Formal Methods in software development
a.y.2016/2017
Prof. Anna Labella
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OBDD
[H-R cap.6]
Representing boolean functions
A formula can be represented by a boolean function,
where its variables are boolean variables, as in circuits.
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OBDD
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[H-R cap.6]
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Exercises
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Binary decision trees
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Exercises
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Exercises
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BDD
Removal of duplicated terminals
Removal of redundant tests
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(they are nomore trees)
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BDD
Removal of duplicated non terminals
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BDD

Eliminate duplicated terminals

Eliminate redundant tests

Eliminate duplicated non terminals
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BDD: how do we introduce
operations? Composing BDD

constants

variables
1
☐
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BDD: sums and products
We can substitute terminals by non
terminals and compose functions
e.g.
 In the conjunction we substitute 1 by the
BDD of the other function
 In the disjunction we substitute 0 by the
BDD of the other function
 In the negation we swap 1 and 0

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Exercises
Let f be represented by
describe
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BDD

Satisfiability: to reach 1 via a coherent
path

Validity: it is not possible to reach 0 via
a coherent path
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OBDD: ordered binary decision
diagrams
Equivalence?
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OBDD
Normal form: order and then reduce
Theorem. We have a unique result
Hence there is a canonical form
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OBDD
An example: the parity function on 4 variables
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Repeated variables
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OBDD
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order without repetitions
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OBDD
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OBDD
•
•
•
•
•
•
Composition is nomore directly allowed
Absence of redundant variables
Test for semanic equivalence with a compatible ordering
Test for validity (reducibility to B1)
Test for implication (reducibility of f. g to B0)
Test for satisfiability (non reducibility to B0)
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Exercises
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OBDD: the algorithm reduce
It identifies equal nodes going bottom up
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OBDD: the algorithm apply
It exploits operations
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OBDD: the algorithm apply
Shannon expansion theorem
ƒ = x ƒ[0/x] + x ƒ[1/x]
In general form
ƒ op g = xi ( ƒ[0/xi] op g[0/xi] + xi ( ƒ[1/xi] op g[1/xi])
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OBDD: the algorithm apply
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OBDD: the control structure for
the algorithm apply
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OBDD: the algorithm apply
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OBDD: the algorithm restrict
restrictig f to f[0/x] and to f[1/x]
restrict(0, x, Bf) works as follows: For each node n labeled
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OBDD: restrict exercise
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OBDD: the algorithm exists
 f = f[0/x] + f[1/x]
 f = f[0/x] . f[1/x]
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OBDD: the algorithm exists
exercise
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OBDD: the algorithm exists
exercise
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Complexity
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