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Enigma 003 : The Four-Square Chessboard

Alice and Bob continue their journey in space. They are challenged by Aïka, who’s hidden a key under one of the four chessboard squares. A coin is then randomly placed on each of the four squares. How will Alice communicate to Bob the exact position of the key by flipping only one coin? You will apply a new concept, the control by the 0-state.

Make sure to watch the following video before getting started with this exercise:

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Note

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Run the cell below to install the necessary packages.

import sys
!{sys.executable} -m pip install qiskit==1.1.1
!{sys.executable} -m pip install qiskit_aer==0.14.2
!{sys.executable} -m pip install pylatexenc==2.10

# Import necessary modules
import numpy as np
from qiskit import QuantumCircuit

Exercise 1 - Quick quiz

×

The enigma uses a modulo 2 addition like this one:

1 0 + 0 1 = 1 1

Such addition has the interesting characteristic that the numbers can be interchanged in any order like this

1 1 + 0 1 = 1 0

Or

1 1 + 1 0 = 0 1

Meaning that adding any two of them gives the third one as an answer (this is true for any numbers). Playing with modulo 2 additions also has other interesting characteristics. In the enigma, adding the first number to the second is done by applying a CNOT between q₄ and q₆ (and q₅ and q₇). Here is the code of the algorithm in the enigma.

#qubits 0 to 3 are the 4 squares
#qubits 4 and 5 is where the key is hidden
#qubits 6 and 7 is where the focus first lands
#qubits 8 and 9 is where the focus lands at the end which is the key location
problem1_qc = QuantumCircuit(10)

#coin distribution on each square
for i in range(4):
    problem1_qc.h(i)

problem1_qc.barrier([4,5])

#hiding the key under one of the 4 squares
problem1_qc.h(4)
problem1_qc.h(5)

problem1_qc.barrier([4,5,6,7])

#finding the parity of 1's on squares for which binary numbers finish by 1 and putting the answer on q5
problem1_qc.cx(1, 6)
problem1_qc.cx(3, 6)
problem1_qc.barrier([6,7])

#finding the parity of 1's on squares for which binary numbers have a 1 as second to last digit and putting the answer on q6
problem1_qc.cx(2, 7)
problem1_qc.cx(3, 7)
problem1_qc.barrier([6,7])

#adding modulo 2 the position of the key and the position of the focus
problem1_qc.cx(4, 6)
problem1_qc.cx(5, 7)
problem1_qc.barrier([6,7])

#turning the right coin
problem1_qc.ccx(7,6,3)
problem1_qc.barrier([6,7])
problem1_qc.x(6)
problem1_qc.ccx(7,6,2)
problem1_qc.x(6)
problem1_qc.barrier([6,7])
problem1_qc.x(7)
problem1_qc.ccx(7,6,1)
problem1_qc.x(7)
problem1_qc.barrier([6,7])
problem1_qc.x(6)
problem1_qc.x(7)
problem1_qc.ccx(7,6,0)
problem1_qc.x(7)
problem1_qc.x(6)
problem1_qc.barrier([6,7,8,9])

#finding the parity of 1's on squares for which binary numbers finish by 1 and putting the answer on q8
problem1_qc.cx(1, 8)
problem1_qc.cx(3, 8)
problem1_qc.barrier([6,7,8,9])

#finding the parity of 1's on squares for which binary numbers have a 1 as second to last digit and putting the answer on q9
problem1_qc.cx(2, 9)
problem1_qc.cx(3, 9)

problem1_qc.draw(output='mpl')

What is the meaning of the values of q₆ and q₇ after all the gates in the circuit have been applied?

Submit Answer

Exercise 2 - Code writing

Can you write the circuit for a 4 by 4 square chess set until you calculate the position of the piece to turn?

Click to reveal HINT 1

Start by drawing a 4 by 4 chess board and number each square from 0 to 15 in decimal and binary numbers starting with the top row.

Click to reveal HINT 2

The trick is now to add (modulo 2) all the squares that end with a 1 and to proceed the same way with all squares that have a 1 on their second bit counting from right to left and so on using four extra squares.

Click to reveal HINT 3

In the following picture, all squares with green numbers must be added modulo 2 and the answer must be placed in the last qubit (q₂₀) of the focus. The same must take place for blue squares with their result on q₂₁, the yellow squares on q₂₂, and the red squares on q₂₃. The zeros and ones on the squares are only there as an example.

nb_coins = 16
nb_key = 4
nb_focus_first = 4
nb_focus_key = 4
nb_qubits = nb_coins + nb_key + nb_focus_first + nb_focus_key

#qubits 0 to 15 are the 16 squares
#qubits 16 to 19 is where the key is hidden
#qubits 20 to 23 is where the focus first lands
#qubits 24 to 27 is where the focus lands at the end which is the key location
problem2_qc = QuantumCircuit(nb_qubits)

# The code for this problem is similar to the one in Exercise 1, simply longer.
### Start your work here ###


problem2_qc.draw(output='mpl')

Click to reveal the answer

nb_coins = 16
nb_key = 4
nb_focus_first = 4
nb_focus_key = 4
nb_qubits = nb_coins + nb_key + nb_focus_first + nb_focus_key

#qubits 0 to 15 are the 16 squares
#qubits 16 to 19 is where the key is hidden
#qubits 20 to 23 is where the focus first lands
#qubits 24 to 27 is where the focus lands at the end which is the key location
problem2_qc = QuantumCircuit(nb_qubits)

#coin distribution on each square
for i in range(16):
    problem2_qc.h(i)

problem2_qc.barrier()

#hiding the key under one of the 16 squares
for i in range(16, 20):
    problem2_qc.h(i)

problem2_qc.barrier()

#finding the parity of 1's on squares for which binary numbers finish by 1 and putting the answer on q20
problem2_qc.cx(1, 20)
problem2_qc.cx(3, 20)
problem2_qc.cx(5, 20)
problem2_qc.cx(7, 20)
problem2_qc.cx(9, 20)
problem2_qc.cx(11, 20)
problem2_qc.cx(13, 20)
problem2_qc.cx(15, 20)
problem2_qc.barrier()

#finding the parity of 1's on squares for which binary numbers have a 1 as second to last digit and putting the answer on q21
problem2_qc.cx(2, 21)
problem2_qc.cx(3, 21)
problem2_qc.cx(6, 21)
problem2_qc.cx(7, 21)
problem2_qc.cx(10, 21)
problem2_qc.cx(11, 21)
problem2_qc.cx(14, 21)
problem2_qc.cx(15, 21)
problem2_qc.barrier()

#finding the parity of 1's on squares for which binary numbers have a 1 as their 3rd digit from the right and putting the answer on q22
problem2_qc.cx(4, 22)
problem2_qc.cx(5, 22)
problem2_qc.cx(6, 22)
problem2_qc.cx(7, 22)
problem2_qc.cx(12, 22)
problem2_qc.cx(13, 22)
problem2_qc.cx(14, 22)
problem2_qc.cx(15, 22)
problem2_qc.barrier()

#finding the parity of 1's on squares for which binary numbers have a 1 as their 4th digit from the right and putting the answer on q23
problem2_qc.cx(8, 23)
problem2_qc.cx(9, 23)
problem2_qc.cx(10, 23)
problem2_qc.cx(11, 23)
problem2_qc.cx(12, 23)
problem2_qc.cx(13, 23)
problem2_qc.cx(14, 23)
problem2_qc.cx(15, 23)
problem2_qc.barrier()

#adding modulo 2 the position of the key and the position of the focus
problem2_qc.cx(16, 20)
problem2_qc.cx(17, 21)
problem2_qc.cx(18, 22)
problem2_qc.cx(19, 23)

problem2_qc.draw(output='mpl')

Exercise 3 - Code writing

Complete the circuit to allow Alice to turn the right coin knowing that the MCX gate is the multi-control X gate.

To avoid using a lot of X gates to control the 0 state, we can specify the control state of each control qubit. Here is the way to use the MCX gate if we wanted to have the control-qubits 20, 21, 22, and 23 control the state 1, 1, 1, and 0 respectively, and qubit 14 as the target-qubit:

problem3_qc.mcx([20, 21, 22, 23], 14, ctrl_state='0111').

ctrl_qubits = [20, 21, 22, 23]
problem3_qc = problem2_qc.copy()
problem3_qc.barrier()

# turning the right coin on the bottom row
problem3_qc.mcx(ctrl_qubits, 15, ctrl_state='1111')

problem3_qc.mcx(ctrl_qubits, 14, ctrl_state='0111')

### Continue writing the code here ###


problem3_qc.draw(output='mpl')

Click to reveal the answer

ctrl_qubits = [20, 21, 22, 23]
problem3_qc = problem2_qc.copy()
problem3_qc.barrier()

#turning the right coin on the bottom row
problem3_qc.mcx(ctrl_qubits, 15, ctrl_state='1111')

problem3_qc.mcx(ctrl_qubits, 14, ctrl_state='0111')

problem3_qc.mcx(ctrl_qubits, 13, ctrl_state='1011')

problem3_qc.mcx(ctrl_qubits, 12, ctrl_state='0011')

#turning the right coin on the 3rd row
problem3_qc.mcx(ctrl_qubits, 11, ctrl_state='1101')

problem3_qc.mcx(ctrl_qubits, 10, ctrl_state='0101')

problem3_qc.mcx(ctrl_qubits, 9, ctrl_state='1001')

problem3_qc.mcx(ctrl_qubits, 8, ctrl_state='0001')

#turning the right coin on the 2nd row
problem3_qc.mcx(ctrl_qubits, 7, ctrl_state='1110')

problem3_qc.mcx(ctrl_qubits, 6, ctrl_state='0110')

problem3_qc.mcx(ctrl_qubits, 5, ctrl_state='1010')

problem3_qc.mcx(ctrl_qubits, 4, ctrl_state='0010')

#turning the right coin on the top row
problem3_qc.mcx(ctrl_qubits, 3, ctrl_state='1100')

problem3_qc.mcx(ctrl_qubits, 2, ctrl_state='0100')

problem3_qc.mcx(ctrl_qubits, 1, ctrl_state='1000')

problem3_qc.mcx(ctrl_qubits, 0, ctrl_state='0000')

#now that the right coin has been turned, it is time to put the focus on the square where the key is located
problem3_qc.barrier()

#finding the parity of 1's on squares for which binary numbers finish by 1 and putting the answer on q24
problem3_qc.cx(1, 24)
problem3_qc.cx(3, 24)
problem3_qc.cx(5, 24)
problem3_qc.cx(7, 24)
problem3_qc.cx(9, 24)
problem3_qc.cx(11, 24)
problem3_qc.cx(13, 24)
problem3_qc.cx(15, 24)
problem3_qc.barrier()

#finding the parity of 1's on squares for which binary numbers have a 1 as second to last digit and putting the answer on q25
problem3_qc.cx(2, 25)
problem3_qc.cx(3, 25)
problem3_qc.cx(6, 25)
problem3_qc.cx(7, 25)
problem3_qc.cx(10, 25)
problem3_qc.cx(11, 25)
problem3_qc.cx(14, 25)
problem3_qc.cx(15, 25)
problem3_qc.barrier()

#finding the parity of 1's on squares for which binary numbers have a 1 as their 3rd digit from the right and putting the answer on q26
problem3_qc.cx(4, 26)
problem3_qc.cx(5, 26)
problem3_qc.cx(6, 26)
problem3_qc.cx(7, 26)
problem3_qc.cx(12, 26)
problem3_qc.cx(13, 26)
problem3_qc.cx(14, 26)
problem3_qc.cx(15, 26)
problem3_qc.barrier()

#finding the parity of 1's on squares for which binary numbers have a 1 as their 4th digit from the right and putting the answer on q27
problem3_qc.cx(8, 27)
problem3_qc.cx(9, 27)
problem3_qc.cx(10, 27)
problem3_qc.cx(11, 27)
problem3_qc.cx(12, 27)
problem3_qc.cx(13, 27)
problem3_qc.cx(14, 27)
problem3_qc.cx(15, 27)

problem3_qc.draw(output='mpl')