Question Discussion & Solution

To find the minimum number of keys in a B-tree of order `m` and height `h`, we need to consider the minimum number of children and keys allowed at each node level. 1. **Order `m=5`**: * Minimum keys for any non-root node: `ceil(m/2) - 1 = ceil(5/2) - 1 = 3 - 1 = 2` keys. * Minimum children for any internal node: `ceil(m/2) = ceil(5/2) = 3` children. * Root node specific: If `h>0`, it must have at least 1 key and at least 2 children. 2. **Height `h=3`**: This means there are 4 levels of nodes (Level 0 to Level 3). 3. **Calculate minimum nodes and keys at each level:** * **Level 0 (Root Node)**: * Number of nodes: 1 * Minimum keys: 1 (since `h=3 > 0`) * **Level 1 (Children of Root)**: * The root must have at least 2 children. So, 2 nodes at Level 1. * Minimum keys per node: 2 * Total keys at Level 1: `2 * 2 = 4` * **Level 2 (Children of Level 1 Nodes)**: * Each of the 2 nodes at Level 1 must have at least `ceil(m/2) = 3` children. * Number of nodes at Level 2: `2 * 3 = 6` * Minimum keys per node: 2 * Total keys at Level 2: `6 * 2 = 12` * **Level 3 (Leaf Nodes - Children of Level 2 Nodes)**: * Each of the 6 nodes at Level 2 must have at least `ceil(m/2) = 3` children. * Number of nodes at Level 3: `6 * 3 = 18` * Minimum keys per node: 2 * Total keys at Level 3: `18 * 2 = 36` 4. **Total Minimum Keys**: Sum of minimum keys from all levels = `1 (Level 0) + 4 (Level 1) + 12 (Level 2) + 36 (Level 3) = 53`. Therefore, the minimum number of keys the B-tree can contain is 53.