| bio | website | |
|---|---|---|
| location | United States | |
| age | ||
| visits | member for | 2 years, 2 months |
| seen | May 19 at 16:00 | |
| stats | profile views | 225 |
"What is now proved was once only imagined." — William Blake
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May 14 |
comment |
When is the right ascension of the mean sun 0? Thanks, this is promising and may explain what I'm seeing in my related question, though I'm not clear right now (it's been a while since I asked) how to modify my "naive" EOT derivation to accommodate this information to confirm that it solves my problem there (help with that would be an answer). |
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May 14 |
revised |
Why is my approach to the equation of time off by a constant? Link starting RA calibration question; fix subscript on "naive" function. |
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May 14 |
revised |
Why is my approach to the equation of time off by a constant? Link starting RA calibration question. |
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Apr 11 |
accepted | Is it ever necessary to extend an analysis of Grover's algorithm beyond $k/N = 1/2$? |
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Apr 10 |
revised |
Is it ever necessary to extend an analysis of Grover's algorithm beyond $k/N = 1/2$? Typos. |
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Apr 10 |
comment |
Is it ever necessary to extend an analysis of Grover's algorithm beyond $k/N = 1/2$? Assume for the moment that I'm following a strategy that I'm happy with up to $k/N = 1/2$. |
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Apr 10 |
comment |
Is it ever necessary to extend an analysis of Grover's algorithm beyond $k/N = 1/2$? Yes, so if whatever strategy I adopt using (unmodified) Grover's algorithm achieves a probability of a hit of $p_h$, then can't I say after measuring $x\in S$ using Grover's, modified as above, that any of the unmeasured values, $y\in S\setminus x$, is a hit with probability $1-p_h$? And if so then whenever I've got a $k$ that gives me an unsatisfactory $p_h$, can't I switch to the modified version? |
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Apr 10 |
revised |
Is it ever necessary to extend an analysis of Grover's algorithm beyond $k/N = 1/2$? Add tag. |
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Apr 9 |
asked | Is it ever necessary to extend an analysis of Grover's algorithm beyond $k/N = 1/2$? |
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Apr 1 |
awarded | Critic |
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Mar 29 |
revised |
Is there a “map” of the interplanetary transport network? Add link to NASA's trajectory browser. |
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Mar 29 |
accepted | Should it be obvious that independent quantum states are composed by taking the tensor product? |
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Mar 21 |
awarded | Necromancer |
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Mar 18 |
awarded | Notable Question |
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Mar 14 |
awarded | Yearling |
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Mar 12 |
comment |
How to apply a Hadamard gate? Do you mean how does one apply $H$ to an individual (e.g. "the first") qubit? |
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Mar 12 |
revised |
How to apply a Hadamard gate? Fix kets LaTeX. |
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Mar 12 |
suggested | suggested edit on How to apply a Hadamard gate? |
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Mar 11 |
comment |
Why the quantum entanglement doesn't break quantum cryptography Doesn't a $CNOT$ gate work for these two special cases? |
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Feb 24 |
comment |
Is “entanglement” unique to quantum systems? Note that I wasn't as clear as I should have been in the original question, where I rely too much on the context of the passage I was citing. There, "entanglement" is effectively defined to mean nothing more than "unfactorable" — i.e., requiring more than $k+l$ bits. Hence my note at the end about what I am seeking: "If one struck out...". If "entanglement" is formally defined differently (e.g. requiring something specific about the substates, such as being physically distant) then my question isn't really about entanglement, but about this specific (factorability and dimensionality) property. |