Enigma 004 : The Monty Hall Exercise

Enigma 004 : The Monty Hall Exercise#

The diamond you won has disappeared. Your search leads you to Monty Hall Manor, where Kettu has hidden your diamond in one of the three safes in front of you. If you choose the right safe, you can retrieve your diamond, otherwise Kettu will keep it forever. Can you determine the best approach to win to solving this enigma?

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

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 - Code writing#

×

Write a circuit that would only use 3 qubits instead of 4 (and still assuming that you initially chose safe number 2) for Enigma 004 - The Monty Hall Exercise.

Use only two qubits to represent the three safes.

Find a way to obtain ¹⁄₃ probability of measuring 00, 01, and 10 (the binary equivalent of 0, 1, and 2).

problem1_qc = QuantumCircuit(3)

### Start your work here ###

problem1_qc.draw("mpl")

Exercise 2 - Code writing#

Write a circuit (using three qubits to hide the diamond exactly like in the enigma) that would randomly determine the chest you choose at the start, and also determine which safe will be opened.

You can use the following circuit that is the equivalent of a multicontrolled Hadamard gate:

problem2_qc.ry(np.pi/4, 2)
problem2_qc.mcx([0, 1], 2)
problem2_qc.ry(-np.pi/4, 2)

You can use q₃, q₄, and q₅ to determine the safe you put your hand on at first and q₆, q₇, and q₈ to determine which safe will be opened (safe 0 is linked to q₀, q₃, and q₆; safe 1 is linked to q₁, q₄, and q₇; safe 2 is linked to q₂, q₅, and q₈).

The circuit to randomly choose the safe you put your hand on at the start is the same as the one used to hide the diamond.

For the circuit to determine which safe will be opened, start with the three cases where the diamond and your hand are on the same safe.

problem2_qc = QuantumCircuit(9)
prob_2on3 = 2 * np.arcsin(np.sqrt(2/3))

### Start your work here ###

problem2_qc.draw("mpl")

Exercise 3 - Code writing#

Time travel

One very important aspect of quantim computing is that all quantum logic gates have an inverse. This means that it is possible to simulate time traveling by going to the end of an algorithm and coming back at the start simply using the inverse of every gate in a backward manner.

The following circuit shows the algorithm seen in the video with an extra qubit used for deciding which door will be opened in the case the diamond is in safe 2 at the beginning (and assuming that you initially chose safe 2). The circuit has been written up to the point in time a safe has been opened by Kettu.

qreg_q = QuantumRegister(5, 'q')
creg_c = ClassicalRegister(1, 'c')
creg_d = ClassicalRegister(1, 'd')
creg_f = ClassicalRegister(1, 'f')
creg_g = ClassicalRegister(1, 'g')
problem3_qc = QuantumCircuit(qreg_q, creg_c, creg_d, creg_f, creg_g)
prob_2on3 = 2 * np.arcsin(np.sqrt(2/3))

problem3_qc.ry(prob_2on3, 0)
""""
## q4 is used to decide which door will be opened in case
   the diamond is in safe 2 at the beginning.
## This is necessary since we don't want this information to be lost
   when measuring q3 again after going back in time.
"""
problem3_qc.h(4)
problem3_qc.ch(0, 1)
problem3_qc.cx(1, 2)
problem3_qc.cx(0, 1)
problem3_qc.x(0)
problem3_qc.barrier()
problem3_qc.cx(0, 3)
problem3_qc.mcx([2, 4], 3)
problem3_qc.measure(3, 0)

problem3_qc.draw("mpl")

Write the rest of the algorithm to travel in time going back to the beginning, choosing a strategy that will allow you to proceed with the rest of the algorithm and win the diamond everytime.

Place the gates in reverse order upto the barrier and choose a safe the diamond is not in.

Use conditional swap to make sure you choose a safe the diamond is not in. For example, here is how you would apply a NOT gate on q₀ with the condition that the classical register g has the value 1: problem3_qc.x(0).c_if(creg_g, 1)

qreg_q = QuantumRegister(5, 'q')
creg_c = ClassicalRegister(1, 'c')
creg_d = ClassicalRegister(1, 'd')
creg_f = ClassicalRegister(1, 'f')
creg_g = ClassicalRegister(1, 'g')
problem3_qc = QuantumCircuit(qreg_q, creg_c, creg_d, creg_f, creg_g)

prob_2on3 = 2 * np.arcsin(np.sqrt(2/3))
problem3_qc.ry(prob_2on3, 0)
problem3_qc.h(4)
problem3_qc.ch(0, 1)
problem3_qc.cx(1, 2)
problem3_qc.cx(0, 1)
problem3_qc.x(0)
problem3_qc.barrier()
problem3_qc.cx(0, 3)
problem3_qc.mcx([2, 4], 3)
problem3_qc.measure(3, creg_g[0])


### Start your work here ###

problem3_qc.barrier()
problem3_qc.measure(0, creg_c[0])
problem3_qc.measure(1, creg_d[0])
problem3_qc.measure(2, creg_f[0])
problem3_qc.measure(3, creg_g[0])

problem3_qc.draw("mpl")

Exercise 4 - Quick quiz#

Let’s run the time travel circuit on a simulator to see the results. Run the cell below.

# Time travel circuit
qreg_q = QuantumRegister(5, 'q')
creg_c = ClassicalRegister(1, 'c')
creg_d = ClassicalRegister(1, 'd')
creg_f = ClassicalRegister(1, 'f')
creg_g = ClassicalRegister(1, 'g')
problem3_qc = QuantumCircuit(qreg_q, creg_c, creg_d, creg_f, creg_g)

prob_2on3 = 2 * np.arcsin(np.sqrt(2/3))
problem3_qc.ry(prob_2on3, 0)
problem3_qc.h(4)
problem3_qc.ch(0, 1)
problem3_qc.cx(1, 2)
problem3_qc.cx(0, 1)
problem3_qc.x(0)
problem3_qc.barrier()
problem3_qc.cx(0, 3)
problem3_qc.mcx([2, 4], 3)
problem3_qc.measure(3, creg_g[0])

problem3_qc.mcx([2, 4], 3)
problem3_qc.cx(0, 3)
problem3_qc.barrier()
problem3_qc.swap(0, 2).c_if(creg_g, 0)
problem3_qc.swap(1, 2).c_if(creg_g, 1)
problem3_qc.cx(0, 3)
problem3_qc.mcx([2, 4], 3)

problem3_qc.barrier()
problem3_qc.measure(0, creg_c[0])
problem3_qc.measure(1, creg_d[0])
problem3_qc.measure(2, creg_f[0])
problem3_qc.measure(3, creg_g[0])

# Executing the circuit on a simulator
simulator = AerSimulator()
result = simulator.run(transpile(problem3_qc, simulator), shots=1000).result()
counts = result.get_counts(problem3_qc)
plot_histogram(counts)

What is the meaning of the result?