Don't understand one step for AVL tree height log n proof

Question

I came across a proof that an AVL tree has $O(\log n)$ height and there's one step which I do not understand.

Let $N_h$ represent the minimum number of nodes that can form an AVL tree of height $h$. Since we're looking for the minimum number of nodes, let its children's number of nodes be $N_{h-1}$ and $N_{h-2}$.

Proof:

$$N_h = N_{h-1} + N_{h-2} + 1 \tag{1}$$ $$N_{h-1} = N_{h-2} + N_{h-3} + 1 \tag{2}$$ $$ N_h = (N_{h-2} + N_{h-3} + 1) + N_{h-2} + 1 \tag{3}$$ $$ N_h > 2N_{h-2} \tag{4}$$ $$N_h > 2^{h/2} \tag{5} $$

I do not understand how we went from (4) to (5). If anyone could explain, that'd be great.

Answer 1

You can continue as same as line 4 the process like that:

$$ N_h > 2N_{h-2}> 2(2 N_{h-4})>2(2(2 N_{h-6}))>\cdots$$ As you can see, the indexs are decreasing by substracting $2$ in each step when you use the inequality. So, the process stops when the index $h$ takes $0$, but from the indexs behavior the half of $h$ (floor) will be the quantity of times that we substract $2$ from $h$.

$$ N_h > 2N_{h-2}>\cdots>2(2(2(2(2h^{h-10}))))> 2^{\frac{h}{2}}$$

Answer 2

Assuming $h$ even, by induction

$$N_h>2N_{h-2}>2^2N_{h-4}>2^3N_{h-6}>\cdots 2^{h/2}N_0$$

because you go up two levels $h/2$ times.

Answer 3

After drawing some trees, I realized that the minimum number of nodes of a tree with height $h-2$ is in fact $2^{h/2}$.

Answer 4

Let's consider an example of an AVL tree with height of 4, then:

$N_8 > 2 N_{6} \ \ \ \Longleftrightarrow \ \ N_8 > 2^1 N_{8 - 2 \cdot 1}$

$N_8 > 4 N_{4} \ \ \ \Longleftrightarrow \ \ N_8 > 2^2 N_{8 - 2 \cdot 2}$

$N_8 > 8 N_{2} \ \ \ \Longleftrightarrow \ \ N_8 > 2^3 N_{8 - 2 \cdot 3}$

$N_8 > 16 N_{0} \ \Longleftrightarrow \ \ N_8 > 2^4 N_{8 - 2 \cdot 4}$

recall, that a tree with only one node (root) has height of 0, so $N_{h=0} = 1$. Hence,

$N_8 > 16 N_{0} \ \ \ \ \Longleftrightarrow \ \ \ \ N_8 > 2^4$

and this is exactly $N_h > 2^{h/2}$.

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The most interesting part of this proof is generalization. Did you notice a pattern in our example?

$N_h > 2^i N_{h - 2 \cdot i}$ $\qquad (1)$

Can we compute the value of $i$ when $N_{h - 2 \cdot i}$ becomes $N_{h=0}$?

Yes, we can get the value of $i$ by solving this equation: $h - 2 \cdot i = 0$. The answer of the equation is $i=h/2$.

The next step is to substitute $N_{h - 2 \cdot i}$ with $N_{h=0}$ and $i$ with $h/2$ in the $(1)$ equation, which is:

$N_h > 2^{h/2} N_0 \ \ \ \ \Longleftrightarrow \ \ N_h > 2^{h/2}$

Proved.