6.896 Sublinear Time Algorithms
May 3, 2007
Lecture 24
Lecturer: Ronitt Rubinfeld
Scribe: Ivaylo Riskov
1
Recap
f
:
{
0
,
1
}
n
→ {
0
,
1
}
x < y
if
∀
i, x
i
≤
y
i
f
is monotone if
∀
x
≤
y, f
(
x
)
≤
f
(
y
)
We define a violating edge (
x, y
) such that
y
=
x
⊕
e
i
, x < y, f
(
x
)
> f
(
y
). We can use the following
tester: Pick
O
(
n
)
edges and reject if any violate.
Finally, we started the proof of the following theorem:
Theorem 1
The above tester passes monotone functions and fails
far from monotone functions (must
chnage
f
on at least
fraction of the inputs) with probability at least
3
/
4
.
2
Proof of Theorem
Claim 2
If
f
is
far from monotone it has at least
2
n
violating edges.
Proof
of claim:
By contrapositive: assume
f
has less than
2
n
violating edges.
Let
f
(
i
)
be
f
after the
i
th phase. That is, phase
i
fixes edges in
i
th direction without ruining lower
direction edges.
Inductive assumption:
f
(
i
−
1)
has no violating edges in directions 1
. . . i
−
1.
Definition 3
Consider the following definitions:
•
H
0
←
nodes
x
such that
x
i
= 0
•
H
1
←
nodes
x
such that
x
i
= 1
•
M
i
←
violating edges in
i
th direction.
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 Fall '04
 RonittRubinfeld
 Algorithms, Following, Christopher Nolan, Adam Copeland

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