GATE 2010
March 13, 2025GATE 2010
March 13, 2025GATE 2010
| Question 18 |
Consider a B+-tree in which the maximum number of keys in a node is 5. What is the minimum number of keys in any non-root node?
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Question 18 Explanation:
― In B+ tree, the root contains minimum two block pointers and maximum ‘p’ block pointers.
― Here,
p = order
key = order – 1 = p – 1
― In the non-root node, the minimum no. of keys = ⌈p/2⌉-1
― So, key = 5
order = 6
― So, minimum no. of keys in root node =⌈6/2⌉ – 1 = 2
― Here,
p = order
key = order – 1 = p – 1
― In the non-root node, the minimum no. of keys = ⌈p/2⌉-1
― So, key = 5
order = 6
― So, minimum no. of keys in root node =⌈6/2⌉ – 1 = 2
Correct Answer: B
Question 18 Explanation:
― In B+ tree, the root contains minimum two block pointers and maximum ‘p’ block pointers.
― Here,
p = order
key = order – 1 = p – 1
― In the non-root node, the minimum no. of keys = ⌈p/2⌉-1
― So, key = 5
order = 6
― So, minimum no. of keys in root node =⌈6/2⌉ – 1 = 2
― Here,
p = order
key = order – 1 = p – 1
― In the non-root node, the minimum no. of keys = ⌈p/2⌉-1
― So, key = 5
order = 6
― So, minimum no. of keys in root node =⌈6/2⌉ – 1 = 2