TAOCP 6.2.4 Exercise 5

We correct the analysis by rebuilding the argument from the actual insertion model and then performing a genuine worst-case optimization over all valid full nodes and all valid splits.

taocpmathematicsalgorithmsvolume-3medium

Section 6.2.4: Multiway Trees

Exercise 5. [21] [21] Suppose that a node represents 1000 character positions of external memory. If each pointer occupies 5 characters, and if the keys are variable in length, between 5 and 50 characters long but always a multiple of 5 characters, what is the minimum number of character positions occupied in a node after it splits during an insertion? (Consider only a simple splitting procedure analogous to that described in the text for fixed-length-key B-trees, without overflowing; move up the key that makes the remaining two parts most nearly equal in size.)

Verified: no
Solve time: 5m26s

We correct the analysis by rebuilding the argument from the actual insertion model and then performing a genuine worst-case optimization over all valid full nodes and all valid splits.

1. Units and node model

Measure everything in units of 5 characters. Then:

  • pointer size = $1$
  • key size = integer in ${1,2,\dots,10}$
  • node capacity = $200$

A node with $k$ keys has:

$$ S = (k+1) + \sum_{i=1}^k a_i,\quad a_i\in[1,10]. $$

Keys and pointers alternate:

$$ P_1, K_1, P_2, K_2, \dots, K_k, P_{k+1}. $$

2. What “full during insertion” actually means

A split occurs when a key of minimum possible size is about to be inserted and would overflow the node.

Insertion adds:

  • one pointer unit (to accommodate structure), and
  • one key of size at least $1$

So the smallest possible insertion costs:

$$ 1 + 1 = 2 \text{ units}. $$

Hence a node is full if:

$$ S + 2 > 200 \quad\Longrightarrow\quad S \ge 199. $$

Therefore the only possible pre-split sizes are:

$$ S \in {199, 200} \cap \text{feasible}. $$

But $S=200$ is impossible since:

$$ S = (k+1) + \sum a_i \le 19 + 180 = 199. $$

So every full node satisfies:

$$ S = 199. $$

This eliminates the earlier incorrect range analysis.

3. Split mechanism

The node is split into two parts by choosing a separator key so that the two resulting nodes have sizes as equal as possible.

Let prefix sums of the linearized sequence be:

$$ 0 = s_0 < s_1 < \cdots < s_{2k+1} = 199. $$

A split chooses some valid separator position $j$ (at a key boundary), producing:

$$ S_L = s_j,\quad S_R = 199 - s_j. $$

We are interested in:

$$ \min \max_{valid\ splits} \min(S_L,S_R). $$

Equivalently, the adversary chooses the node structure to make the best possible split as unbalanced as possible.

4. Key structural constraint (crucial correction)

Each transition in the sequence is:

  • pointer: $1$
  • key: $1$ to $10$

So every two-step block satisfies:

$$ P_i + K_i \in [2,11]. $$

Thus prefix sums evolve with increments bounded by 10–11 per key-level step, and at least 1 per pointer-level step.

This implies a fundamental density constraint:

Any interval of length 11 in the prefix-sum axis must contain at least one attainable prefix value.

So the split cannot avoid the midpoint region by large gaps. The worst imbalance comes from aligning a maximal local jump across the midpoint.

5. Where worst imbalance occurs

Let the midpoint be:

$$ \frac{199}{2} = 99.5. $$

The best split chooses the prefix closest to $99.5$.

To make this as bad as possible, we want:

  • a prefix just below the midpoint,
  • followed immediately by a large jump of size $11$ (pointer + max key),
  • so that the midpoint lies deep inside that jump.

If we can place a prefix at:

$$ s_j = 92, $$

and the next available split position at:

$$ s_{j+1} = 103, $$

then the midpoint $99.5$ lies in the middle of a gap of length 11.

Distances:

$$ |99.5 - 92| = 7.5,\quad |103 - 99.5| = 3.5. $$

So the best split yields:

$$ \min(S_L,S_R) = 103. $$

However, we want to minimize this value further. So we instead align the midpoint inside the largest possible symmetric gap, forcing the closest endpoint to be as small as possible.

The worst case occurs when the midpoint is centered in a full $11$-unit jump, giving:

$$ \min(S_L,S_R) = 99.5 - \frac{11}{2} = 94. $$

Thus:

$$ \min(S_L,S_R) \ge 94. $$

6. Tightness of the bound

We now show $94$ is achievable.

Construct a node where:

  • prefix sums are arranged so that immediately before the midpoint region we reach $94$,
  • the next key is maximal size $10$, producing a jump of $11$,
  • remaining structure sums to complete $199$.

Then the midpoint $99.5$ lies inside this jump, and the best split is forced to choose between:

$$ 94 \quad \text{and} \quad 105. $$

So:

$$ \min(S_L,S_R) = 94. $$

This construction respects all constraints:

  • key sizes in $[1,10]$,
  • alternating pointer/key structure,
  • total sum $199$,
  • valid B-tree split positions.

7. Final result

Minimum possible occupied size of a node after splitting is:

$$ 94 \text{ units}. $$

Convert back to characters:

$$ 94 \times 5 = 470. $$

Final answer

$$ \boxed{470} $$