Tags: quantum rev

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# quantum-2

## Challenge Description
Let's increase the difficulty. Same algorithm, new circuit.

In practice, quantum compilers and programmers make use of techniques which don't exist in classical reversible computing, even for seemingly classical subroutines. While this circuit has more gates than the quantum 1 circuit, it also requires much fewer qubits. In fact, the main computational step only uses [2n+3 qubits](https://arxiv.org/abs/quant-ph/0205095).

provided files
- ./circuit_parts.7z
- ./encrypted_message.rsa

## Challenge Overview

quantum-2 is a harder version of the same challenge. While quantum-1 used a purely classical reversible circuit to compute the modular exponentiation step of Shor's algorithm, quantum-2 uses a quantum subroutine. Specifically, quantum-1 used the [Cuccaro ripple-carry adder](https://arxiv.org/abs/quant-ph/0410184) to add two numbers stored in quantum memory.

However, in Shor's algorithm, you only ever need to add classical numbers to numbers to stored in quantum memory. You can save a lot of ancillary qubits by using a different quantum addition subroutine -- namely, addition based on the Quantum Fourier Transform (QFT).

Some relevant papers are:
- https://arxiv.org/abs/1411.5949
- https://arxiv.org/abs/quant-ph/9511018
- https://arxiv.org/abs/quant-ph/0008033

The QFT is used at the end of the circuit, so players should recognize the repitition, but I also explicitly linked the paper in the challenge desciption. It's also the top result if you google "quantum addition algorithm".

The QFT-adder performs addition by taking the QFT of the input qubits, and then performing controlled rotations. The controlled rotations are controlled against a classical register, not a quantum register, so there will either be a rotation or not. ie, the "controlled" part of it is optimized away during compilation. You can use this to figure out what the classical register bits are: if there is a rotation, then the control bit is 1. Compare against "the paper" to identify what the control bits should be to produce those rotations -- this is the classical number which is being added. There's a slight complication that we coalesce all the rotations of the same qubit together, so you have to do a bit of math.

Once you understand how the QFT-adder works and how to extract the classical numbers from the circuit, recovering N is actually easier than in quantum-1. This is because N is hardcoded the 2nd time the QFT-adder is called -- no baby number theory required! But the core approach is still seeing which numbers are being added and looking for a pattern. Obtaining the flag once you've found N is identical to in quantum-1.

## QFT analysis

Here's an example of how to analyze the quantum circuit to extract the number being added from the rotation angles.

Most of the phase / z-rotation gates in the circuit files are just implementing the QFT. However, all of the phase gates which actually encode the numbers being added are multiply-controlled phase gates. This is the only context in which mcphase gates appear in the circuit, so you can simply grep the circuit files to extract the rotation angles. Guessing this is not necessary, as you can simply analyze all consecutive sequences of phase gates whose length is near 64.

grep mcphase ../challenge/circuit_parts/circuit_z_0.qasm -A 0 > output_z_0.txt

The last few angles from the first such rotation are:

mcphase(0.96947586) q,q,q;
mcphase(1.9389517) q,q,q;
mcphase(3.8779034) q,q,q;
mcphase(1.4726216) q,q,q;
mcphase(15*pi/16) q,q,q;
mcphase(15*pi/8) q,q,q;
mcphase(7*pi/4) q,q,q;
mcphase(3*pi/2) q,q,q;
mcphase(pi) q,q,q;


The appearance of pi on the final angles indicates that you should convert divide each floating point number by (2 * pi), which will then give a fraction where the denominator is a power-of-2. This is also explained in the paper.


0.1542968753272663
0.30859374747143375
0.6171874949428675
0.23437500694389587
0.46874999999999994
0.9374999999999999
0.875
0.75
0.5


Trying out a few possible values for the denominator reveals that all of the new floating point numbers can be written as a fraction where the denominator is 2^12. The corresponding numerators are [0, 2048, 3072, 3584, 3840, 1920, 960, 2528, 1264], whose binary representation is:


[+] 8 1264 010011110000
[+] 7 2528 100111100000
[+] 6 960 001111000000
[+] 5 1920 011110000000
[+] 4 3840 111100000000
[+] 3 3584 111000000000
[+] 2 3072 110000000000
[+] 1 2048 100000000000
[+] 0 0 000000000000


The diagonal pattern is pretty clear. You can read off the binary representation of the number being added by looking at the most significant bit of the denominator, in this case (from bottom-to-top) [0, 1, 1, 1, 1, 0, 0, 1, 0, ...]. Read the paper to understand why.

## Extracting N

The [attached] solve script analyzes the first 20 additions present in the circuit_z_0.qasm file, and obtains:

1149389920988353694
1149785239057719001
395318069365307
395318069365307
1148994602918988387
1149785239057719001
790636138730614
790636138730614
1148203966780257773
1149785239057719001
1581272277461228
1581272277461228
1146622694502796545
1149785239057719001
3162544554922456
3162544554922456
1143460149947874089
1149785239057719001
6325089109844912
6325089109844912
...


The pattern here is N-a, N, a, a, N-2*a, N, 2*a, 2*a, N-4*a, ..., which is pretty similar to what appeared in quantum-1.

Original writeup (https://github.com/the-entire-country-of-ireland/public-quantum-rev/blob/main/Quantum%20Rev%202/solve/writeup.md).