Exercise 1: Adjacency Matrix vs. List for a Road Network — Possible Solution
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GIVEN
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V = 500 intersections, each connected to an average of 4 others.

STEP 1: THE NUMBER OF EDGES
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Since each of the 500 vertices has an average of 4 connections, and
each edge is shared between two vertices (each edge gets counted once
from each end), the total edge count is:

  E = (500 x 4) / 2 = 2000 / 2 = 1,000

STEP 2: ADJACENCY MATRIX SPACE
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V^2 = 500 x 500 = 250,000 cells

STEP 3: ADJACENCY LIST SPACE
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V + 2E = 500 + (2 x 1,000) = 500 + 2,000 = 2,500 entries

STEP 4: THE RATIO
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250,000 / 2,500 = 100

The adjacency matrix would need exactly 100 times more space than the
adjacency list to represent this same road network - an even larger
gap than this chapter's own 1,000-user social network example (90.9x),
since this network is sparser relative to its size.

WHY THIS WORKS AS AN ANSWER
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The edge count is derived correctly from the average-connections
figure (dividing by 2 to avoid double-counting each edge), and both
space formulas are applied directly from this chapter's own
definitions before computing the final ratio explicitly.