Exercise 4. [M24] [M24] (A. D. Booth and A. J. T. Colin.) Given N keys in random order, suppose that we use the first 2”, 1 to construct a perfectly balanced tree, placing 2" keys on level k for 0 < k <n; then we use Algorithm T to insert the remaining keys. What is the average number of comparisons in a successful search? [Hint: Modify Eq. (2).]

Verified: yes
Solve time: 4m26s

The failure in the previous solution is the incorrect introduction of an inhomogeneous “deviation dynamics.” Once that is removed, the structure becomes a standard first-order linear recurrence with a known particular solution and a one-parameter family of homogeneous corrections.

1. Correct expectation recurrence

Let $a_s = \mathbb{E}[I_s]$, the expected internal path length after $s$ keys.

For any BST shape of size $s$, inserting a random key among the remaining keys gives the standard identity (TAOCP 6.2.2, Eq. (2)):

$$ a_{s+1}

a_s + \frac{a_s + 2s}{s+1}. $$

Multiply by $s+1$:

$$ (s+1)a_{s+1} = (s+2)a_s + 2s. $$

This holds for all $s \ge m$, independent of the initial shape.

2. Solve the recurrence

We solve

$$ (s+1)a_{s+1} - (s+2)a_s = 2s. $$

2.1 Known particular solution

A standard solution for random BSTs is:

$$ b_s = 2s(H_s - 1), $$

and it satisfies the same recurrence:

$$ (s+1)b_{s+1} = (s+2)b_s + 2s. $$

2.2 Homogeneous equation

The homogeneous recurrence is:

$$ (s+1)x_{s+1} = (s+2)x_s. $$

So:

$$ \frac{x_{s+1}}{s+2} = \frac{x_s}{s+1} = C, $$

hence

$$ x_s = C(s+1). $$

2.3 General solution

Therefore,

$$ a_s = 2s(H_s - 1) + C(s+1). $$

3. Determine the constant from the initial balanced tree

Let

$$ m = 2^n - 1, \qquad r = N - m. $$

The initial perfectly balanced tree has internal path length

$$ a_m = I_m = \sum_{k=0}^{n-1} k 2^k = (n-2)2^n + 2. $$

Substitute into the general form:

$$ a_m = 2m(H_m - 1) + C(m+1). $$

$$ C = \frac{a_m - 2m(H_m - 1)}{m+1}. $$

4. Expected internal path length at $N$

$$ a_N = 2N(H_N - 1) + \frac{N+1}{m+1}\Big(a_m - 2m(H_m - 1)\Big). $$

Substitute $m = 2^n - 1$:

$$ a_N

2N(H_N - 1) + \frac{N+1}{2^n} \left( (n-2)2^n + 2 - 2(2^n - 1)(H_{2^n-1} - 1) \right). $$

5. Average number of comparisons in a successful search

A successful search visits one node per key on the search path, so:

$$ C_N = 1 + \frac{a_N}{N}. $$

Thus,

$$ C_N

1 + 2(H_N - 1) + \frac{N+1}{N(m+1)} \left(a_m - 2m(H_m - 1)\right). $$

Substituting $m = 2^n - 1$:

$$ \boxed{ C_N

2H_N - 1 + \frac{N+1}{N \cdot 2^n} \left( (n-2)2^n + 2 - 2(2^n - 1)(H_{2^n-1} - 1) \right) }. $$

6. Structural interpretation

The term $2H_N - 1$ is the universal mean for random BST insertion.
The second term is a pure homogeneous correction induced by the non-random initial condition.
No harmonic sum or accumulated forcing terms appear, because both processes share identical inhomogeneous dynamics; the difference evolves only through the homogeneous equation.

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TAOCP 6.2.2 Exercise 4