### SPOJ: 11409. Fibonacci With a Twist

After passing hours on it, finally I was able to solve it. Problem statement is pretty clear, given a recursive function, find its nth term.

f(0) = 0

f(1) = 1

f(n) = f(n-1) + f(n-2) + (n-1)

The given range on n makes it obvious that this solution must have a logarithmic algorithm or constant close form. However, I don't know about the close form, mathematics was not my thing. I am also not sure if this can be solved directly using matrix exponentiation.

Looking at the recurrence, it is obvious that the equation can be written in the following form for higher n:

f(n) = a_{n}f(0) + b_{n}f(1) + c_{n}

Here, a_{n}, b_{n} and c_{n} are coefficients for nth term. We can also discard all the terms containing f(0) which is actually 0. So, if we continue writing this, we can find a really nice pattern:

f(0) = 0

f(1) = 1

f(2) = f(1) + (2-1) = 1 + (1)

f(3) = f(2) + f(1) + (3-1) = 2 + (1) + (2)

f(4) = f(3) + f(2) + (4-1) = 3 + 2(1) + (2) + (3)

similarly...

f(5) = 5 + 3(1) + 2(2) + (3) + (4)

f(6) = 8 + 5(1) + 3(2) + 2(3) + (4) + (5)

f(7) = 13 + 8(1) + 5(2) + 3(3) + 2(4) + (5) + (6)

f(8) = 21 + 13(1) + 8(2) + 5(3) + 3(4) + 2(5) + (6) + (7)

I think some Fibonacci numbers caught our eyes already. So, now we try to generalize. Just to note: we can do this because the coefficients of similar terms are consecutive fibonacci numbers.

f(n) = fib(n) + fib(n-1)*1 + fib(n-2)*2 + fib(n-3)*3 + ... ... + fib(1)*(n-1)

f(n+1) = fib(n+1) + fib(n)*1 + fib(n-1)*2 + fib(n-2)*3 + ... ... + fib(1) * n

subtracting f(n) from f(n+1), we can get:

f(n+1)-f(n) = fib(n+1) + {fib(n-1) + ... ... + fib(1)}

f(n+1)-f(n) = fib(n+1) + fib(n+1) - 1 [sum(fib(n)) = fib(n+2)-1]

f(n+1) = f(n) + 2fib(n+1) - 1

This can be solved directly using matrix exponentiation. However, we will need a 4x4 matrix to do that. But, we can actually reduce this equation to a non recursive version:

f(n+1) = f(n) + 2fib(n+1) - 1

f(n+1) = f(n-1) 2fib(n) + 2fib(n+1) - 2

f(n+1) = f(n-2) + 2fib(n-1) + 2fib(n) + 2fib(n+1) - 3

f(n+1) = f(0) + 2fib(1) + 2fib(2) + 2fib(3) + ... ... + 2fib(n+1) - (n+1)

f(n+1) = 2(fib(n+3) - 1) - (n+1)

So, rewriting for f(n):

f(n) = 2(fib(n+2) - 1) - n

Now, this is no more a hard recursive function, we just need to know (n+2)th fibonacci term, which can be computed using only a 2x2 matrix. This post shows you how, if you need to know.

However, for this specific problem, be careful about the final result, you may need modulus correction.

excellent

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DeleteBrilliant mind!!

DeleteVery good observation !

ReplyDeletenice explanation...!!

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