Reducing BDDsExample 2

isomorphic subgraphs

aT

f

bET

cET

1 0

cET

1 0

E

cET

0 1

aT

f

bET

cET

1 0

E

cET

0 1

aT

f

E

cET

0 1

cET

1 0

Lets try to reduce and OBDDIterative apply

Identify isomorphic subgraphsRemove redundant nodes

Fb = F b(redundant node)

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Summary, and then some

Binary Decision Diagrams (BDDs)Efficient mechanism to representation of Boolean functions in terms of memory andCPU

Translating function BDD and BDD functionImportance of variable orderingMethod to reduced OBDDs, getting a ROBDD

We said that BDDs can be efficiently stored and manipulated HOW? Refer to handwritten notes Supplem ental materials attached


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ITE Operator Two argument operators expressed in terms of ITE

FFF0011

ITE(F, G, 0)FG F > G0010

ITE(F, G, 0)FG AND(F, G)0001

0000000

GGG0101

ITE(F, 0, G)FGF < G0100

ITE(F, 1, G)F + GOR(F, G)0111

ITE(F, G, G)F GXOR(F, G)0110

ITE(F, G, G)(F G) XNOR(F, G)1001

ITE(F, 0, G)(F + G) NOR(F, G)1000

ITE(F, 1, G)F + G F G1011

ITE(G, 0, 1)G NOT(G)1010

ITE(F, G, 1)F+ GF G1101

ITE(F, 0, 1)F NOT(F)1100

ITE(F, G, 1)(FG) NAND(F, G)1110

1111111

Equivalent FormExpressionNameTable


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ITE Algorithm

ITE(F, G, H) = FG + FH

= v(FG+ FH) v + v(FG+ FH) v

= v(F vGv + F vHv) + v(F v Gv + F v Hv )

= ITE( v, ITE(F v, G v, H v), ITE(F v , G v , H v ) )

Terminal cases are as follows

Most standard manipulation of BDDs can be done with ITE Algorithm recursive, based on formulation where v is the top variable of F, G, H

ITE(1, F, G) = ITE(0, G, F) = ITE(F, 1, 0) = ITE(G, F, F) = F

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Pseduo-code of the ITE Algorithm

ITE(F, G, H){

(result , terminal_case ) = TERMINAL_CASE(F, G, H)

if (terminal_case ) return ( result )

(result , in_computed_table ) = COMPUTED_TABLE_HAS_ENTRY(F, G, H)

if (in_computed_table ) return ( result )

v = TOP_VARIABLE(F, G, H)

T = ITE(F v, Gv, Hv)

E = ITE(F v , Gv , H v )

R = FIND_OR_ADD_UNIQUE_TABLE(v, T, E)

INSERT_COMPUTED_TABLE((F, G, H), R)

return (R)

}

// did we find a terminal case?

// have we already calculated this value?

// recursively calculate this value

// see if subtreealready present

// record the calculated value

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