# 8.5. Faster Computer, or Faster Algorithm?¶

## 8.5.1. Faster Computer, or Faster Algorithm?¶

Imagine that you have a problem to solve, and you know of an algorithm whose running time is proportional to \(n^2\) where \(n\) is a measure of the input size. Unfortunately, the resulting program takes ten times too long to run. If you replace your current computer with a new one that is ten times faster, will the \(n^2\) algorithm become acceptable? If the problem size remains the same, then perhaps the faster computer will allow you to get your work done quickly enough even with an algorithm having a high growth rate. But a funny thing happens to most people who get a faster computer. They don’t run the same problem faster. They run a bigger problem! Say that on your old computer you were content to sort 10,000 records because that could be done by the computer during your lunch break. On your new computer you might hope to sort 100,000 records in the same time. You won’t be back from lunch any sooner, so you are better off solving a larger problem. And because the new machine is ten times faster, you would like to sort ten times as many records.

If your algorithm’s growth rate is linear (i.e., if the equation that
describes the running time on input size \(n\) is
\(\mathbf{T}(n) = cn\) for some constant \(c\)),
then 100,000 records on the new machine will be sorted in the same
time as 10,000 records on the old machine.
If the algorithm’s growth rate is greater than \(cn\),
such as \(c_1n^2\), then you will *not* be able to do a
problem ten times the size in the same amount of time on a machine
that is ten times faster.

How much larger a problem can be solved in a given amount of time by a faster computer? Assume that the new machine is ten times faster than the old. Say that the old machine could solve a problem of size \(n\) in an hour. What is the largest problem that the new machine can solve in one hour? The following table shows how large a problem can be solved on the two machines for five running-time functions.

Table 8.5.1

The increase in problem size that can be run in a fixed period of time on a computer that is ten times faster. The first column lists the right-hand sides for five growth rate equations. For the purpose of this example, arbitrarily assume that the old machine can run 10,000 basic operations in one hour. The second column shows the maximum value for \(n\) that can be run in 10,000 basic operations on the old machine. The third column shows the value for \(n'\), the new maximum size for the problem that can be run in the same time on the new machine that is ten times faster. Variable \(n'\) is the greatest size for the problem that can run in 100,000 basic operations. The fourth column shows how the size of \(n\) changed to become \(n'\) on the new machine. The fifth column shows the increase in the problem size as the ratio of \(n'\) to \(n\).

This table illustrates many important points.
The first two equations are both linear; only the value of the
constant factor has changed.
In both cases, the machine that is ten times faster gives an increase
in problem size by a factor of ten.
In other words, while the value of the constant
does affect the absolute size of the problem that can be solved in a
fixed amount of time, it does not affect the *improvement* in
problem size (as a proportion to the original size) gained by a faster
computer.
This relationship holds true regardless of the algorithm’s growth
rate:
Constant factors never affect the relative improvement gained
by a faster computer.

An algorithm with time equation \(\mathbf{T}(n) = 2n^2\) does not
receive nearly as great an improvement from the faster machine as an
algorithm with linear growth rate.
Instead of an improvement by a factor of ten, the improvement
is only the square root of that: \(\sqrt{10} \approx 3.16\).
Thus, the algorithm with higher growth rate not only solves a smaller
problem in a given time in the first place, it *also*
receives less of a speedup from a faster computer.
As computers get ever faster, the disparity in problem sizes becomes
ever greater.

The algorithm with growth rate \(\mathbf{T}(n) = 5 n \log n\) improves by a greater amount than the one with quadratic growth rate, but not by as great an amount as the algorithms with linear growth rates.

Note that something special happens in the case of the algorithm whose running time grows exponentially. If you look at its plot on a graph, the curve for the algorithm whose time is proportional to \(2^n\) goes up very quickly as \(n\) grows. The increase in problem size on the machine ten times as fast is about \(n + 3\) (to be precise, it is \(n + \log_2 10\)). The increase in problem size for an algorithm with exponential growth rate is by a constant addition, not by a multiplicative factor. Because the old value of \(n\) was 13, the new problem size is 16. If next year you buy another computer ten times faster yet, then the new computer (100 times faster than the original computer) will only run a problem of size 19. If you had a second program whose growth rate is \(2^n\) and for which the original computer could run a problem of size 1000 in an hour, than a machine ten times faster can run a problem only of size 1003 in an hour! Thus, an exponential growth rate is radically different than the other growth rates shown in the table. The significance of this difference is an important topic in computational complexity theory.

Instead of buying a faster computer, consider what happens if you replace an algorithm whose running time is proportional to \(n^2\) with a new algorithm whose running time is proportional to \(n \log n\). In a graph relating growth rate functions to input size, a fixed amount of time would appear as a horizontal line. If the line for the amount of time available to solve your problem is above the point at which the curves for the two growth rates in question meet, then the algorithm whose running time grows less quickly is faster. An algorithm with running time \(\mathbf{T}n=n^2\) requires \(1024 \times 1024 = 1,048,576\) time steps for an input of size \(n=1024\). An algorithm with running time \(\mathbf{T}(n) = n \log n\) requires \(1024 \times 10 = 10,240\) time steps for an input of size \(n = 1024\), which is an improvement of much more than a factor of ten when compared to the algorithm with running time \(\mathbf{T}(n) = n^2\). Because \(n^2 > 10 n \log n\) whenever \(n > 58\), if the typical problem size is larger than 58 for this example, then you would be much better off changing algorithms instead of buying a computer ten times faster. Furthermore, when you do buy a faster computer, an algorithm with a slower growth rate provides a greater benefit in terms of larger problem size that can run in a certain time on the new computer.