.Problem 1 Show, by means of a counterexample, that the following "greedy" strategy does not always determine an optimal way to cut rods. Define the density of a rod of length i to be pi/i, that is, its value per inch. The greedy strategy for a rod of length n cuts off a first piece of length i, where 1 ≤ i ≤ n, having maximum density. It then continues by applying the greedy strategy to the remaining piece of length n-i.
.Problem 1 Show, by means of a counterexample, that the following "greedy" strategy does not always determine an optimal way to cut rods. Define the density of a rod of length i to be pi/i, that is, its value per inch. The greedy strategy for a rod of length n cuts off a first piece of length i, where 1 ≤ i ≤ n, having maximum density. It then continues by applying the greedy strategy to the remaining piece of length n-i.
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Please solve the following problem and show all work (this class is analysis of algorithms we use cpp psuedocode mostly)

Transcribed Image Text:.Problem 1
Show, by means of a counterexample, that the following "greedy" strategy does
not always determine an optimal way to cut rods. Define the density of a rod of
length i to be pi/i, that is, its value per inch. The greedy strategy for a rod of
length n cuts off a first piece of length i, where 1 ≤ i ≤ n, having maximum
density. It then continues by applying the greedy strategy to the remaining piece of
length n-i.
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