Timeline for Complexity analysis on a parameterized recurrence relation
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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May 4, 2015 at 9:35 | answer | added | Leo | timeline score: 0 | |
Apr 30, 2015 at 7:53 | vote | accept | Leo | ||
Apr 25, 2015 at 17:06 | answer | added | Emil Jeřábek | timeline score: 4 | |
Apr 25, 2015 at 8:54 | comment | added | Emil Jeřábek | @Aravind: That's just not true. You seem to start from the assumption that $k$ is constant, and derive that the optimal $k$ is constant, but that's circular reasoning. For constant $k$, the recurrence grows as $O(\alpha_k^n)$ for certain constants $\alpha_k$ that decrease towards $1+\phi$ as $k$ increases. From this, it's clear that the optimal choice of $k=g(n)$ must be an unbounded function. | |
Apr 24, 2015 at 14:06 | comment | added | Aravind | The generating function for $T(n)$ depends only on $k$ and not on $n$; it should be a rational function rather than a polynomial in this case; the optimal value of $k$ depends only on this rational function and is not a function of $n$. | |
Apr 24, 2015 at 10:20 | history | edited | Leo | CC BY-SA 3.0 |
Add an idea of simplification
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Apr 24, 2015 at 10:10 | history | edited | Leo | CC BY-SA 3.0 |
Recurrence formula updated.
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Apr 24, 2015 at 9:24 | comment | added | R B | It seems that no matter which $k$ you choose you'll end up with $T(n)=2^{\Theta(n)}$, probably somewhere between $2^n$ and $3^n$. | |
Apr 24, 2015 at 8:56 | review | First posts | |||
Apr 27, 2015 at 10:16 | |||||
Apr 24, 2015 at 8:51 | history | asked | Leo | CC BY-SA 3.0 |