1. Read Once Branching Programs: a model of computation used in CAD.x1 1
xi: query nodes x3 x2 1 1 x3 x2 1 1
Output nodes 1

2. What is the ROBP for OR function with 4 variables?

3. Represent a BP with a polynomial:
the polynomial for the output node 1. p x 1
(1-x)p xp 1 x1
(1- x1) 1 x1 x1
(1- x1) x3 x2 1
p1 …. pn 1
x2(1- x1) x3 x2 1 x
p1+…+pn 1 1 1

4. Relationship between a ROBP B and the assigned polynomial p:
For any Boolean assignment to B’s variables, all polynomials assigned
to its nodes evaluate to 0 or 1. Polynomials evaluating to 1 are those on
the computation path for that assignment. Hence B and p agree on
Boolean input.
Since B is read once, we may write p as a sum of product terms:
y1y2….ym, where yi is xi , (1- xi) or 1 and each term correspond a path
from the start node to the output node 1.
If the path doesn’t contain xi then yi =1 . Take each term of p
with yi =1 and split it into xi and (1- xi). Doing this yields the same
polynomial. Continue until all yi is either xi or (1- xi).
The end result is an equivalent polynomial q that contains a product term
for each assignment on which B evaluates to 1.

5. EQROBP = { <B1, B2> | B1 and B2 are equivalent ROBP }.
Thm: EQROBP is in BPP.
Proof:
Suppose there are m variables in both ROBP’s.
Transform both ROBP’s into polynomials as above.
Let F be a field with at least 3m elements.
D = “on input <B1, B2>:
1. Select randomly a1,…, am from F.
2. Evaluate the assigned polynomials p1 and p2 at ai’s.
3. If p1(a1,…, am ) = p2(a1,…, am ) accept; o/w reject.”
If B1 and B2 are equivalent, then D always accepts, since they evaluate
to 1 on exactly the same assignments. Else D rejects with a probability
at least 2/3 by the following lemmas.

6. Lemma:
For every d 0, a degree d polynomial p on a single variable x
either has at most d roots, or it is a zero polynomial.
Proof:
By induction on d.
It is obvious for d = 0, since degree-0 polynomial either has no root or is
identical to zero..
Suppose it is true for d-1 and prove true for d.
If p != 0 and of degree d with a root at a, then (x-a)|p and p/(x-a) is a
non-zero polynomial of degree d-1 and has at most d-1 roots
by induction hypothesis.

7. Lemma:
Let F be a finite field with f elements and let p be a nonzero polynomial on the variables x1 through xm, where each variable has degree at most d. If a1 through am are selected randomly from F, then Pr[p(a1,…, am ) = 0]  md/f.
Proof: By ind on m.
For m=1, by the previous lemma, p has at most d roots, so the probability is at most d/f.
Assume it is true for m-1 and prove for m. Represent p as
p = p0+x1p1+…+x1d pd .
If p(a1,…, am ) = 0, then either all pi evaluate to 0 or some p1 doesn’t evaluate to 0 and a1 is a root of the single variable poly obtained by evaluating p0 through pd on a2 through am.

8. p = p0+x1p1+…+x1d pd
Since p is nonzero, there is a pj is nonzero.
Pr[ all pi evaluate to 0 ]  Pr[ pj evaluate to 0]  (m-1)d/f,
by induction hypothesis.
If some pi doesn’t evaluate to 0, then on the assignment of a2 through am, p reduces to a nonzero single variable polynomial. The basis already shows that a1 is a root of such a poly with probability of at most d/f.
Therefore, Pr[p(a1,…, am ) = 0]  (m-1)d/f +d/f = md/f.