5
$\begingroup$

Here is an exercise (4.27) from Nielsen and Chuang and I found the answer (given in the figure below) online without any explanation. The question was to construct a circuit by seeing a matrix (given in the figure).

So I want to know what should be the thought process after seeing the matrix?

enter image description here

$\endgroup$
1
  • 3
    $\begingroup$ This thread that is about the same exercise might be interesting. $\endgroup$ Commented Sep 16, 2020 at 9:48

2 Answers 2

1
$\begingroup$

This exercise has been asked about a few times: like here and here.

I had posted a ‘semi-mechanical’ way that essentially eliminates guessing: here. The idea is to use the fact that any boolean function can be written as a boolean polynomial, and use the resulting polynomial to synthesize a circuit.

$\endgroup$
0
$\begingroup$

First, your transformation is not correct. Imagine multiplying this matrix with $|1\rangle$, and the result would be $|2\rangle$ instead of $|7\rangle$.

Personally, I prefer the method provided in this slide. I do not understand Japanese, but hopefully, I have gotten its general idea. Next, I will explain that idea with a detailed circuit drawing.

The circuit $U$ transforms the computational basis $$|0\rangle\rightarrow|0\rangle\quad|1\rangle\rightarrow|2\rangle\quad|2\rangle\rightarrow|3\rangle\quad|3\rangle\rightarrow|4\rangle\quad|4\rangle\rightarrow|5\rangle\quad|5\rangle\rightarrow|6\rangle\quad|6\rangle\rightarrow|7\rangle\quad|7\rangle\rightarrow|1\rangle$$ Let $P_{mn}$ represent $|m\rangle\leftrightarrow|n\rangle$. Then, $$U=P_{12}P_{23}P_{34}P_{45}P_{56}P_{67}$$

We can implement each one based on their binary representation. Note that Toffoli=$P_{67}$, and the Fredkin gate in Exercise 4.25 is $P_{56}$. Most $P$ here can be implemented similarly with either Toffoli or Fredkin.

$P_{34}$ is tricky, though. Nevertheless, we can replace it with $P_{35}P_{45}P_{35}$. To see that, we can verify with $|3\rangle\rightarrow|5\rangle\rightarrow|4\rangle\rightarrow|4\rangle$, $|4\rangle\rightarrow|4\rangle\rightarrow|5\rangle\rightarrow|3\rangle$, and $|5\rangle\rightarrow|4\rangle\rightarrow|3\rangle\rightarrow|5\rangle$.

The final circuit is

enter image description here

Although there are simpler circuits for this question, this idea is nice because it is general in some sense. Some theory probably says that any permutation can be decomposed into pair-wise permutations $P_{mn}$. So, maybe this method can also solve some other permutation problems.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, it is known that any permutation can be decomposed into transpositions ($P_{mn}$ type permutations). In fact, any permutation can be decomposed into simple transpositions ($P_{i,i+1}$). $\endgroup$
    – AYun
    Commented Oct 26, 2023 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.