FDD2BDD algorithm doubt

Question

I want to convert this FDD2BDD, i have done this using by FDD by set->RMNF->SNF to get BDD. That worked perfectly. 

This one I'm bit confused . Not sure why all the nodes got reduced.  I took the FDD (answer ) then try to find the BDD for this(question) . 

I was following the step, but node C is 0 for me. Which means , I cant find node C in the BDD, it reduced to 0.  Could you please help on this?

Answer 1

The algorithm always takes the cofactors, regardless which variable is their root. So, in that case, you just have the leaf node 0.

Comments

So ..what I have done is correct?

FDD2BDD(n6) = Fdd2Bdd(n4) xor  Fdd2Bdd(n3) split by a
FDD2BDD(n4) = 0 xor  1 split by b
FDD2BDD(n3) = 0 xor  Fdd2Bdd(n2) split by b
FDD2BDD(n2) = 1 xor  1 split by c

If I draw here, both -ve and +ve cofactors of c , are 1. Thats not correct in the given question.

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No, it is not correct, since you also got the wrong BDD. The correct computation is as follows:

FDD2BDD(n6)
    = a ? FDD2BDD(n4)⊕FDD2BDD(n3) : FDD2BDD(n4)
    = a ? FDD2BDD(n4)⊕FDD2BDD(n3) : FDD2BDD(n4)
    = a ? (b ? 1 : 0)⊕(b ? (c ? 0 : 1) : 0) : (b ? 1 : 0)
    = a ? (b ? 1⊕(c ? 0 : 1) : 0⊕0) : (b ? 1 : 0)
    = a ? (b ? (c ? 1⊕0 : 1⊕1) : 0⊕0) : (b ? 1 : 0)
    = a ? (b ? (c ? 1 : 0) : 0) : (b ? 1 : 0)


FDD2BDD(n4)
    = b ? FDD2BDD(0)⊕FDD2BDD(1) : FDD2BDD(0)
    = b ? 0⊕1 : 0
    = b ? 1 : 0

FDD2BDD(n3)
    = b ? FDD2BDD(0)⊕FDD2BDD(n2) : FDD2BDD(0)
    = b ? 0⊕(c ? 0 : 1) : 0
    = b ? (c ? 0 : 1) : 0


FDD2BDD(n2)
    = c ? FDD2BDD(1)⊕FDD2BDD(1) : FDD2BDD(1)
    = c ? 1⊕1 : 1
    = c ? 0 : 1

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Thanks for the suggestion . I did computed the +ve cofactor directly, like negative. Now it's perfect. For BDD2FDD we are computing -ve and xor of -ve and +ve. Now we need to revert those steps. Thanks for the clarification.