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One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $T \to \infty$ which is impractical. In addition if $T$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $\tau$, using quantum annealing on a 5-qubit system, you only likely to find 2 of the 3 possible states.

[arXiv:0808.0365v3] Ground-state statistics from annealing algorithms: Quantum vs classical approaches - Matsuda et al.

One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $T \to \infty$ which is impractical. In addition if $T$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $\tau$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states.

[arXiv:0808.0365v3] Ground-state statistics from annealing algorithms: Quantum vs classical approaches - Matsuda et al.

One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $T \to \infty$ which is impractical. In addition if $T$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $\tau$, using quantum annealing on a 5-qubit system, you only likely to find 2 of the 3 possible states.

[arXiv:0808.0365v3] Ground-state statistics from annealing algorithms: Quantum vs classical approaches - Matsuda et al.

2 added 109 characters in body
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One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $T \to \infty$ which is impractical. In addition if $T$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $\tau$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states.

Matsuda et al. https[arXiv://arxiv.org/abs/0808.0365v30365v3] Ground-state statistics from annealing algorithms: Quantum vs classical approaches - Matsuda et al.

One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $T \to \infty$ which is impractical. In addition if $T$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $\tau$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states.

Matsuda et al. https://arxiv.org/abs/0808.0365v3

One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $T \to \infty$ which is impractical. In addition if $T$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $\tau$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states.

[arXiv:0808.0365v3] Ground-state statistics from annealing algorithms: Quantum vs classical approaches - Matsuda et al.

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One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $T \to \infty$ which is impractical. In addition if $T$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $\tau$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states.

Matsuda et al. https://arxiv.org/abs/0808.0365v3