Read Once Branching Programs:

1
Read Once Branching Programs
a model of computation used in CAD.
x1
1
0
xi query nodes
x3
x2
0
0
1
1
x2
x3
1
0
0
1
Output nodes
0
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.
x
p
xp
1
(1-x)p
0
1
x1
1
(1- x1)
x1
0
x1
x3
(1- x1)
x2
0
0
1
p1 . pn
1
x2(1- x1)
x2
x3
1
0
x
p1pn
0
1
0
1
0
1
4
Relationship between a ROBP B and the assigned
polynomial p For any Boolean assignment to Bs
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 doesnt 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 ltB1, B2gt B1 and B2 are equivalent
ROBP .
Thm EQROBP is in BPP. Proof
Suppose there are m variables in both
ROBPs. Transform both ROBPs into polynomials as
above. Let F be a field with at least 3m
elements. D on input ltB1, B2gt
1. Select randomly a1,, am from F.
2. Evaluate the assigned polynomials p1 and p2
at ais. 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 Prp(a1,, am ) 0 ?
md/f. Proof By ind on m. For m1, 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
p0x1p1x1d pd . If p(a1,, am ) 0, then
either all pi evaluate to 0 or some p1 doesnt
evaluate to 0 and a1 is a root of the single
variable poly obtained by evaluating p0 through
pd on a2 through am.
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p p0x1p1x1d 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 doesnt 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, Prp(a1,, am ) 0 ? (m-1)d/f
d/f md/f.