22 votes
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Has there been any truly ground breaking advance in quantum algorithms since Grover and Shor?

Have there been any truly ground breaking algorithms besides Grover's and Shor's? It depends on what you mean by "truly ground breaking". Grover's and Shor's are particularly unique because they ...
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17 votes
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How many logical qubits are needed to run Shor's algorithm efficiently on large integers ($n > 2^{1024}$)?

The question is about how many logical qubits it takes to implement Shor's algorithm for factoring an integer $N$ of bit-size $n$, i.e., a non-negative integer $N$ such that $1 \leq N \leq 2^n{-}1$. ...
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16 votes
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What integers have been factored with Shor's algorithm?

The prime factorization of 21 (7x3) seems to be the largest done to date with Shor's algorithm; it was done in 2012 as detailed in this paper. It should be noted, however, that much larger numbers, ...
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  • 3,331
10 votes
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Does Shor's algorithm end the search for factoring algorithms in the quantum world of computation?

Asymptotically, Shor's algorithm is really efficient. Basically it's just: superposition, modular exponentiation (the slowest step), and a fourier transform. Modular exponentiation is what you do to ...
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  • 1,661
9 votes

Why should we use inverse QFT instead of QFT in Shor's algorithm?

For Shor's algorithm, it actually doesn't matter which one you use. If you apply the QFT twice, it is equivalent to a classical multiplication by -1 modulo $2^n$ where $n$ is the size of the register....
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8 votes

Do the probability amplitudes of the superposition state produced by the QFT transform convey useful information?

You probably shouldn't be thinking of the Quantum Fourier Transform as being something where you want to extract the outcoming probability amplitudes. As you say, when you start measuring, you destroy ...
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8 votes
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When increase the shot, why the result is different?

In general, the number of shots does not increase the accuracy of an experiment. Rather it gives a more precise answer. Attached is a figure showing the distance (in terms of Hellinger distance) for ...
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  • 2,159
8 votes
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How to show that amount of qubits needed to crack the RSA-2048 protocol using Shor's algorithm?

I assume you mean the result from this paper, where the authors (including 'our very own' Craig Gidney) have estimated that if you have $\sim20$ million noisy qubits it would take you around $8$ hours ...
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  • 4,728
8 votes
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Why doesn't Shor's algorithm output a solution for some numbers?

If $f(x) = a^x \pmod{N}$ passes through $-1$, that value of $a$ won't work. For example, $a=2$ fails for $N=33$ because $2^5 = 32 \equiv -1 \pmod{33}$. This should have been mentioned in whatever ...
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7 votes
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How do quantum computers prevent "quantum noise"?

How do we prevent quantum noise in a quantum computer? Well, technically the answer is (at least for most systems): we use ridiculously low temperatures (much colder than space), we shield everything ...
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7 votes
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Measuring ancillas in Shor's algorithm

Is that correct? Is it [not] necessary to measure the ancilla qubits in Shor's algorithm? Correct, it is not necessary to measure the ancillae. This is easily seen by appealing to the no-...
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7 votes

Where exactly does entanglement appear in Shor's algorithm?

Your question contains the answer, as you mention the controlled-U gate which is an entangling gate. You will see in the page I linked, that the action of c-U on $|+\rangle|0\rangle$ for example can ...
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7 votes
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In Shor's factorization algorithm for $N$, why can we always find $n$ such that $N^2\le 2^n\le 2N^2$?

Let's represent $N^2$ as $2^a+b$, where $a$ is the greatest power of 2 that not exceeds $N^2$, and $b \ge 0$ (which is always possible to do - $a$ is just the number of bits in binary representation ...
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7 votes
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Simplified explanation of Shor/QFT transformation as thumbtack

Let me attempt to give a rather unconventional answer to this question: ...
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7 votes
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Why is quantum Fourier transform required in Shor's algorithm?

The essential feature of this problem is that while both the quantum and classical algorithms can make use of the efficient classical function of calculating $a^k\text{ mod }N$, the issue is how many ...
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7 votes
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Why is the size of the top register for Shor's algorithm chosen as it is?

In a nutshell, the number of qubits in the top register directly corresponds to the number of bits of precision to which $x/T$ approximates $s/r$, and we need enough precision to be able to determine $...
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  • 4,538
7 votes
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Does the quantum Fourier transform have many applications beyond period finding?

Given that the QFT is exponentially faster than the FFT, The problem with quantum computing is that they are not actually parallel computers: One is tweaking the qubits in such a way that when ...
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6 votes

What integers have been factored with Shor's algorithm?

For Shor's algorthm: State of the art is still 15. In order to "factor" 21 in the paper Heather mentions, they had to use the fact that $21=7\times 3$ to choose their base $a$. This was explained in ...
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6 votes
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Expected repetitions of the quantum part of Shor's algorithm

The number of runs required is arbitrarily close to 1, using the correct post-processing. See "On the success probability of quantum order finding" by Martin Ekerå from Jan 2022: We prove a ...
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  • 22.4k
6 votes

Expected repetitions of the quantum part of Shor's algorithm

This self-answer gives a not-very-good worst case analysis. I'd really rather have a proper distribution of repetition counts. Probability of a period resulting in factoring In Shor's original paper,...
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  • 22.4k
6 votes
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Confusion about random sampling of integers in Shor's algorithm

You skipped a step in the algorithm. First check if $N$ is even. $35$ is not even. Next determine if $N=a^b$ for $a \geq 1$ and $b \geq 2$. It's not. Randomly choose $x$ in the range $1$ to $N-1$. If ...
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  • 1,679
6 votes
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What happens with first phase factor in QFT?

If you have a quantum state like $$|\Psi\rangle_n = a_0|0\rangle_n+a_1|1\rangle_n+...+a_n|2^n-1\rangle_n$$ and you measure it in the $\{|0\rangle_n,...,|2^{n-1}\rangle_n\}$ basis, then the probability ...
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6 votes
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Why do we use the quantum superposition for a period instead of factors in Shor's algorithm?

A QFT can't arbitrarily raise the probability of any state you want to any value you want. Once you create a superposition, you need to find some way to make destructive interference occur between the ...
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  • 1,661
6 votes
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Can I access device specifications of IBMQ paying devices?

The first thing that comes to my mind with your problem is the fake devices available in Qiskit. Here is how to use it, the main idea is that Qiskit stored device properties in Terra and with it, you ...
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  • 2,374
6 votes
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Why is the superposition of all states an eigenvector, with eigenvalue 1, of the QFT?

Perhaps a simpler example will help. Let the unitary $U$ permute $\vert 0\rangle$ and $\vert 1 \rangle$. That means $U\vert 0\rangle = \vert 1\rangle$ and $U\vert 1\rangle = \vert 0\rangle$. Then ...
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  • 518
6 votes
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Classical algorithm with complexity similar to Shor's discovered: Are there more efficient quantum algorithms than Shor's?

Shor's algorithm is often quoted as having a cost of $\mathcal{O}(\log^3 n)$ quantum operations, though it's actually a bit smaller than that: $\mathcal{O}\left((\log n)^{2} (\log \log n) (\log \log \...
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5 votes

What integers have been factored with Shor's algorithm?

The size of the number factored is not a good measure for the complexity of the factorization problem, and correspondingly the power of a quantum algorithm. The relevant measure should rather be the ...
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5 votes
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Shor's algorithm effectiveness

Check out Thoerem 5.3 in Nielsen and Chuang. It conveys that we are almost guaranteed to get a good value of $x$ (the probability is stated to be at least $1-\frac{1}{2^m}$ where there are $m$ unique ...
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  • 47.1k
5 votes
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Is there a simple, formulaic way to construct a modular exponentiation circuit?

Here's, surely, a very non-optimal way of doing it. Imagine we have a unitary $V$ which performs the operation $$ V|x\rangle|y\rangle=|x\rangle|xy\text{ mod }N\rangle. $$ We can deal with how $V$ ...
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