CODEWITHSUNDEEP
binary-decision-diagramcudd
husna
husna

Reputation: 187

CUDD sum of products Boolean expression

I would like to create BDD for the following Boolean function:

F = (A'B'C'D') OR (A'B C) OR (C' D') OR (A)

I managed to create only F = (A'B'C'D') with the following code but how to add other product terms to the existing BDD?

 int main (int argc, char *argv[])
{
    char filename[30];
    DdManager *gbm; /* Global BDD manager. */
    gbm = Cudd_Init(0,0,CUDD_UNIQUE_SLOTS,CUDD_CACHE_SLOTS,0); /* Initialize a new BDD manager. */
    DdNode *bdd, *var, *tmp_neg, *tmp;
    int i;
    bdd = Cudd_ReadOne(gbm); /*Returns the logic one constant of the manager*/
    Cudd_Ref(bdd); /*Increases the reference count of a node*/

    for (i = 3; i >= 0; i--) {
        var = Cudd_bddIthVar(gbm,i); /*Create a new BDD variable*/
        tmp_neg = Cudd_Not(var); /*Perform NOT boolean operation*/
        tmp = Cudd_bddAnd(gbm, tmp_neg, bdd); /*Perform AND boolean operation*/
        Cudd_Ref(tmp);
        Cudd_RecursiveDeref(gbm,bdd);
        bdd = tmp;
    }

    bdd = Cudd_BddToAdd(gbm, bdd); 
    print_dd (gbm, bdd, 2,4);   
    sprintf(filename, "./bdd/graph.dot"); 
    write_dd(gbm, bdd, filename);  
    Cudd_Quit(gbm);
    return 0;
}

Upvotes: 1

Views: 602

Answers (2)

Johannes Thorn
Johannes Thorn

Reputation: 932

Build every conjunction independently so that you get conj0 to conj3 make sure to only negate the correct literals. I'm not particularly versed in C and don't have a development environment setup right now so you will need to make some corrections.

I will use the following mapping

A <=> BDD(0)
B <=> BDD(1)
C <=> BDD(2)
D <=> BDD(3)

Build conj0 the way you do it now in your for loop. Make sure conj0 = bdd afterwards.

For conj1 which will encode (A' B C) use

bdd = Cudd_IthVar(gbm, 0);
bdd = Cudd_Not(bdd);
tmp = Cudd_And(gbm, bdd, Cudd_IthVar(gbm, 1));
Cudd_Ref(tmp);
Cudd_Deref(gbm, bdd);
bdd = tmp;
tmp = Cudd_And(gbm, bdd, Cudd_IthVar(gbm, 2));
Cudd_Ref(tmp);
Cudd_Deref(gbm, bdd);
bdd = tmp;
conj1 = bdd;

Do the same for conj2 and conj3.

After you've got all the conjunctions build build the top level disjunction by using Cudd_bddOr(). Also make sure that you get the Cudd_Ref() and Cudd_Deref() right otherwise you'll leak memory.

Upvotes: 2

Fabio Somenzi
Fabio Somenzi

Reputation: 203

If you are only interested in that particular function, here's a way to build it and inspect it:

#include <stdio.h>
#include <stdlib.h>
#include "cudd.h"

int main(void) {
  /* Get set. */
  DdManager * mgr = Cudd_Init(4,0,CUDD_UNIQUE_SLOTS,CUDD_CACHE_SLOTS,0);
  DdNode *a = Cudd_bddIthVar(mgr, 0);
  DdNode *c = Cudd_bddIthVar(mgr, 1);
  DdNode *b = Cudd_bddIthVar(mgr, 2);
  DdNode *d = Cudd_bddIthVar(mgr, 3);
  char const * const inames[] = {"a", "c", "b", "d"};
  /* Build BDD. */
  DdNode * tmp = Cudd_bddIte(mgr, c, b, Cudd_Not(d));
  Cudd_Ref(tmp);
  DdNode * f = Cudd_bddOr(mgr, a, tmp);
  Cudd_Ref(f);
  Cudd_RecursiveDeref(mgr, tmp);
  /* Inspect it. */
  printf("f");
  Cudd_PrintSummary(mgr, f, 4, 0);
  Cudd_bddPrintCover(mgr, f, f);
  char * fform = Cudd_FactoredFormString(mgr, f, inames);
  printf("%s\n", fform);
  /* Break up camp and go home. */
  free(fform);
  Cudd_RecursiveDeref(mgr, f);
  int err = Cudd_CheckZeroRef(mgr);
  Cudd_Quit(mgr);
  return err;
}

Note the choice of (optimal) variable order. You should see this output:

f: 5 nodes 1 leaves 12 minterms
1--- 1
-11- 1
-0-0 1

a | (c & b | !c & !d)

Upvotes: 1

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