The Lazy Colleague | PennyLane Challenges

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Challenge statement

This challenge is included in the QHack 2023 Flashback Badge Challenge event.

The project Zenda's team is working on can only be completed if at least three people in the team are working. Let's model this situation in a circuit:

drawing

In this diagram, the qubit $e_i$ refers to the $i$ -th employee, which will take the value 1 if this employee is chosen to work on the project. The output state labelled result will take the value 0 if the project was not completed and 1 if it was. Let us imagine that employee $e_1$ is the one who does not work. Then, if we apply the operator to the state $|1\rangle|1\rangle|0\rangle|1\rangle|0\rangle$ (that is, we select $e_1$ , $e_2$ and $e_4$ for the project), the output will be $|1\rangle|1\rangle|0\rangle|1\rangle|0\rangle$ . As we can see, the last qubit is still $\lvert 0\rangle$ , i.e. the project has not been carried out. This is because there are only two employees that actually work on the project, and a minimum of three is required.

Zenda wants to know who the lazy worker is, executing as few projects as possible. For this reason, she asks you to help her with your quantum skills. You are asked to discover who the lazy employee is, using a single shot and a single call to the "Project execution" operator.

Challenge code

On one hand, you are asked to complete circuit (you only need to apply gates). You can only call the project_execution operator once, which is already incorporated in the template. On the other hand, you must complete process_output, which will take the output of your circuit and will return who the lazy guy is.

The project_execution function will be generated when testing the solution; if you want to experiment with the function output in the notebook, you can temporarily replace project_execution with an operator of the form qml.MultiControlledX(wires=['e1', 'e2', 'e4', 'result']). In this case, the absence of "e3" on the wires means that in this project, "e3" will be the lazy employee. Just remember to switch it back to project_execution before submitting, so that your function uses the correct project_execution during testing!

Output

To judge this challenge, we will arbitrarily generate 5000 different projects (project_execution), which we will send one by one to the circuit to check that your prediction is correct ("e1", "e2", "e3" or "e4"). Therefore, in this case, there will be no public and private test cases. Good luck!