# Is my solution to a mathematics puzzle I created the most efficient solution there is to it?

Here's a puzzle of mine that I created around 2 hours ago:

Let $$f(x):=x^2$$ and $$g(x):=x-4$$. Starting with x=0, what is the least amount of times you need to apply the functions $$f$$ and $$g$$ so that at some point, the output is $$1024$$?

#### Restrictions

Negative numbers less than $$-8$$ are not allowed.

I wanted to try my hand at getting the most efficient solution to this, so here is my most efficient attempt so far (my second attempt, first attempt used up a total of 60 functions):

Total functions: 14$$f(g(g(g(g(g(g(g(g(f(g(g(f(g(0))))))))))))))$$

1. Apply $$g(x)$$ and then $$f(x)$$ to get $$16$$ (total of $$2$$ functions right now)
2. Apply $$g(x)$$ twice to get $$8$$ and then apply $$f(x)$$ to get $$64$$ (total of $$5$$ functions right now)
3. Apply $$g(x)$$ a total of $$8$$ times to get $$32$$ and then apply $$f(x)$$ to get $$1024$$ (total of $$14$$ functions)

However, my question is: Is this truly the most efficient solution to my puzzle, or can YOU find a solution that is more efficient?

unsure if I need more tags or not or if tags need to be changed/removed

Of course I know the most efficient solution, I'm just seeing if you can find it :)

Our number will always be a multiple of $$4$$. If the result of the last $$f$$ is greater than $$1024$$, it's at least $$36^2=1296$$, and we'll save many, many applications by reducing the argument of the last $$f$$ to $$32$$ instead of reducing its output to $$1024$$. (It can't be $$-32$$, because that's forbidden.)
The result of the f prior to that has to be the square of another multiple of $$4$$. $$16$$ is too small, so the same argument tells us that the second-last $$f$$ must output $$64$$. Your path used $$f(8)$$, but $$f(-8)$$ is allowed, and this is where the reduction lies.
$$f(g^8) fgg(0) = 1024$$ in $$12$$ applications.