Consider a B-tree of height h, minimum degree t ≥ 2 that contains any n-key, where n ≥ 1. Which of the following is correct?

Question

Consider a B-tree of height h, minimum degree t &ge; 2 that contains any n-key, where n &ge; 1. Which of the following is correct?

Prepp Team · Accepted Answer

Understanding B-Tree Height, Minimum Degree, and Keys

A B-tree is a self-balancing tree data structure that maintains sorted data and allows searches, sequential access, insertions, and deletions in logarithmic time. A key characteristic of a B-tree is its minimum degree, denoted by $t$.

The minimum degree $t \ge 2$ governs the number of keys and children each node must contain:

Every node except the root must have at least $t-1$ keys.
Every node except the root must have at least $t$ children.
The root node must have at least 1 key (if it's not a leaf) and at least 2 children (if it's not a leaf).
A node can hold at most $2t-1$ keys.
A node can have at most $2t$ children.

The height of a B-tree is the number of edges on the path from the root to the deepest leaf. A B-tree with only a root node has a height of 0.

Relating Height to the Minimum Number of Keys

To find the relationship between the height $h$ and the number of keys $n$, let's consider the minimum number of keys a B-tree of height $h$ can possibly hold. This occurs when every node has the minimum required number of keys and children.

Let $n_{min}(h)$ be the minimum number of keys in a B-tree of height $h \ge 0$ with minimum degree $t \ge 2$.

Height $h = 0$: The tree consists only of the root node, which is also a leaf. The root must have at least 1 key. So, $n_{min}(0) = 1$.
Height $h \ge 1$: The root is an internal node.
- Level 0 (Root): The root has at least 1 key and must have at least 2 children.
- Level 1: The root's children are non-root nodes. There are at least 2 nodes at this level. Each must have at least $t-1$ keys. Total keys at level 1: $2 \times (t-1)$. Each must also have at least $t$ children.
- Level 2: The nodes at level 1 each have at least $t$ children. So, there are at least $2 \times t$ nodes at level 2. Each is a non-root node (since $h \ge 2$), so must have at least $t-1$ keys. Total keys at level 2: $2t \times (t-1)$.
- Level $i$ ($1 \le i < h$): There are at least $2 \times t^{i-1}$ nodes at this level. Each is a non-root internal node, with at least $t-1$ keys. Total keys at level $i$: $2t^{i-1} \times (t-1)$.
- Level $h$ (Leaves): There are at least $2 \times t^{h-1}$ nodes at this level. These are non-root leaf nodes, each with at least $t-1$ keys. Total keys at level $h$: $2t^{h-1} \times (t-1)$.

Summing the minimum keys across all levels for $h \ge 1$:

$n_{min}(h) = (\text{keys at level 0}) + \sum_{i=1}^{h} (\text{min keys in nodes at level i})$

$n_{min}(h) = 1 + \sum_{i=1}^{h} (\text{min nodes at level i}) \times (t-1)$

$n_{min}(h) = 1 + \sum_{i=1}^{h} (2t^{i-1}) \times (t-1)$

$n_{min}(h) = 1 + 2(t-1) \sum_{i=1}^{h} t^{i-1}$

The sum $\sum_{i=1}^{h} t^{i-1}$ is a geometric series $t^0 + t^1 + \dots + t^{h-1}$, which equals $\frac{t^h - 1}{t-1}$ for $t \ne 1$. Since $t \ge 2$, we can use this formula:

$n_{min}(h) = 1 + 2(t-1) \left(\frac{t^h - 1}{t-1}\right)$

$n_{min}(h) = 1 + 2(t^h - 1)$

$n_{min}(h) = 1 + 2t^h - 2$

$n_{min}(h) = 2t^h - 1$

This formula holds even for $h=0$, as $2t^0 - 1 = 2(1) - 1 = 1$, which matches $n_{min}(0)$.

Deriving the Height Inequality based on Number of Keys

The question states the B-tree contains $n$ keys, where $n \ge 1$. The number of keys $n$ must be at least the minimum number of keys possible for that height $h$. Therefore:

$n \ge n_{min}(h)$

$n \ge 2t^h - 1$

We need to isolate $h$ from this inequality:

$n + 1 \ge 2t^h$

$\frac{n+1}{2} \ge t^h$

Now, take the logarithm base $t$ of both sides. Since $t \ge 2 > 1$, the logarithm function is increasing, and the inequality direction is preserved:

$\log_t\left(\frac{n+1}{2}\right) \ge \log_t(t^h)$

$\log_t\left(\frac{n+1}{2}\right) \ge h$

Rewriting the inequality, we get:

$h \le \log_t\left(\frac{n+1}{2}\right)$

This inequality gives an upper bound on the height of a B-tree with $n$ keys and minimum degree $t$.

Comparing with Options

Let's compare our derived inequality with the given options:

$h \ge \log_t \left(\frac{n+1}{2}\right)$
$h \le \log_t \left(\frac{n+1}{2}\right)$
$h \ge \log_t \left(\frac{n-1}{2}\right)$
$h \le \log_t \left(\frac{n-1}{2}\right)$

Our derivation leads to $h \le \log_t \left(\frac{n+1}{2}\right)$, which matches option 2.

Expression Derived	Corresponds to Option	Inequality
$n \ge 2t^h - 1$		This represents the minimum number of keys for height h.
$\frac{n+1}{2} \ge t^h$		Rearranged inequality.
$h \le \log_t \left(\frac{n+1}{2}\right)$	2	Taking log base t on both sides.

Therefore, the correct inequality relating the height $h$, the number of keys $n$, and the minimum degree $t$ of a B-tree is $h \le \log_t \left(\frac{n+1}{2}\right)$.

Revision Table: B-Tree Properties and Height Analysis

Property	Description	Relevance to Height
Minimum Degree $t$	Minimum number of children for non-root nodes ($\ge 2$). Dictates min/max keys per node.	Larger $t$ means more keys per node, potentially leading to shorter trees for a given $n$.
Keys per Node	Each non-root node has $[t-1, 2t-1]$ keys. Root has $[1, 2t-1]$ keys.	Minimum keys per node determines the minimum number of keys a tree can hold for a given structure, impacting the minimum height bound.
Children per Node	Each non-root internal node has $[t, 2t]$ children. Root has $[2, 2t]$ children if not leaf.	Minimum children per node dictates the minimum branching factor, crucial for minimum key calculation per level.
Height $h$	Number of edges from root to deepest leaf.	The relationship $h \le \log_t \left(\frac{n+1}{2}\right)$ provides an upper bound on height for a tree with $n$ keys and minimum degree $t$. This is related to the minimum number of keys a tree can hold.
Number of Keys $n$	Total number of keys stored in the tree ($\ge 1$).	Larger $n$ generally implies a greater height, but the increase is logarithmic with base related to $t$.

Additional Information: B-Tree Height Bounds

The height of a B-tree is logarithmic with respect to the number of keys $n$. We derived the inequality $h \le \log_t \left(\frac{n+1}{2}\right)$, which represents the maximum possible height for a given number of keys $n$ and minimum degree $t$. This maximum height occurs when the tree is minimally filled (each node has the minimum number of keys).

We can also consider the minimum possible height for a B-tree containing $n$ keys. This occurs when the tree is maximally filled, i.e., every node contains the maximum number of keys ($2t-1$).

Level 0 (Root): At most $2t-1$ keys. If it has children, at most $2t$ children.
Level 1: At most $2t$ nodes. Each has at most $2t-1$ keys. Total keys: $(2t)(2t-1)$.
Level 2: At most $(2t)^2$ nodes. Each has at most $2t-1$ keys. Total keys: $(2t)^2(2t-1)$.
Level $i$: At most $(2t)^i$ nodes. Each has at most $2t-1$ keys. Total keys: $(2t)^i(2t-1)$.

The total maximum number of keys $n_{max}(h)$ in a tree of height $h$:

$n_{max}(h) = (\text{max keys level 0}) + \sum_{i=1}^{h} (\text{max keys in nodes at level i})$

$n_{max}(h) = (2t-1) + \sum_{i=1}^{h} (2t)^i (2t-1)$

$n_{max}(h) = (2t-1) \left(1 + \sum_{i=1}^{h} (2t)^i\right)$

$n_{max}(h) = (2t-1) \sum_{i=0}^{h} (2t)^i$

This is a geometric series with common ratio $2t$. The sum is $\frac{(2t)^{h+1}-1}{2t-1}$.

$n_{max}(h) = (2t-1) \left(\frac{(2t)^{h+1}-1}{2t-1}\right)$

$n_{max}(h) = (2t)^{h+1} - 1$

If the tree contains $n$ keys, then $n \le n_{max}(h)$:

$n \le (2t)^{h+1} - 1$

$n + 1 \le (2t)^{h+1}$

Taking $\log_{2t}$ on both sides (since $2t \ge 4 > 1$):

$\log_{2t}(n + 1) \le h + 1$

$h \ge \log_{2t}(n+1) - 1$

This gives a lower bound on the height of a B-tree with $n$ keys. Combining the bounds:

$\log_{2t}(n+1) - 1 \le h \le \log_t\left(\frac{n+1}{2}\right)$

The question specifically asks for one of the provided inequalities which relates to the maximum possible height, which is determined by the minimum number of keys property, leading to the derivation $h \le \log_t \left(\frac{n+1}{2}\right)$.

Prepp Team · Answer

h &ge; log t $\rm\frac{n+1}{2}$

Prepp Team · Answer

h &ge; log t $\rm\frac{n&minus;1}{2}$

Prepp Team · Answer

h &le; log t $\rm\frac{n&minus;1}{2}$

Property	Description	Relevance to Height
Minimum Degree \(t\)	Minimum number of children for non-root nodes (\(\ge 2\)). Dictates min/max keys per node.	Larger \(t\) means more keys per node, potentially leading to shorter trees for a given \(n\).
Keys per Node	Each non-root node has \([t-1, 2t-1]\) keys. Root has \([1, 2t-1]\) keys.	Minimum keys per node determines the minimum number of keys a tree can hold for a given structure, impacting the minimum height bound.
Children per Node	Each non-root internal node has \([t, 2t]\) children. Root has \([2, 2t]\) children if not leaf.	Minimum children per node dictates the minimum branching factor, crucial for minimum key calculation per level.
Height \(h\)	Number of edges from root to deepest leaf.	The relationship \(h \le \log_t \left(\frac{n+1}{2}\right)\) provides an upper bound on height for a tree with \(n\) keys and minimum degree \(t\). This is related to the minimum number of keys a tree can hold.
Number of Keys \(n\)	Total number of keys stored in the tree (\(\ge 1\)).	Larger \(n\) generally implies a greater height, but the increase is logarithmic with base related to \(t\).