Introduction to Quantum Algorithms Part I: Quantum Gates and Simon s Algorithm

Transcription

1 Part I: Quantum Gates and Simon s Algorithm Martin Rötteler NEC Laboratories America, Inc. 4 Independence Way, Suite 00 Princeton, NJ 08540, U.S.A. International Summer School on Quantum Information, Max-Planck-Institut für Physik komplexer Systeme Dresden, August 3, 005

2 Overview Today: Introduction to qubits, quantum gates, and circuits. Appetizer: Two-bit problem where quantum beats classical! The power of quantum computing: Simon s algorithm Basic principles used: Computing in superposition Constructive/destructive interference Tomorrow: Shor s algorithm and Grover s algorithm

3 Basics: Quantum Information Quantum bit (qubit) A qubit is a normalized state in C : α 0 + β, α, β C, where α + β =. Quantum register A quantum register of length n is a collection of qubits q,..., q n. Possible operations / dynamics A quantum computer can perform unitary operations on quantum registers measure single qubits

4 Basics: Quantum Information Quantum bit (qubit) A qubit is a normalized state in C : α 0 + β, α, β C, where α + β =. Quantum register A quantum register of length n is a collection of qubits q,..., q n. Possible operations / dynamics A quantum computer can perform unitary operations on quantum registers measure single qubits

5 Basics: Quantum Information Quantum bit (qubit) A qubit is a normalized state in C : α 0 + β, α, β C, where α + β =. Quantum register A quantum register of length n is a collection of qubits q,..., q n. Possible operations / dynamics A quantum computer can perform unitary operations on quantum registers measure single qubits

6 Bits vs Qubits One classical bit One quantum bit (qubit) is in one of the basis states = High = = Low = 0 is in a coherent superposition Ψ = α + β of the basis states, Boolean state space: IF = {, } = {0, } consisting of elements Quantum state space H = C : { } α + β : α + β = of dimensions

7 Bits vs Qubits One classical bit One quantum bit (qubit) is in one of the basis states = High = = Low = 0 is in a coherent superposition Ψ = α + β of the basis states, Boolean state space: IF = {, } = {0, } consisting of elements Quantum state space H = C : { } α + β : α + β = of dimensions

8 Two Bits vs Two Qubits Two classical bits Two bits can hold either of the 4 states in the state space {0, } {0, } = {0, } = {,,, }. Two coupled qubits A register of two coupled qubits can hold any of the states Ψ = α + β + γ + δ in the state space H H = C C. Two separate qubits Two separate qubits can hold any of the product states Ψ Ψ = (α + β ) (α + β ) in the state space H H C C.

9 Two Bits vs Two Qubits Two classical bits Two bits can hold either of the 4 states in the state space {0, } {0, } = {0, } = {,,, }. Two coupled qubits A register of two coupled qubits can hold any of the states Ψ = α + β + γ + δ in the state space H H = C C. Two separate qubits Two separate qubits can hold any of the product states Ψ Ψ = (α + β ) (α + β ) in the state space H H C C.

10 Two Bits vs Two Qubits Two classical bits Two bits can hold either of the 4 states in the state space {0, } {0, } = {0, } = {,,, }. Two coupled qubits A register of two coupled qubits can hold any of the states Ψ = α + β + γ + δ in the state space H H = C C. Two separate qubits Two separate qubits can hold any of the product states Ψ Ψ = (α + β ) (α + β ) in the state space H H C C.

11 Many Bits vs Many Qubits Classical register of n bits holds one of the n states ɛ = ɛ ɛ n of the state space {0, } n = {0, } {0, }. Quantum register of n qubits can hold any coherent superposition Ψ = α ɛ ɛ n ɛ ɛ ɛ n ɛ {0,} n in the n dimensional space H n = C C C = C n. Product states of n qubits can only hold a product state Ψ Ψ Ψ n = (α + β ) (α n + β n ) (thus, only linear scaling of system and dimension)

12 Many Bits vs Many Qubits Classical register of n bits holds one of the n states ɛ = ɛ ɛ n of the state space {0, } n = {0, } {0, }. Quantum register of n qubits can hold any coherent superposition Ψ = α ɛ ɛ n ɛ ɛ ɛ n ɛ {0,} n in the n dimensional space H n = C C C = C n. Product states of n qubits can only hold a product state Ψ Ψ Ψ n = (α + β ) (α n + β n ) (thus, only linear scaling of system and dimension)

13 Many Bits vs Many Qubits Classical register of n bits holds one of the n states ɛ = ɛ ɛ n of the state space {0, } n = {0, } {0, }. Quantum register of n qubits can hold any coherent superposition Ψ = α ɛ ɛ n ɛ ɛ ɛ n ɛ {0,} n in the n dimensional space H n = C C C = C n. Product states of n qubits can only hold a product state Ψ Ψ Ψ n = (α + β ) (α n + β n ) (thus, only linear scaling of system and dimension)

14 Our Model of Measurements von Neumann measurements of one qubit First, specify a basis B for C, e. g. { 0, }. The outcome of measuring the state α 0 + β is described by a random variable X. The probabilities to observe 0 or are given by Pr(X = 0) = α, Pr(X = ) = β. Measuring a state in C n in an orthonormal basis B Recall: Orthonormal Basis of C N B = { ψ i : i =,..., N}, where ψ i ψ j = δ i,j Let ϕ = N i= α i i, where N i= α i =. Then measuring ϕ in the basis B gives random variable X B taking values,..., N: Pr(X B = ) = ψ ϕ,..., Pr(X B = N) = ψ N ϕ.

15 Our Model of Measurements von Neumann measurements of one qubit First, specify a basis B for C, e. g. { 0, }. The outcome of measuring the state α 0 + β is described by a random variable X. The probabilities to observe 0 or are given by Pr(X = 0) = α, Pr(X = ) = β. Measuring a state in C n in an orthonormal basis B Recall: Orthonormal Basis of C N B = { ψ i : i =,..., N}, where ψ i ψ j = δ i,j Let ϕ = N i= α i i, where N i= α i =. Then measuring ϕ in the basis B gives random variable X B taking values,..., N: Pr(X B = ) = ψ ϕ,..., Pr(X B = N) = ψ N ϕ.

16 Two Important Types of Operations Local operations N U = U U.,.. where U U(). U Conditioned operation: the controlled NOT (CNOT) = 0 = 0

17 Two Important Types of Operations Local operations N U = U U.,.. where U U(). U Conditioned operation: the controlled NOT (CNOT) = 0 = 0

18 Notation for Quantum Gates Gate in Feynman notation Corresponding transformation H U u u u u

19 Notation for Quantum Gates Gate in Feynman notation Corresponding transformation H U u u u u

20 Two Constructions for Matrices Tensor product ( ) ( ) T T = (T I)(I T ) = I T classically parallel T T quantum parallel = independent dynamics classical: expensive, quantum: easy, since local operations Direct sum T T = (T I)(I T ) = ( ) I T conditional operation T T CASE operator = coupled dynamics classical: easy, quantum: difficult, since entangled

21 Two Constructions for Matrices Tensor product ( ) ( ) T T = (T I)(I T ) = I T classically parallel T T quantum parallel = independent dynamics classical: expensive, quantum: easy, since local operations Direct sum T T = (T I)(I T ) = ( ) I T conditional operation T T CASE operator = coupled dynamics classical: easy, quantum: difficult, since entangled

22 The Hadamard Transform The Hadamard transform on one qubit H = ( ) = ( ) xy x y. x,y IF The Hadamard transform on n qubits The n-fold tensor product of H is given by H n = n/ where x y := n i= x iy i IF. x,y IF n ( ) x y x y,

23 The Hadamard Transform The Hadamard transform on one qubit H = ( ) = ( ) xy x y. x,y IF The Hadamard transform on n qubits The n-fold tensor product of H is given by H n = n/ where x y := n i= x iy i IF. x,y IF n ( ) x y x y,

24 The Hadamard Transform Properties of the Hadamard transform By definition H is the following map: 0 ( 0 + ) ( 0 ) Identity involving the Pauli matrices: H σ x H = σ z Identity involving the CNOT gate: H H H H =

25 The Hadamard Transform Properties of the Hadamard transform By definition H is the following map: 0 ( 0 + ) ( 0 ) Identity involving the Pauli matrices: H σ x H = σ z Identity involving the CNOT gate: H H H H =

26 The Hadamard Transform Properties of the Hadamard transform By definition H is the following map: 0 ( 0 + ) ( 0 ) Identity involving the Pauli matrices: H σ x H = σ z Identity involving the CNOT gate: H H H H =

27 Quantum Gates and Circuits Elementary quantum gates.. U (i) = U.. i CNOT (i,j) =..... i. j Universal set of gates Theorem (Barenco et al., 995): U( n ) = U (i), CNOT (i,j) : i, j =,..., n, i j Quantum gates: main problem Find efficient factorizations for given U U( n )!

28 Quantum Gates and Circuits Elementary quantum gates.. U (i) = U.. i CNOT (i,j) =..... i. j Universal set of gates Theorem (Barenco et al., 995): U( n ) = U (i), CNOT (i,j) : i, j =,..., n, i j Quantum gates: main problem Find efficient factorizations for given U U( n )!

29 Quantum Gates and Circuits Elementary quantum gates.. U (i) = U.. i CNOT (i,j) =..... i. j Universal set of gates Theorem (Barenco et al., 995): U( n ) = U (i), CNOT (i,j) : i, j =,..., n, i j Quantum gates: main problem Find efficient factorizations for given U U( n )!

30 Different Levels of Abstraction Unitary matrix U = 0 C A Factorized unitary matrix U= (I H ) (I σ x ) (H I) 0 ( )0 I B@ CA B@ CA = σ x Quantum circuit H H

31 Different Levels of Abstraction Unitary matrix U = 0 C A Factorized unitary matrix U= (I H ) (I σ x ) (H I) 0 ( )0 I B@ CA B@ CA = σ x Quantum circuit H H

32 Different Levels of Abstraction Unitary matrix U = 0 C A Factorized unitary matrix U= (I H ) (I σ x ) (H I) 0 ( )0 I B@ CA B@ CA = σ x Quantum circuit H H

33 Basic Facts About Boolean Functions Boolean functions Any function f : {0, } n {0, } m, where n is the number of input bits and m the number of output bits is said to be a Boolean function. Any Boolean function can be represented by a truth table. If f = (f,..., f m ), this is a matrix of size n m where in column i we have the list of values f i (x,..., x n ), where x j {0, } for j =,..., n. The number of Boolean functions There are n Boolean functions f : {0, } n {0, }, i. e., functions with n inputs and one output (since we can assign an arbitrary value for each of the n inputs). There are mn Boolean functions f : {0, } n {0, } m with n inputs and m outputs.

34 Basic Facts About Boolean Functions Boolean functions Any function f : {0, } n {0, } m, where n is the number of input bits and m the number of output bits is said to be a Boolean function. Any Boolean function can be represented by a truth table. If f = (f,..., f m ), this is a matrix of size n m where in column i we have the list of values f i (x,..., x n ), where x j {0, } for j =,..., n. The number of Boolean functions There are n Boolean functions f : {0, } n {0, }, i. e., functions with n inputs and one output (since we can assign an arbitrary value for each of the n inputs). There are mn Boolean functions f : {0, } n {0, } m with n inputs and m outputs.

35 Universal Gates: Classical Computing Connectives x x x x NOT gate: x x 0 0 AND gate: x x x x x x x x OR gate: NAND gate: Theorem Any deterministic classical circuit and thereby any deterministic classical computation can be realized by using NAND gates only. Any probabilistic computation can be realized using NAND gates and in addition one gate which realizes a fair coin flip.

36 Universal Gates: Classical Computing Connectives x x x x NOT gate: x x 0 0 AND gate: x x x x x x x x OR gate: NAND gate: Theorem Any deterministic classical circuit and thereby any deterministic classical computation can be realized by using NAND gates only. Any probabilistic computation can be realized using NAND gates and in addition one gate which realizes a fair coin flip.

37 Loading More Structure Onto Bits Finite field of two elements The set {0, } can be equipped with a multiplication and an addition such that the field axioms hold. Truth tables for these operations: x y x y x y x y Then ({0, },, ) becomes a field consisting of two elements only, also denoted by IF. A finite field with n elements exists if and only if n is a prime power. Important identity: ( ) x y = ( ) x ( ) y We can also rewrite CNOT: x y x x y

38 Loading More Structure Onto Bits Finite field of two elements The set {0, } can be equipped with a multiplication and an addition such that the field axioms hold. Truth tables for these operations: x y x y x y x y Then ({0, },, ) becomes a field consisting of two elements only, also denoted by IF. A finite field with n elements exists if and only if n is a prime power. Important identity: ( ) x y = ( ) x ( ) y We can also rewrite CNOT: x y x x y

39 Loading More Structure Onto Bits Finite field of two elements The set {0, } can be equipped with a multiplication and an addition such that the field axioms hold. Truth tables for these operations: x y x y x y x y Then ({0, },, ) becomes a field consisting of two elements only, also denoted by IF. A finite field with n elements exists if and only if n is a prime power. Important identity: ( ) x y = ( ) x ( ) y We can also rewrite CNOT: x y x x y

40 Loading More Structure Onto Bits Finite field of two elements The set {0, } can be equipped with a multiplication and an addition such that the field axioms hold. Truth tables for these operations: x y x y x y x y Then ({0, },, ) becomes a field consisting of two elements only, also denoted by IF. A finite field with n elements exists if and only if n is a prime power. Important identity: ( ) x y = ( ) x ( ) y We can also rewrite CNOT: x y x x y

41 Loading More Structure Onto Bits Finite field of two elements The set {0, } can be equipped with a multiplication and an addition such that the field axioms hold. Truth tables for these operations: x y x y x y x y Then ({0, },, ) becomes a field consisting of two elements only, also denoted by IF. A finite field with n elements exists if and only if n is a prime power. Important identity: ( ) x y = ( ) x ( ) y We can also rewrite CNOT: x y x x y

42 Reversible Computation of Boolean Functions Basic issue of reversible computing Suppose, we want to compute a function f : {0, } n {0, } m that is not reversible. How can we do this? One possible solution Define a new Boolean function which takes n + m inputs and n + m outputs as follows: F(x, y) := (x, y f (x)) Properties of F(x, y) On the special inputs (x, 0), where x {0, } n we obtain that F(x, 0) = (x, f (x)). Furthermore, F is reversible. Theorem (Bennett): If f can be computed using K gates, then F can be computed using K + m gates.

43 Reversible Computation of Boolean Functions Basic issue of reversible computing Suppose, we want to compute a function f : {0, } n {0, } m that is not reversible. How can we do this? One possible solution Define a new Boolean function which takes n + m inputs and n + m outputs as follows: F(x, y) := (x, y f (x)) Properties of F(x, y) On the special inputs (x, 0), where x {0, } n we obtain that F(x, 0) = (x, f (x)). Furthermore, F is reversible. Theorem (Bennett): If f can be computed using K gates, then F can be computed using K + m gates.

44 Reversible Computation of Boolean Functions Basic issue of reversible computing Suppose, we want to compute a function f : {0, } n {0, } m that is not reversible. How can we do this? One possible solution Define a new Boolean function which takes n + m inputs and n + m outputs as follows: F(x, y) := (x, y f (x)) Properties of F(x, y) On the special inputs (x, 0), where x {0, } n we obtain that F(x, 0) = (x, f (x)). Furthermore, F is reversible. Theorem (Bennett): If f can be computed using K gates, then F can be computed using K + m gates.

45 Universal Gates: Reversible Computing The Toffoli gate TOF x y z x y z x y z x y z x y Theorem (Toffoli, 98) Any reversible computation can be realized by using TOF gates and ancilla (auxiliary) bits which are initialized to 0.

46 Universal Gates: Reversible Computing The Toffoli gate TOF x y z x y z x y z x y z x y Theorem (Toffoli, 98) Any reversible computation can be realized by using TOF gates and ancilla (auxiliary) bits which are initialized to 0.

47 Reversible Computation of Boolean Functions More features of the reversible embedding F(x, y) Possibly, there are reversible functions that compute G(x, 0) = (f (x), junk(x)) and use use fewer than n + m bits. Actually, only log (max c y ) many extra bits are needed to make f reversible, where c y = {x : f (x) = y} are the sizes of the collisions of f. The advantage of F is its uniform definition. Computing a function by a quantum circuit Any function f : {0, } n {0, } m can be computed by means of the reversible function F : {0, } n+m {0, } n+m. Hence we can compute f also by a quantum circuit U f x y = x y f (x).

48 Reversible Computation of Boolean Functions More features of the reversible embedding F(x, y) Possibly, there are reversible functions that compute G(x, 0) = (f (x), junk(x)) and use use fewer than n + m bits. Actually, only log (max c y ) many extra bits are needed to make f reversible, where c y = {x : f (x) = y} are the sizes of the collisions of f. The advantage of F is its uniform definition. Computing a function by a quantum circuit Any function f : {0, } n {0, } m can be computed by means of the reversible function F : {0, } n+m {0, } n+m. Hence we can compute f also by a quantum circuit U f x y = x y f (x).

49 A Two-Bit Problem The Deutsch Jozsa problem (most simple case) Given a Boolean function f : {0, } {0, }. Decide whether the property f (0) = f () holds or not. The four possible functions x f (x) x f (x) x f (x) x f (x) Observation From querying the function only on one input, we cannot determine whether f (0) = f () with certainty. E. g., if the answer is f (0) = 0, it could be the first or third function. On the other hand, two queries determine the function completely. What is the quantum query complexity?

50 A Two-Bit Problem The Deutsch Jozsa problem (most simple case) Given a Boolean function f : {0, } {0, }. Decide whether the property f (0) = f () holds or not. The four possible functions x f (x) x f (x) x f (x) x f (x) Observation From querying the function only on one input, we cannot determine whether f (0) = f () with certainty. E. g., if the answer is f (0) = 0, it could be the first or third function. On the other hand, two queries determine the function completely. What is the quantum query complexity?

51 A Two-Bit Problem The Deutsch Jozsa problem (most simple case) Given a Boolean function f : {0, } {0, }. Decide whether the property f (0) = f () holds or not. The four possible functions x f (x) x f (x) x f (x) x f (x) Observation From querying the function only on one input, we cannot determine whether f (0) = f () with certainty. E. g., if the answer is f (0) = 0, it could be the first or third function. On the other hand, two queries determine the function completely. What is the quantum query complexity?

52 The Phase Kick-Back Trick Computing the function into the phase x 0? U f? Computing the effect for different inputs ( ) 0 f (x) f (x) x x ( ) 0 = x ( ) f (x) = ( ) f (x) x ( ) 0

53 The Phase Kick-Back Trick Computing the function into the phase x 0? U f? Computing the effect for different inputs ( ) 0 f (x) f (x) x x ( ) 0 = x ( ) f (x) = ( ) f (x) x ( ) 0

54 The Phase Kick-Back Trick Computing the function into the phase x 0? U f? Computing the effect for different inputs ( ) 0 f (x) f (x) x x ( ) 0 = x ( ) f (x) = ( ) f (x) x ( ) 0

55 The Phase Kick-Back Trick Computing the function into the phase x 0? U f? Computing the effect for different inputs ( ) 0 f (x) f (x) x x ( ) 0 = x ( ) f (x) = ( ) f (x) x ( ) 0

56 The Phase Kick-Back Trick Computing the function into the phase x 0 U f ( ) f (x) x 0 Computing the effect for different inputs ( ) 0 f (x) f (x) x x ( ) 0 = x ( ) f (x) = ( ) f (x) x ( ) 0

57 The Phase Kick-Back Trick The quantum algorithm 0 H H Phase kick-back in superposition H U f ( ) f (0) 0 = U f H? H? 0 + ( ) f () ( ) f (0) 0 + ( ) f (0) f () H ( ) f (0) f (0) f () 0 0

58 The Phase Kick-Back Trick The quantum algorithm 0 H H Phase kick-back in superposition H U f ( ) f (0) 0 = U f H? H? 0 + ( ) f () ( ) f (0) 0 + ( ) f (0) f () H ( ) f (0) f (0) f () 0 0

59 The Phase Kick-Back Trick The quantum algorithm 0 H H Phase kick-back in superposition H U f ( ) f (0) 0 = U f H? H? 0 + ( ) f () ( ) f (0) 0 + ( ) f (0) f () H ( ) f (0) f (0) f () 0 0

60 The Phase Kick-Back Trick The quantum algorithm 0 H H Phase kick-back in superposition H U f ( ) f (0) 0 = U f H? H? 0 + ( ) f () ( ) f (0) 0 + ( ) f (0) f () H ( ) f (0) f (0) f () 0 0

61 The Phase Kick-Back Trick The quantum algorithm 0 H H Phase kick-back in superposition H U f ( ) f (0) 0 = U f H? H? 0 + ( ) f () ( ) f (0) 0 + ( ) f (0) f () H ( ) f (0) f (0) f () 0 0

62 The Phase Kick-Back Trick The quantum algorithm 0 H H f (0) f () H U f H Phase kick-back in superposition H U f ( ) f (0) 0 = 0 + ( ) f () ( ) f (0) 0 + ( ) f (0) f () H ( ) f (0) f (0) f () 0 0 Hence, we from measuring the first qubit in the computational basis, we obtain the answer f (0) f () which reveals whether f (0) = f () or not.

63 The Phase Kick-Back Trick The quantum algorithm 0 H H f (0) f () H U f H Phase kick-back in superposition H U f ( ) f (0) 0 = 0 + ( ) f () ( ) f (0) 0 + ( ) f (0) f () H ( ) f (0) f (0) f () 0 0 Hence, we from measuring the first qubit in the computational basis, we obtain the answer f (0) f () which reveals whether f (0) = f () or not.

64 What can be computed? The strong Church-Turing thesis Any reasonable algorithmic process carried out by a physical machine can be efficiently simulated by a probabilistic Turing machine. The slow-down for this is at most polynomial. Efficient computations Given a problem of size n, an algorithm is said to have a polynomial running time if the number of steps it needs to find a solution is bounded by p(n), where p is a polynomial. How do quantum algorithms fit in? They can solve problems which are believed to be intractable for classical computers. There are physically reasonable algorithmic processes carried which seem to be hard to simulate for any classical probabilistic Turing machine.

65 What can be computed? The strong Church-Turing thesis Any reasonable algorithmic process carried out by a physical machine can be efficiently simulated by a probabilistic Turing machine. The slow-down for this is at most polynomial. Efficient computations Given a problem of size n, an algorithm is said to have a polynomial running time if the number of steps it needs to find a solution is bounded by p(n), where p is a polynomial. How do quantum algorithms fit in? They can solve problems which are believed to be intractable for classical computers. There are physically reasonable algorithmic processes carried which seem to be hard to simulate for any classical probabilistic Turing machine.

66 What can be computed? The strong Church-Turing thesis Any reasonable algorithmic process carried out by a physical machine can be efficiently simulated by a probabilistic Turing machine. The slow-down for this is at most polynomial. Efficient computations Given a problem of size n, an algorithm is said to have a polynomial running time if the number of steps it needs to find a solution is bounded by p(n), where p is a polynomial. How do quantum algorithms fit in? They can solve problems which are believed to be intractable for classical computers. There are physically reasonable algorithmic processes carried which seem to be hard to simulate for any classical probabilistic Turing machine.

67 Computational Model: Quantum Circuits Example: quantum circuit Quantum circuit and corresponding directed acyclic graph. x 4 G G 6 U U 6 x 3 U 3 U x U 5 U 4 x G3 G G 5 G4 Uniform families of quantum circuits A family F := {C n U( n ) n IN} of quantum circuits is called uniform if there exists a polynomial-time deterministic Turing machine which computes n C n, where n is the problem size. Theorem (Yao 93) Uniform families of quantum circuits and quantum Turing machines (see Bernstein/Vazirani 93) are equivalent. y 4 y 3 y y

68 Computational Model: Quantum Circuits Example: quantum circuit Quantum circuit and corresponding directed acyclic graph. x 4 G G 6 U U 6 x 3 U 3 U x U 5 U 4 x G3 G G 5 G4 Uniform families of quantum circuits A family F := {C n U( n ) n IN} of quantum circuits is called uniform if there exists a polynomial-time deterministic Turing machine which computes n C n, where n is the problem size. Theorem (Yao 93) Uniform families of quantum circuits and quantum Turing machines (see Bernstein/Vazirani 93) are equivalent. y 4 y 3 y y

69 Computational Model: Quantum Circuits Example: quantum circuit Quantum circuit and corresponding directed acyclic graph. x 4 G G 6 U U 6 x 3 U 3 U x U 5 U 4 x G3 G G 5 G4 Uniform families of quantum circuits A family F := {C n U( n ) n IN} of quantum circuits is called uniform if there exists a polynomial-time deterministic Turing machine which computes n C n, where n is the problem size. Theorem (Yao 93) Uniform families of quantum circuits and quantum Turing machines (see Bernstein/Vazirani 93) are equivalent. y 4 y 3 y y

70 Asymptotics of Functions Landau notation We use Landau notation to compare the asymptotics of two functions f, g : IN C. Typically, f is the running time for an algorithm for input of size n and g is another function. f (n) = O(g(n)) means that for some m there exists a constant c > 0, such that f (n) cg(n) for all n m. f (n) = Ω(g(n)) means that for some m there exists a constant c > 0, such that f (n) cg(n) for all n m. f (n) = Θ(g(n)) means that both f (n) = O(g(n)) and f (n) = Ω(g(n)) hold. Example Let f (n) be number of operations needed to compute a classical FFT of a vector of length n. Then f (n) = O(n log(n)).

71 Asymptotics of Functions Landau notation We use Landau notation to compare the asymptotics of two functions f, g : IN C. Typically, f is the running time for an algorithm for input of size n and g is another function. f (n) = O(g(n)) means that for some m there exists a constant c > 0, such that f (n) cg(n) for all n m. f (n) = Ω(g(n)) means that for some m there exists a constant c > 0, such that f (n) cg(n) for all n m. f (n) = Θ(g(n)) means that both f (n) = O(g(n)) and f (n) = Ω(g(n)) hold. Example Let f (n) be number of operations needed to compute a classical FFT of a vector of length n. Then f (n) = O(n log(n)).

72 Computational Complexity Measuring the problem size Usually algorithms can use different kinds of resources. Typical examples for resources which can be measured are time, space, depth, and queries. The resources needed are always measured as a function of the input size. Examples Consider the problem of multiplying two n-bit numbers. Clearly, this can be done in O(n ) additions and multiplications using the school method. The best known method uses O(n log n log log n) operations. The converse problem of factoring an n-bit number N into its prime factors is much harder. The currently best known algorithm needs exp ( c(log N) /3 (log log N) /3) operations.

73 Computational Complexity Measuring the problem size Usually algorithms can use different kinds of resources. Typical examples for resources which can be measured are time, space, depth, and queries. The resources needed are always measured as a function of the input size. Examples Consider the problem of multiplying two n-bit numbers. Clearly, this can be done in O(n ) additions and multiplications using the school method. The best known method uses O(n log n log log n) operations. The converse problem of factoring an n-bit number N into its prime factors is much harder. The currently best known algorithm needs exp ( c(log N) /3 (log log N) /3) operations.

74 Query Problems Query complexity and black boxes Instead of counting the number of operations we can count the number of queries to f in order to solve the problem. Often upper and lower bounds can be shown for the query complexity model only. Black-box model: we assume that we are given a function f but cannot analyze how f is actually implemented. Most real-world problems are actually white-box, for example FACTORING, GRAPH-ISO, etc. Lower bounds in the white-box problems are typically very weak. Examples Black box problems: Simon s algorithm, Grover s algorithm. White-box problems: Factoring and dlog, phase estimation.

75 Query Problems Query complexity and black boxes Instead of counting the number of operations we can count the number of queries to f in order to solve the problem. Often upper and lower bounds can be shown for the query complexity model only. Black-box model: we assume that we are given a function f but cannot analyze how f is actually implemented. Most real-world problems are actually white-box, for example FACTORING, GRAPH-ISO, etc. Lower bounds in the white-box problems are typically very weak. Examples Black box problems: Simon s algorithm, Grover s algorithm. White-box problems: Factoring and dlog, phase estimation.

76 Simon s Problem The XOR bit mask problem We consider only functions f : {0, } n {0, } n for which there exists s {0, } n such that x {0, } n we have f (x) = f (x s), x, y {0, } n we have that if x y s, then f (x) f (y). The task is to find the hidden string s. Example where f : {0, } 3 {0, } Here s = (0,, ).

77 Simon s Problem The XOR bit mask problem We consider only functions f : {0, } n {0, } n for which there exists s {0, } n such that x {0, } n we have f (x) = f (x s), x, y {0, } n we have that if x y s, then f (x) f (y). The task is to find the hidden string s. Example where f : {0, } 3 {0, } Here s = (0,, ).

78 Simon s Problem The XOR bit mask problem We consider only functions f : {0, } n {0, } n for which there exists s {0, } n such that x {0, } n we have f (x) = f (x s), x, y {0, } n we have that if x y s, then f (x) f (y). The task is to find the hidden string s. Example where f : {0, } 3 {0, } Here s = (0,, ).

79 Simon s Problem Some comments about Simon s problem There is some similarity to the Deutsch-Jozsa problem, however, here the task is not to distinguish between two cases (constant vs balanced) but between exponentially many cases (one for each unknown string s). Problem might seem artificial, but Shor s algorithm has the same underlying idea. Note: This is a promise problem! Problem gave the first strong (polynomial vs exponential) separation between quantum and classical computing. Before that only super-polynomial separations were known. Literature D. R. Simon. On the Power of Quantum Computation. Proc. FOCS 94, 6 3. IEEE, 994.

80 Simon s Problem Some comments about Simon s problem There is some similarity to the Deutsch-Jozsa problem, however, here the task is not to distinguish between two cases (constant vs balanced) but between exponentially many cases (one for each unknown string s). Problem might seem artificial, but Shor s algorithm has the same underlying idea. Note: This is a promise problem! Problem gave the first strong (polynomial vs exponential) separation between quantum and classical computing. Before that only super-polynomial separations were known. Literature D. R. Simon. On the Power of Quantum Computation. Proc. FOCS 94, 6 3. IEEE, 994.

81 Simon s Problem Some comments about Simon s problem There is some similarity to the Deutsch-Jozsa problem, however, here the task is not to distinguish between two cases (constant vs balanced) but between exponentially many cases (one for each unknown string s). Problem might seem artificial, but Shor s algorithm has the same underlying idea. Note: This is a promise problem! Problem gave the first strong (polynomial vs exponential) separation between quantum and classical computing. Before that only super-polynomial separations were known. Literature D. R. Simon. On the Power of Quantum Computation. Proc. FOCS 94, 6 3. IEEE, 994.

82 Simon s Problem Some comments about Simon s problem There is some similarity to the Deutsch-Jozsa problem, however, here the task is not to distinguish between two cases (constant vs balanced) but between exponentially many cases (one for each unknown string s). Problem might seem artificial, but Shor s algorithm has the same underlying idea. Note: This is a promise problem! Problem gave the first strong (polynomial vs exponential) separation between quantum and classical computing. Before that only super-polynomial separations were known. Literature D. R. Simon. On the Power of Quantum Computation. Proc. FOCS 94, 6 3. IEEE, 994.

83 Simon s Problem: Specification How the function f is given We are given f as a black box in form of a quantum circuit computing U f : x y U f x y f (x) How can we query f? Classical algorithm: makes queries x,..., x k resulting in the answers f (x ),..., f (x k ) (counts as k queries). Quantum algorithm: can query in superposition, i. e., starting with the state ϕ = k i= x i results in U f ϕ 0 = k U f x i 0 = i= k x i f (x i ). i= This counts as one query!

84 Simon s Problem: Specification How the function f is given We are given f as a black box in form of a quantum circuit computing U f : x y U f x y f (x) How can we query f? Classical algorithm: makes queries x,..., x k resulting in the answers f (x ),..., f (x k ) (counts as k queries). Quantum algorithm: can query in superposition, i. e., starting with the state ϕ = k i= x i results in U f ϕ 0 = k U f x i 0 = i= k x i f (x i ). i= This counts as one query!

85 Simon s Problem: Specification How the function f is given We are given f as a black box in form of a quantum circuit computing U f : x y U f x y f (x) How can we query f? Classical algorithm: makes queries x,..., x k resulting in the answers f (x ),..., f (x k ) (counts as k queries). Quantum algorithm: can query in superposition, i. e., starting with the state ϕ = k i= x i results in U f ϕ 0 = k U f x i 0 = i= k x i f (x i ). i= This counts as one query!

86 Measuring the State of the System Partial measurements / non-demolition measurements Suppose we are given a Boolean function f : {0, } n {0, } and have prepared the state ψ = n/ α x x f (x) x {0,} n = n/ α x x 0 + α x x. x:f (x)=0 x:f (x)= Measuring the last qubit gives a random variable X. Pr(X = 0) = α x, Pr(X = ) = α x. x:f (x)=0 x:f (x)= This talk: We only consider von Neumann measurements of some (or all) of the qubits.

87 Measuring the State of the System Partial measurements / non-demolition measurements Suppose we are given a Boolean function f : {0, } n {0, } and have prepared the state ψ = n/ α x x f (x) x {0,} n = n/ α x x 0 + α x x. x:f (x)=0 x:f (x)= Measuring the last qubit gives a random variable X. Pr(X = 0) = α x, Pr(X = ) = α x. x:f (x)=0 x:f (x)= This talk: We only consider von Neumann measurements of some (or all) of the qubits.

88 Measuring the State of the System Partial measurements / non-demolition measurements Suppose we are given a Boolean function f : {0, } n {0, } and have prepared the state ψ = n/ α x x f (x) x {0,} n = n/ α x x 0 + α x x. x:f (x)=0 x:f (x)= Measuring the last qubit gives a random variable X. Pr(X = 0) = α x, Pr(X = ) = α x. x:f (x)=0 x:f (x)= This talk: We only consider von Neumann measurements of some (or all) of the qubits.

89 Measuring the State of the System Why partial/ non-demolition measurements We can still work with the collapsed state! For instance if measuring the last qubit of ψ = n/ α x x 0 + α x x x:f (x)=0 x:f (x)= yields the result X =, then the collapsed state is ψ = α x x, where s = {x : f (x) = }. s x:f (x)= Note: cannot be used to find solutions to f (x) =. Why? Further reading: the most general operation which can be applied to a quantum system in order to obtain classical information is a POVM (positive operator valued measure).

90 Measuring the State of the System Why partial/ non-demolition measurements We can still work with the collapsed state! For instance if measuring the last qubit of ψ = n/ α x x 0 + α x x x:f (x)=0 x:f (x)= yields the result X =, then the collapsed state is ψ = α x x, where s = {x : f (x) = }. s x:f (x)= Note: cannot be used to find solutions to f (x) =. Why? Further reading: the most general operation which can be applied to a quantum system in order to obtain classical information is a POVM (positive operator valued measure).

91 Measuring the State of the System Why partial/ non-demolition measurements We can still work with the collapsed state! For instance if measuring the last qubit of ψ = n/ α x x 0 + α x x x:f (x)=0 x:f (x)= yields the result X =, then the collapsed state is ψ = α x x, where s = {x : f (x) = }. s x:f (x)= Note: cannot be used to find solutions to f (x) =. Why? Further reading: the most general operation which can be applied to a quantum system in order to obtain classical information is a POVM (positive operator valued measure).

92 Simon s Problem: Preparing Useful States Creating the uniform superposition The basic idea is to prepare 0 0 H n n U f n x 0 x IF n x f (x). x IF n Collapsing the uniform superposition Now, measuring the second register will yield a random s IF n in the image of f. The state collapses to ϕ x0,s = ( x 0 + x 0 s ).

93 Simon s Problem: Preparing Useful States Creating the uniform superposition The basic idea is to prepare 0 0 H n n U f n x 0 x IF n x f (x). x IF n Collapsing the uniform superposition Now, measuring the second register will yield a random s IF n in the image of f. The state collapses to ϕ x0,s = ( x 0 + x 0 s ).

94 Computing the Hadamard Transform We apply H n to the collapsed states ϕ x0,s : H n ϕ x0,s = ( ) n/ ( ) x y x y ( x 0 + x 0 s ) = = n+ n+ x,y IF n ( ) x x 0 x + x IF n x IF n ( ) x (x0 s) x ( ( ) x x 0 x + ( ) x x 0 x s) ) x = = n+ n+ x IF n ( ( ) x x 0 x + ( ) x x 0 ( ) x s) x x IF n ( ) x x ( 0 + ( ) x s) x x IF n

95 Computing the Hadamard Transform We apply H n to the collapsed states ϕ x0,s : H n ϕ x0,s = ( ) n/ ( ) x y x y ( x 0 + x 0 s ) = = n+ n+ x,y IF n ( ) x x 0 x + x IF n x IF n ( ) x (x0 s) x ( ( ) x x 0 x + ( ) x x 0 x s) ) x = = n+ n+ x IF n ( ( ) x x 0 x + ( ) x x 0 ( ) x s) x x IF n ( ) x x ( 0 + ( ) x s) x x IF n

96 Computing the Hadamard Transform We apply H n to the collapsed states ϕ x0,s : H n ϕ x0,s = ( ) n/ ( ) x y x y ( x 0 + x 0 s ) = = n+ n+ x,y IF n ( ) x x 0 x + x IF n x IF n ( ) x (x0 s) x ( ( ) x x 0 x + ( ) x x 0 x s) ) x = = n+ n+ x IF n ( ( ) x x 0 x + ( ) x x 0 ( ) x s) x x IF n ( ) x x ( 0 + ( ) x s) x x IF n

97 Computing the Hadamard Transform We apply H n to the collapsed states ϕ x0,s : H n ϕ x0,s = ( ) n/ ( ) x y x y ( x 0 + x 0 s ) = = n+ n+ x,y IF n ( ) x x 0 x + x IF n x IF n ( ) x (x0 s) x ( ( ) x x 0 x + ( ) x x 0 x s) ) x = = n+ n+ x IF n ( ( ) x x 0 x + ( ) x x 0 ( ) x s) x x IF n ( ) x x ( 0 + ( ) x s) x x IF n

98 Computing the Hadamard Transform We apply H n to the collapsed states ϕ x0,s : H n ϕ x0,s = ( ) n/ ( ) x y x y ( x 0 + x 0 s ) = = n+ n+ x,y IF n ( ) x x 0 x + x IF n x IF n ( ) x (x0 s) x ( ( ) x x 0 x + ( ) x x 0 x s) ) x = = n+ n+ x IF n ( ( ) x x 0 x + ( ) x x 0 ( ) x s) x x IF n ( ) x x ( 0 + ( ) x s) x x IF n

99 Simon s Problem: Destructive Interference What have we gained by doing this? H n ϕ x0,s = = n+ n+ ( ) x x ( 0 + ( ) x s) x x IF n ( ) x x 0 x x IF n x s=0 Hence measuring this state yields a random element in What we really want... s = {x IF n x s = 0}.... are elements from s itself (there is only 0 and s itself since s is one-dimensional). How can we compute s from s?

100 Simon s Problem: Destructive Interference What have we gained by doing this? H n ϕ x0,s = = n+ n+ ( ) x x ( 0 + ( ) x s) x x IF n ( ) x x 0 x x IF n x s=0 Hence measuring this state yields a random element in What we really want... s = {x IF n x s = 0}.... are elements from s itself (there is only 0 and s itself since s is one-dimensional). How can we compute s from s?

101 Simon s Problem: Destructive Interference What have we gained by doing this? H n ϕ x0,s = = n+ n+ ( ) x x ( 0 + ( ) x s) x x IF n ( ) x x 0 x x IF n x s=0 Hence measuring this state yields a random element in What we really want... s = {x IF n x s = 0}.... are elements from s itself (there is only 0 and s itself since s is one-dimensional). How can we compute s from s?

102 Simon s Problem: Destructive Interference What have we gained by doing this? H n ϕ x0,s = = n+ n+ ( ) x x ( 0 + ( ) x s) x x IF n ( ) x x 0 x x IF n x s=0 Hence measuring this state yields a random element in What we really want... s = {x IF n x s = 0}.... are elements from s itself (there is only 0 and s itself since s is one-dimensional). How can we compute s from s?

103 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

104 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

105 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

106 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

107 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

108 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

109 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

110 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

111 Solution to Simon s Problem Quantum algorithm Given: Function f : {0, } n {0, } n with Simon promise, i. e., preimages of a fixed image have the form x 0 and x 0 s. Task: Find the unknown bit-string s IF n. Repeat the following steps n times:. Initialize two quantum registers: 0 0. Equal distribution on first register: 3. Compute f in superposition: 4. Measure second register: 5. Compute H n on first register: n n P x {0,} x 0 P n x {0,} x f (x) n ( x 0 + x 0 s ) = ϕ x0,s n+ X x IF n x s=0 ( ) x x 0 x 6. Measure first register: Sample y IF n with y s = 0. Further classical post-processing is necessary!

112 Classical Postprocessing in Simon s Algorithm Classical post-processing After n iterations: y,..., y n IF n with y i s = 0. We have to infer s by a purely classical computation. Show high probability of success over the choice of y i. Linear algebra over IF We are given the linear system of equations A s = y. y n s = 0 Hence, we have to compute the kernel of A IF (n ) n. If the kernel is one-dimensional, then s is uniquely determined.

113 Classical Postprocessing in Simon s Algorithm Classical post-processing After n iterations: y,..., y n IF n with y i s = 0. We have to infer s by a purely classical computation. Show high probability of success over the choice of y i. Linear algebra over IF We are given the linear system of equations A s = y. y n s = 0 Hence, we have to compute the kernel of A IF (n ) n. If the kernel is one-dimensional, then s is uniquely determined.

114 Classical Post-Processing in Simon s Algorithm The probability of success Since dim( s ) = we have that dim( s ) = n. Hence we have to bound the probability that n random vectors in IF n are linear independent: n n n n Pr(rk(A) = n ) = n n n = Complexity of the quantum algorithm n n n We have used n iterations and each individual run uses one query, n Hadamard transforms, and n + single qubit measurements Postprocessing: Computing the kernel of a matrix of size n n is linear algebra and can be solved in time O(n 3 ). Overall we have found a polynomial-time quantum algorithm.

115 Classical Post-Processing in Simon s Algorithm The probability of success Since dim( s ) = we have that dim( s ) = n. Hence we have to bound the probability that n random vectors in IF n are linear independent: n n n n Pr(rk(A) = n ) = n n n = Complexity of the quantum algorithm n n n We have used n iterations and each individual run uses one query, n Hadamard transforms, and n + single qubit measurements Postprocessing: Computing the kernel of a matrix of size n n is linear algebra and can be solved in time O(n 3 ). Overall we have found a polynomial-time quantum algorithm.

116 Classical Post-Processing in Simon s Algorithm The probability of success Since dim( s ) = we have that dim( s ) = n. Hence we have to bound the probability that n random vectors in IF n are linear independent: n n n n Pr(rk(A) = n ) = n n n = Complexity of the quantum algorithm n n n We have used n iterations and each individual run uses one query, n Hadamard transforms, and n + single qubit measurements Postprocessing: Computing the kernel of a matrix of size n n is linear algebra and can be solved in time O(n 3 ). Overall we have found a polynomial-time quantum algorithm.

117 Classical Post-Processing in Simon s Algorithm The probability of success Since dim( s ) = we have that dim( s ) = n. Hence we have to bound the probability that n random vectors in IF n are linear independent: n n n n Pr(rk(A) = n ) = n n n = Complexity of the quantum algorithm n n n We have used n iterations and each individual run uses one query, n Hadamard transforms, and n + single qubit measurements Postprocessing: Computing the kernel of a matrix of size n n is linear algebra and can be solved in time O(n 3 ). Overall we have found a polynomial-time quantum algorithm.

118 Simon s Problem: Classical Lower Bound Theorem Let A be a classical probabilistic algorithm which determines s using k queries to the function f. Then k = Ω( n/ ). Computational complexity theory lingo Hence there exists an oracle O with respect which we have a separation between classical and quantum computation: BPP O BQP O This is a so-called relativized result, i. e., it holds in a world in which calls to the oracle cost only one query. Whether this also holds for our world, i. e., without oracles, is a major open problem in theoretical computer science. Looking ahead: Why does the fact that FACTORING is in BQP, whereas no classical algorithm is known for it, does not imply that BPP BQP?

119 Simon s Problem: Classical Lower Bound Theorem Let A be a classical probabilistic algorithm which determines s using k queries to the function f. Then k = Ω( n/ ). Computational complexity theory lingo Hence there exists an oracle O with respect which we have a separation between classical and quantum computation: BPP O BQP O This is a so-called relativized result, i. e., it holds in a world in which calls to the oracle cost only one query. Whether this also holds for our world, i. e., without oracles, is a major open problem in theoretical computer science. Looking ahead: Why does the fact that FACTORING is in BQP, whereas no classical algorithm is known for it, does not imply that BPP BQP?

120 Simon s Problem: Classical Lower Bound Theorem Let A be a classical probabilistic algorithm which determines s using k queries to the function f. Then k = Ω( n/ ). Computational complexity theory lingo Hence there exists an oracle O with respect which we have a separation between classical and quantum computation: BPP O BQP O This is a so-called relativized result, i. e., it holds in a world in which calls to the oracle cost only one query. Whether this also holds for our world, i. e., without oracles, is a major open problem in theoretical computer science. Looking ahead: Why does the fact that FACTORING is in BQP, whereas no classical algorithm is known for it, does not imply that BPP BQP?

121 Simon s Problem: Classical Lower Bound Theorem Let A be a classical probabilistic algorithm which determines s using k queries to the function f. Then k = Ω( n/ ). Computational complexity theory lingo Hence there exists an oracle O with respect which we have a separation between classical and quantum computation: BPP O BQP O This is a so-called relativized result, i. e., it holds in a world in which calls to the oracle cost only one query. Whether this also holds for our world, i. e., without oracles, is a major open problem in theoretical computer science. Looking ahead: Why does the fact that FACTORING is in BQP, whereas no classical algorithm is known for it, does not imply that BPP BQP?

122 Simon s Problem: Classical Lower Bound Theorem Let A be a classical probabilistic algorithm which determines s using k queries to the function f. Then k = Ω( n/ ). Computational complexity theory lingo Hence there exists an oracle O with respect which we have a separation between classical and quantum computation: BPP O BQP O This is a so-called relativized result, i. e., it holds in a world in which calls to the oracle cost only one query. Whether this also holds for our world, i. e., without oracles, is a major open problem in theoretical computer science. Looking ahead: Why does the fact that FACTORING is in BQP, whereas no classical algorithm is known for it, does not imply that BPP BQP?

123 Simon s Problem: Classical Lower Bound Sketch of proof of the Ω( n/ ) lower bound Suppose A makes k queries x,..., x k, where x i x j. Let F k := {f (x i ) : i =,..., k} and E k := {x i x j : i j}. If F k < k then we have found a collision, i. e. a pair (i 0, j 0 ) with f (x i0 ) = f (x j0 ). Then s = x i0 x j0. Suppose there was no collision. Then s E k and E k = ( k ) candidates have been eliminated. However, there are n ( k ) candidates for s. We show that they are equally likely for a given F k. Then k = Ω( n ). Bayes rule: Pr(s = s 0 F k = k) = Pr( F k = k s = s 0 ) Pr(s = s 0 ). Pr( F k = k) The a priori probabilities Pr(s = s 0 ) are equal and by symmetry Pr( F k = k s = s0 ) is independent of s 0.

124 Simon s Problem: Classical Lower Bound Sketch of proof of the Ω( n/ ) lower bound Suppose A makes k queries x,..., x k, where x i x j. Let F k := {f (x i ) : i =,..., k} and E k := {x i x j : i j}. If F k < k then we have found a collision, i. e. a pair (i 0, j 0 ) with f (x i0 ) = f (x j0 ). Then s = x i0 x j0. Suppose there was no collision. Then s E k and E k = ( k ) candidates have been eliminated. However, there are n ( k ) candidates for s. We show that they are equally likely for a given F k. Then k = Ω( n ). Bayes rule: Pr(s = s 0 F k = k) = Pr( F k = k s = s 0 ) Pr(s = s 0 ). Pr( F k = k) The a priori probabilities Pr(s = s 0 ) are equal and by symmetry Pr( F k = k s = s0 ) is independent of s 0.

125 Simon s Problem: Classical Lower Bound Sketch of proof of the Ω( n/ ) lower bound Suppose A makes k queries x,..., x k, where x i x j. Let F k := {f (x i ) : i =,..., k} and E k := {x i x j : i j}. If F k < k then we have found a collision, i. e. a pair (i 0, j 0 ) with f (x i0 ) = f (x j0 ). Then s = x i0 x j0. Suppose there was no collision. Then s E k and E k = ( k ) candidates have been eliminated. However, there are n ( k ) candidates for s. We show that they are equally likely for a given F k. Then k = Ω( n ). Bayes rule: Pr(s = s 0 F k = k) = Pr( F k = k s = s 0 ) Pr(s = s 0 ). Pr( F k = k) The a priori probabilities Pr(s = s 0 ) are equal and by symmetry Pr( F k = k s = s0 ) is independent of s 0.

126 Simon s Problem: Classical Lower Bound Sketch of proof of the Ω( n/ ) lower bound Suppose A makes k queries x,..., x k, where x i x j. Let F k := {f (x i ) : i =,..., k} and E k := {x i x j : i j}. If F k < k then we have found a collision, i. e. a pair (i 0, j 0 ) with f (x i0 ) = f (x j0 ). Then s = x i0 x j0. Suppose there was no collision. Then s E k and E k = ( k ) candidates have been eliminated. However, there are n ( k ) candidates for s. We show that they are equally likely for a given F k. Then k = Ω( n ). Bayes rule: Pr(s = s 0 F k = k) = Pr( F k = k s = s 0 ) Pr(s = s 0 ). Pr( F k = k) The a priori probabilities Pr(s = s 0 ) are equal and by symmetry Pr( F k = k s = s0 ) is independent of s 0.

127 Simon s Problem: Classical Lower Bound Sketch of proof of the Ω( n/ ) lower bound Suppose A makes k queries x,..., x k, where x i x j. Let F k := {f (x i ) : i =,..., k} and E k := {x i x j : i j}. If F k < k then we have found a collision, i. e. a pair (i 0, j 0 ) with f (x i0 ) = f (x j0 ). Then s = x i0 x j0. Suppose there was no collision. Then s E k and E k = ( k ) candidates have been eliminated. However, there are n ( k ) candidates for s. We show that they are equally likely for a given F k. Then k = Ω( n ). Bayes rule: Pr(s = s 0 F k = k) = Pr( F k = k s = s 0 ) Pr(s = s 0 ). Pr( F k = k) The a priori probabilities Pr(s = s 0 ) are equal and by symmetry Pr( F k = k s = s0 ) is independent of s 0.

128 For Further Reading G. Alber, Th. Beth, M. Horodecki, P. Horodecki, R. Horodecki, M. Roetteler, H. Weinfurter, and A. Zeilinger. Quantum Information: An Introduction to Basic Theoretical Concepts and Experiments. Springer, 00. J. Gruska Quantum Computing. McGraw-Hill, 999. A. Yu. Kitaev, A. H. Shen, and M. N. Vyalyi. Classical and Quantum Computation. Graduate Studies in Mathematics, vol. 47, AMS, 00. M. Nielsen und I. Chuang. Quantum Computation and Information. Cambridge University Press, 000.

129 Conclusions Elementary quantum gates: Controlled NOT gate Local unitary transformations A simple quantum algorithm on two qubits which distinguishes constant from balanced functions. Separation: query (quantum) vs queries (classical) Quantum algorithm for Simon s problem based on: Computing with superpositions Interference of computational paths

130 Part II: The Algorithms of Shor and Grover Martin Rötteler NEC Laboratories America, Inc. 4 Independence Way, Suite 00 Princeton, NJ 08540, U.S.A. International Summer School on Quantum Information, Max-Planck-Institut für Physik komplexer Systeme Dresden, September, 005

131 Overview Today: Shor s algorithm Outlook: Modular exponentiation Period extraction via Quantum Fourier Transform Classical post-processing Generalizations of Shor s algorithm Grover s algorithm for searching a list Universal quantum gates On September 9, Markus Grassl will continue with an introduction to quantum error-correcting codes.

132 The Integer Factorization Problem Basic problem Given a natural number N. Find a (prime) factor of N. Best known classical algorithm The number field sieve has a complexity of exp ((.93 + o())(log N) /3 (log log N) /3) which is (sub)exponential in the number n = log N of bits of N. Making money with factoring The company RSA offers $ for anybody who can factor a certain 67 digit number N. This number is known to be of the form N = pq but finding p and q is infeasible using the best known classical algorithms.

133 The Integer Factorization Problem Basic problem Given a natural number N. Find a (prime) factor of N. Best known classical algorithm The number field sieve has a complexity of exp ((.93 + o())(log N) /3 (log log N) /3) which is (sub)exponential in the number n = log N of bits of N. Making money with factoring The company RSA offers $ for anybody who can factor a certain 67 digit number N. This number is known to be of the form N = pq but finding p and q is infeasible using the best known classical algorithms.

134 The Integer Factorization Problem Basic problem Given a natural number N. Find a (prime) factor of N. Best known classical algorithm The number field sieve has a complexity of exp ((.93 + o())(log N) /3 (log log N) /3) which is (sub)exponential in the number n = log N of bits of N. Making money with factoring The company RSA offers $ for anybody who can factor a certain 67 digit number N. This number is known to be of the form N = pq but finding p and q is infeasible using the best known classical algorithms.

135 The RSA Factoring Challenge RSA-00 is factored! In May 005 a small team of people has factored RSA-00. At 663 bits, this is the largest RSA Challenge Number factored. The classical effort undertaken Sieving equivalent of 55 years on a single. GHz Opteron CPU. The matrix step took about 3 months on a cluster of 80. GHz Opterons. Computed from late 003 to May 005. RSA-00 and its factors N = p = q =

136 The RSA Factoring Challenge RSA-00 is factored! In May 005 a small team of people has factored RSA-00. At 663 bits, this is the largest RSA Challenge Number factored. The classical effort undertaken Sieving equivalent of 55 years on a single. GHz Opteron CPU. The matrix step took about 3 months on a cluster of 80. GHz Opterons. Computed from late 003 to May 005. RSA-00 and its factors N = p = q =

137 The RSA Factoring Challenge RSA-00 is factored! In May 005 a small team of people has factored RSA-00. At 663 bits, this is the largest RSA Challenge Number factored. The classical effort undertaken Sieving equivalent of 55 years on a single. GHz Opteron CPU. The matrix step took about 3 months on a cluster of 80. GHz Opterons. Computed from late 003 to May 005. RSA-00 and its factors N = p = q =

138 Shor s Algorithm: Reduction to Order Finding Reformulating the factoring problem We can factor N if the following problem can be solved: Input: A number a with < a < N. Output: The order r of a modulo N, i. e., the smallest integer r > 0 such that a r mod N. Why is this a reduction? Suppose we want to find a divisor of N different from + or. Pick a random a with < a < N and find its order r Suppose that r is even (happens with high probability): 0 = (a r ) = (a r/ )(a r/ + ) mod N. If a r/ ± then gcd(a r/, N) and gcd(a r/ +, N) yield at least one nontrivial divisor of N.

139 Shor s Algorithm: Reduction to Order Finding Reformulating the factoring problem We can factor N if the following problem can be solved: Input: A number a with < a < N. Output: The order r of a modulo N, i. e., the smallest integer r > 0 such that a r mod N. Why is this a reduction? Suppose we want to find a divisor of N different from + or. Pick a random a with < a < N and find its order r Suppose that r is even (happens with high probability): 0 = (a r ) = (a r/ )(a r/ + ) mod N. If a r/ ± then gcd(a r/, N) and gcd(a r/ +, N) yield at least one nontrivial divisor of N.

140 Shor s Algorithm: Reduction to Order Finding Remarks about this reduction Note that gcd(a, b) of two n-bit integers can be computed in poly(log(n)) time. There were two events in which the reduction fails: (i) we pick a with an odd order r and (ii) a r/ = ±. We have to bound the probability for one of these events to occur. Theorem Let N = p µ... pµm m with m and p i >. Then The big question Pr(neither (i) nor (ii) occurs) m How can we efficiently determine the multiplicative order of a random element a modulo N?

141 Shor s Algorithm: Reduction to Order Finding Remarks about this reduction Note that gcd(a, b) of two n-bit integers can be computed in poly(log(n)) time. There were two events in which the reduction fails: (i) we pick a with an odd order r and (ii) a r/ = ±. We have to bound the probability for one of these events to occur. Theorem Let N = p µ... pµm m with m and p i >. Then The big question Pr(neither (i) nor (ii) occurs) m How can we efficiently determine the multiplicative order of a random element a modulo N?

142 Shor s Algorithm: Reduction to Order Finding Remarks about this reduction Note that gcd(a, b) of two n-bit integers can be computed in poly(log(n)) time. There were two events in which the reduction fails: (i) we pick a with an odd order r and (ii) a r/ = ±. We have to bound the probability for one of these events to occur. Theorem Let N = p µ... pµm m with m and p i >. Then The big question Pr(neither (i) nor (ii) occurs) m How can we efficiently determine the multiplicative order of a random element a modulo N?

143 Defining a Period Function The modular exponentiation map Let N be an integer and let a Z N. Let M be an integer. The modular exponentiation is the map f : x (a x mod N) from Z M to Z N. Result: The map f can be implemented efficiently using standard arithmetic in O(poly(log N)) operations. Hence also the map U f : x y x a x mod N can be implemented efficiently. Recall that the order of a is defined as the smallest integer r such that a r = mod N. Observation The function f : x (a x mod N) is periodic and has period length r, i. e., f (x) = f (x + r) for all inputs x.

144 Defining a Period Function The modular exponentiation map Let N be an integer and let a Z N. Let M be an integer. The modular exponentiation is the map f : x (a x mod N) from Z M to Z N. Result: The map f can be implemented efficiently using standard arithmetic in O(poly(log N)) operations. Hence also the map U f : x y x a x mod N can be implemented efficiently. Recall that the order of a is defined as the smallest integer r such that a r = mod N. Observation The function f : x (a x mod N) is periodic and has period length r, i. e., f (x) = f (x + r) for all inputs x.

145 Defining a Period Function The modular exponentiation map Let N be an integer and let a Z N. Let M be an integer. The modular exponentiation is the map f : x (a x mod N) from Z M to Z N. Result: The map f can be implemented efficiently using standard arithmetic in O(poly(log N)) operations. Hence also the map U f : x y x a x mod N can be implemented efficiently. Recall that the order of a is defined as the smallest integer r such that a r = mod N. Observation The function f : x (a x mod N) is periodic and has period length r, i. e., f (x) = f (x + r) for all inputs x.

146 Defining a Period Function The modular exponentiation map Let N be an integer and let a Z N. Let M be an integer. The modular exponentiation is the map f : x (a x mod N) from Z M to Z N. Result: The map f can be implemented efficiently using standard arithmetic in O(poly(log N)) operations. Hence also the map U f : x y x a x mod N can be implemented efficiently. Recall that the order of a is defined as the smallest integer r such that a r = mod N. Observation The function f : x (a x mod N) is periodic and has period length r, i. e., f (x) = f (x + r) for all inputs x.

147 Defining a Period Function The modular exponentiation map Let N be an integer and let a Z N. Let M be an integer. The modular exponentiation is the map f : x (a x mod N) from Z M to Z N. Result: The map f can be implemented efficiently using standard arithmetic in O(poly(log N)) operations. Hence also the map U f : x y x a x mod N can be implemented efficiently. Recall that the order of a is defined as the smallest integer r such that a r = mod N. Observation The function f : x (a x mod N) is periodic and has period length r, i. e., f (x) = f (x + r) for all inputs x.

148 Setting up a Periodic State Observation The function f : x a x mod N is periodic and has period length r, i. e., f (x) = f (x + r) for all inputs x. The graph of the function f (x) = x mod 65 y = f (x) x

149 Shor s Problem: Preparing Useful States Creating the graph of f Let f (x) = a x mod N be the modular exponentiation and let U f : x y x y f (x) be as usual. We compute (letting M = m ) 0 0 H m x 0 Uf M x Z M M Collapsing the graph of f x Z M x f (x). Now, measuring the second register will yield a random s Z N in the image of f. The state collapses to N/r x 0 + k r. N/r k=0

150 Shor s Problem: Preparing Useful States Creating the graph of f Let f (x) = a x mod N be the modular exponentiation and let U f : x y x y f (x) be as usual. We compute (letting M = m ) 0 0 H m x 0 Uf M x Z M M Collapsing the graph of f x Z M x f (x). Now, measuring the second register will yield a random s Z N in the image of f. The state collapses to N/r x 0 + k r. N/r k=0

151 An Application of the DFT: Period Extraction Motivation We would like to apply the trick from Simon s algorithm: ϕ x0,s = ( x 0 + x 0 s ) n+ ( ) x x 0 x. x IF n x s=0 The unknown offset x 0 is transfered into the phases. The analogue of ϕ x0,s in case of the cyclic group Z N is ψ x0,r = N/r x 0 + k r N/r k=0 r {}}{ Again, we would like to transfer x 0 into the phases. x 0

152 An Application of the DFT: Period Extraction Motivation We would like to apply the trick from Simon s algorithm: ϕ x0,s = ( x 0 + x 0 s ) n+ ( ) x x 0 x. x IF n x s=0 The unknown offset x 0 is transfered into the phases. The analogue of ϕ x0,s in case of the cyclic group Z N is ψ x0,r = N/r x 0 + k r N/r k=0 r {}}{ Again, we would like to transfer x 0 into the phases. x 0

153 The Discrete Fourier Transformation (DFT) Definition of the DFT DFT N := [ ] ωn k l, N k,l=0...n ω N = e πi/n Example DFT

154 Estimating Character Sums Useful trick in quantum computing (Character Lemma) Lemma: For all i = 0,..., n the following holds: n ωn ij = n δ i,0 j=0 Proof: Let S := n j=0 ωij n. Then n ωns i = ωnω i n ij n = ω i(j+) n n = ωn ij = S j=0 j=0 j=0 If i 0 then ω i n, i. e., ( ω i n) 0. Hence S = 0. If i = 0 then ω i n = which implies that S = n.

155 Estimating Character Sums Useful trick in quantum computing (Character Lemma) Lemma: For all i = 0,..., n the following holds: n ωn ij = n δ i,0 j=0 Proof: Let S := n j=0 ωij n. Then n ωns i = ωnω i n ij n = ω i(j+) n n = ωn ij = S j=0 j=0 j=0 If i 0 then ω i n, i. e., ( ω i n) 0. Hence S = 0. If i = 0 then ω i n = which implies that S = n.

156 Estimating Character Sums Useful trick in quantum computing (Character Lemma) Lemma: For all i = 0,..., n the following holds: n ωn ij = n δ i,0 j=0 Proof: Let S := n j=0 ωij n. Then n ωns i = ωnω i n ij n = ω i(j+) n n = ωn ij = S j=0 j=0 j=0 If i 0 then ω i n, i. e., ( ω i n) 0. Hence S = 0. If i = 0 then ω i n = which implies that S = n.

157 Estimating Character Sums Useful trick in quantum computing (Character Lemma) Lemma: For all i = 0,..., n the following holds: n ωn ij = n δ i,0 j=0 Proof: Let S := n j=0 ωij n. Then n ωns i = ωnω i n ij n = ω i(j+) n n = ωn ij = S j=0 j=0 j=0 If i 0 then ω i n, i. e., ( ω i n) 0. Hence S = 0. If i = 0 then ω i n = which implies that S = n.

158 Extracting the Period of a Function Theorem (Fourier duality) Let N IN and let r Z N be a divisor of N, and let x 0 Z N. Then DFT N ψ x0,r = DFT N X N/r p x 0 + k r A = N/r k=0 X r r l=0 ω lx 0 N r N l N r Proof: DFT N ψ x0,r = = = r N r N 0 N X ω ij N i j A@ N/r X p x 0 + k r N N/r i,j=0 0 X X N/r ω i(x 0+k r) N i=0 k=0 N X N/r X ω ix 0 N ωn ikr i=0 k=0 {z } =:α i A i i k=0 A

159 Extracting the Period of a Function Theorem (Fourier duality) Let N IN and let r Z N be a divisor of N, and let x 0 Z N. Then DFT N ψ x0,r = DFT N X N/r p x 0 + k r A = N/r k=0 X r r l=0 ω lx 0 N r N l N r Proof: DFT N ψ x0,r = = = r N r N 0 N X ω ij N i j A@ N/r X p x 0 + k r N N/r i,j=0 0 X X N/r ω i(x 0+k r) N i=0 k=0 N X N/r X ω ix 0 N ωn ikr i=0 k=0 {z } =:α i A i i k=0 A

160 Extracting the Period of a Function Theorem (Fourier duality) Let N IN and let r Z N be a divisor of N, and let x 0 Z N. Then DFT N ψ x0,r = DFT N X N/r p x 0 + k r A = N/r k=0 X r r l=0 ω lx 0 N r N l N r Proof: DFT N ψ x0,r = = = r N r N 0 N X ω ij N i j A@ N/r X p x 0 + k r N N/r i,j=0 0 X X N/r ω i(x 0+k r) N i=0 k=0 N X N/r X ω ix 0 N ωn ikr i=0 k=0 {z } =:α i A i i k=0 A

161 Extracting the Period of a Function Theorem (Fourier duality) Let N IN and let r Z N be a divisor of N, and let x 0 Z N. Then DFT N ψ x0,r = DFT N X N/r p x 0 + k r A = N/r k=0 X r r l=0 ω lx 0 N r N l N r Proof: DFT N ψ x0,r = = = r N r N 0 N X ω ij N i j A@ N/r X p x 0 + k r N N/r i,j=0 0 X X N/r ω i(x 0+k r) N i=0 k=0 N X N/r X ω ix 0 N ωn ikr i=0 k=0 {z } =:α i A i i k=0 A

162 Constructive / Destructive Interference Computing the coefficients α i For each i = 0,..., N we have to compute α i = N/r k=0 ω ikr N. Case : i = N r l for some l = 0,..., r. Then α i = N/r k=0 ω N r lkr N/r N = k=0 = N r. Case : i N r l for all l = 0,..., r. Then α i = N/r k=0 by the Character Lemma. ( ) k ω N r lkr N = 0

163 Constructive / Destructive Interference Computing the coefficients α i For each i = 0,..., N we have to compute α i = N/r k=0 ω ikr N. Case : i = N r l for some l = 0,..., r. Then α i = N/r k=0 ω N r lkr N/r N = k=0 = N r. Case : i N r l for all l = 0,..., r. Then α i = N/r k=0 by the Character Lemma. ( ) k ω N r lkr N = 0

164 Constructive / Destructive Interference End of proof (Fourier duality) DFT N ψ x0,r = = r N r N N i=0 r l=0 ω ix 0 N ( N/r ω l N r x 0 N N r k=0 ) N r ωn ikr i = N ln = r r r l=0 i=0 ω l N r x 0 N ω ix 0 N α i i ln r What happens if r does not divide N? In this case the state can be approximated very accurately by ) DFT N ( k x 0 + k r k ω lµx 0 N lµ with an element µ Z N such that µr N.

165 Constructive / Destructive Interference End of proof (Fourier duality) DFT N ψ x0,r = = r N r N N i=0 r l=0 ω ix 0 N ( N/r ω l N r x 0 N N r k=0 ) N r ωn ikr i = N ln = r r r l=0 i=0 ω l N r x 0 N ω ix 0 N α i i ln r What happens if r does not divide N? In this case the state can be approximated very accurately by ) DFT N ( k x 0 + k r k ω lµx 0 N lµ with an element µ Z N such that µr N.

166 Constructive / Destructive Interference End of proof (Fourier duality) DFT N ψ x0,r = = r N r N N i=0 r l=0 ω ix 0 N ( N/r ω l N r x 0 N N r k=0 ) N r ωn ikr i = N ln = r r r l=0 i=0 ω l N r x 0 N ω ix 0 N α i i ln r What happens if r does not divide N? In this case the state can be approximated very accurately by ) DFT N ( k x 0 + k r k ω lµx 0 N lµ with an element µ Z N such that µr N.

167 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

168 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

169 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

170 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

171 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

172 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

173 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

174 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

175 Solution to the Period Extraction Problem Quantum algorithm Given: Modular exponentiation function U a : x 0 x a x mod N. Task: Find the order r of a modulo N. Repeat the following steps one time: (w/o normalizations, M = m >> N). Initialize two quantum registers: 0 0. Equal distribution on first register: P M x=0 x 0 3. Compute f in superposition: P M x=0 x ax mod N P M/r 4. Measure second register: P k=0 x 0 + k r r 5. Compute DFT M on first register: N l=0 ωl r x 0 M l N r 6. Measure first register: Sample a rational number p q which is very close to l 0 r. How can we classically reconstruct r from p q?

176 Classical Post-Processing How to obtain r from p q? N Since we know N and l 0, we could easily compute r from l 0 However, we are just given p q for which we only know that the following holds p q l 0 r < r Diophantine approximation We apply the continued fractions algorithm to p q. This will lead to several principal fractions and actually l 0 r will be one of them. Note that we can check whether a candidate r is indeed the order. r. Theorem (Shor 94) FACTORING BQP.

177 Classical Post-Processing How to obtain r from p q? N Since we know N and l 0, we could easily compute r from l 0 However, we are just given p q for which we only know that the following holds p q l 0 r < r Diophantine approximation We apply the continued fractions algorithm to p q. This will lead to several principal fractions and actually l 0 r will be one of them. Note that we can check whether a candidate r is indeed the order. r. Theorem (Shor 94) FACTORING BQP.

178 Classical Post-Processing How to obtain r from p q? N Since we know N and l 0, we could easily compute r from l 0 However, we are just given p q for which we only know that the following holds p q l 0 r < r Diophantine approximation We apply the continued fractions algorithm to p q. This will lead to several principal fractions and actually l 0 r will be one of them. Note that we can check whether a candidate r is indeed the order. r. Theorem (Shor 94) FACTORING BQP.

179 Diophantine Approximation The continued fractions algorithm Input: x IR. Output: Sequence of b i Z (possibly infinite) which represents x. Let b 0 := x, x :=, b x b := x, x :=,.... Then 0 x b x = b 0 + b + b + Example with rational input Suppose that x = The algorithm results in (vector of b i s): [,,,, 5, 3,,, 3,,,,,,,, 5,,,,,,,,,, 3] Consider the convergents C n which are obtained by truncating after n steps: n C n n C n

180 Diophantine Approximation The continued fractions algorithm Input: x IR. Output: Sequence of b i Z (possibly infinite) which represents x. Let b 0 := x, x :=, b x b := x, x :=,.... Then 0 x b x = b 0 + b + b + Example with rational input Suppose that x = The algorithm results in (vector of b i s): [,,,, 5, 3,,, 3,,,,,,,, 5,,,,,,,,,, 3] Consider the convergents C n which are obtained by truncating after n steps: n C n n C n

181 Diophantine Approximation Lagrange s Theorem Let x Q and assume that we are given p Q such that q x p q q Then x is a convergent C n of p q, namely that for which Cn p q < q. Example (cont d) Let y = 345 be the inverse period. Since denominator of y is we can work with precision 33. Suppose we measure p = q : n C n n C n

182 Diophantine Approximation Lagrange s Theorem Let x Q and assume that we are given p Q such that q x p q q Then x is a convergent C n of p q, namely that for which Cn p q < q. Example (cont d) Let y = 345 be the inverse period. Since denominator of y is we can work with precision 33. Suppose we measure p = q : n C n n C n

183 Parallels between DSP and QC w w w w Digital Signal Processing Quantum Computing signal f (x) quantum information sampling X f = f (x)δ x x transform modelling X x α x x f (x) unitary transform & measurement result probabilities

184 Example: Fast Fourier Transform (FFT) Cooley-Tukey FFT The matrix DFT N = [ ] ω k l N N k,l=0...n, where ω N = e πi/n can be written as a short product of sparse matrices. DFT 4 = Π rev ( DFT ) diag (DFT ) [ ] [ ] [ ] [ ] [ ] i i = i i i FFT Theorem Multiplication with DFT N can be performed classically in O(N log N) elementary operations. We can do much better on a quantum computer!

185 Example: Fast Fourier Transform (FFT) Cooley-Tukey FFT The matrix DFT N = [ ] ω k l N N k,l=0...n, where ω N = e πi/n can be written as a short product of sparse matrices. DFT 4 = Π rev ( DFT ) diag (DFT ) [ ] [ ] [ ] [ ] [ ] i i = i i i FFT Theorem Multiplication with DFT N can be performed classically in O(N log N) elementary operations. We can do much better on a quantum computer!

186 Example: Fast Fourier Transform (FFT) Cooley-Tukey FFT The matrix DFT N = [ ] ω k l N N k,l=0...n, where ω N = e πi/n can be written as a short product of sparse matrices. DFT 4 = Π rev ( DFT ) diag (DFT ) [ ] [ ] [ ] [ ] [ ] i i = i i i FFT Theorem Multiplication with DFT N can be performed classically in O(N log N) elementary operations. We can do much better on a quantum computer!

187 Example: Fast Fourier Transform (FFT) Cooley-Tukey FFT The matrix DFT N = [ ] ω k l N N k,l=0...n, where ω N = e πi/n can be written as a short product of sparse matrices. DFT 4 = Π rev ( DFT ) diag (DFT ) [ ] [ ] [ ] [ ] [ ] i i = i i i FFT Theorem Multiplication with DFT N can be performed classically in O(N log N) elementary operations. We can do much better on a quantum computer!

188 Fast Fourier Transform Cooley-Tukey Formula Π n DFT n = ( DFT n DFT n DFT n D n DFT n D n ) = ( DFT n ) ( n D n ) (DFT n ) Factorization of the twiddle factors D n := ω n ω n... ω n n = ( ω n n ) ( )... ω n

189 Fast Fourier Transform Cooley-Tukey Formula Π n DFT n = ( DFT n DFT n DFT n D n DFT n D n ) = ( DFT n ) ( n D n ) (DFT n ) Factorization of the twiddle factors D n := ω n ω n... ω n n = ( ω n n ) ( )... ω n

190 Cooley-Tukey Realization of DFT Quantum circuit for DFT N.. H H D 4 D N D N H D 4. Cost Classical Computer Quantum Computer T (N) = T (N/) + O(N) T (N) = T (N/) + O(logN)) T (N) = O(NlogN) T (N) = O(log N)

191 Cooley-Tukey Realization of DFT Quantum circuit for DFT N.. H H D 4 D N D N H D 4. Cost Classical Computer Quantum Computer T (N) = T (N/) + O(N) T (N) = T (N/) + O(logN)) T (N) = O(NlogN) T (N) = O(log N)

192 Shor s Algorithm: Gate Counts Modular exponentiation U : x 0 x a x mod N, where N is the k-bit number to be factored, and x is a k-bit number. Implementation using 396k(k + O(k)) elementary operations [Beckman et al.]. Quantum Fourier Transform: DFT n needs n(n ) two-qubit gates and n one-qubit gates. An upper bound on the resources for k-bit number N About 400k 3 operations are needed (can be improved to O(k log k log log k))). The space needed is 5k + qubits. Example: Factoring 8-bit and 04-bit numbers For 8-bit we need operations and 64 qubits. For 04-bit number operations and 5 qubits.

193 Shor s Algorithm: Gate Counts Modular exponentiation U : x 0 x a x mod N, where N is the k-bit number to be factored, and x is a k-bit number. Implementation using 396k(k + O(k)) elementary operations [Beckman et al.]. Quantum Fourier Transform: DFT n needs n(n ) two-qubit gates and n one-qubit gates. An upper bound on the resources for k-bit number N About 400k 3 operations are needed (can be improved to O(k log k log log k))). The space needed is 5k + qubits. Example: Factoring 8-bit and 04-bit numbers For 8-bit we need operations and 64 qubits. For 04-bit number operations and 5 qubits.

194 Shor s Algorithm: Gate Counts Modular exponentiation U : x 0 x a x mod N, where N is the k-bit number to be factored, and x is a k-bit number. Implementation using 396k(k + O(k)) elementary operations [Beckman et al.]. Quantum Fourier Transform: DFT n needs n(n ) two-qubit gates and n one-qubit gates. An upper bound on the resources for k-bit number N About 400k 3 operations are needed (can be improved to O(k log k log log k))). The space needed is 5k + qubits. Example: Factoring 8-bit and 04-bit numbers For 8-bit we need operations and 64 qubits. For 04-bit number operations and 5 qubits.

195 Shor s Algorithm: Gate Counts Modular exponentiation U : x 0 x a x mod N, where N is the k-bit number to be factored, and x is a k-bit number. Implementation using 396k(k + O(k)) elementary operations [Beckman et al.]. Quantum Fourier Transform: DFT n needs n(n ) two-qubit gates and n one-qubit gates. An upper bound on the resources for k-bit number N About 400k 3 operations are needed (can be improved to O(k log k log log k))). The space needed is 5k + qubits. Example: Factoring 8-bit and 04-bit numbers For 8-bit we need operations and 64 qubits. For 04-bit number operations and 5 qubits.

196 HSP: History of the Problem D. Simon, 994 Hidden Subgroups in (Z ) n. P. Shor, 994 Factoring Discrete Logarithm } Hidden Subgroups in Z M resp. Z M Z M Kitaev 95, Brassard & Høyer 97, Mosca & Ekert 98 Generalization to arbitrary abelian groups. Open Problems Can the Hidden Subgroups Problem be solved for Any non-abelian group? (yes!) All non-abelian groups? (don t know) Interesting non-abelian groups? (progress, but still open)

197 HSP: History of the Problem D. Simon, 994 Hidden Subgroups in (Z ) n. P. Shor, 994 Factoring Discrete Logarithm } Hidden Subgroups in Z M resp. Z M Z M Kitaev 95, Brassard & Høyer 97, Mosca & Ekert 98 Generalization to arbitrary abelian groups. Open Problems Can the Hidden Subgroups Problem be solved for Any non-abelian group? (yes!) All non-abelian groups? (don t know) Interesting non-abelian groups? (progress, but still open)

198 HSP: History of the Problem D. Simon, 994 Hidden Subgroups in (Z ) n. P. Shor, 994 Factoring Discrete Logarithm } Hidden Subgroups in Z M resp. Z M Z M Kitaev 95, Brassard & Høyer 97, Mosca & Ekert 98 Generalization to arbitrary abelian groups. Open Problems Can the Hidden Subgroups Problem be solved for Any non-abelian group? (yes!) All non-abelian groups? (don t know) Interesting non-abelian groups? (progress, but still open)

199 HSP: History of the Problem D. Simon, 994 Hidden Subgroups in (Z ) n. P. Shor, 994 Factoring Discrete Logarithm } Hidden Subgroups in Z M resp. Z M Z M Kitaev 95, Brassard & Høyer 97, Mosca & Ekert 98 Generalization to arbitrary abelian groups. Open Problems Can the Hidden Subgroups Problem be solved for Any non-abelian group? (yes!) All non-abelian groups? (don t know) Interesting non-abelian groups? (progress, but still open)

200 HSP: History of the Problem D. Simon, 994 Hidden Subgroups in (Z ) n. P. Shor, 994 Factoring Discrete Logarithm } Hidden Subgroups in Z M resp. Z M Z M Kitaev 95, Brassard & Høyer 97, Mosca & Ekert 98 Generalization to arbitrary abelian groups. Open Problems Can the Hidden Subgroups Problem be solved for Any non-abelian group? (yes!) All non-abelian groups? (don t know) Interesting non-abelian groups? (progress, but still open)

201 HSP: History of the Problem D. Simon, 994 Hidden Subgroups in (Z ) n. P. Shor, 994 Factoring Discrete Logarithm } Hidden Subgroups in Z M resp. Z M Z M Kitaev 95, Brassard & Høyer 97, Mosca & Ekert 98 Generalization to arbitrary abelian groups. Open Problems Can the Hidden Subgroups Problem be solved for Any non-abelian group? (yes!) All non-abelian groups? (don t know) Interesting non-abelian groups? (progress, but still open)

202 HSP: History of the Problem D. Simon, 994 Hidden Subgroups in (Z ) n. P. Shor, 994 Factoring Discrete Logarithm } Hidden Subgroups in Z M resp. Z M Z M Kitaev 95, Brassard & Høyer 97, Mosca & Ekert 98 Generalization to arbitrary abelian groups. Open Problems Can the Hidden Subgroups Problem be solved for Any non-abelian group? (yes!) All non-abelian groups? (don t know) Interesting non-abelian groups? (progress, but still open)

203 The Hidden Subgroup Problem (HSP) Definition of the problem Given: Finite group G, finite set S, map f : G S Promise: There exists H G where f constant on G/H, g H g H implies f (g ) f (g ). Problem: Find generators for H. Note This is a natural generalization of Simon s problem. There G = F n and in addition we know H = s =. In fact, the HSP for any subgroup H IF n can be solved efficiently. Mathematically G f S π i G/H

204 The Hidden Subgroup Problem (HSP) Definition of the problem Given: Finite group G, finite set S, map f : G S Promise: There exists H G where f constant on G/H, g H g H implies f (g ) f (g ). Problem: Find generators for H. Note This is a natural generalization of Simon s problem. There G = F n and in addition we know H = s =. In fact, the HSP for any subgroup H IF n can be solved efficiently. Mathematically G f S π i G/H

205 HSP: Separation of Pre-images Visualization of the cosets of G f H f g H. g H n f G R

206 Examples for Hidden Subgroup Problems Simon s Problem Find hidden subgroup in G = Z n of a black-box function f : G {0, } m, where m n and x y + H f (x) = f (y). Factoring Find hidden subgroup in G = Z M with respect to the function Discrete logarithm problem f (x) := a x mod N. Find hidden subgroup in G = Z M Z M with respect to the function f (x, y) := a x b y mod p. The graph isomorphism problem Can be reduced to the problem of finding certain hidden subgroups of order in the non-abelian group G = S n S.

207 Examples for Hidden Subgroup Problems Simon s Problem Find hidden subgroup in G = Z n of a black-box function f : G {0, } m, where m n and x y + H f (x) = f (y). Factoring Find hidden subgroup in G = Z M with respect to the function Discrete logarithm problem f (x) := a x mod N. Find hidden subgroup in G = Z M Z M with respect to the function f (x, y) := a x b y mod p. The graph isomorphism problem Can be reduced to the problem of finding certain hidden subgroups of order in the non-abelian group G = S n S.

208 Examples for Hidden Subgroup Problems Simon s Problem Find hidden subgroup in G = Z n of a black-box function f : G {0, } m, where m n and x y + H f (x) = f (y). Factoring Find hidden subgroup in G = Z M with respect to the function Discrete logarithm problem f (x) := a x mod N. Find hidden subgroup in G = Z M Z M with respect to the function f (x, y) := a x b y mod p. The graph isomorphism problem Can be reduced to the problem of finding certain hidden subgroups of order in the non-abelian group G = S n S.

209 Examples for Hidden Subgroup Problems Simon s Problem Find hidden subgroup in G = Z n of a black-box function f : G {0, } m, where m n and x y + H f (x) = f (y). Factoring Find hidden subgroup in G = Z M with respect to the function Discrete logarithm problem f (x) := a x mod N. Find hidden subgroup in G = Z M Z M with respect to the function f (x, y) := a x b y mod p. The graph isomorphism problem Can be reduced to the problem of finding certain hidden subgroups of order in the non-abelian group G = S n S.

210 Duality of the Fourier Transform Basic identity DFT A ( U x U+c Geometric interpretation ) x = U y U ϕ c,y y Amplitude Shifted Line DFT Phase

211 Duality of the Fourier Transform Basic identity DFT A ( U x U+c Geometric interpretation ) x = U y U ϕ c,y y Amplitude Shifted Line DFT Phase