IT代考 # Part 3: Visualizing Order of Growth – cscodehelp代写
# Part 3: Visualizing Order of Growth
In this part, you’ll get to visually experiment with different C
values and lower bounds on N in order to show that a function fits a
certain order of growth. To start, do some imports:
“`python
import pandas as pd
import matplotlib
from matplotlib import pyplot as plt
from math import log2, log10, ceil
“`
Also remember to do `%matplotlib inline`
Now paste the following code in a cell (you don’t need to understand it for the purposes of this lab):
“`python
matplotlib.rcParams[“font.size”] = 20
def get_ax():
fig, ax = plt.subplots(figsize=(8,6))
ax.spines[“right”].set_visible(False)
ax.spines[“top”].set_visible(False)
ax.set_xlim(1, 10)
return ax
def scale_ax():
ax = get_ax()
ax.set_xlabel(“N (data size)”)
ax.set_ylabel(“Steps”)
return ax
def plot_func(ax, f, C=1, color=”k”, label=”work”):
start = ax.get_xlim()[0]
width = ax.get_xlim()[1] – ax.get_xlim()[0]
s = pd.Series([],dtype=float)
for i in range(100):
N = start + width * (i+1)/100
s[N] = eval(f)
s.sort_index().plot(ax=ax, color=color, linewidth=3, label=label)
plt.text(s.index[-1], s.iloc[-1], f, verticalalignment=’center’)
def upper_bound(ax, order, C=1, minN=None):
f = order
if C != 1:
f = “C * (%s)” % order
plot_func(ax, f, C=C, color=”r”, label=”upper bound”)
if minN != None:
ax.axvspan(minN, ax.get_xlim()[1], color=’0.85′)
ax.legend(frameon=False)
“`
## Exercise 1: show `N+100` is in `O(N)`.
You need to show that some multiple of `N` is an upper bound on
`N+100` for large values of `N`. Paste+run this code:
“`python
ax = scale_ax()
ax.set_xlim(0, 10) # TODO: change upper bound
plot_func(ax, “N + 100″)
upper_bound(ax, order=”N”) # TODO: pass C and minN
“`
It should look like this:
Clearly `g(N)=N` is not an upper bound on `f(N)=N+100`, but remember
that you are allowed to do the following:
1. multiple `g(N)` by a constant `C`
2. require that `N` be large
To use your first capability, pass in a `C` value of of 25, changing the call to `upper_bound(…)` to be like this:
“`python
upper_bound(ax, order=”N”, C=25)
“`
It should look like this:
Better, but now the red line is only an upper bound for large `N`
values. Let’s set a lower bound on `N`, like this:
“`python3
upper_bound(ax, order=”N”, C=25, minN=6)
“`
It should look like this:
Great!
In general for all of these exercises, your job is to choose `C` and
`minN` values so that the red line is above the black line in the
shaded portion.
Finally, you should make sure the upper bound keeps holding for large
N values. If this were a math course, you would prove this is true
for all large N values. But since this is a programming course, let’s
just make the `xlim` really big and make sure there are no obvious
problems, like this:
“`python
ax.set_xlim(0, 1e6) # 1 million
“`
Looks good!
## Exercise 2: show `100*N` is in `O(N)`.
Start with this:
“`python
ax = scale_ax()
ax.set_xlim(0, 10)
plot_func(ax, “100*N”)
upper_bound(ax, order=”N”)
“`
Do you need to set `minN` for this one, or is choosing the right `C` good enough?
## Exercise 3: show `N**2 + 5*N + 25` is in `O(N**2)`
Start with this:
“`python
ax = scale_ax()
ax.set_xlim(0, 10)
plot_func(ax, “N**2 + 5*N + 25″)
upper_bound(ax, order=”N**2”)
“`
## Exercise 4: *try* to show `2*N + (N/15)**4` is in `O(N)`
Start with this:
“`python
ax = scale_ax()
ax.set_xlim(0, 10)
plot_func(ax, “2*N + (N/15)**4″)
upper_bound(ax, order=”N”)
“`
Use `C=3` and `minN=1`.
What happens when you increase the x limit to 75?
“`python
ax.set_xlim(0, 75)
“`
It turns out `2*N + (N/15)**4` is NOT in `O(N)` after all. What
Big O order of growth does `2*N + (N/15)**4` have?
## Exercise 5: show `log2(N)` is in `O(log10(N))`
Start with this:
“`python
ax = scale_ax()
ax.set_xlim(1, 100)
plot_func(ax, “log2(N)”)
upper_bound(ax, order=”log10(N)”)
“`
Any `C` value that is at least `log2(10)` will work here. In general,
any log curve with base M is just a constant multiple of another log
curve with base N. Thus, it is common to just say an algorithm is
`O(log N)`, without bothering to specify the base.
## Exercise 6: show `ceil(log2(N))` is in `O(log N)`
Start with this:
“`python
ax = scale_ax()
ax.set_xlim(1, 100)
plot_func(ax, “ceil(log2(N))”)
upper_bound(ax, order=”log2(N)”)
“`
## Exercise 7: show `N * ceil(log2(N))` is in `O(N log N)`
Start with this:
“`python
ax = scale_ax()
ax.set_xlim(1, 100)
plot_func(ax, “N * ceil(log2(N))”)
upper_bound(ax, order=”N * log2(N)”)
“`
## Exercise 8: show `F(N) = 0+1+2+…+(N-1)` is in `O(N**2)`
Replace `????` in the following:
“`python
ax = scale_ax()
ax.set_xlim(1, 100)
plot_func(ax, “sum(range(int(N)))”)
upper_bound(ax, order=????)
“`
