Assume that you run one of the executables for various N values, time them and record a table of values such as the following:
|N value||Time in seconds|
Say you decide to test these values to see if they're closer to O(N) or to O(N2). If they're O(N) then for large enough N they should obey Time = b*N where b is some constant (whose exact value is unimportant). Then the ratio of two times should obey Time1/Time2 = b*N1/b*N2 = N1/N2. For example we should have Time(N=32)/Time(N=16) = 32/16 = 2, whereas the actual ratio from the table you've collected is 144/40 = 3.6, which is a lot larger than 2.
So it appears as if the complexity is worse than O(N) so you try O(N2). In this case you should eventually see Time1/Time2 = b*N12/b*N22 = N12/N22. For the case we looked at before, comparing N=32 to N=16 gave us a ratio of 3.6 whereas 322/162 = (32/16)2 = 22 = 4. Now 3.6 is a lot closer to 4 than 2 is to 4. If you're patient enough to run the N=64 case and wait for about 9 minutes for it to complete, the ratio of the two times is 3.8 or so, even closer to 4.
In doing all these computations, you might want to add a few columns to your table and fill in some ratios. Also note that it is not necessary to compare the ratios of times for N values that are just a factor of 2 apart. In fact, for O(NlgN) behavior it is better to compare times for N values that are 4x or 8x or even 16x apart to see the difference between O(N) and O(NlgN) behavior. But of course don't use such small values of N that an interfering term could be deceiving you.
In conclusion, in this example we've found that as N gets large, the ratio of successive times for doubled N values is approaching 4, which would suggest O(N2) behavior.
Now I'll reveal the actual function that was used to create the values above. It is Time = (1/8)*(N2 + 4*N). This function eventually is dominated by the N2 term so it truly is O(N2). But for small values of N the linear 4*N factor contributes significantly. In fact, the ratio of the two contributions is N2/(4*N) = N/4, so for N=4 they contribute equally, whereas for N = 64, the second term is only 1/16 the first, or about 6% of the dominate term.
I think you'll find this lab interesting and instructive and not that difficult or time-consuming to do.