Announcement: Microsoft Q# Coding Contest – Winter 2019
By Nickolas, 6 years ago, In English

Microsoft’s Quantum team is excited to announce the Q# Coding Contest – Winter 2019! In this contest you can put your quantum programming skills to the test, solving quantum computing tasks in Q#. Winners will receive a Microsoft Quantum T-shirt!

Quantum computing is a radically different computing paradigm compared to classical computing. Indeed, it is so different that some tasks that are believed to be classically intractable (such as factoring integers or simulating physical systems) can be performed efficiently on a quantum computer. In 2017 Microsoft introduced the Quantum Development Kit which includes the Q# programming language. Q# can be used with Visual Studio, Visual Studio Code or the command line, on Windows, macOS, and Linux.

In summer of 2018 we hosted the first quantum programming contest, which included problems on introductory topics in quantum computing: superposition, measurement, quantum oracles and simple algorithms. This contest will offer harder problems on some of these topics as well as introduce some new topics.

The contest will run from March 1 to March 4. The rules of the contest are:

  • The contest will have 12 tasks of various complexity levels.
  • To solve each task, you will write Q# code to implement the described transformation on the given set of qubits or to perform a more challenging task. Solutions are accepted in Q# only.
  • The solution is correct if it passes all tests from a predefined test set. You will know whether the solution is correct soon after submitting it.
  • Participants are ranked according to the number of correctly solved tasks.
  • Ties are resolved based on lowest penalty time for all tasks, which is computed as the latest submission time (the time since the start of the contest) for any of the correctly solved tasks. There is no penalty for failed submissions.
  • The top 50 ranked participants will receive a Microsoft Quantum T-shirt.
  • NO PURCHASE NECESSARY. Must be 16 years of age or older. Game ends 3/4/19. For details, see Official Rules.

We will offer a warmup round the weekend before the contest, from February 22 to February 25. Participation in the warmup round is entirely optional. The warmup round includes simpler tasks on the topics covered in the main contest and gives you an opportunity to get familiar with the contest environment and submission system beforehand, as well as refresh or learn the basics of quantum computing and the Q# programming language. During the warmup round everybody is encouraged to discuss the tasks and the solutions. Once the warmup round is over, we will publish the editorials explaining both the quantum computing logic behind the solution and the Q# implementation on the contest page.

Another great way to prepare for the contest is to solve some of the Quantum Katas. They offer problems on a variety of quantum programming topics, and they are very similar to the ones used in the contest. In fact, the participants of the summer Q# contest will recognize the contest problems in some of the kata tasks :-)

Good luck! We hope you enjoy the contest!

For first time Codeforces users:

  1. Create user account here.
  2. Register for the warmup round here.
  3. Register for the contest here.
  4. Once the warmup round starts on February 22, access the problems here.
  5. Once the contest starts on March 1, access the problems here.

Quantum computing and Q# materials:

Note that this contest will use Q# 0.4, while the previous contest used Q# 0.2. A lot of code written in 0.2 will still work in 0.4; for details on breaking changes and new features please see release notes.

Update. The warmup round is over; here are the explanations for the problems. See you in the main contest this weekend!

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Comments Comments (126)
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6 years ago, # |
  Vote: I like it +8 Vote: I do not like it

this is so cool. i love that you bring this new innovative stuff on codeforces. i have read a lot on quantum computing and it is definitely the future. moving bits around and all that stuff is awesome. now my only hope is that you won’t mix math and quantum sharp in this contest / this series of contests because Q sharp is very interesting and i really wouldn’t like it to be dirtied up with useless math. i don’t think i’m the only one.

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6 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Do you expect higher difficulty than on the first quantum contest, i.e. the T-shirts not being decided on time only?

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    6 years ago, # ^ |
      Vote: I like it +48 Vote: I do not like it

    We are definitely aiming to have more difficult problems in this contest! I asked one of our researchers to write several problems, and I still have to figure out how to solve them myself :-)

    We're also going to change scoring to be based on the last solved problem submission time, not on the sum of submission times, so the impact of the time the competitor joins the contest should be reduced.

    That being said, Codeforces community has surprised us once already, and I'm not willing to bet that you won't do it again :-)

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      6 years ago, # ^ |
        Vote: I like it -28 Vote: I do not like it

      yeah but math is useless and makes everything unnecessarily harder. so don’t bullshit us with it

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6 years ago, # |
  Vote: I like it +29 Vote: I do not like it

Can we expect the t-shirts to be different from the Summer round ones? ;)

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6 years ago, # |
← Rev. 2   Vote: I like it +12 Vote: I do not like it

Well, time to finish the syntax highlighting I started half a year ago. It's going to take a long time because I'm a lazy shit.

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I tried solving some of the questions of the previous contest today, but I was getting a lot of "denial of judgment" and "compilation error" errors, even when trying to use the official solution provided. Does anyone know what is the problem?

Specifically for problems B3 and C1.

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    6 years ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    The checkers of the previous contest have not been updated to the new version of Q# yet. I'm investigating this, but this might take a while. For B3 and B4, for example, the only change I had to do in the reference solution was switch from ";" to "," as a separator between array elements — in Polygon it just worked, but I'm not sure why this gives me "denial of judgement" on the website...

    If you want to practice on those problems, you can solve them in the Quantum Katas — they are migrated to 0.4, and as a bonus you get error messages on your machine, not via Codeforces interface.

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      6 years ago, # ^ |
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      I checked the Quantum Katas, and they are amazing, but I think I found a mistake, and I don't know how else to report it.

      In the 4th Joint Measurement task, it says the output should be "0 if qubits were in W state, 1 if they were in the second superposition."

      But the solution given has the reverse, returning 1 for W and 0 for the other.

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        6 years ago, # ^ |
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        Nice catch, thank you! I've opened a PR to fix this.

        (If you have a GitHub account, you can report possible issues with the katas directly in Issues)

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    6 years ago, # ^ |
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    The checkers have been updated now. The problems that were affected are B3-B4, C1-C2 and E1-E2.

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6 years ago, # |
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Is there a more extensive / detailed documentation for Q#? At this moment one can only guess what adjoint controlled distribute means at the end of an operation. I've seen it in the katas for Superposition. A little more explanation in the docs or somewhere about functors would be nice. Or is there a place and i didn't manage to find it?

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6 years ago, # |
  Vote: I like it +5 Vote: I do not like it

https://codeforces.net/contest/1001/submission/50058667

I wonder if the checker is updated since the error message is from the checker... Also, I tried a submission that has passed the test before and it got a CE verdict on the same problem.

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    6 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    Nice catch, indeed I missed the problem from warmup round which needed the same update. Should be fixed now.

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6 years ago, # |
  Vote: I like it +55 Vote: I do not like it

Is it allowed to stream solving the warm-up round on Youtube?

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is this contest rated?

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6 years ago, # |
← Rev. 2   Vote: I like it +36 Vote: I do not like it

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

So, the penalty is based on latest time rather than sum of times?

That's good to hear! Last time, if you were one hour late starting the contest, you would get 10 hours added to your penalty...

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    6 years ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    Yes. This was one of the big issues mentioned in the discussion of the first contest, so we switched to calculating time penalty this way.

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      6 years ago, # ^ |
      ← Rev. 4   Vote: I like it 0 Vote: I do not like it

      Looks like wrong tries are not counted in the total penalty. Also the number of wrong tries is not correctly calculated. It's -1 (one 'free' wrong try) for some reason. EDIT: Oh, I was wrong. It looks like you have free wrong tries if you fail test#1, but higher test fails are counted.

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        6 years ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        Yes, we've removed the penalty for failed submissions as well — I don't think any of the problems are easy to bruteforce, and racking up penalty for learning the language is just frustrating. We'll see if we need to put it back it for the main contest.

        I think forgiving failure on test 1 is a feature in all Codeforces contests, it's not special to this one.

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6 years ago, # |
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Am I the only one facing trouble to ResetAll qubits? " The called operation does not support the necessary functor(s) for the requested auto-generation of a specialization."

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    6 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    If you specified the operation to have an automatically generated adjoint (as the oracle tasks do with adjoint auto or adjoint invert clause), it can not use measurements (and ResetAll just measures the qubits and flips them if 1 was measured). Automatically generated adjoint reverses the order of operations and performs their adjoints, and you can not reverse measurement.

    You can either specify adjoint of an operation that uses measurements manually (if it is still a unitary transformation and has an adjoint — for example, implementing rotation using repeat-until-success loop as described here), or you need to "uncompute" the qubits you're going to release instead of measuring them (see this example).

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      6 years ago, # ^ |
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      Good to know this before the contest! I kept getting compilation error for ResetAll, as well, and I got really frustrated.

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        6 years ago, # ^ |
          Vote: I like it +15 Vote: I do not like it

        This is exactly why we're running a warmup round before the main event :-)

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6 years ago, # |
  Vote: I like it +16 Vote: I do not like it

This is how warmup round went :)
https://cv1.pikabu.ru/video/2019/02/07/1549525232275164028_640x640.webm

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6 years ago, # |
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For one of the problems, I used ApplyToEach(X,qs) to flip a bunch of qubits. The compiler didn't like it, because for some reason it doesn't play nice with "adjoint auto". What's the reason for that?

After I changed it to a "for" loop, the compiler error disappeared.

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    6 years ago, # ^ |
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    adjoint auto requires that each operation in the body has an adjoint. ApplyToEach does not have an adjoint even if the operation it applies to the qubits does have an adjoint. You can use ApplyToEachA if you need an adjointable version of that, it will work.

    This is an unfortunate quirk of the language which is going away in one of the future releases. For now a lot of the library functions have to exist in four versions — one with no functors (*), one with adjoint version (*A), one with controlled version (*C) and one with all (*CA).

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6 years ago, # |
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Is there any limit on the number of extra qubits that are used in the code? I am getting Memory Limit exceeded on Test 8 in both G1, G2. Can anyone help?

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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yes, the memory limit on the problem implicitly defines the limit on the total number of qubits that can be used. However, for tasks G1 and G2 you don't need to allocate any extra qubits! I'll publish the editorial for G1 shortly, meanwhile a hint: take a look at task 1.3 from the Grover search kata.

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      6 years ago, # ^ |
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      Thank you. I didn't know that there was a library function analogous to n-bit Toffoli. So, I was implementing n-bit Toffoli gate and it required extra qubits.

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        6 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        On a real quantum computer the gate controlled on multiple qubits would have to be decomposed into 1- and 2-qubit gates, so it would indeed require extra qubits — but you'll still be able to use a library and let it take care of the decomposition. These problems run on a simulator, and it can do controlled gates directly, without extra qubits.

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        6 years ago, # ^ |
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        I actually implement a 8-bit Toffoli, with only CNOT gate and CCNOT gate, and using 4 extra qubits. It passed the test.

        The official solution is definitely much easier, though.

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      6 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      I keep on getting "Invalid initializer expression. Possible initializers are "Qubit()", "Qubit[expr]", or tuples thereof" when I try to use some extra qubits with 'using(qs = Qubit[number])' which is the same declaration as I saw in one of the References. Please correct me. Thanks.

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        6 years ago, # ^ |
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        Your code has using (qs = Qubits[8]) (plural), and the type name is Qubit singular, so you have to use using (qs = Qubit[8]).

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6 years ago, # |
  Vote: I like it +21 Vote: I do not like it
A hint for problem G1
Full solution for problem G1
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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it
    Question about solution
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      6 years ago, # ^ |
        Vote: I like it +6 Vote: I do not like it
      You can not compare qubit states directly
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6 years ago, # |
  Vote: I like it +9 Vote: I do not like it
A hint for problem U1
Full solution for problem U1
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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Could you explain how to use the DumpUnitary tool?

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      6 years ago, # ^ |
        Vote: I like it +11 Vote: I do not like it

      1)Clone the quantum katas repo https://github.com/Microsoft/QuantumKatas.git 2)Go to path /QuantumKatas/utilities/DumpUnitary 3)There add your Solve function to the file Operations.qs 4)Change line number 50 to call your function Solve 5)now do $dotnet run in bash 6)After running a file named DumpUnitaryPattern.txt is formed. Verify your solution from it. 7) To change the number of qubits change N in Driver.cs

      Hope this helps you

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      6 years ago, # ^ |
      ← Rev. 3   Vote: I like it +12 Vote: I do not like it

      In Operations.qs file make a new operation that takes a Qubit array. Here I called it "YourCode" then modify the existing CallDumpUnitary so it passes your new function to DumpUnitaryToFiles like below. 

      // make a new operation to put your code in
operation YourCode (qs : Qubit[]) : Unit {
    // put your code here
		
}

// The operation which is called from C# code, change it so it looks like this
operation CallDumpUnitary (N : Int) : Unit {
     DumpUnitaryToFiles(N, YourCode);
}

      Then run it. Look in your output folder, there are 2 files, DumpUnitary.txt, and DumpUnitaryPattern.txt, that is your output, that is how your code changed the qubits that got passed into it. If you want to change N, the number of qubits, change the N value in the Driver.cs file's static Main method, run it and now look at your files, they are now a 2^N x 2^N matrix of whatever you changed N to.

      Then when you want to submit basically copy all the code inside YourCode and paste it in their solve template where it says "//your code here".

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6 years ago, # |
  Vote: I like it -9 Vote: I do not like it

Custom Invocation is not expected to be working, is it?

It looks like lack of Driver and Main leads to compilation error, whatever is sent as Source and Input.

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    6 years ago, # ^ |
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    It should be working, same as in the previous contest. Here is the signature of the code you should use to run it (note that the namespace and operation name have to match this code exactly, since this is what is called from C#)

    namespace Solution {
    open Microsoft.Quantum.Primitive;
    open Microsoft.Quantum.Canon;

    // ------------- Operation which is called from C# -------------------
    operation RunQsharp () : Bool
    {
        Message("Hi!");
        return true;
    }
}
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      6 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      This is what is returned to output if I copy this code to Source:

      Invocation failed [COMPILATION_ERROR]
Can't compile file:
Microsoft (R) Build Engine version 15.7.177.53362 for .NET Core
Copyright (C) Microsoft Corporation. All rights reserved.

  Restoring packages for K:\invoker-prod\work\codeforces6\e884e6915636a650cf7f602c6c25726a\compile-7e62e93193bbb36402d44e52e1ee29a2\qsharp.csproj...
  Generating MSBuild file K:\invoker-prod\work\codeforces6\e884e6915636a650cf7f602c6c25726a\compile-7e62e93193bbb36402d44e52e1ee29a2\obj\qsharp.csproj.nuget.g.props.
  Generating MSBuild file K:\invoker-prod\work\codeforces6\e884e6915636a650cf7f602c6c25726a\compile-7e62e93193bbb36402d44e52e1ee29a2\obj\qsharp.csproj.nuget.g.targets.
  Restore completed in 210.02 ms for K:\invoker-prod\work\codeforces6\e884e6915636a650cf7f602c6c25726a\compile-7e62e93193bbb36402d44e52e1ee29a2\qsharp.csproj.
  
  ____________________________________________
  
  Q#: Success! (0 errors, 0 warnings) 
CSC : error CS5001: Program does not contain a static 'Main' method suitable for an entry point [K:\invoker-prod\work\codeforces6\e884e6915636a650cf7f602c6c25726a\compile-7e62e93193bbb36402d44e52e1ee29a2\qsharp.csproj]

=====
Used: 0 ms, 0 KB
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6 years ago, # |
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I try to solve the Chessboard Unitary problem and have the correct matrix output by DumpUnitary. But in the contest, I get WA 1. What could have gone wrong?

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    6 years ago, # ^ |
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    Comes out, my matrix was of the desired pattern but not unitary. I guess it would be a good idea to add a unitarity test to the DumpUnitary.

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6 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Im not very familiar with quantum mechanics, so i have been trying to solve the Unitary problems by manipulating vectors. I keep coming up against the wall of the 1 vector (a n*n vector with all weights being equal). For one single qubit a set(One, x) plus H(x) gets the desired result, but I am clueless as how to extend this to more than one qubit.

(For context, the 3rd problem is very "easily" solved by a controlled 1 gate followed by negating the upper half, and the second by splitting the qubits into even and odd, and applying the 1 gate to both sets individually)

Edit: its very posible this way of approaching the problems is just incorrect;

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6 years ago, # |
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Quantum computers are fairy tales and will never be constructed due to Quantum Zeno Effect. Better give us more practical special contests related to web design.

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    6 years ago, # ^ |
      Vote: I like it +37 Vote: I do not like it

    Quantum Zeno Effect isn't a problem for quantum computing, as it requires constant measuring. The usual quantum computation scheme is to prepare the system, then let it evolve on its own, and only at the end there is a measurement. If you solved the problems, you maybe noticed that the problems don't require making measurements, until the moment you check if the answer is correct (except last year's problem that explicitly asked for measurement).

    However, it is true that we are still very far from creating fully-scalable universal quantum computer, and with current state of knowledge it seems impossible. The existing "quantum computers" are either very small (~15 qubits) or not fully entangled. The problem is, there are two competing requirements: control and isolation from environment. If the qubits are too weakly coupled with the environment, it is impossible to control them. On the other hand, if they are too strongly coupled with the environment, the information flows to the environment and results in errors in the computation. There are many candidate systems for physical qubit implementation, but all of them suffer in at least one of these issues.

    Nevertheless, developing the theoretical base of quantum algorithms can be made even without a working quantum computer and will certainly become useful one day. Some of them don't require full capability of a perfect quantum computer and can be implemented on the easier defect ones. The theoretical quantum informatics also results in quantum error correction protocol, which loosens the criteria for errors.

    Forgive me for replying to (I suppose) a trolling attempt, but the information from that post isn't that easy to verify (apart from being rude, which is obvious) so I thought I can clarify a little.

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6 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

Why is the sequence of operations

Spoiler

Giving me Wrong Answer in G2?. I just wanted to do the Quantum analog of De Morgan, but apparently I am failing :)

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    6 years ago, # ^ |
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    You should do MultiX(x) again, cause you shouldn't change the state of given input.

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      6 years ago, # ^ |
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      Thank you very much! Although the problem doesn't say so anywhere (the Microsft docs "encourage" you to do so, though). In any case, should'nt it just throw a "Runtime error" or something in cases like this?

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        6 years ago, # ^ |
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        Not completely sure, but I remember seeing the author saying that they turn all RE into WA.

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        6 years ago, # ^ |
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        The problem does say so: "... performs a transformation " — this means that x is left unchanged after applying the operation.

        I tried to draw a better line between Wrong Answer and Runtime Error failures; currently any solution that fails the problem-specific checks (such as returning qubits in a state other than expected or using measurements in a task which prohibits them) should be judged as WA, and any solution that fails more general checks (such as releasing allocated qubits in non-zero state or trying to access array elements outside of array bounds) should be judged as RE. In this case the qubits are returned in the state other than expected, so the judge gives WA.

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          6 years ago, # ^ |
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          Oh ok , cool! Didn't thought you would verify the input qubits too.

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6 years ago, # |
  Vote: I like it +8 Vote: I do not like it

U4: the same form of the matrix, but the top left 2 × 2 square is anti-unitary and the bottom right (2N - 2) × (2N - 2) is all 'X'-s.

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    6 years ago, # ^ |
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    I can't think of a way a unitary can enforce that the OR of the last N-1 qubits remains unchanged! How to solve this?

    My failed attempt
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      6 years ago, # ^ |
      ← Rev. 2   Vote: I like it +8 Vote: I do not like it

      I have no idea how, just throwing it out there. Maybe I should've written U4e+3.

      If it was solvable, I'm guessing the solution would be based on the 1e-5 threshold — creating a matrix with not exactly zero coefficient for , but something close enough.

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        6 years ago, # ^ |
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        Let's apply controlled Hadamard gate on the Suffix using the Or of the Suffix. We can then choose randomly an index from the Suffix and enforce its state to be One. This should follow the pattern you described but I'm unsure if it's a unitary.

        implementation
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          6 years ago, # ^ |
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          Reset uses measurement, so you can't do that.

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            6 years ago, # ^ |
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            We can enforce One using another temp QuBit, which have the same state as the qubit in the random index (using CNOT and X), and applying Controlled X on the random index using both the temp and the Or but I don't think treating QuBits this way is correct.

            CNOT(qs[idx + 1], quTmp);
X(quTmp);
Controlled X([quOr, quTmp], qs[idx + 1]);
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              6 years ago, # ^ |
              ← Rev. 2   Vote: I like it +8 Vote: I do not like it

              When you use extra qubits and Controlled operations, keep in mind:

              • the extra qubits are going to end up entangled with others
              • you have to make sure they're zero at the end

              I've thought about this for long enough and it really isn't a warmup problem — if it has a solution, it's not a simple elegant solution.

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                6 years ago, # ^ |
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                Xellos — you seem to have a much better grasp on this stuff than I do. I asked Nickolas during the warmup round if there existed a general method for devising an arbitrary matrix. Nickolas answered right after the regular contest ended and said finally can answer this question now contest has ended and yes it is possible, here is link to paper:

                https://arxiv.org/pdf/1306.3200.pdf

                I'm going through it but don't understand any of it, certainly not enough to implement it. Does it make sense to you?

                Could you use this to implement U4?

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                  6 years ago, # ^ |
                    Vote: I like it 0 Vote: I do not like it

                  It's approximation, so you can't use a fixed number of gates to get an exact matrix you want (in general, maybe it works for U4 but there's a specific solution for U4 that doesn't use this anyway). On the other hand, it uses a much more limited set of gates than we have.

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    6 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Have you tried incorporating the R transformations? Like Rx, Ry, Rz? According to the dump tool the imaginary component counts just as much as the real component. I think in order to pull off your proposed U4 maybe working in Y and Z as well H and X.

    I got it so the left top 2x2 matrix were unique values unseen in the rest of the matrix and I feel like I can exploit that to turn that upper left 2x2 into an anti-unitary but not sure how.

    But at this point I have no idea what I'm doing, I just toss up X's, H's, Controlled X's, Controlled H's, Y, Z, T, S, etc. Until it finally does what I want. So don't take my word for it.

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      6 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      Yeah, R should be useful. For example, we can convert the last qubit into "pure if and only if everything is " easily. Still, more complex operations (no pun intended) are more complex to use, so the question becomes less "which operations?" and more "how?".

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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    What do you mean by anti-unitary? In mathematics anti-unitary is anti-linear map, and you can't merge matrix that represent linear map and matrix that represent anti-linear map as blocks of the bigger matrix. This is just ill-defined.

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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Did you even read U3?

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        6 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Ah, you wanted to say "anti-diagonal". Should've guess that.
        Anti-unitary is a common notion and it's very different from "anti-diagonal".

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          6 years ago, # ^ |
            Vote: I like it -10 Vote: I do not like it

          Yeah, that's what I meant. I pay less attention to precision when it's just a rough idea in the context of the previous problem.

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    6 years ago, # ^ |
    ← Rev. 3   Vote: I like it +10 Vote: I do not like it

    Well, I've gotten close. 2 cells missing. I can't seem to get rid of them and get them above epsilon, though I can push them around in neat ways with CNOT and SWAP, just no matter what I tried yet they stick around. Its pretty clean as well, the zeroes are pure zeroes, and the 2 upper left are pure 1.0. The 2 problem cells are pure zeroes as well, not sure how to mix them in and swirl them around to pick up some value without disturbing the other zeroes.

    open Microsoft.Quantum.Primitive;
open Microsoft.Quantum.Canon;
open Microsoft.Quantum.Extensions.Diagnostics;
open Microsoft.Quantum.Extensions.Math;
open Microsoft.Quantum.Extensions.Convert;

operation Solve (x : Qubit[]) : Unit {
   let n = Length(x);
   let Hn = ApplyToEachC(H, _);
   let theta = ArcSin(Sqrt(PI())/Sqrt(7.0));
   for(i in 1..n-1) {
      Controlled Hn([x[i]], x[0..i-1]);
      Controlled Hn([x[i]], x[i+1..n-1]);
   }
   X(x[0]);
   for(j in 0..n-1){
      (ControlledOnInt(1, ApplyToEachCA(Ry(2.0*theta,_), _)))(x[j .. n - 1], x[0 .. j - 1]);
   }
}


.X......
X.......
..XX..XX
..XX..XX
..XXXXXX
..XXXXXX
..XXXXXX
..XXXXXX
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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it +10 Vote: I do not like it

    Well I have N = 2 or 3 working, I feel like the pattern is in the alternation of CNOT and Ry applications, but I have no idea what the pattern is just yet for N>3. 

    open Microsoft.Quantum.Primitive;
open Microsoft.Quantum.Canon;
open Microsoft.Quantum.Extensions.Diagnostics;
open Microsoft.Quantum.Extensions.Math;
open Microsoft.Quantum.Extensions.Convert;

operation Solve (x : Qubit[]) : Unit {
   let n = Length(x);
   let Hn = ApplyToEachC(H, _);
   let theta = ArcSin(Sqrt(2.0)/Sqrt(3.0));
   for(i in 1..n-1) {
      Controlled Hn([x[i]], x[0..i-1]);
      Controlled Hn([x[i]], x[i+1..n-1]);
   }
   X(x[0]);
   for(i in 1..n-1){
      CNOT(x[i],x[i-1]);
   }
   for(j in 0..n-1){
      (ControlledOnInt(1, ApplyToEachCA(Ry(2.0*theta,_), _)))(x[j .. n - 1], x[0 .. j - 1]);
   }
   for(i in 1..n-1){
      CNOT(x[i],x[i-1]);
   }
   for(i in 2..n-1){
      CNOT(x[i-1],x[i]);
   }
   for(j in 0..n-1){
      (ControlledOnInt(1, ApplyToEachCA(Ry(2.0*theta,_), _)))(x[j .. n - 1], x[0 .. j - 1]);
   }
}

N=2
-------------
.X..
X...
..XX
..XX

N=3
--------------
.X......
X.......
..XXXXXX
..XXXXXX
..XXXXXX
..XXXXXX
..XXXXXX
..XXXXXX

But at N=4 it starts to break down ...
------------------------
.X..............
X...............
..XXXXXX....XX..
..XXXXXX....XX..
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
..XXXXXXXXXXXX.X
..XXXXXXXXXXXXX.
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
..XXXXXXXXXXXXXX
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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Xellos — Well here is your proposed U4 working for 2 <= N <= 5. Top left is 2 × 2 square is anti-unitary and the bottom right (2^N - 2) × (2^N - 2) is all 'X'-s. 

    open Microsoft.Quantum.Primitive;
open Microsoft.Quantum.Canon;
open Microsoft.Quantum.Extensions.Diagnostics;
open Microsoft.Quantum.Extensions.Math;
open Microsoft.Quantum.Extensions.Convert;

operation Solve(x : Qubit[]) : Unit {
   let n = Length(x);
   let Hn = ApplyToEachC(H, _);
   let theta = ArcSin(Sqrt(2.0)/Sqrt(3.0));
   for(i in 1..n-1) {
      Controlled Hn([x[i]], x[0..i-1]);
      Controlled Hn([x[i]], x[i+1..n-1]);
   }
   X(x[0]);
   for(k in 1..n-1){
      for(j in 0..n-1){
         for(i in k+j+1..n-1){
            CNOT(x[i-1],x[i]);
         }
         for(i in k+j..n-1){
            CNOT(x[i],x[i-1]);
         }
      }
      for(i in 0..n-1){
         (ControlledOnInt(1, ApplyToEachCA(Ry(2.0*theta,_), _)))(x[i .. n - 1], x[0 .. i - 1]);
      }
      for(j in 0..n-1){
         for(i in k+j+1..n-1){
            CNOT(x[i-1],x[i]);
         }
         for(i in k+j..n-1){
            CNOT(x[i],x[i-1]);
         }
      }
      for(i in 0..n-1){
         (ControlledOnInt(1, ApplyToEachCA(Ry(2.0*theta,_), _)))(x[i .. n - 1], x[0 .. i - 1]);
      }
      for(i in k..n-1){
         (ControlledOnInt(1, ApplyToEachCA(Ry(2.0*theta,_), _)))(x[i .. n - 1], x[0 .. i - 1]);
      }
   }
}

N=5, works for 2,3,4 as well.
--------------------------------

.X..............................
X...............................
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is this sequence of operations giving wrong answer on U3. I checked with DumpUnitaryTool. I know what littleEndian means. Can someone help me?

Spoiler
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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Nope. Its wrong for every N, according to DumpUnitary.

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      6 years ago, # ^ |
      ← Rev. 3   Vote: I like it 0 Vote: I do not like it

      DumpUnitary's Output follows littleEndian. So for n=2, should I get

      .X..
X...
..XX
..XX

      or this?

      ..X.
.X.X
X...
.X.X
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        6 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        DumpUnitary follows the same convention as the statement (and the checked in UnitaryPatterns kata), so you should get the first pattern, but your solution produces the second one.

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6 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Here is the editorial for the round: https://assets.codeforces.com/rounds/1115/warmup-editorial.pdf

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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Is there a general algorithm to make the matrix look anyway you want, like a Fast Fourier Transform equivalent or something similar?

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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      The problem is that the pattern itself doesn't specify the elements of the matrix. There is a Quantum Fourier Transform that decomposes a unitary matrix into simpler unitary matrices.

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      6 years ago, # ^ |
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      Now that the main contest is over, I can finally answer this question :-)

      There are ways to implement a unitary once you have figured out the coefficients it has; one such way is described here. This method, however, is fairly slow — it times out on the largest test in each of the harder D problems.

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        6 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Do you (or anyone, feel free to answer) know what Z[i,1/sqrt(2)] means in the paper? I thought Z was integers like https://en.wikipedia.org/wiki/Integer

        But the hollow Z seems to mean something different in this paper.

        So if I understand the paper correctly, but vaguely, you need to devise a U unitary operation that will give you what you want, then this algorithm will construct the arbitrary U operation out of Clifford + T gates for you so when it is applied it changes the identity matrix to the output that you want. Is that the main idea of it? I could be completely wrong, the paper seems above my head at this point.

        So like the problem D2, I got a matrix 2^3 x 2^3 for N = 3 like 

        let N = Length(x);
for(i in 0..N-1){
	H(x[i]);
	Controlled H(x[i+1..N-1], x[i]);
}

XXXX....
XXXX....
XXXX....
XXXX....
....XX..
....XX..
......X.
.......X

        And I could flip it so the small this part pointed to the upper right, but not top left.

        let N = Length(x);
for(i in 0..N-1){
	X(x[i]);
	H(x[i]);
	Controlled H(x[i+1..N-1], x[i]);
}

.......X
......X.
....XX..
....XX..
XXXX....
XXXX....
XXXX....
XXXX....

        But I couldn't figure out how to flip it like this, its gotta be elementary because it looks like an "anti-diagonal matrix transpose" to flip it from pointing to lower right to pointing to upper left. 

        X.......
.X......
..XX....
..XX....
....XXXX
....XXXX
....XXXX
....XXXX

        Not sure what the proper term is, I figure transpose flips along main diagonal, what I was looking for is a flip along the anti-diagonal. Probably reference implementation just constructs it easily in one shot, but I felt so close when I had it pointing down and right.:

        But instead, using the ideas in the paper, I could formulate coefficients for a U that gives me the correct output. Then once I have devised that, the algorithm will tell me how to construct U from Clifford and T gates and possibly up to two ancilla in order to create that arbitrary U operation?

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      6 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      There is a way to decompose any unitary matrix to a product of CNOT gates and a single qubit gates. It's in the famous Nielsen and Chuang book, page 189. I used it in my solution of D6. Though not in a straightforward way.

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    6 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    When will the problems be added to practice session?

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    6 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    For problem G3, do you really need the ancillary qubits?

    You can just CNOT(x[0],x[N-1]), CNOT(x[1],x[N-2]), etc. Then the second half of the array will be all 0s iff the original array was a palindrome. Then revert the CNOTs at the end,

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      6 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      It is possible to solve this problem the way you described as well. The solution I described felt more natural to me, I guess. It has the additional benefit of allowing me to explain in more detail the use of ancillas with uncomputing them in the end instead of measuring, which is a very frequent source of confusion.

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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      This is a really neat solution! I was wondering how I could solve it without using extra memory, since I kept running out of memory for the last test case when using n ancillary qubits.

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        6 years ago, # ^ |
        ← Rev. 4   Vote: I like it 0 Vote: I do not like it

        Yeah, that is how I solved it, I never thought of using extra qubits, at the time I didn't think it was allowed, but I guess it is allowed as long as you don't measure them and you also undo your operations on the ancilla at the end. But this is what I came up with, took me 13 tries since I struggled more with the fact that ranges are inclusive and the arrays are zero based, kind of a rare pairing of paradigms. And no while loop is available I could find, so I couldn't do 

        mutable i=0, j=N-1; 
while (i<j) {
  //do they match
  i++, j--;
}

        Which seems like the most natural control structure for palindromes to me. So my failed submissions were largely from off by one errors dealing with these inclusive ranges. It would be a nice addition to Q# to have regular for loops and while loops, so you can do for(mutable i=0;i<N;i++) or while(i<j), etc. Most people are just so used to zero-based, right exclusive thinking that when we, well at least I, have to think about inclusive ranges its like learning to ride a bike again, I actually have to think about it. But anyway I digress.

        Main idea is looping i through 0 .. N/2 and CNOTing the palindromic partners, turns out this operation is easily reversible and so legal in these oracle questions.

        So CNOT has a truth table like this:

        CNOT(qs[i], qs[N-1-i])
-----------------------
0,0 -> 0,0
0,1 -> 0,1
1,0 -> 1,1
1,1 -> 1,0

        So left, qs[i] always stays the same, while qs[N-1-i], the right, the target, is assigned a zero if and only if qs[i] and qs[N-1-i] are the same, and its assigned a 1 if they are different, almost like classical XOR except your output is the second operand, almost like take some unknowns A,B in {0,1} and you are doing B = A XOR B;

        So lets take a string, "01011011010", loop through i in 0 .. N/2-1, and CNOT qs[i] with its antipodal partner qs[N-1-i], if they are the same then qs[N-1-i] will be 0, so also after that flip it with X(qs[N-1-i]) because later we will re-use our bitwise AND solution from the problem before.

        So applying this to my example "01011011010" it will end up as "01011011111", then Controlled X on the last N/2 qubits with y as target to test for a bitwise AND on the N/2..N-1 range, if they are all 1s its was a palindrome, if not a palindrome then there will at least be one zero in the range and y won't be flipped. Then take care of edge case where Length(qs)==1, since length 1 is always a palindrome but there is nothing to CNOT it with so if length is 1 just flip y with X(y) and return. Unless maybe Controlled X on an empty range is always true? Wasn't sure so I handled as a special case.

        Then how to undo it? Well look at the truth table, well if left is 0 and right is 1 then after a second CNOT right stays 1, then X flipped again and right is back to 0, its original value. Remember if right is 1 then the original was a match, and left is unchanged, and its a 0, so right must have been a 0 as well. And so going through the truth table you see just applying the same operation again reverts the array back to its original values.

        Take left is 0 and right is 0 then they weren't a match originally, because right == 0, so applying second CNOT leaves right == 0, then X flipped and its a 1 again, like we expect, since right is 0 it wasn't a match originally and the unchanged 0 on the left means right must have been a 1 originally.

        I'm probably rambling at this point but you can go through the truth table and see its undoable just by reversing the operation along the same right half range.

        Spoiler
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    6 years ago, # ^ |
      Vote: I like it -10 Vote: I do not like it

    I think Q# partial application technique is a bit terrifying in the QM context. Unitary operator is not unitary anymore if you look just on some subset of qubits.

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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      They still are unitary operators; they're just tensor products of other unitary operators, with the identity for the qubits you're not looking at.

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        6 years ago, # ^ |
          Vote: I like it -10 Vote: I do not like it

        No. Try to describe what is SWAP(q,_) where q is some fixed qubit.

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          6 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Nothing. It doesn't exist. There's no such operation defined in the language.

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            6 years ago, # ^ |
              Vote: I like it -10 Vote: I do not like it

            I'm pretty sure that 

            let WTF = SWAP(q1, _);   
WTF(q2);

            is a legal code :)

            If you are trying to be sarcastic, then don't :)
            There are a lot of novices that can't catch that.

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              6 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              That's the same as SWAP(q1, q2). q1 isn't fixed, it's a mutable qubit. Unless by "fixed" you mean "a variable", i.e. fixed symbol instead of fixed value.

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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Are there any plans to publish editorial to contest itself?

    And/or include the contest tasks into Katas along with Reference?

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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Yes, of course, we've published the editorial here.

      We'll definitely publish the tasks as part of the Katas (and a couple of other tasks which were not a great fit for the contest but will be great for the Katas). It will just take some time — the code is mostly the same but the presentation differs, so contest tasks need a little work to be ported to the kata format. The first PR is already up :-)

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6 years ago, # |
  Vote: I like it +20 Vote: I do not like it

I'm looking at the final standings and I'm afraid I have no clue how the scoring works. What does '+' or '+2' mean? What does '-3' mean? How is the penalty calculated? Does it depend on the times, and/or these '+' and '-'? If so, how?

A detailed explanation would be much appreciated.

Cheers.

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    6 years ago, # ^ |
      Vote: I like it +20 Vote: I do not like it

    Any submission with a '+' is a solved problem. Number means how many incorrect attempts it took before solving the problem (no number = solved at first try). The time below is when the problem was accepted. Problems marked with a '-' are not solved, and again the number is how many incorrect attempts there were.

    In this contest, the first thing that counts is the number of problems solved. Then, the tie-breaker (marked 'penalty') is the time of the last correctly solved problem.

    In some other contests, each wrong try on the correctly solved problems gives additional penalty (usually 20 minutes), but here there is no rule. The incorrect tries for unsolved problems don't matter (but they are useful to show during the contest because they will be important if the problem gets solved).

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      6 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      Ah, and I see now the penalty number is literally the number of minutes since the start. Thank you very much for the clarification!

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    6 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    Why are people disliking a genuine question like this?

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I can't get the T-shirt, greens never get T-shirt!

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Hi, I'm not able to understand why this code gives compilation error. Please help

namespace Solution { open Microsoft.Quantum.Primitive; open Microsoft.Quantum.Canon;

operation Solve (x : Qubit[], y : Qubit) : Unit {
    body (...) {
        mutable f = -1;
        set f = 0;
    }
    adjoint auto;
}

} Link: https://codeforces.net/contest/1115/submission/50589565

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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Mutable variables don't allow adjoint variant of the operation to be generated automatically.

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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I still get compilation error after removing adjoint auto. For some reason it compiles if I remove the set f = 0; line.

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        6 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        If you remove adjoint auto, the problem G1 will fail compilation because this specific problem requires the operation to have an adjoint specified. If you remove set f = 0, you still have a mutable variable but it's never updated, so it's the same as an immutable variable, and adjoint can get generated automatically.

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          6 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          thanks :) I have one more doubt, suppose I want to set a mutable variable to some value and adjoint auto is mentioned how do I do this ?

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            6 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            You can specify adjoint in a different way (for example, manually — the problem requires an adjoint but doesn't restrict how it has to be specified), or you can extract the mutable variable into a function (if it is part of a purely classical processing, for example, in this task).

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        6 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I just tried it without adjoint auto in custom invocation. No compilation error.

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5 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Given the fact 9 months has passed, I still haven't received my T-shirt. sad... Can I track the package in someway?

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