2
$\begingroup$

I am working on a paper about Continuous Variable QKD. (https://arxiv.org/abs/1711.08500v2)

I read about direct and reverse reconciliation in this paper. I don't understand what exactly Reconciliation is? Reconciliation algorithms, reconciliation protocols and so on.

I am really confused. Can any one explain reconciliation to me?

$\endgroup$
5
  • $\begingroup$ Please provide a link to the paper where you encountered this so that we can better help you. $\endgroup$
    – JSdJ
    Nov 17, 2020 at 10:12
  • 1
    $\begingroup$ I believe the term "reconciliation" or "information reconciliation" refers to the error correction step that comes after Alice and Bob have generated their raw key. Their raw keys $K_A$ and $K_B$ will not be exactly equal so Alice and Bob have to communicate some classical information to allow them to error correct $K_A$ and $K_B$ so that after the procedure they have new keys $J_A$, $J_B$ such that $J_A = J_B$ (with high probability). $\endgroup$
    – Rammus
    Nov 17, 2020 at 10:26
  • 1
    $\begingroup$ @Rammus Exactly what I thought, I was writing up an answer:) $\endgroup$
    – JSdJ
    Nov 17, 2020 at 10:34
  • $\begingroup$ @JSdJ thanks for your comment. I added the paper link. $\endgroup$ Nov 19, 2020 at 6:58
  • $\begingroup$ @Rammus as I understand, reverse reconciliation and post selection are two different method to reach significant transmission distances. Am I right? $\endgroup$ Nov 19, 2020 at 7:12

1 Answer 1

4
$\begingroup$

I assume the paper you are reading is referring to information reconciliation.

Information reconciliation is a vital part of post-processing in QKD, to limit (or erase in the best-case scenario) the amount of errors/differences between the key of Alice and Bob.

In that sense, it is a form of (classical) error-correction, and, broadly speaking, it works like this:

  • Alice and Bob run a QKD protocol together and both obtain a raw key $k_{a}$ and $k_{b}$ of whatever length.
  • In real world scenarios, $k_{a} \not = k_{b}$ due to measurement errors, imperfect channels etc. Alice and Bob want the exact same key, so they need to fix this.
  • However, $k_{a} \simeq k_{b}$; we can write $k_{b} = k_{a} + \epsilon$, where $\epsilon$ is the error/difference between Alice and Bob, and therefore $\epsilon \simeq \overrightarrow{0}$ (i.e. $\epsilon$ is a bitstring with almost all entries equal to $0$ because there are not that many errors).
  • Alice and Bob have (in advance) agreed to use a certain error code to perform error correction. This error code is some linear function $f$ that calculates the syndrome $s$ of a key. Note that the error code may or may not be secret to Alice and Bob.
  • It is hard to reverse this function in general. $s$ is of considerably shorter length than $k$. So, there are multiple $k$ with the same $s$.
  • Alice calculates $s_{a} = f(k_{a})$ and publicly communicates $s_{a}$. Any Eve cannot do anything with this because the function is in general hard to reverse.
  • Bob receives $s_{a}$ and calculates $s_{b}$. The function is linear, so $s_{b} = f(k_{b}) = f(k_{a} + \epsilon) = f(k_{a}) + f(\epsilon) = s_{a} + s_{\epsilon}$. Therefore, the 'error syndrome' $s_{\epsilon} = s_{a} - s_{b}$ encodes only information about $\epsilon$.
  • Only Bob can calculate $s_{\epsilon}$. Because the code was designed properly (and for this exact function), and because $\epsilon \simeq \overrightarrow{0}$, Bob can decode the error syndrome to estimate $\epsilon$: $\hat{\epsilon} = f^{-1}(s_{\epsilon})$.
  • Finally, Bob calculates $\hat{k_{a}} = k_{b} - \hat{\epsilon}$. If the code is any good, $\hat{k_{a}} - k_{a} \rightarrow 0$, i.e. Alices key and Bobs corrected key are very close/almost always the same.

There are a lot of important details I have grossed over, but this is the general idea. For a nice introduction of information reconciliation/error correction in the scope of QKD, check this paper.

Note that the above is error-correction for general QKD, not necessarily only for continuous variable QKD. I suspect that there are small discrepancies between the two, but the main goal should be the same.

$\endgroup$

Your Answer

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

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