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Assume that an O(log2N) algorithm runs for 10 milliseconds when the input size (N) is 32. What input size makes the algorithm run for 14 milliseconds

Sagot :

Answer:

An input size of N = 128 makes the algorithm run for 14 milliseconds

Explanation:

O(log2N)

This means that the running time for an algorithm of length N is given by:

[tex]F(N) = c\log_{2}{N}[/tex]

In which C is a constant.

Runs for 10 milliseconds when the input size (N) is 32.

This means that [tex]F(32) = 10[/tex]

So

[tex]F(N) = c\log_{2}{N}[/tex]

[tex]10 = c\log_{2}{32}[/tex]

Since [tex]2^5 = 32, \log_{2}{32} = 5[/tex]

Then

[tex]5c = 10[/tex]

[tex]c = \frac{10}{5}[/tex]

[tex]c = 2[/tex]

Thus:

[tex]F(N) = 2\log_{2}{N}[/tex]

What input size makes the algorithm run for 14 milliseconds

N for which [tex]F(N) = 14[/tex]. So

[tex]F(N) = 2\log_{2}{N}[/tex]

[tex]14 = 2\log_{2}{N}[/tex]

[tex]\log_{2}{N} = 7[/tex]

[tex]2^{\log_{2}{N}} = 2^7[/tex]

[tex]N = 128[/tex]

An input size of N = 128 makes the algorithm run for 14 milliseconds