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:
In which C is a constant.
Runs for 10 milliseconds when the input size (N) is 32.
This means that
So
Since
Then
Thus:
What input size makes the algorithm run for 14 milliseconds
N for which . So
An input size of N = 128 makes the algorithm run for 14 milliseconds
Answer:
The definition of the subject in question is presented in section elsewhere here.
Explanation:
The total value of almost all of the quantities mostly on unique die features will indeed be:
⇒
⇒
The number of the 4 lateral sides throughout the first tossing will be 15, therefore determines the total of both top and bottom features would be:
⇒
⇒
The compositions including its {top and bottom} current valuations that give a total of Six include {1, 5}, {5, 1}, {2, 4}, as well as {4, 2}. Consequently, because neither of those variations includes Six, it indicates that six doesn't really feature on the very first toss mostly on high or low hand.
And then let's look at second tossed situation. Throughout this particular instance, the total amount obtained would be 12.
Which means it has to be the number of top and bottom:
⇒
⇒
{4, 5},{5, 4}, {6, 3}, and {3, 6} seem to be the {top, bottom} facial variations that provide a total of Nine.
Currently regarding all the tosses, {6, 3} and perhaps {3, 6} have been the only situations wherein the 6 emerge towards top or bottom. Then the value of the unique die mostly on face identical to 6 must've been 3.