Is the number of nodes in an ORBDD 2n + 2?

Question

Wikipedia says that the number of nodes in a ORBDD (Order Reduced Binary Decision Diagrams) of order $x_1 < x_2 < \dots < x_{2n}$ is $2n + 2$.

But I can't find proof. Anyone?

Answer 1

It doesn't say that. To quote:

Let us consider the Boolean function $f(x_1,\ldots, x_{2n}) = x_1x_2 + x_3x_4 + \cdots + x_{2n-1}x_{2n}$. Using the variable ordering $x_1 < x_3 < \cdots < x_{2n-1} < x_2 < x_4 < \cdots < x_{2n}$, the BDD needs $2^{n+1}$ nodes to represent the function. Using the ordering $x_1 < x_2 < x_3 < x_4 < \cdots < x_{2n-1} < x_{2n}$, the BDD consists of $2n + 2$ nodes.

It is only talking about that particular function, and demonstrating the difference that different variable orderings can make.

As it says a few lines later:

There are functions for which the graph size is always exponential — independent of variable ordering.

Bryant [1] proves several functions have exponentially sized OBDDs, regardless of the variable ordering (see for example Theorems 2 and 4 in the paper).

References

1. Randal E. Bryant, "On the Complexity of VLSI Implementations and Graph Representations of Boolean Functions with Application to Integer Multiplication", IEEE Trans. on Computers, 40(2):205-226, 1991.