Algorithmic run time is a bit different from the run time of a program. Since an algorithm is simply an idea, there's no limit to the processing speed for evaluating the algorithm. This means that an expression of algorithmic run time in minutes or seconds is meaningless.

Without factors such as processor speed and architecture, the important unknown for an algorithm is input size. A sorting algorithm running on 1,000 elements will certainly take longer than the same sorting algorithm running on 10 elements. The input size is generally denoted by n, and each atomic step can be expressed as a number. The run time of a simple algorithm, such as the one that follows, can be expressed in terms of n.

for(i = 1 to n) {

Do something;

Do another thing;

}

Do one last thing;

This algorithm loops n times, each time doing two actions, and then does one last action, so the time complexity for this algorithm would be 2n + 1. A more complex algorithm with an additional nested loop tacked on, shown below, would have a time complexity of n^{2} + 2n + 1, since the new action is executed n^{2} times.

for(x = 1 to n) {

for(y = 1 to n) {

Do the new action;

}

}

for(i = 1 to n) {

Do something;

Do another thing;

}

Do one last thing;

But this level of detail for time complexity is still too granular. For example, as n becomes larger, the relative difference between 2n + 5 and 2n + 365 becomes less and less. However, as n becomes larger, the relative difference between 2n^{2} + 5 and 2n + 5 becomes larger and larger. This type of generalized trending is what is most important to the run time of an algorithm.

Consider two algorithms, one with a time complexity of 2n + 365 and the other with 2n^{2} + 5. The 2n^{2} + 5 algorithm will outperform the 2n + 365 algorithm on small values for n. But for n = 30, both algorithms perform equally, and for all n greater than 30, the 2n + 365 algorithm will outperform the 2n^{2} + 5 algorithm. Since there are only 30 values for n in which the 2n^{2} + 5 algorithm performs better, but an infinite number of values for nin which the 2n + 365 algorithm performs better, the 2n + 365 algorithm is generally more efficient.

This means that, in general, the growth rate of the time complexity of an algorithm with respect to input size is more important than the time complexity for any fixed input. While this might not always hold true for specific real-world applications, this type of measurement of an algorithm's efficiency tends to be true when averaged over all possible applications.

#### Asymptotic Notation

Asymptotic notation is a way to express an algorithm's efficiency. It's called asymptotic because it deals with the behavior of the algorithm as the input size approaches the asymptotic limit of infinity.

Returning to the examples of the 2n + 365 algorithm and the 2n^{2} + 5 algorithm, we determined that the 2n + 365 algorithm is generally more efficient because it follows the trend of n, while the 2n^{2} + 5 algorithm follows the general trend of n^{2}. This means that 2n + 365 is bounded above by a positive multiple of n for all sufficiently large n, and 2n^{2} + 5 is bounded above by a positive multiple of n^{2} for all sufficiently large n.

This sounds kind of confusing, but all it really means is that there exists a positive constant for the trend value and a lower bound on n, such that the trend value multiplied by the constant will always be greater than the time complexity for all n greater than the lower bound. In other words, 2n^{2} + 5 is in the order of n^{2}, and 2n + 365 is in the order of n. There's a convenient mathematical notation for this, called big-oh notation, which looks like O(n^{2}) to describe an algorithm that is in the order of n^{2}.

A simple way to convert an algorithm's time complexity to big-oh notation is to simply look at the high-order terms, since these will be the terms that matter most as n becomes sufficiently large. So an algorithm with a time complexity of 3n^{4} + 43n^{3} + 763n + log n + 37 would be in the order of O(n^{4}), and 54n^{7} + 23n^{4} + 4325 would be O(n^{7}).