quantum cryptography christian schaffner illc, university of amsterdam centrum wiskunde &...
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Quantum Cryptography
Christian SchaffnerILLC, University of Amsterdam
Centrum Wiskunde & Informatica
Logic, Language and ComputationMonday, 21 September 2015
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2
1969: Man on the Moon
NASA
The Great Moon-Landing Hoax?
How can you prove that you are at a specific location?http://www.unmuseum.org/moonhoax.htm
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3 What will you Learn from this Talk?
Classical Cryptography Quantum Computation & Teleportation Position-Based Cryptography Garden-Hose Model
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4Classical Cryptography 3000 years of fascinating history until 1970: private communication was the only goal
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5Modern Cryptography is everywhere! is concerned with all settings where people do not trust
each other
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6Secure Encryption
k = ?
Alice
Bob
Goal: Eve does not learn the message Setting: Alice and Bob share a secret key k
Evek = 0101 1011 k = 0101
1011
m = 0000 1111m = “I love you”
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eXclusive OR (XOR) Functionx y x © y0 0 01 0 10 1 11 1 0
Some properties: 8 x : x © 0 = x 8 x : x © x = 0
7
) 8 x,y : x © y © y = x
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One-Time Pad Encryption
Alice
Bob
Goal: Eve does not learn the message Setting: Alice and Bob share a key k Recipe:
Is it secure?
Eve
x y x © y0 0 0
0 1 1
1 0 1
1 1 0k = 0101 1011
m = 0000 1111
c = m © k = 0101 0100 c © k = 0000 1111c © k = m © k © k = m © 0 = m
c = 0101 0100
k = 0101 1011
m = 0000 1111 c = m © k = 0101 0100 m = c © k = 0000 1111
k = 0101 1011 k = 0101 1011
k = ?
8
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Perfect Security
Alice
Bob
Given that c = 0101 0100, is it possible that m = 0000 0000 ?
Yes, if k = 0101 0100. is it possible that m = 1111 1111 ?
Yes, if k = 1010 1011. it is possible that m = 0101 0101 ?
Yes, if k = 0000 0001 In fact, every m is possible. Hence, the one-time pad is perfectly secure!
Eve
x y x © y0 0 0
0 1 1
1 0 1
1 1 0
m = ?
k = ?k = ?
c = m © k = 0101 0100 m = c © k = ?
k = ?
9
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Problems With One-Time Pad
Alice
Bob
The key has to be as long as the message (Shannon’s theorem) The key can only be used once.
Eve
m = 0000 1111
k = 0101 1011 k = 0101 1011
c = m © k = 0101 0100 m = c © k = 0000 1111
k = ?
10
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11Information Theory 6 ECTS MoL course, given in 2nd block: Nov/Dec 2015 mandatory for Logic & Computation track first lecture: Tuesday, 28 October 2015, 9:00, G3.13 http://homepages.cwi.nl/~schaffne/courses/inftheory/2015/
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Problems With One-Time Pad
Alice
Bob
The key has to be as long as the message (Shannon’s theorem) The key can only be used once. In practice, other encryption schemes (such as AES) are used
which allow to encrypt long messages with short keys. One-time pad does not provide authentication:
Eve can easily flip bits in the message
Eve
m = 0000 1111
k = 0101 1011 k = 0101 1011
c = m © k = 0101 0100 m = c © k = 0000 1111
k = ?
12
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Symmetric-Key Cryptography
Alice
Encryption insures secrecy: Eve does not learn the message, e.g. one-time pad
Authentication insures integrity: Eve cannot alter the message
General problem: players have to exchange a key to start with
Eve
Bob
13
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14Introduction to Modern Cryptography 6 ECTS MoL course, usually given in Feb/March 2016: probably not http://homepages.cwi.nl/~schaffne/courses/crypto/2015/
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15 What to Learn from this Talk?
Classical Cryptography Quantum Computing & Teleportation Position-Based Cryptography Garden-Hose Model
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16Quantum Bit: Polarization of a Photonqubit as unit vector in C2
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17Qubit: Rectilinear/Computational Basis
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18Detecting a Qubit
Bob
no photons: 0
Alice
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19Measuring a Qubit
Bob
no photons: 0photons: 1
with prob. 1 yields 1measurement:
0/1
Alice
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20Diagonal/Hadamard Basis
with prob. ½ yields 0
with prob. ½ yields 1
Measurement:
0/1=
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21Illustration of a Superposition
with prob. ½ yields 0
with prob. ½ yields 1
Measurement:
0/1=
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22Illustration of a Superposition
==
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23Quantum Mechanics
with prob. 1 yields 1Measurements:
+ basis
£ basis
with prob. ½ yields 0
with prob. ½ yields 1
0/1
0/1
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Wonderland of Quantum Mechanics
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25Quantum is Real!
25
generation of random numbers
(diagram from idQuantique white paper)
no quantum computation, only quantum communication required
50%
50%
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26 Possible to build in theory, no fundamental theoretical obstacles have been found yet.
Canadian company “D-Wave” claims to have build a quantum computer with 1024 qubits. Did they?
2014: Martinis group “acquired” by Google 2014/15: 135+50 Mio € investment in QuTech centre in
Delft 2015: QuSoft center in Amsterdam
Can We Build Quantum Computers?
Martinis group (UCSB)9 qubits
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27No-Cloning Theorem
??
?
quantum operations: U
Proof: copying is a non-linear operation
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Quantum Key Distribution (QKD)Alice
Bob
Eve security against unrestricted eavesdroppers:
quantum states are unknown to Eve, she cannot copy them honest players can check whether Eve interfered
technically feasible: no quantum computation required, only quantum communication
[Bennett Brassard 84]
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29EPR Pairs
prob. ½ : 0 prob. ½ : 1
prob. 1 : 0
[Einstein Podolsky Rosen 1935]
“spukhafte Fernwirkung” (spooky action at a distance) EPR pairs do not allow to communicate
(no contradiction to relativity theory) can provide a shared random bit
EPR magic!
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30Quantum Teleportation[Bennett Brassard Crépeau Jozsa Peres Wootters 1993]
does not contradict relativity theory teleported state can only be recovered
once the classical information ¾ arrives
?
[Bell]
? ?
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31 What to Learn from this Talk?
Classical Cryptography
Quantum Computing & Teleportation Position-Based Cryptography Garden-Hose Model
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32
How to Convince Someone of Your Presence at a Location
The Great Moon Landing Hoax
http://www.unmuseum.org/moonhoax.htm
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33Position-Based Cryptography
Possible Applications: Launching-missile command comes
from within the military headquarters Talking to the correct country Pizza-delivery problem /
avoid fake calls to emergency services …
Can the geographical location of a player be used as sole cryptographic credential ?
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34Position-Based Cryptography
http://nos.nl/op3/artikel/692138-gamer-krijgt-swatteam-in-zn-nek-swatting.html
https://youtu.be/TiW-BVPCbZk?t=117
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35
Basic task: Position Verification
Prover wants to convince verifiers that she is at a particular position
no coalition of (fake) provers, i.e. not at the claimed position, can convince verifiers
assumptions: communication at speed of light instantaneous computation verifiers can coordinate
Verifier1 Verifier2Prover
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36
Position Verification: First Try
Verifier1 Verifier2Prover
time
distance bounding [Brands Chaum ‘93]
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37
Position Verification: Second Try
Verifier1 Verifier2Prover
position verification is classically impossible ![Chandran Goyal Moriarty Ostrovsky: CRYPTO ’09]
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38
Equivalent Attacking Game
independent messages mx and my copying classical information this is impossible quantumly
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39
Position Verification: Quantum Try[Kent Munro Spiller 03/10]
Let us study the attacking game
?
?
?
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40
?
Attacking Game
impossible but possible with entanglement!!
?? ?
?
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41
?
Entanglement attack
done if b=1
[Bell]
?
?
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42
?
Entanglement attack
the correct person can reconstruct the qubit in time! the scheme is completely broken
[Bell]
?
?[Bell]
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43more complicated schemes?
Different schemes proposed by Chandran, Fehr, Gelles, Goyal, Ostrovsky [2010] Malaney [2010] Kent, Munro, Spiller [2010] Lau, Lo [2010]
Unfortunately they can all be broken! general no-go theorem [Buhrman, Chandran,
Fehr, Gelles, Goyal, Ostrovsky, S 2014]
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46No-Go Theorem
Any position-verification protocol can be broken using an exponential number of EPR-pairs
Question: is this optimal? Does there exist a protocol such that:
any attack requires many EPR-pairs honest prover and verifiers efficient
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47
Single-Qubit Protocol: SQPf[Kent Munro Spiller 03/10]
if f(x,y)=0?
?
?
if f(x,y)=1
efficiently computable
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48
?
Attacking Game for SQPf
Define E(SQPf) := minimum number of EPR pairs required for attacking SQPf
? ?
if f(x,y)=0 if f(x,y)=1
x y
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49 What to Learn from this Talk?
Classical Cryptography
Quantum Computing & Teleportation
Position-Based Cryptography Garden-Hose Model
http://arxiv.org/abs/1109.2563Buhrman, Fehr, S, Speelman
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share s waterpipes
50
The Garden-Hose Model
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The Garden-Hose Model
based on their inputs, players connect pipes with pieces of hose Alice also connects a water tap
51
if water exits @ Alice if water exits @ Bob
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Garden-Hose complexity of f:GH(f) := minimum number of pipes needed to compute f
52
if water exits @ Alice if water exits @ Bob
The Garden-Hose Model
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Demonstration: Inequality on Two Bits53
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GH( Inequality ) · demonstration: 3n challenge: 2n + 1 (first student to email me solution wins)
world record: ~1.359n [Chiu Szegedy et al 13]GH( Inequality ) ¸ n [Pietrzak ‘11]
n-Bit Inequality Puzzle54
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Relationship betweenE(SQPf) and GH(f)
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GH(f) ¸ E(SQPf)Garden-Hose Attacking Game
teleport teleportteleport
teleport
?
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GH(f) ¸ E(SQPf)Garden-Hose Attacking Game
teleport teleportteleport
teleport
?
y, Bob’s telep. keys
x, Alice’s telep. keys
using x & y, can follow the water/qubit correct water/qubit using all
measurement outcomes
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58
last slide: GH(f) ¸ E(SQPf) The two models are not equivalent:
exists f such that GH(f) = n , but E(SQPf) · log(n)
Quantum garden-hose model: give Alice & Bob also entanglement research question: are the models now equivalent?
GH(f) = E(SQPf) ?
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59Garden-Hose Complexity Theory every f has GH(f) · 2n+1
if f in logspace, then GH(f) · polynomial efficient f & no efficient attack ) P L
exist f with GH(f) exponential (counting argument) for g 2 {equality, IP, majority}: GH(g) ¸ n / log(n)
techniques from communication complexity
Many open problems!
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60
What Have You Learned from this Talk?
Classical Cryptography
Quantum Computing & Teleportation
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61
What Have You Learned from this Talk?
No-Go Theorem Impossible unconditionally, but attack requires
unrealistic amounts of resources
Garden-Hose Model model of communication complexity
Position-Based Cryptography
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62 Take on the crypto challenge! GH( Inequality ) = 2n + 1 pipes
the first person to tell me ([email protected]) the protocol wins:
course “Information Theory” see you in the next block on 28 October 2015!