# A very primitive implementation of two basic Binary decision diagrams

Binary decision diagrams are awesome. Read about them. Knuth has written about them. You should be able to locate some material on the net. In a nutshell, BDD is a way to represent a Boolean function uniquely.

Now lets talk about this primitive implementation. We have an array called ‘mem’ which keeps all the nodes of a BDD.

In each BDD, there are two sink nodes which are fixed at the bottom of an array ‘mem’ which keeps all of our bdd-nodes. These two nodes, sink0 and sink1 are the last nodes in a BDD to which all other nodes must have a path. All nodes in any BDD must be unique, in a sense that no two nodes in a BDD should have equal triple value of (tag, lo, hi).

A data-structure called ‘bddNode’ has been type-defined which has four fields. These are tag, index, lo, hi (actually there is a fifth one also which we are not using). The field, tag, is the number of nodes in a particular BDD; index, is the variable index; lo, is the address of its LO node; and hi, is the address of HI node. Field ‘tag’ is used for making debugging easier. THIS IS NOT THE MOST EFFICIENT DATA-STRUCTURE.

We are constructing two already reduced BDDs. First is  $bdd1 = x_1 + x_3$ and second one is $bdd2 = x_2 \oplus x_3$. Since there are only two variables in each bdd, the question of ordering does not arise. However these two functions are defined over 3 variables, .

On these line, many primitive BDD’s can be built over a collection of variables.

Since file was getting too longer to understand, I have not coded how to compose two bdd’s such as $bdd3 = bdd1 \& bdd2$ etc.. One can write a generic function to do this. HINT : Think of a BDD node as an if-then-else structure e.g. if variable 3 then goto node 2 else sink0.

A simple printNode function is provided. It will print a single BDD node. It would be good if you can write printBDD function which can print a full BDD. HINT : Graph traversal.

Great many things can be done with/on BDD’s. Have fun.

```
/*
* =====================================================================================
*
*       Filename:  demo_bdd.c
*
*    Description:  A demo program to show how operations on BDD are
*    implemented.
*
*        Version:  1.0
*        Created:  Thursday 18 August 2011 01:15:29  IST
*       Revision:  none
*       Compiler:  gcc
*
*         Author:  Dilawar Singh (Graduate Student, EE IITB), dilawar@ee.iitb.ac.in
*      Institute:  IIT Bombay
*
* =====================================================================================
*/

#include    <stdio.h>
#include    <stdlib.h>
#include    <string.h> /*  for length function. */

int totalNodes = 0; /* this many nodes are currently in use. */

typedef struct node
{
unsigned int *tag; /* Useful in debugging.*/
unsigned int index;
int vRef; /*  reference count minus 1. NOT USING. */
} bddNode;

int i;

void printNode(bddNode*);
void printBdd(bddNode*);
bddNode* constructBdd(char* fVar[], char* var[], char* binOp);

bddNode mem[100];

/*
*  These sink-nodes never change. Let's define them and we'll put them in 'mem' at bottom. There,
*  they will rest in peace all the time.
*/
bddNode* sink0;
bddNode* sink1;

int main()
{
/*  Let's define variables with ordering. There are algorithms available to
*  find a suitable ordering such that BDD size is minimum. For now, we
*  assume that ordering is know to us */

/*  for simplicity sake, char 1 means x_1 and so on. */
char *var[] = {"1", "2", "3"}; /*  Note that these are char and not numbers. */

/*  Now we should have some binary operations. What is a use of BDDs if we
*  can not compose them. */
char *binOp[] = {"&", "|", "^", "\$", "#"}; /* I am not using all of them.  */

/*  now we need to construct some bdd so that we can have some fun! Ideally,
*  there should be a function to which we can pass a boolean function and
*  presto, we get a bdd. Why don't you do that and I hardcode two BDD's.*/

/*  I am hardcoding two bdds, bdd1 and bdd2. Later we will combine them
*  using a binary operation and produce a new BDD.
*
* let bdd1 = x_1 | x_3, read x_1 'or' x_3.
* NOTE : Sink nods are labelled as 000 (0) and 011. DO NOT CONFUSE THEM
* WITH OTHER NODE LEVELS which are variable names.
*
*              1
*             / \
*           LO   \
*           /     HI
*          3       \
*        /  \      011
*       LO   HI
*      /      \
*    000      011
*/

/*  initialize sink nodes and put them at the bottom of mem*/
sink0 = mem;
sink1 = mem + 1;
totalNodes += 2; /*  two sink nodes. */

sink0->tag = 0;
sink0->index = 0;
sink0->lo = 0;
sink0->hi = 0;

sink1->tag = 1;
sink1->index = 0;
sink1->lo = 1;
sink1->hi = 1;

/*  let's construct bdd1 */
char* var1[] = {"1", "3"}; /*  these are variables bdd needs. */
bddNode* bdd1 = constructBdd(var1, var, "&");

/*  and now bdd2 which is x_2 xor x_3
*
*            2
*         /    \
*        LO      HI
*       /         \
*      3           3
*     / \         / \
*    LO  HI     HI  LO
*     |    \   /    /
*     |     \/     /
*      \   /  \   /
*       000    011
*
*/
char* var2[] = {"2", "3"};
bddNode* bdd2 = constructBdd(var2, var, "^");
return 0;
}

void printNode(bddNode* p)
{
printf("Node : ");
}

void printBdd(bddNode* p)
{

}

bddNode* constructBdd(char* fVar[], char* var[], char* binOp)
{
if (strncmp(binOp, "&", 1) == 0)
{
int nodeInThisBDD = 0;
printf("Building a BDD with AND operation\n");
bddNode* r = mem + totalNodes + 1; nodeInThisBDD++; totalNodes++;
bddNode* rl = mem + totalNodes + 1; nodeInThisBDD++; totalNodes++;
nodeInThisBDD += 2; /*  two sink nodes. */
r->tag = nodeInThisBDD;
r->index = (*fVar[0] - 48) ;
rl->tag = --nodeInThisBDD;
rl->index = (*fVar[1] - 48);
printNode(r); printNode(rl);
return r;
}
if (strncmp(binOp, "^", 1) == 0)
{
int nodeInThisBDD = 0;
printf("Building a BDD with XOR operation.\n");
bddNode* r = mem + totalNodes + 1; nodeInThisBDD++;  totalNodes++;
bddNode* rl = mem + totalNodes + 1; nodeInThisBDD++; totalNodes++;
bddNode* rh = mem + totalNodes + 1; nodeInThisBDD++; totalNodes++;

nodeInThisBDD += 2; /*  two sink nodes */
/*  name it */
r->tag = nodeInThisBDD;
r->index = (*fVar[0] - 48);

rl->tag = nodeInThisBDD -1 ; rh->tag = nodeInThisBDD - 2;
rl->index = rh->index = *fVar[1] - 48;
rl->lo = rh->hi = sink0;
rl->hi = rh->lo = sink1;
printNode(r);
return r;
}
else
{
fprintf(stderr, "This operation is not implemented yet.");
}
}

```