Question

Assume that for some binary tree node implementation, a pointer requires 4 bytes and a data object requires 24 bytes. Further assume that we are storing a Full binary tree, that the internal nodes are implemented to store two pointers and a data field, and that the leaf nodes store only a data field. Type a fraction (like "1/2") that represents the fraction of the total space taken up by overhead. Give your fraction in lowest terms.

Answer 1

The fraction of the total space taken up by overhead for a full binary tree with given node implementation is 1/7. This calculation assumes two pointers of 4 bytes each and a data object of 24 bytes within internal nodes, and only a data object within leaf nodes. As the tree grows large, the overhead of 8 bytes per internal node coalesces into a constant ratio of the total used memory.

Step-by-step explanation:

We are given that pointers take up 4 bytes and a data object takes up 24 bytes. A full binary tree has internal nodes with two pointers and a data field, while leaf nodes contain only a data field. To find the fraction representing the overhead space taken, we must consider the space taken by the pointers in relation to the total space.

In a full binary tree, each level has twice the number of nodes as the previous level, except for the last level, which is filled with leaf nodes. Since it is a full binary tree, the number of leaf nodes is equal to the number of internal nodes plus one. The two pointers in each internal node represent the overhead since they do not store actual data.

Each internal node takes 4 bytes for one pointer, another 4 bytes for the second pointer, plus 24 bytes for the data object, totaling 32 bytes. Leaf nodes take 24 bytes, as they store only a data object and have no pointers. The overhead for each internal node is then 8 bytes for both pointers.

Assuming n internal nodes, the total overhead would be 8n bytes. The total memory used by the internal nodes would be 32n bytes, and the total memory used by the leaf nodes would be 24(n+1) bytes. The total memory consumption is therefore 32n bytes + 24(n+1) bytes. The fraction of the total space taken up by the overhead is the total overhead memory divided by the total memory consumption:

Fraction = \( \frac{8n}{32n + 24(n+1)} \)

To simplify, we get:

Fraction = \( \frac{8n}{56n + 24} \)

Assuming the tree has more than one node, we can ignore the constant (24) when looking for the ratio since the number of nodes n will dominate as it grows large (this is like the concept of 'big O' notation in computer science), and the fraction simplifies to about:

Fraction = \( \frac{8}{56} \) = \( \frac{1}{7} \)

The fraction of the total space taken up by overhead, in lowest terms, is 1/7.