Applying BigO

• We analyzed squares.py and determined that it's 'ideal' running time is T(n)=1+1+8n+1+n+7n^2+1+1.
• Exercise: Prove that T(n)=1+1+8n+1+n+7n^2+1+1 is in O(n^2).
• We let t(n) be the number of milliseconds that squares takes to solve a problem of size n (that is, with n points). Based on our above analysis, we conclude that (using Maple notation) t := n -> a2*n^2+a1*n+a0 where a2, a1, a0 are to be determined.
• We ran squares.py on a few inputs generated by writePoints.py (see squares.data) points:=[ [1000,78.9041519165], [3000,671.78606987], [5000,1709.86890793], [10000,6951.59602165], [20000,26423.099041] ] That is, t(1000)=78.9041519165 t(3000)=671.78606987 t(5000)=1709.86890793 t(10000)=6951.59602165 t(20000)=26423.099041
• We solve the linear system... solve({t(1000)=78.9041519165, t(3000)=671.78606987, t(5000)=1709.86890793}, {a2, a1, a0}) to get {a2 = 0.5565011501e-4, a1 = 0.7384049892e-1, a0 = -50.58646202}
• Now if we set a2:= 0.5565011501e-4; a1:= 0.7384049892e-1; a0:= -50.58646202; we can evaluate t for other input sizes... t(1000)= 78.90415188 t(3000)= 671.7860699 t(5000)= 1709.868908 t(10000)= 6252.830028 t(20000)= 23686.26952 t(40000)= 91943.21752 or plot them both together...(see plotting with maple) with(plots, textplot, display); plot1 := plot(points); plot2 := plot(t, 1000 .. 40000); display(plot1, plot2);