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I have been trying to build myself an example of Shor's computations for the discrete log. I started out with this objective and I realized I should understand the factorization first, which I did and did. Having a clue about the factorization, perhaps now I can tackle the discrete log, but it's still not that easy.

Problem. Find $a$ such that $2^a = 7 \bmod 29$. (We know $2$ is a generator of $Z_{29}$.)

Peter Shor tells us to find $q$, a power of $2$ that is close to $29$, that is, $29 < q < 2\times29$. So $q = 32 = 2^5$ suffices. Next he tells us to put two register |$a$> and |$b$> in uniform superposition $\bmod 28$. (Why $\kern-0.4em\bmod 28$? Why not $\kern-0.4em\bmod 29$?) Then in a third register compute |$2^a 7^{-b} \bmod 29$>.

This will produce a periodic sequence in superposition. Applying the QFT to this register, we should be able to extract this period. When I look at the sequence for this concrete case (which is $2^a 7^{-b} \bmod 29$), I find $[1, 25, 16, 23, 24, 20, 7, 1, 25, 16, 23, 24, 20, 7, 1, ...]$ So, I can see the period is $7$.

What is the calculation that I must do now to extract the solution $a = 12$?

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  1. Fermat's little theorem says that for a prime $p$, $a^{p-1}\equiv 1\mod p$ for all $a$ co-prime to $p$. This means the order of the group of powers of $a$ will divide $p-1$, rather than $p$, hence why the algorithm uses $28$.

  2. The "input" is two-dimensional here; it should be all pairs $(a,b)$ for $0\leq a\leq 27$ and $0\leq b\leq 27$. The period you're looking for is a pair of integers $(r_1,r_2)$ such that $2^a7^{-b}\equiv 2^{a+r_1}7^{-b-r_2}\mod p$.

Let $x$ be the discrete log, i.e., $7\equiv 2^x\mod p$. Then we can re-write this as

$$ 2^a2^{-bx}\equiv 2^{a+r_1}2^{-x(b+r_2)}\mod p$$ Since we know that $2^{28}\equiv 1\mod p$, we can treat the exponents as integers modulo $28$, i.e.,

$$a-bx \equiv a+r_1-x(b+r_2)\mod 28$$ We can cancel out terms with $a$ and $b$:

$$\begin{align}0\equiv &r_1-xr_2\mod 28\\ -r_1r_2^{-1}\equiv &x\mod 28\end{align}$$

So the discrete log is $-r_1r_2^{-1}\mod 28$.

This means what you really want to do is take a two-dimensional array of results for different values of $a$ and $b$, and find a period in that.

It looks like what you've actually found is that the order of $7$ is $7$, i.e., $7^7\equiv 1\mod p$. I don't think this has any impact on Shor's algorithm, but I'm not sure. Usually you would only apply Shor's algorithm to a prime-order group, since you can use the Pohlig-Hellman algorithm to reduce composite-order discrete logs to smaller prime-order discrete logs.

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  • $\begingroup$ Your answer looks very interesting. I might take some time digesting that. More to follow! $\endgroup$
    – user14021
    Dec 9, 2020 at 0:18
  • $\begingroup$ How can you be sure that r2 has an inverse mod 28? That's only true if r2 and 28 are coprime. $\endgroup$
    – Omeglac
    Nov 14, 2023 at 10:24
  • $\begingroup$ You're right, but notice that if $r_1$ has an inverse mod 28 and so does $x$, then necessarily $r_2$ does as well. For real-world cryptography we also work only in prime-order subgroups, so the issue doesn't come up. $\endgroup$
    – Sam Jaques
    Nov 15, 2023 at 19:32

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