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    <p>So I was thinking about this problem and I think that it's possible to do this in O(lg n) time with O(lg n) memory usage. This is based on the fact that</p> <p>F(n) = (1 / &radic;5) (&Phi;<sup>n</sup> - &phi;<sup>n</sup>)</p> <p>Where &Phi; = (1 + &radic;5)/2 and &phi; = 1 - &Phi;.</p> <p>The first observation is that &phi;<sup>n</sup> &lt; 1 for any n > 1. This means that for any n > 2, we have that</p> <p>F(n) = &lfloor; &Phi;<sup>n</sup> / &radic;5 &rfloor;</p> <p>Now, take n and write it in binary as b<sub>k-1</sub>b<sub>k-2</sub>...b<sub>1</sub>b<sub>0</sub>. This means that</p> <p>n = 2<sup>k-1</sup> b<sub>k-1</sub> + 2<sup>k-2</sup> b<sub>k-2</sub> + ... + 2<sup>1</sup> b<sub>1</sub> + 2<sup>0</sup> b<sub>0</sub>.</p> <p>This means that</p> <p>F(n) = &lfloor; &Phi;<sup>2<sup>k-1</sup> b<sub>k-1</sub> + 2<sup>k-2</sup> b<sub>k-2</sub> + ... + 2<sup>1</sup> b<sub>1</sub> + 2<sup>0</sup> b<sub>0</sub></sup> / &radic;5 &rfloor;</p> <p>Or, more readably, that</p> <p>F(n) = &lfloor; &Phi;<sup>2<sup>k-1</sup> b<sub>k-1</sub></sup>&Phi;<sup>2<sup>k-2</sup> b<sub>k-2</sub></sup> ... &Phi;<sup>2<sup>1</sup> b<sub>1</sub></sup>&Phi;<sup>2<sup>0</sup> b<sub>0</sub></sup> / &radic;5 &rfloor;</p> <p>This suggests the following algorithm. First, start computing &Phi;<sup>2<sup>k</sup></sup> for all k until you compute a number &Phi;<sup>z</sup> such that &lfloor; &Phi;<sup>z</sup> / &radic;5 &rfloor; that's greater than your number F(n). Now, from there, iterate backwards across all of the powers of &Phi; you generated this way. If the current number is bigger than the indicated power of &Phi;, then divide it by that power of &Phi; and record that the number was divided by this value. This process essentially recovers one bit of n at a time by subtracting out the largest power of 2 that you can at a time. Consequently, once you're done, you'll have found n.</p> <p>The runtime of this algorithm is O(lg n), since you can generate &Phi;<sup>2<sup>i</sup></sup> by repeated squaring, and we only generate O(lg n) terms. The memory usage is O(lg n), since we store all of these values.</p>
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    1. POAn inverse Fibonacci algorithm?
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    1. UStemplatetypedef
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    1. COYou can escape floating point computations if you use 2x2 matrixes instead. It should be faster and somewhat simpler.
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      1. USNikita Rybak
    2. CODon't agree with this. Compute phi^2^k itself is a problem. How precise? Then you need to take the floors etc. What is wrong with a simple binary search, computing Fibonacci using matrix multiplication? :-P
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    3. CO@Moron, @Nikita Rybak- I like the idea to use the matrix representation. However, I don't see how to recover individual bits out of those representations. Could you clarify that step? I definitely would like something numerically stable!
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