BDD

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BDD
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Binary Decision Diagrams

• Boolean variable
• Boolean function
• Examples x, xy, xy, represent others in terms
  of these
• Can represent a boolean function by truth table
  
• B function compact, determining satisfaction
  difficult checking equivalence is hard
• Truth table inefficient space satisfaction easy
  comparison is difficult
• Could give canonical representations as DNF, CNF
  complexity of algorithms is high

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Binary Decision Tree
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BDD

• OBDD are a cononical form representation for b
  formulas
• more compagt, manipulated efficiently
• Used widely in variety of areas eg CAD and
  verification

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Binary Decision Trees

• Nonterminal vertex v labeled by a variable var(v)
  and has 2 successors low v (var (v) assigned 0)
  and high v (var (v) assigned 1).
• Each terminal vertex is labelled by 0 0r 1.
• Each BDD determined a b function
• Much like truth table

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Simplify

• Remove duplicate terminals
• Remove duplicate nonterminals (if var(u) var(v)
  low(u) low(v) and high(u) high(v)
• Remove redundant tests (if nonterminal v has
  low(v) high (v) has

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• Boolean variable and expression
• grammar t x 0 1 -t t t t V t
  t gt t
• - and v suffice
• Truth assignments 0/x1, 1/x2
• If ordering of variables is established a b exp t
  over n varables can be viewed as a function
  f(x1,xn) Bn -gt B
• Ordering plays crucial role in considering
  compact representations of b. expressions
• Satisfiability, tautology, equivalence
• Normal forms DNF, CNF
• Satisfiability of b. expressions is NP complete

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DNF, CNF
Similarly, a Conjunctive Normal Form (CNF) is an
expression that can be written as conjunction of
disjunction. Like,
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• if then else operator x-gt y0, y1 defined as
• (x y0) V (-x y1)
• Call x the test expression
• All operators can be expressed using the if then
  else operator and constants 0 and 1, in such a
  way as all tests occur only on un-negated
  variables and variables occur nowhere else
• Eg x is (x -gt 0,1)
• Since variables must occur only in tests the B
  exp x is (x -gt 1,0)
• A new Normal Form If-then-else Normal Form (INF)
• Some Predefined Boolean expressions
• 111, 100, 1v11, 1v01, 1?11, 1?00, 0?11,
  0?01
• 1XX, X00, 1vX1, Xv0X, 1?XX, X?11, 0?X1

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• Shannon expansion of term (expression) t with
  respect to variable x x -gt t1/x, t0/x
• t is equivalent to its Shannon expansion along
  any variable
• Proposition Any b exp is equivalent to an
  expression in INF
• Proof
• Use Shannon expansion to generate the INF for
  any expression t
• If t contains no variables it is equivalent to 0
  or 1 which is in INF
• Else form Shannon expansion of t with wrt one of
  its variables,x
• Since terms t0/x and t1/x have one less
  variable we recursively find INFs for both of
  these call them t0 and t1
• An INF for t is x -gt t0, t1

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• Example Find INF for t (x1? y1) (x2?y2)
• Lets assume the ordering x1 lt y1 lt x2 lt y2
• Now for x1,
• t x1?t1,t0 means, (x1t1)v (x1t0)
• where t1t1/ x1, t0t0/ x1
• evaluating,t1(1?y1)(x2?y2)
• t0(0? y1)(x2?y2)
• For y1,
• t1 y1?t11,t10 where t11t11/ y1, t10t10/
  y1
• evaluating,t11(1?1)(x2?y2)(x2?y2)
• t10(1? 0)(x2?y2)0
• So, t1 y1?t11,0

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• t0 y1?t01,t00 where t01t011/ y1, t00t10/
  y1
• evaluating,t01(0?1)(x2?y2)0
• t00(0? 0)(x2?y2)(x2?y2)
• So, t0 y1?0, t00
• For x2,
• t11 x2?t111,t110 where t111t111/ x2,
  t110t110/ x2
• evaluating,t111 1?y2
• t110 0?y2
• t00 x2?t001, t000 where t001t001/ x2,
  t000t000/ x2
• evaluating,t001 1?y2
• t000 0?y2

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• For y2 ,
• t001 y2? t0011, t0010 where t0011t0011/ y2,
  t0010t0010/ y2
• evaluating,t0011(1?1)1
• t0010(1? 0) 0
• So, t001 y2?1,0
• t000 y2? t0001, t0000 where t0001t0001/ y2,
  t0000t0000/ y2
• evaluating,t0001(0?1)0
• t0010(0? 0) 1
• So, t000 y2?0,1

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• Similarly show,
• t111 y2? t1110, t1111
• y2?1,0
• t110 y2? t1101, t1100
• y2?0,1
• See figure 2.
• Here,
• Low edge else part dotted
• high edge then part solid

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• Decision tree for above b expression
• Show INF as graph it is decision tree (somewhat
  reduced from the above)
• Can reduce even further some expressions are
  identical
• t110 identical to t000 and t111 identical to
  t001
• If we identify all equal subexpressions we end up
  with Binary decision diagram (BDD) no longer a
  tree but a directed acyclic graph (DAG)
• Substitute t000 for t110 in RHS of t11 and also
  t001 for t11, see that t00 and t11 are identical
  and in t 1 replace t11 with t00

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• Figure 2 with caption

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• Figure 3 with caption

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• Figure 4 with caption

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• A BDD is a rooted DAG such that
• One or more terminal nodes of out degree zero
  labelled 0 or 1
• A set of variable nodes of out degree 2. (Out
  going edges are given by functions low and high
  in next figure, low(u) - dashed line high
  (u)-solid line). A variable var(u) is associated
  with each variable node
• A BDD is ordered iff on all paths through the
  graph the variables respect a given linear order
  x1 lt x2 lt x3 lt xn An (O)BDD is reduced R(O)BDD
  iff
• Uniqueness no two distinct nodes u and v have the
  sam variable name and low and high successor
  (var(u)var(v), low(u) low(v) and
  high(u)high(v) gt u v
• Non-redundant tests no variable nodeu has
  identical low and high successor (low(u) is not
  equal t high (u) )

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• Interesting properties of ROBDDs
• Compact representation of b expressions
• There are efficient algorithms for performing all
  kinds of logical operations on ROBDDs
• Based on fact that for any function f Bn -gt B
  there is exactly one ROBDD representing it
• In particular there is exactly one ROBDD for the
  constant T function on Bn (and one for the
  constantly F function)
• Can test in constant time if an ROBDD is
  constantly T (or constantly F) Compare with b
  expressions - this problem is NP complete!

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• An ROBDD represents a function
• Nodes u inductively define b expression tu
  terminal node is a b constant nonterminal node
  marked x is an if-then-else expression where
  condition is x and the two branches are the b
  expressions given by low and high son
• t0 0 t1 1 tu var(u) -gt t high(u), t
  low(u)
• If x1lt x2lt lt xn, is var ordering associate with
  each node u the function fu that maps (b1,,bn)
  to tub1/x1,bn/xn
• Lemma For any function f Bn -gt B there is
  exactly one ROBDD u with variable ordering
  x1ltx2lt,,lt xn such that fu f(x1,,xn)
• Proof Induction on number of arguments of f
  for induction step use two function f0 and f1 of
  n arguments by fixing the first argument of f to
  0 and 1 resp then
• f(x1,,xn1 ) x1 -gt f1(x2,xn1),
  f0(x2,,xn1).
• Buy induction result holds for f0 and f1 -
  and there are unique ROBDD nodes u0 and u1 with
  fu0 f0 and fu1 f1. Two cases to consider to
  show result holds for f (Case u0 u1 and case
  u0 not u1) See reading for remainder of proof.

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• The ordering determines the ROBDD and in
  particular the number of nodes varies see figure
  6.
• Difficult to find minimal ROBDD some heuristics
  for particular classes of functions (see text and
  reading)

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• Figure 6 and caption

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• Rather than construct OBDD and then reduce it
  there is algorithm to reduce the OBDD during
  construction ( see reading)

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Represent Kripke Structures

• If Q is an nary relation over 0,1 then Q can be
  represented by the ROBDD for its characteristic
  function
• fQ(x1,,xn) 1 iff (x1,xn) is in Q
• Let Q be n-ary relation over D (wlog let number
  of elements of D be 2n).
• Let g be bijection g 0,1m -gt D. g takes
  m-tuple x (of 0s and 1s) to element of D --
  d is rep by x
• Q is rep as the char function of the boolean
  relation Q of arity m x n where Q(x1,,xn)
  Q(g(x1) , g(xn)).

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• If M (S, R, L)
• Assume 2n states let g be mapping g 0,1n -gt
  S each assignment is in S so the characteristic
  function representing S is the OBDD representing
  the characteristic function for 1 (all s in S are
  in relation 1),denoted as f1
• For R use same encoding g need 2 sets of
  variables one to represent start state and one to
  represent final state of a transition. If
  transition relation R is encoded by R(x, x),
  then R is rep by char function for R denoted as
  fR
• Let L AP -gt 2S. For each p we have set L(p)
  use encoding g to find char function of that set.