Question

Recall that the dynamic programming algorithm for the traveling salesman problem uses O(n^2 \cdot 2^n)O(n

2

⋅2

n

) time and O(n \cdot 2^n)O(n⋅2

n

) space (as usual, nn is the number of vertices). You are going to run this algorithm on a graph with 50 vertices. Roughly how much space is needed for this assuming that each cell of the dynamic programming table occupies 8 bytes? (See How much is 1 megabyte, gigabyte, etc?)

Answer 1

Answer:

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