A limit on nonlocality in any
world in which communication
complexity is not trivial
IFT6195
Alain Tapp
In collaboration with…
Gilles Brassard
 Harry Buhrman
 Naoh Linden
 André Allan Methot
 Falk Unger


Quant-ph/0508042
Motivation

What would be the consequences if the
non local collerations in our world were
stronger than the one given by quantum
mechanics?
Theoretical computer science?
 Foundation of physics?
 Philosophy?

Perfect Non Local Boxes
Bob
Alice
a
x
NLB
ab  x y
b
y
NLB and communication
One bit of communication is enough to
implement a NLB.


Alice sends a to Bob and output x=0
Bob outputs y  a  b
NLB and communication
NLBs does not allow for communication.

We can have a perfect box for which x and y
are uniformly distributed and independent of
(respectively) a and b.
NLB, classical deterministic strategies
0 0 a 0 0 b a b
00
01
10
11
yes
yes
yes
no
yes
yes
no
yes
yes
no
yes
yes
no
yes
yes
yes
NLB classical implementation
There is a probabilistic strategy with
succes probability ¾ on all input.
 There is no classical déterministic strategy
with success proportion greater than ¾.
 There is no probabilistic strategy with
success probability greater than ¾.

¾
NLB quantum strategy
1
 00  11 
 
2
 cos( )  sin(  ) 

R( )  
 sin(  ) cos( ) 
Alice and Bob have the same strategy.
If input=0 applies R (  / 16) otherwise R (3 / 16)
Measure and output the result.
This strategy works on all inputs with probability:
cos ( / 8)  85%
2
NLB quantum strategy
cos ( / 8)  85%
2
Tsirelson proved in 1980 that this is
optimal whatever the entanglement
shared by the players.
Bell theorem
85%  cos ( / 8)  3 / 4
2
The classical upper bound and the
quantum lower bound do not match.
 We can derive an inequality from this that
provides a Bell theorem proof.
 This is known as the CHSH inequality.

Classical Communication Complexity
Alice
Bob
z R {0,1}k
x
f ( x, y )
y
Quantum Communication Complexity
Alice
x
f ( x, y )
1
 1

00

11 

2
 2

k
Bob
y
Communication Complexity
The classical/quantum probabilistic
communication complexity of f, C(f)/Q(f)
is the amount of classical communication
required by the best protocol that
succeeds on all input with probability at
least 1 / 2   when the players have
unlimited prior classical/quantum
correlation.
Inner product (IP)
IP ( x, y )  x  y
x  x1 x2  xn
y  y1 y2  yn
x  y  ( x1  y1 )  ( x2  y3 )    ( xn  yn )
n
x  y   xi yi (mod 2)
i 1
Inner product (IP)
C ( IP )  n  O(1)
Q( IP )  (n)
Most functions are difficult
For most functions f
C ( f )  n  O(1)
Q( f )  n  O(1)
Equality
x  y  EQ( x, y )  1
x  y  EQ( x, y )  0
Alice and Bob each have a very large file and
they want to know if it is exactly the same.
How much do they need to communicate?
Equality
z R 0,1
n
Alice x
ma  x  z
Bob y
mb  y  z
mb
Output ma  mb
Equality
x  y  Pz  x  z  y   1
1
x  y  Pz  x  z  y  
2
By repeating the protocol twice we have
success probability of at least ¾.
C (EQ )  2
Scheduling
Alice and Bob want to find a time where
they are both available for a meeting.
x  x1 x2  xn
y  y1 y2  yn
S ( x, y )  ( x1  y1 )  ( x2  y3 )    ( xn  yn )
 n

S ( x, y )    xi yi  0 
 i 1

Scheduling
C ( S )  ( n )
Q ( S )  ( n )
Raz separation
There exists a problem such that:
 n


C ( S )  
 log( n) 
1/ 4
Q ( S )  O (log( n))
IP using NLB
( Ai , Bi )  NLB( xi , yi )
Ai  Bi  xi  yi
x  y  ( x1  y1 )  ( x2  y3 )    ( xn  yn )
x  y  ( A1  B1 )  ( A2  B2 )    ( An  Bn )
x  y  ( A1  A2   An )  ( B1  B2    Bn )
Perfect NLB implies trivial CC
Any function can be computed with a serie of
AND gates and negations.
Distributed bit
x  xa  xb
Input bit
xa  x1 , xb  0
Negation:
AND
Outcome
x  xa  xb
Two NLBs
y  ya  yb
Bob sends
yb to Alice
AND
( A1 , B1 )  NLB( xa , yb )
( A2 , B2 )  NLB( ya , xb )
A1  B1  xa  yb
A2  B2  ya  xb
x  y  ( xa  xb )  ( ya  yb )
x  y  ( xa  ya )  ( xa  yb )  ( xb  ya )  ( xb  yb )
x  y  (( xa  ya )  A1  A2 )  ( B1  B2  ( xb  yb ))
Main result
In any world where non local boxes can be
implemented with accuracy larger than 0.91
communication complexity is trivial.
1 1
NLB  
2
6

f , C* ( f )  1
CC with a bias

We say that a function f can be computed
with a bias if Alice and Bob can produce a
distributed bit z such that
z  z a  zb
1
P[ f ( x, y )  z ] 
2
CC with a bias
Every function can be computed with a bias.
Alice’s input: x
Bob’s input: y
Alice and Bob share z
Alice outputs a=f(x,z)
Bob outputs b=0 if y=z and a random bit otherwise.
1 
1 1
1
P[ f ( x, y )  a  b]  n  1  n  
2  2 2
2
Idea

We want a bounded bias.

Let’s amplify the bias.

Repetition and majority?
Idea
Maj
Maj
Maj
~
~
~
f ( x) f ( x) f ( x)
Maj
~
~
~
f ( x) f ( x) f ( x)
Maj
Maj
~
~
~
f ( x) f ( x) f ( x)
Maj
~
~
~
f ( x) f ( x) f ( x)
~
f ( x)
Maj
~
~
~
f ( x) f ( x) f ( x)
Maj
Maj
~
~
~
f ( x) f ( x) f ( x)
Maj
Maj
Maj
~
~
~
f ( x) f ( x) f ( x)
~
~
~
f ( x) f ( x) f ( x)
~
~
~
f ( x) f ( x) f ( x)
Non local majority
y  Maj ( x1 , x2 , x3 )  1 iff
x1  x2  x3  2
x1  x  x
1
a
1
b
x2  x  x
2
a
2
b
x3  x  x
3
a
3
b
y  y a  yb
ya  yb  Maj ( x  x , x  x , x  x )
1
a
1
b
2
a
2
b
3
a
3
b
NLM > 5/6

If NLM can be computed with probability
stricly greather than 5/6 than every
fonction can be computed with a bounded
bias.

Below that treshold NLM makes things
worst.
NLM > 5/6
p  1/ 2  
(  0)
q  5/ 6
(  0)
h( p)  q( p 3  3 p 2 (1  p))  (1  q)(3 p(1  p) 2  (1  p)3 )
1
3 
1
s 

2 2 1  3 2
1 / 2  p  s  h( p )  p
Non local equality
y  NLE ( x1 , x2 , x3 )  0 iff
x1  x2  x3
x1  x  x
1
a
1
b
x2  x  x
2
a
2
b
x3  x  x
3
a
3
b
y  y a  yb
ya  yb  NLE ( x  x , x  x , x  x )
1
a
1
b
2
a
2
b
3
a
3
b
NLE implies NLM
ya  yb  NLE ( x  x , x  x , x  x )
1
a
1
b
z a  ya  x  x  x
1
a
2
a
2
a
2
b
3
a
3
b
3
a
z b  yb  x  x  x
1
b
2
b
3
b
z a  zb  Maj ( x  x , x  x , x  x )
1
a
1
b
2
a
2
b
3
a
3
b
2 NLB implies NLE
x1  x1a  xb1
x 2  xa2  xb2
x 3  xa3  xb3
z1a  zb1  NLB( x1a  xa2 , xb2  xb3 )
z a2  zb2  NLB(1  xa2  xa3 , xb1  xb2 )
NLE ( x1 , x 2 , x 3 )
 ( x1  x 2 )  ( x 2  x 3 )
 ( x1a  xb1  xa2  xb2 )  ( xa2  xb2  3a  xb3 )
 (( x1a  xa2 )  ( xa2  xa3 ))  z1a  z a2
 (( xb1  xb2 )  ( xb2  xb3 ))  zb1  zb2
To conclude the proof
1 1
5
5
NLB  
 NLE   Maj 
2
6
6
6
•Compute f several times with a bias
•Use a tree of majority to improve the bias.
•Bob sends his share of the outcome to Alice.
Open question
Show some unacceptable
consequences of correlations
epsilon-stronger than the one
predicted by quantum mechanics.