Praktek Algoritma Deutsch-Jozsa dengan Matlab / Octave

Download pdf dari materi tutorial praktek Algoritma Deutsch-Jozsa dengan Matlab / Octave = Deutsch-JozsaAlgorithm

Deutsch-Jozsa Algorithm:
Deutsch, D., & Jozsa, R. (1992). Rapid solution of problems by quantum computation. Proceedings of the Royal Society of London. Series A: Mathematical and Physical Sciences, 439(1907), 553-558.
https://royalsocietypublishing.org/doi/10.1098/rspa.1992.0167

Deutsch-Jozsa Algorithm: n inputs –> balanced or constant function?
Dalam komputer tradisional, skenario terburuk membutuhkan evaluasi fungsi 2^{n − 1} + 1.
Bisakah kita melakukan yang lebih baik menggunakan Algoritma Quantum?

Kode Program Matlab/Octave:

qubit_0 = [1; 0]
qubit_1 = [0; 1]
qubits_00_0 = kron( kron(qubit_0, qubit_0) , qubit_0)
qubits_01_0 = kron( kron(qubit_0, qubit_1) , qubit_0)
qubits_10_0 = kron( kron(qubit_1, qubit_0) , qubit_0)
qubits_11_0 = kron( kron(qubit_1, qubit_1) , qubit_0)
Ufunction_gate = [0 1 0 0 0 0 0 0; 1 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0; 0 0 1 0 0 0 0 0; 0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 1]
input00_output1 = Ufunction_gate * qubits_00_0
input01_output1 = Ufunction_gate * qubits_01_0
input10_output0 = Ufunction_gate * qubits_10_0
input11_output0 = Ufunction_gate * qubits_11_0

Deutsch-Jozsa: kasus fungsi seimbang dengan 2 input.
Jumlah input teratas = 2 qubit.

Kode Program Matlab/Octave:

qubit_0 = [1; 0]
qubit_1 = [0; 1]
Nx = 2
input_x = [1]
for k=1:Nx
  input_x = kron(input_x, qubit_0);
end
input_xy = kron(input_x, qubit_1);
checkpoint_0 = input_xy
Hadamard_gate = [ 1/sqrt(2) 1/sqrt(2); 1/sqrt(2) -1/sqrt(2) ]
Hx = [1]
for k=1:Nx
  Hx = kron(Hx, Hadamard_gate);
end
Hxy = kron(Hx, Hadamard_gate);
checkpoint_1 = Hxy * checkpoint_0

Kode Program Matlab/Octave:

qubit_0 = [1; 0]
qubit_1 = [0; 1]
Nx = 2

input_x = [1]
for k=1:Nx
  input_x = kron(input_x, qubit_0);
end
input_xy = kron(input_x, qubit_1);
checkpoint_0 = input_xy

Hadamard_gate = [ 1/sqrt(2) 1/sqrt(2); 1/sqrt(2) -1/sqrt(2) ]
Hx = [1]
for k=1:Nx
  Hx = kron(Hx, Hadamard_gate);
end
Hxy = kron(Hx, Hadamard_gate);
checkpoint_1 = Hxy * checkpoint_0

Ufunction_gate = [0 1 0 0 0 0 0 0; 1 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0; 0 0 1 0 0 0 0 0; 0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 1]

checkpoint_2 = Ufunction_gate * checkpoint_1

Kode Program Matlab/Octave:

qubit_0 = [1; 0]
qubit_1 = [0; 1]
Nx = 2

input_x = [1]
for k=1:Nx
  input_x = kron(input_x, qubit_0);
end
input_xy = kron(input_x, qubit_1);
checkpoint_0 = input_xy

Hadamard_gate = [ 1/sqrt(2) 1/sqrt(2); 1/sqrt(2) -1/sqrt(2) ]
Hx = [1]
for k=1:Nx
  Hx = kron(Hx, Hadamard_gate);
end
Hxy = kron(Hx, Hadamard_gate);
checkpoint_1 = Hxy * checkpoint_0

Ufunction_gate = [0 1 0 0 0 0 0 0; 1 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0; 0 0 1 0 0 0 0 0; 0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 1]
checkpoint_2 = Ufunction_gate * checkpoint_1
Identity_gate = eye(2)
HxIy = kron(Hx, Identity_gate);
checkpoint_3 = HxIy * checkpoint_2

Deutsch-Jozsa Algorithm: –> Percepatan eksponensial!
Kita telah memecahkan masalah dengan cukup SATU KALI evaluasi fungsi, yang bertentangan dengan evaluasi fungsi yang diperlukan dalam perhitungan klasik.
Kita telah mendemonstrasikan algoritma, menggunakan case Balanced Function A.
Sekarang, kita cukup mengukur qubit teratas.
Jika itu dalam keadaan | 0>, maka kita tahu bahwa f adalah fungsi konstan, jika tidak maka itu adalah fungsi seimbang.

Exponential Speedup: 

Kode Program Matlab / Octave:

qubit_0 = [1; 0]
qubit_1 = [0; 1]
Identity_gate = eye(2)
Hadamard_gate = [ 1/sqrt(2) 1/sqrt(2); 1/sqrt(2) -1/sqrt(2) ]

balance_A_function = [0 1 0 0 0 0 0 0; 1 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0; 0 0 1 0 0 0 0 0; 0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 1]
balance_B_function = [1 0 0 0 0 0 0 0; 0 1 0 0 0 0 0 0; 0 0 1 0 0 0 0 0; 0 0 0 1 0 0 0 0; 0 0 0 0 0 1 0 0; 0 0 0 0 1 0 0 0; 0 0 0 0 0 0 0 1; 0 0 0 0 0 0 1 0]
constant_D_function = [0 1 0 0 0 0 0 0; 1 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0; 0 0 1 0 0 0 0 0; 0 0 0 0 0 1 0 0; 0 0 0 0 1 0 0 0; 0 0 0 0 0 0 0 1; 0 0 0 0 0 0 1 0]

% Change this as you like!
Nx = 2
SecretFunction_gate = balance_A_function

input_x = [1]
for k=1:Nx
  input_x = kron(input_x, qubit_0);
end
input_xy = kron(input_x, qubit_1);

Hx = [1]
for k=1:Nx
  Hx = kron(Hx, Hadamard_gate);
end

Hxy = kron(Hx, Hadamard_gate);

DeutschJozsaAlgorithm = kron(Hx,Identity_gate) * SecretFunction_gate * Hxy

result = DeutschJozsaAlgorithm * input_xy

Untuk proof pembuktian matematikanya, langsung lihat di buku sumber = Yanofsky, N. S., & Mannucci, M. A. (2008). Quantum computing for computer scientists. Cambridge University Press.
http://www.sci.brooklyn.cuny.edu/~noson/qctext.html

Kode program Algoritma Deutsch-Jozsa Matlab/Octave di Github =

https://github.com/keamanansiber/qiskit/blob/master/Matlab_Octave/Deutsch_Jozsa_Algorithm_Matlab.m