Formal Verification Lab 3

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Formal Verification Lab 3

• Dan Goldwasser
• dgoldwas_at_cs.haifa.ac.il

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Binary Decision Tree

• A directed rooted tree with
• Non Terminal Nodes
• Labeled by a variable var(v)
• Two successors high(v), low(v)
• Terminal Nodes
• Labeled by value(v) 0/1
• Binary Decision Tree are used to represent a
  Boolean function

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Binary Decision Tree

• Deciding the truth
• value of a formula is
• done by traversing
• the tree according to
• a truth assignment

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Binary Decision Diagram

• There is a lot of redundancy in this
  representation a smaller representation could
  be achieved by merging isomorphic sub-trees.
• The result is a BDD

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Ordered Binary Decision Diagram

• A canonical representation can be obtained by
  following two restrictions
• Variable ordering
• No isomorphic sub trees

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Ordered Binary Decision Diagram

• Assign arbitrary total ordering to variables
• e.g., x1 lt x2 lt x3
• Variables must appear in ascending order along
  all paths

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Ordered Binary Decision Diagram
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Ordered Binary Decision Diagram
Merge isomorphic nodes
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Ordered Binary Decision Diagram
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Ordered Binary Decision Diagram
Initial Graph
Reduced Graph

• Canonical representation of Boolean function
• For given variable ordering
• Two functions equivalent if and only if graphs
  isomorphic
• Can be tested in linear time
• Desirable property simplest form is canonical.

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Logical Operations

• Logical operations can be implemented efficiently
  ( O(n) , n being the size of the OBDD )
• Key Idea Shannons expansion
• f ( ? x ? fx?0) ? (x ? fx?1)

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APPLY

• Definitions
• Let be an arbitrary two argument logical
  operation,
• Let f and f be two Boolean functions
• Let v and v be the roots of the OBDDS of f and
  f
• Let var(v)x and var(v)x

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APPLY

• if v and v are both terminal vertices, then ff
  is value(v) value(v)
• If xx, then we use the Shannon expansionf f
  ( ? x ? (fx?0 f x?0 )) ?
  ( x ? (fx?1 f x?1 ))

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APPLY

• If x lt x (according to the order of the
  variables), thenf x?0 f x?1 f
• since f does not depend on x, so that
• f f ( ? x ? (fx?0 f )) ?
  ( x ? (fx?1 f ))

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Apply Operation - AND
a AND c
ac
a


c
AND