1 / 65

Quantum Cryptography

Quantum Cryptography. 1e29 = 29996224275833 x 3475385758524527 (LOCK) (MESSAGE/CODE) (KEY) Classical computer doing 20 GFLOPS needs 10 15 /20x10 9 ~ 0.5 day to solve this Largest prime has 22,338,618 digits  4000 years !.

marano
Download Presentation

Quantum Cryptography

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Quantum Cryptography 1e29 = 29996224275833 x 3475385758524527 (LOCK) (MESSAGE/CODE) (KEY) Classical computer doing 20 GFLOPS needs 1015/20x109 ~ 0.5 day to solve this Largest prime has 22,338,618 digits  4000 years!

  2. Classical Computation

  3. Inverter/NOT gate in out out in NOT 0 1 1 0

  4. 2-input Logic Gates 0 0 0 0 0 0 0 0 1 1 1 0 AND OR 1 1 0 0 0 1 1 1 1 1 1 1

  5. 2-input Logic Gates 0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 NAND NOR XOR 1 1 1 0 0 0 1 1 0 1 1 1 1 1 1 0 0 0 SUM

  6. NAND is universal A A A B NAND NOT AND = NOT(NAND) OR made of NAND and NOT A + B = NOT[NOT(A+B)] = NOT[NOT(A) and NOT(B)]

  7. NAND is universal A.B A A.B B NAND NOT A A A.B = A + B NOT B B NOT

  8. 2 Bit Half-Adder A 0 0 0 CARRY = AND B 0 1 0 SUM = XOR 1 0 0 A B CARRY SUM 1 1 1 0 0 1 1 1 1 0 2

  9. Full-Adder A Sum S B Carry Cout Cin S = A + B + Cin Cout = A.B + B.Cin + Cin.A

  10. Encoder: n inputs, 2n outputs (Identifies each A-B possibility) Enable D0 A D1 B D2 D0 = A NOR B = NOT(A) AND NOT(B) D1 = NOT(A) AND B D2 = A AND NOT(B) D3 = A AND B D3

  11. Decoder: 2n inputs, n outputs Collapses possibilities Enable D0 A D1 B D2 A = D2 + D3 B = D1 + D3 D3

  12. 2x1 and 4x 1 Multiplexer (Selector) S1 S0 Control S A A B F F C B D S=0, F=A S=1, F=B F = S.A + S.B (S0,S1) = (0,0), F=A (0,1), F=B (1,0), F=C (1,1), F=D F = S0.S1.A + S0.S1.B + S0.S1.C + S0.S1.D

  13. 8 x 1 Multiplexer DEMUX S1 S0 S2 S1 S0 S2 D0 D1 D0 D2 D1 D3 D2 F D4 D3 F D5 D4 D6 D5 D7 D6 D7

  14. 8 x 1 DEMUX

  15. STORAGE/MEMORY (FLIP-FLOPS) Need feedback S Q R Q

  16. How do we realize these gates?

  17. Build a “NOT” with CMOS inverters PMOS NMOS • voltage turns • PMOS on, shorts • to power supply voltage + voltage turns NMOS on, shorts to ground ON Gain OFF

  18. Build a NAND with CMOS inverters 0 0 1 0 1 1 NAND 1 0 1 1 1 0

  19. Build a NOR with CMOS inverters 0 0 1 0 1 0 NOR 1 0 0 1 1 0

  20. Build an oscillator with CMOS inverters

  21. Build memory with CMOS (DRAM, SRAM, Flash)

  22. Energetics of Computation

  23. Dissipation in CMOS PMOS Pdiss = IOFFVDD + aCVDD2f NMOS • Leakage high with scaling IOFF • Reduce VDD Steep devices • Reduce C/ION ratio  FinFETs • F2D = 0.5√L • F3D = 0.5√(L/Nlayers)

  24. Dissipation in binary switching “0” Barrier large enough to prevent spontaneous backflip “1”

  25. Dissipation when electron falls downhill eVAB A VAB B N els: (N+2)kBTln2 We’re already pretty efficient for on-chip dissipation! (Zhirnov) 3kBTln2 Perr = e-qDV/kT < ½  qDV > kBTln2 kBTln2

  26. Reversible Switching (“Moonwalking potential”) When we know where electron sits SLIDE When we don’t know where electron sits OOPS !! KNOWLEDGE CONNECTED TO DISSIPATION !

  27. Information entropy DF = -TDS • S = kBlnW • = <kBln(1/p) > • = -SikB(pilnpi)

  28. Dissipation in CMOS DF = -TDS S = -SikB(pilnpi) When we know where electron sits (read and copy) Si = 0 Sf = 0 When we don’t know where electron sits Sf = 0 Si = -kBln1/2 = kBln2

  29. Dissipation in AND AND 0 0 0 DF = -TDS 0 1 0 S = -SikB(pilnpi) 1 0 0 1 1 1 Si = -4kB(1/4ln1/4) Sf = -kB(3/4ln3/4 + 1/4ln1/4) Squeezing of phase space (2 inputs, 1 output)

  30. No Dissipation (CNOT) 0 0 0 A A A B Aout Bout B Flip B iff A = 1 0 1 0 0 1 0 1 1 Preserve phase space Bout is XOR ! 1 1 1 1 0 SUM

  31. CNOT is reversible ! Aout A A Bout B B

  32. CSWAP (Fredkin gate) A A B Flip B and C if A = 1 C A B Aout Bout C Cout 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 1 0 0 1 0 0 0 1 0 1 0 1 1 0 1 1 1 0 0 1 1 1 1 1

  33. CCNOT (Toffoli gate) A Flip A iff B = 1 and C = 1 B B A B Aout Bout C Cout C C 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 Set C = 1 Get A XOR B ! (Same as SUM bit) 0 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 1 1 1

  34. Quantum Computation

  35. PRIME FACTORIZATION Use Superposition + Entanglement + reversible gates for one-shot computation • Shor’s Algorithm • Grover’s Algorithm • Kane’s Nuclear spin QC

  36. What is Entanglement ?

  37. EPR Paradox: Pion decay  two entangled states in a singlet (e and e+ with opposite spins) |00> + |11>  Move far far away Measure bit 1  0 Bit 2 must collapse to 0 instantly !! (Nonlocality since light needs finite time)

  38. Is entanglement real? Could uncertainty accommodate a hidden variable? Is uncertainty just in knowledge or fundamental ??

  39. Bell’s Inequality |00> + |11>  Separate particles Run each through a rotation Uq1 x Uq2 Uq1= Uq2 = U+(p/2-q2) cosq1 -sinq1 sinq1 cosq1 1 0 |0> = |1> = 0 1

  40. Result -sin(q1-q2) [|00> + |11>]  Coincidence + cos(q1-q2) [ |10> - |01>]  Non-coinc P(AC = 1) = sin2(q1-q2) P(AC = -1) = cos2(q1-q2) P(AC = -1) – P(AC = 1) = P(01) + P(10) – P(00) – P(11) = correlation = cos[2(q1-q2)]

  41. Result (Alain Aspect etc)

  42. Bell

  43. Proof: Bell’s inequality Own vs Rent: O and O Car vs No Car: C and C TV vs No TV: T and T OCT + OCT + OCT + OCT + OCT + OCT + OCT + OCT = 1 If only two questions on a survey – coalesce info thus O O T T C C C C O O T T ? ? ? ? ? ? ? ? ? ? ? ? Jay Sulzberger

  44. Proof: Bell’s inequality Test for accuracy: OCT + OCT + OCT + OCT + OCT + OCT + OCT + OCT = 1 TO + TO + OC + OC + CT + CT ≤ 2 Proof: O O T T C C C C O O T T ? ? ? ? ? ? TOC + TOC + TOC + TOC + OCT + OCT + OCT + OCT + CTO + CTO + CTO + CTO = 2 [ 1 – OCT – OCT] ≤ 2 LHS: ? ? ? ? ? ?

  45. Bell

  46. P ( A & [not B] ) + P ( B & [not C] ) ≥ P ( A & [not C] ) TO + TO + OC + OC + CT + CT ≤ 2 http://arxiv.org/pdf/0704.2529v2.pdf

  47. Entanglement is real Electrons are in superposition A classical measurement decouples them by entangling with an apparatus. Decoherence kills phase information

  48. Entanglement is real How can we use this for QC ?

  49. Prime Factorization

  50. Useful for Quantum Cryptography 1e29 = 29996224275833 x 3475385758524527 (LOCK) (MESSAGE/CODE) (KEY) Classical computer doing 20 GFLOPS needs 1015/20x109 ~ 0.5 day to solve this Largest prime has 22,338,618 digits  4000 years! Create superposition of all states related to key as input Entangle with gateable output Measure output  input collapses (don’t peek yet !) Use gates to do Fourier Transform on input  input further collapses Now Read input From input + output reads, we can get factor !!!

More Related