For equation **F(x) = F(x-1) + F(x-2) + ( F(x-1) * F(x-2) )** given ** F(0)** and ** F(1)** can we calculate ** F(n)** in

**time?**

*log(n)***F(x) = F(x-1) + F(x-2) + ( F(x-1) * F(x-2) )** given ** F(0)** and ** F(1)** can we calculate ** F(n)** in

We can easily found f(n) in O(n) time,but not sure about O(logn).

yeah O(n) is pretty simple but the problem required log(n) or something similar . 0 < n < 10^8â€¦

This problem is similar of finding nth Fibonacci in O(logn) time. I found this, https://www.geeksforgeeks.org/program-for-nth-fibonacci-number/ , in which we can find fibonnaci in O(logn) time. Hope this might help.

Well, you can just define G(x) = F(x)+1, which has the recurrence relation G(x) = G(x-1)G(x-2), which you can calculate in \log(n) time.

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Wowâ€¦ How did you got this logic ??

Awesomeâ€¦

I see itâ€™s available on math. stack exchange

I found it online that F(n) = 2^{Fib(n)}-1 when f(0)=0 and f(1)=1. And 2^{fib(n)} can be calculated in logn.

But it is special case. For the generalized version, @theoneyouwant wrote the perfect answer.

@theoneyowwant how to calculate G(x) = G(x-1)G(x-2) in log(n) time?

Itâ€™s just Fibonacci in powerâ€¦ Take log and see the patternâ€¦ Then use matrix exponentiation as shared by @sagar2405

I hope itâ€™s not a question for an ongoing contestâ€¦ Otherwise you can be banned from discuss siteâ€¦ Please delete this post in case itâ€™s for an ongoing contestâ€¦

Well, I just thought of this myself, since itâ€™s somewhat similar to think of it as a functional equation, where such transforms are quite common. In general, you can try such substitutions on various reccurences to get a better form/ to solve it faster. You may have also solved some dynamic programming problems where redefinitions give a faster solution.

Its not from a ongoing contest , its from a past contest i cam across â€¦

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