quantum computing mas 725 hartmut klauck ntu 26.3.2012
TRANSCRIPT
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Quantum ComputingMAS 725Hartmut KlauckNTU26.3.2012
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Order finding over ZN
We are given x, N, x<N Order r(x) of x in ZN:
min. r0: xr =1 mod N „Period“ of the powers x
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Order finding over ZN
Is there a quantum algorithm to find r(x)? Shor‘s algorithm finds r(x) in time poly(log N) trivial approach: compute xi for i=1,...,r(x)
• this is inefficient, could be that r(x)=N-1
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Application
Factorization problem: Given a natural number N, find some nontrivial prime factor (or even all of them)
Factorization can be reduced to order finding!• Purely classical reduction
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Shor‘s algorithm
We follow the general outline of Simon‘s algorithm Start with Hadamard transform, query the black box But then we need another transformation, the
quantum Fourier transform
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Fourier Transform
Fourier transform: g is a function ZL ! C
[or a vector with L entries]
Let w=e2 i/L . Then the Fourier transform is a linear map with matrix FTL(i,j)=wij; 0· i,j· L-1
The trivial algorithm to compute the Fourier transform takes time O(L2)
Fast Fourier Transform [FFT] takes times O(L log L)
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Quantum Fourier Transform
Set L=2n. Consider the state |i=j=0,...,L-1 j |ji . The Fourier transform of |i is
|i =j=0,...,L-1 j |ji, with
This is just the Fourier transform on the superposition Also called QFT Can we implement the QFT efficiently? Efficient means here:
polynomial in n=log L
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Quantum Fourier Transform
Let L=2n. Consider |i=j=0,...,L-1 j |ji Write j=j1 jn; j = j12n-1 ++jn20
Set 0.jt jt+1 ... jn = jt/2++jn/2n-t+1
QFT has the following product representation: |j1...jni maps to
1/2n/2 ¢ t=n,...,1 (|0i+ e2i 0. jt...jn |1i)
=1/2n/2 ¢ t=1,...,n (|0i+ e2ij/2t |1i)
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Quantum Fourier Transform
|j1...jni is mapped to 1/2n/2 ¢ t=n,...,1 (|0i+ e2i 0. jt... jn |1i)
Let Rk be the following gate/unitary operator
Apply H to j1. Result: 1/21/2 ¢ (|0i+ e2i 0. j1 |1i) |j2,...,jni Now apply the Rt gate controlled by jt for t=2,...,n to the first
qubit. Result: 1/21/2 ¢ (|0i+ e2i 0. j1,...,j
n |1i) |j2,...,jni
First qubit is now correct (corresponds to last desired qubit)
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Quantum Fourier Transform
This is the circuit for QFT (up to changing the order of qubits)Number of gates: n+(n-1)++1=O(n2)=O(log2 L)
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Quantum Fourier Transform
Caveat: The result of the QFT is a superposition, there is no exponential speedup of computing the Fourier transform in the classical sense (computing the whole vector)
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Properties of the QFT
Computes in time O(n2), ie. can als be approximated by standard gates quickly
QFT is unitary Set w=e2i/L, then FT-1
L(i,j)=w-ij;0· i,j· L-1
Translation invariance: Let QFT j=0,...,L-1 j |ji = j=0,...,L-1 j |ji
Tk: |ji |j+k mod Li. QFT Tk j=0,...,L-1 j |ji= QFT j=0,...,L-1 j |j+k mod Li
= j=0,...L-1 e2 ijk/L j |ji
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Period finding
Function f: ZL!ZN given as black boxPromise: there is a r<N: f(i)=f(i+r) for all i2ZL
i j+kr ) f(i)f(j) Find r Try to solve this for arbitrary f Black box:
Uf: |ji |yi |ji |f(j) yi; j2ZL; f(j)y 2 ZN
Note that Order finding is an instance of Period finding with f(i)=xi
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Shor‘s Algorithm
log L+log N work space log L qubits in |0i ; 02ZL
log N qubits in |1i; 12ZN
Apply Hadamard on the first register Apply Uf Result:
Measure second register Result:
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Shor‘s Algorithm
Result:
0 · j0 · r-1; L-r · j0+(A-1)r · L-1 A-1 < L/r < A+1
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Shor‘s Algorithm
Result:
Now apply QFT Result:
i.e. the probability of k is independent of j0 (translation invariance)
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Shor‘s Algorithm
Result:
Measurement now: Probability of k is
Assumption : r is a divisor of L, i.e. A=L/r, then
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Shor‘s Algorithm
Assumption : r is a divisor of L, i.e. A=L/r, then
If A is a divisor of k, then =1/r If A is no divisor of k, then = 0
(because there are r values k that are multiples of A, each contributing probability 1/r)
I.e. we receive a multiple of A=L/r, say, cL/r with 0· c· r-1 With high probability: c and L/r have no common divisor Then gcd(cL/r,L)=L/r, L is known, hence we learn r.
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Shor‘s Algorithm
In general: the probability of k is
„favorizes“ values of k with kr/L close to an integer Geometric sum
with k=2kr (mod L)/ L
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Shor‘s Algorithm
with k=2kr (mod L))/ L There are exactly r values k2ZL with
-r/2· kr (mod L) · r/2 For those also - r/L· k· r/L
i.e. with 0· j· A-1<L/r the angles jk all lie in the same halfspace ) constructive interference!
Call such a k good
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Shor‘s Algorithm
Some bounds: |1-eik|· |k|
[direct distance „1“ to „eik“ is smaller than the length of the arc] |1-eiAk|¸ 2A|k|/, if A|k|·
Set dist(0,)=|1-ei|,then dist(0,)/||¸ dist(0,)/=2/
A < (L/r)+1,hence Ak · A r/L < (1+r/L) use that kr· r/2 for a good k
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Shor‘s Algorithm
|1-eik|· |k| ; |1-eiAk|¸ 2A|k|/, if A|k|·
Ak · A r/L < (1+r/L)
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Shor‘s Algorithm
Each of the r good values of k has probability close to 1/r, hence with constant probability we get a k with-r/2· kr (mod L) · r/2 [Success]
|kr-cL|· r/2 for some c Then:|k/L-c/r|· 1/(2L), i.e. k/L is approximation of c/r We know k and L. Consider k/L as rational number (reduced). c is uniformly random from 0,...,r-1 c and r have no common divisor with probability at least 1/log r Then: computing c/r (as a rational number in reduced form) gives us also r Choose L large enough to get a good approximation
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Shor‘s Algorithm
With constant probability we get k with |k/L-c/r|· 1/(2L) With probability 1/log r > 1/log L we have gcd(c,r)=1 Let r<N, L=N2
c/r is a rational number with denominator <N Any two such numbers are not closer than 1/N2=1/L > 1/(2L) The interval contains only one rational number c/r with
denominator < N Find the rational number with denominator < N that is close to
k/L Use the continued fractions algorithm to do that
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Continued fractions
The continued fractions algorithm computed for a real its representation as continued fraction
If |c/r-|· 1/(2r2), then one of the steps computes the pair c,r , after at mostO(t3) Operations for t-bit numbers
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Total running time/success probability k is good with constant probability With probability 1/log N also c is good (i.e. no common divisor
with r) Need to repeat only O(log N) times
For order finding in ZN choose L=N2,i.e. 2 log N +log N qubits are used
Fourier transform in O(log2 L) Continued fractions finds r from k/L in time O(log3 L) Can check r for correctness using the black box
Total time is O(log4 N), can be reduced to O(log3 N)
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Continued fractions
Given: real Approximate by
Take integer part as a0, invert remaining number, iterate Theorem: |p/q-|· 1/(2q2), then p/q appears after at most
O(log (p+q)) steps