Recursion on staircase

Question

I\'m trying to understand the solution provided in a book to the following question:

\"A child is running up a staircase with n steps and can hop either 1 step, 2 st

Answer 1

To answer your first question, it turns it is the beauty of mathematics: if there is 1 step for the staircase, there is 1 way to solve it. If there is 0 steps, there is also 1 way to solve it, which is to do nothing.

It is like, for an n-step staircase, for m times, you can either walk 1, 2, or 3 steps to finish it. So if n is 1, then m is 1, and there is 1 way. If n is 0, m is 0, and there is also 1 way -- the way of not taking any step at all.

If you write out all the ways for a 2-step staircase, it is [[1, 1], [2]], and for 1-step staircase, it is [[1]], and for 0-staircase, it is [[]], not []. The number of elements inside of the array [[]] is 1, not 0.

This will become the fibonacci series if the problem is that you can walk 1 step or 2 steps. Note that fib(0) = 1 and fib(1) = 1, and it corresponds to the same thing: when staircase is 1 step, there is 1 way to solve it. When there is 0 steps, there is 1 way to solve it, and it is by doing nothing. It turns out the number of ways to walk a 2-step staircase is fib(2) is 2 and it equals fib(1) + fib(0) = 1 + 1 = 2, and it wouldn't have worked if fib(0) were equal to 0.