Suppose that we have an algorithm that takes as input a string of n bits. We are told that the e

Suppose that we have an algorithm that takes as input a...

Key Concepts

Markov's Inequality

Markov's inequality is a probabilistic tool that provides an upper bound on the probability that a non-negative random variable exceeds a certain multiple of its expected value. Specifically, for any non-negative random variable X and any a > 0, it states that P(X ? a) ? E[X]/a. This inequality is fundamental when deriving tail bounds from expected values.

Tail Bounds

Tail bounds quantify the probability that a random variable deviates significantly from some central measure, like its mean. In the context of algorithm running times, tail bounds such as those obtained from Markov’s inequality help in assessing the likelihood that the running time will be much larger than the average, even if the worst-case may be much higher.

Expected Value

The expected value is a measure of the average or mean outcome of a random variable. In algorithm analysis, it gives the average running time over all possible inputs drawn from a certain distribution. This concept is critical for understanding the typical performance of an algorithm when the input is random, even though it might not reflect the worst-case scenario.

Worst-case Running Time

Worst-case running time is the maximum running time an algorithm can require for any input of a given size. While expected running time gives a useful average-case measure, worst-case analysis ensures that no input will cause the algorithm to use an unreasonable amount of resources. Markov's inequality, however, does not provide a deterministic worst-case bound; it only offers a probabilistic guarantee that the running time exceeds a constant multiple of the expected value with a correspondingly small probability.

Transcript

00:01 Okay, so we want to find the largest end for which one can solve within one second a algorithm which requires f to the end operations so basically what they say is how many operations can be performed? given that each bit operation is carried out in 10 to minus nine seconds so first what we want to figure out is how many bit operations can be performed? so we have one bit operation one operation one operation can be performed in 10 to the minus nine seconds and nine seconds.

00:33 So what we want to go instead is we want to go from one second.

00:36 So if we're going from 10 to the minus 9 to 1 second, we multiply each side by 10 to the minus 10 to 9, sorry, so 10 to the 9 operations.

00:48 So then to find the largest n for a, we have log n.

00:54 So yeah, log n.

00:56 And we want to solve log n is equal to 10 to 9, or really log place 2.

01:02 So to solve this, you just have to use your log clause.

01:05 So n is just 2 to the 10 to the 9.

01:08 So where i got this from is if you have log base a, b is equal to c, you just have a, c is equal to b.

01:18 So b will be your n, a will be your 2, c will be 10 to 9 and you have this.

01:24 So for d, you have n and this is just equal to 10 to the 9 and that's really just it.

01:30 This is b and for c we have n log n is equal to 10 to the 9 now this this you can try to solve this it's very difficult to solve this um all i can tell you is that there's just there's just a function that does this for you and it gives you n is equal to 3 .96 times like 10 to 7 and to get this number need to use the lambert w function.

02:07 And this is just a very complicated question...

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