Final answer:
To prove an algorithm's running time is theta (g(n)), it must have its worst-case running time as O(g(n)) and its best-case running time as omega (g(n)), implying the running time grows at the same rate as g(n).
Step-by-step explanation:
To prove theta (g(n)) for an algorithm's running time, we must show both:
The algorithm's worst-case running time is O(g(n)), meaning that for sufficiently large n, there exists a constant c1 such that the running time does not exceed c1 * g(n).
The algorithm's best-case running time is omega (g(n)), meaning that there exists a constant c2 such that for sufficiently large n, the running time is at least c2 * g(n).
For an algorithm to have a running time of theta (g(n)), these two conditions must be met simultaneously. This means the running time is tightly bound above and below by g(n), making g(n) an asymptotically tight bound. In simpler terms, the algorithm's running time grows at the same rate as g(n) as n increases.
The statement that the running time of an algorithm is theta (g(n)) if and only if its worst-case running time is O(g(n)) and its best-case running time is omega(g(n)) is known as the tight bound theorem.
To prove this, we can start by assuming that the running time of the algorithm is theta(g(n)). This means that the algorithm's running time is both O(g(n)) and omega(g(n)).
If the running time is O(g(n)), it means that the worst-case running time of the algorithm is bounded above by g(n). And if the running time is omega(g(n)), it means that the best-case running time of the algorithm is bounded below by g(n).