Rating:

# Crunchy

We are given a small Python program which computes the flag:

```

def crunchy(n):

if n < 2: return n

return 6 * crunchy(n - 1) + crunchy(n - 2)

g = ... # a _really_ big number

print("Your flag is: INSA{%d}"%(crunchy(g)%100000007))

```

Running this, of course, won't get us anywhere with the exponentially

slow implementation of `crunchy`. A simple improvement would be to

cache computed vaues of `crunchy(n)`, but that reduces the complexity

to linear, which is still not nearly enough with the hundred-digit

parameter.

Since `crunchy` is a linear recurrence, we can compute it in

logarithmic time. We can express the mapping from $(c(n), c(n-1))$ to

$(c(n+1), c(n))$ (where $c$ is the function) as a matrix multiplication:

$$ \begin{pmatrix}c(n+1)\\c(n)\end{pmatrix} =

\begin{pmatrix}6&1\\1&0\end{pmatrix}

\begin{pmatrix}c(n)\\c(n-1)\end{pmatrix} $$

If we expand the right-hand side of this equation $n-1$ times, we will

get:

$$ \begin{pmatrix}c(n+1)\\c(n)\end{pmatrix} =

\begin{pmatrix}6&1\\1&0\end{pmatrix}^n

\begin{pmatrix}c(1)\\c(0)\end{pmatrix} $$

Since we know the values of $c(1)$ and $c(0)$, all that remains is to

compute the exponentiated matrix. We can do this in

$\mathcal{O}(\log{n})$ multiplications by [repeated

squaring](https://en.wikipedia.org/wiki/Exponentiation_by_squaring),

which is completely reasonable for the input value of `g` (a few

hundreds of multiplications).

Note that all multiplications should be done modulo $10^9 + 7$ instead

of only doing so at the end -- otherwise, the intermediate values will

be to large to work with (they grow exponentially).

Original writeup (https://de298.user.srcf.net/writeups/insa/crunchy.html).