4.4: Interpreting results
- Page ID
- 12746
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Based on our understanding of how ArrayList works, we expect the add method to take constant time when we add elements to the end. So the total time to add n elements should be linear.
Figure \(\PageIndex{1}\): Profiling results: run time versus problem size for adding n elements to the end of an ArrayList.
To test that theory, we could plot total run time versus problem size, and we should see a straight line, at least for problem sizes that are big enough to measure accurately. Mathematically, we can write the function for that line:
\[ runtime = a + bn \nonumber \]
where a is the intercept of the line and b is the slope.
On the other hand, if add is linear, the total time for n adds would be quadratic. If we plot run time versus problem size, we expect to see a parabola. Or mathematically, something like:
\[ runtime = a + bn + cn^2 \nonumber \]
With perfect data, we might be able to tell the difference between a straight line and a parabola, but if the measurements are noisy, it can be hard to tell. A better way to interpret noisy measurements is to plot run time and problem size on a log-log scale.
Why? Let’s suppose that run time is proportional to \( n^k \), but we don’t know what the exponent k is. We can write the relationship like this:
\[ runtime - a + bn + \dots + cn^k \nonumber \]
For large values of n, the term with the largest exponent is the most important, so:
\[ runtime \approx cn^k \nonumber \]
where \( \approx \) means “approximately equal”. Now, if we take the logarithm of both sides of this equation:
\[ \log(runtime) \approx \log(c) + k\log(n) \nonumber \]
This equation implies that if we plot runtime versus n on a log-log scale, we expect to see a straight line with intercept \( \log(c) \) and slope k. We don’t care much about the intercept, but the slope indicates the order of growth: if \( k = 1\), the algorithm is linear; if \( k = 2 \), it’s quadratic.
Looking at the figure in the previous section, you can estimate the slope by eye. But when you call plotResults it computes a least squares fit to the data and prints the estimated slope. In this example:
Estimated slope = 1.06194352346708
which is close to 1; and that suggests that the total time for n adds is linear, so each add is constant time, as expected.
One important point: if you see a straight line on a graph like this, that does not mean that the algorithm is linear. If the run time is proportional to \( n^k \) for any exponent k, we expect to see a straight line with slope k. If the slope is close to 1, that suggests the algorithm is linear. If it is close to 2, it’s probably quadratic.