Provided by: alliance_5.0-20110203-4_amd64 bug

NAME

       implybddnode - computes a bdd that implies a conjonction of two bdd nodes.

SYNOPSYS

       #include "bdd101.h"
       bddnode ∗implybddnode( BddSystem, BddNode1, BddNode2 )
         bddsystem ∗BddSystem;
         bddnode   ∗BddNode1;
         bddnode   ∗BddNode2;

PARAMETERS

       BddSystem           The bdd system.

       BddNode1            The first bdd node.

       BddNode2            The second bdd node.

DESCRIPTION

       implybddnode  tests  if  the intersection of BddNode1 and not BddNode2, exists, in the bdd
       system BddSystem.  If a null pointer is given, the default bdd system is used.

RETURN VALUE

       implybddnode returns the bdd node zero if there is no intersection,  and  a  computed  bdd
       node otherwise.

EXAMPLE

       #include "bdd101.h"
          bddsystem  ∗BddSystem;
          bddcircuit ∗BddCircuit;
          bddnode    ∗BddNode1;
          bddnode    ∗BddNode2;
          bddnode    ∗BddImply;
          chain_list ∗Expr;
          BddSystem  = createbddsystem( 100, 1000, 100, 50000 );
          BddCircuit = createbddcircuit( "hello_world", 10, 10, BddSystem );
          Expr = createablbinexpr( ABL_OR,
                                   createablatom( "i0" ),
                                   createablatom( "i1" ) );
          BddNode1 = addbddcircuitabl( BddCircuit, Expr );
          freeablexpr( Expr );
          Expr = createablbinexpr( ABL_AND,
                                   createablatom( "i0" ),
                                   createablatom( "i1" ) );
          BddNode1 = addbddcircuitabl( BddCircuit, Expr );
          freeablexpr( Expr );
          BddImply = implybddnode( (bddsystem ∗)0, BddNode1, BddNode2 );
          Expr = convertbddcircuitabl( BddCircuit, BddNode );
          /* displays (i0 and (not i1)) */
          viewablexpr( Expr, ABL_VIEW_VHDL );
          freeablexpr( Expr );
          destroybddsystem( (bddsystem ∗)0 );
          destroybddcircuit( (bddcircuit ∗)0 );

SEE ALSO

       bdd(1)