1

我想为以下布尔函数创建 BDD：

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

我设法仅F = (A'B'C'D')使用以下代码创建，但如何将其他产品术语添加到现有 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;
}
4

2 回答 2

2

独立构建每个连词，这样您就conj0可以conj3确保只否定正确的文字。我不是特别精通 C 语言，现在也没有开发环境设置，因此您需要进行一些更正。

我将使用以下映射

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

在你的 for 循环中构建conj0你现在做的方式。conj0 = bdd事后确定。

conj1将对其进行编码(A' B C)使用

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;

conj2对和做同样的事情conj3。

在你构建完所有的连词之后，使用Cudd_bddOr(). 还要确保你得到正确的Cudd_Ref()和，Cudd_Deref()否则你会泄漏内存。

于 2016-04-08T10:43:40.890 回答
1

如果您只对该特定功能感兴趣，这里有一种构建和检查它的方法：

#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;
}

注意（最佳）变量顺序的选择。你应该看到这个输出：

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

a | (c & b | !c & !d)
于 2016-04-14T22:52:59.977 回答