# quantum-1
## Challenge DescriptionWe've found some sort of advanced codebreaker machine, and some encrypted messages it was trying to break. Can you crack it first?
Hint: This circuit is implementing a famous quantum algorithm to factor the public modulus. This modulus *should* be stored in the classical memory before evaluating the circuit (and immediately copied via `cnot` to the quantum memory during `circuit_initialization.qasm`), but is not provided in the challenge. However, the modulus was used to derive many numbers which are used by subroutines in the quantum algorithm, and all of those numbers are hardcoded in the circuit.
Hint: modular exponentiation is just repeated conditional modular multiplication, and modular multiplication is just repeated conditional modular addition, and modular addition is just several regular additions.
provided files- ./circuit_parts.7z- ./encrypted_message.rsa
## Challenge OverviewWe're provided with a quantum circuit implementing Shor's algorithm to factor a number N, as well as a collection of ciphertexts encrypted with N. The goal of this challenge is to analyze the quantum circuit and extract the modulus N, which is sufficiently small (< 64 bits) that it can be factored classically. Then you can use this to decrypt the ciphertexts.
Shor's algorithm consists of 4 stages on the quantum computer:1) hadamards / prepare a uniform superposition2) modular exponentiation3) quantum fourier transform4) measurement
There's classical post-processing on extracting the period and using that to factor N, but we'll ignore that and focus on the quantum component.
The key point is that most of Shor's algorithm is just computing a modular exponentiation step `a^x mod N`, which is implemented via repeated squaring. So the exponentiation step is broken down into repeated controlled modular multiplication. Modular multiplication is then implemented via repeated additions and negations. In this challenge, addition is performed via the [Cuccaro ripple-carry adder](https://arxiv.org/abs/quant-ph/0410184). A very brief overview is presented in [section II.B of this paper](https://arxiv.org/abs/1905.09749). But understanding how it works in-detail isn't too necessary, as long as you get that its implementing addition.
While `N` itself is never explicitly loaded (the quantum circuit expects it to be provided as input, which is then copied over to quantum memory), many values which depend on `N` are hardcoded into the circuit.  Specifically, `pow(a, pow(2,i)) * pow(2,j) % N for j in range(64) for i in range(128)`.
Players should first get a [high-level familiarity with Shor's algorithm](https://en.wikipedia.org/wiki/Shor%27s_algorithm), and then identify which qubits are being used as the exponent qubits via the Hadamard gates at the beginning of the circuit.
## static analysis solution
When analyzing the circuit, players should notice that while the circuit itself is large, it is massively repetitive, hence why it compresses down to under a MB. Thus, they should look at trying to identify repeated chunks of code as functions. This will lead them to the addition function. All of stage 2) can be efficiently simulatable, so players can emulate the function with concrete inputs and see that it performs addition in little-endian. However, this is not necessary.
Next, players should look at the arguments being passed to the function. A lot of the arguments are directly loadly immediately beforehand, via the `ccx` instruction where the first argument is an exponent qubit, example:
```ccx q[0],q[128],q[384];ccx q[0],q[128],q[386];...ADDITION_CIRCUIT;```
In classical reversible computing, `ccx` is often used as a conditional-conditional-write. The above is flipping registers `q[384]` and `q[386]` if and only if registers `q[0]` and `q[128]` are `1`. Because `384 = 6 * 64`, this indicates that these destination registers are part of a single 64-bit variable, and the circuit snippet is really performing a conditional write of `1010000000000...` to that variable. With little-endian encoding, this is writing `5`.
Continue analyzing the circuit and see what other values are being written to this variable, conditioned on an exponent qubit, right before the addition function is called. Players should notice that the numbers loaded are being computed `mod N`, and then do some basic number theory to recover `N`.
A possible approach is in the provided `solve_static.py` python script, with the output for the 1st circuit_part file saved in `rewritten_circuit.txt`. A short selection of the rewritten code is below:```ccx q[0],q[128], `74583193119493` -> #6add #5, #6negate #5add #5, #3negate #5ccx q[449], #3, #4add #4, #5negate #5add #6, #5negate #5add #5, #6x q[449];ccx q[0],q[128], `74583193119493` -> #6ccx q[0],q[129], `149166386238986` -> #6add #5, #6negate #5add #5, #3negate #5ccx q[449], #3, #4add #4, #5negate #5add #6, #5negate #5add #5, #6x q[449];ccx q[0],q[129], `149166386238986` -> #6ccx q[0],q[130], `298332772477972` -> #6...```Examining the disassembled code reveals that it is loading repeated doublings of `a = 74583193119493` into register #6, but after a certain point the numbers plateau off and stop monotonically increasing. This is because the doublings are happening mod N, which lets you recover N from a single circuit_part file.
Once `N` has been recovered, you can factor it in sage in under a minute. Using the public exponent `e = 65537 (0x10001)`, you can then compute `d` and decrypt the ciphertext to get the flag.
##  dynamic analysis solution
The circuit_part files only use SWAP, NOT, CNOT, and CCNOT gates, which are all classical gates, so you can emulate the circuit. At the very start of each file, it loads in `a, a^2, a^4, a^8, ...`, so pausing the emulation after the first values have been loaded will reveal the modular powers of `a`.
We know that `a^2 - (a^2 mod N)` is a multiple of `N`, so you can recover (a small multiple of) `N` via:```gcd( a^2 - (a^2 mod N), a^4 - (a^4 mod N) )```
Unfortunately, as `N` is not loaded during the circuit (it is expected to be provided in classical memory and then copied over to quantum memory during the initialization routine), the entire circuit cannot be correctly emulated.
## hindsight 2021
When designing the challenge, I had made the decision to not hardcode `N` into the circuit, as I thought that this would make it too easy to solve. Instead, I wanted teams to recover the modular powers and doublings of `a` and use that to recover `N`.
This turned out to be a bad call and made the challenge frustrating, as few teams were able to recover any of the integers hardcoded into the circuit. Additionally, not having `N` loaded meant that teams could not successfully emulate the entire modular exponentiation subroutine, which would have been a very elegant solution.
As few teams were able to solve the first quantum challenge, and the entire series could only be solved sequentially, we chose to not release the third and final quantum challenge `='(`.