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CS151
Complexity Theory
Lecture 7
April 19, 2011
April 19, 2011
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Outline
• 3 examples of the power of
randomness
– communication complexity
– polynomial identity testing
– complexity of finding unique solutions
• randomized complexity classes
April 19, 2011
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1. Communication complexity
• Goal
: compute
f(x, y)
while communicating as
few bits as possible between Alice and Bob
• count number of bits exchanged (computation free)
• at each step: one party sends bits that are a
function of held input and received bits so far
two parties: Alice and Bob
function f:{0,1}
n
x {0,1}
n
{0,1}
Alice holds x
{0,1}
n
; Bob holds y
{0,1}
n
April 19, 2011
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Communication complexity
• simple function (equality):
EQ(x, y) = 1 iff x = y
• simple protocol:
– Alice sends x to Bob (n bits)
– Bob sends EQ(x, y) to Alice (1 bit)
– total: n + 1 bits
– (works for any predicate f)
April 19, 2011
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Communication complexity
• Can we do better?
– deterministic protocol?
– probabilistic protocol
?
• at each step: one party sends bits that are
a function of held input and received bits so
far
and the result of some coin tosses
• required to output f(x, y)
with high
probability
over all coin tosses
April 19, 2011
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Communication complexity
Theorem
: no deterministic protocol can
compute
EQ(x, y)
while exchanging fewer
than n+1 bits.
• Proof:
– “input matrix”:
X = {0,1}
n
Y = {0,1}
n
f(x,y)
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April 19, 2011
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Communication complexity
– assume 1 bit sent at a time, alternating (same
proof works in general setting)
– A sends 1 bit depending only on x:
X = {0,1}
n
Y = {0,1}
n
inputs x causing
A to send 1
inputs x causing
A to send 0
April 19, 2011
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Communication complexity
– B sends 1 bit depending only on y and
received bit:
X = {0,1}
n
Y = {0,1}
n
inputs y causing
B to send 1
inputs y causing
B to send 0
April 19, 2011
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Communication complexity
– at end of protocol involving
k bits of
communication
, matrix is partitioned into at
most
2
k
combinatorial rectangles
– bits sent in protocol are the same for every
input (x, y) in given rectangle
– conclude: f(x,y) must be constant on each
rectangle
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Communication complexity
– any partition into combinatorial rectangles with
constant
f(x,y) must have
2
n
+ 1
rectangles
– protocol that exchanges ≤ n bits can only create 2
n
rectangles, so must exchange at least n+1 bits.
X = {0,1}
n
Y = {0,1}
n
1
1
1
1
0
0
Matrix for EQ:
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Communication complexity
• protocol for EQ employing randomness?
– Alice picks
random prime p
in {1.
..4n
2
}, sends:
• p
• (x mod p)
– Bob sends:
• (y mod p)
– players output 1 if and only if:
(x mod p) = (y mod p)
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Communication complexity
– O(log n)
bits exchanged
– if x = y, always correct
– if x ≠ y, incorrect if and only if:
p divides x – y
– # primes in range is ≥ 2n
– # primes dividing x – y is ≤ n
– probability incorrect ≤
1/2
Randomness gives an exponential advantage!!
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