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Praktek Algoritma Deutsch-Jozsa Cybersecurity dengan Matlab / Octave

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<sup>n ? 1</sup> + 1.
Bisakah kita melakukan yang lebih baik menggunakan Algoritma Quantum?


Kode Program Matlab/Octave:

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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:

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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:

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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:

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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:

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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

https://keamanansiber.id/blog/2020/07/09/praktek-algoritma-deutsch-jozsa-dengan-matlab-octave/

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