Analysis of Algorithms¶

Code examples from Think Complexity, 2nd edition.

In [1]:

%matplotlib inline

import matplotlib.pyplot as plt
import numpy as np
import seaborn as sns

import os
import string

from utils import decorate, savefig

Empirical order of growth¶

Sometimes we can figure out what order of growth a function belongs to by running it with a range of problem sizes and measuring the run time.

To measure runtimes, we'll use etime, which uses os.times to compute the total time used by a process, including "user time" and "system time". User time is time spent running your code; system time is time spent running operating system code on your behalf.

In [2]:

def etime():
    """Measures user and system time this process has used.

    Returns the sum of user and system time."""
    user, sys, chuser, chsys, real = os.times()
    return user+sys

time_func takes a function object and a problem size, n, runs the function, and returns the elapsed time.

In [3]:

def time_func(func, n):
    """Run a function and return the elapsed time.
    
    func: function
    n: problem size
    
    returns: user+sys time in seconds
    """
    start = etime()
    func(n)
    end = etime()
    elapsed = end - start
    return elapsed

Here's an example: a function that makes a list with the given length using append.

In [4]:

def append_list(n):
    t = []
    for i in range(n):
        t.append(i)
    return t

time_func(append_list, 1000000)

Out[4]:

0.07999999999999985

run_timing_test takes a function, runs it with a range of problem sizes, and returns two lists: problem sizes and times.

In [5]:

def run_timing_test(func, max_time=1):
    """Tests the given function with a range of values for n.
    
    func: function object

    returns: list of ns and a list of run times.
    """
    ns = []
    ts = []
    for i in range(10, 28):
        n = 2**i
        t = time_func(func, n)
        print(n, t)
        if t > 0:
            ns.append(n)
            ts.append(t)
        if t > max_time:
            break

    return ns, ts

Here's an example with append_list

In [6]:

ns, ts = run_timing_test(append_list)

1024 0.0
2048 0.0
4096 0.0
8192 0.0
16384 0.010000000000000009
32768 0.0
65536 0.010000000000000009
131072 0.0
262144 0.020000000000000018
524288 0.040000000000000036
1048576 0.07000000000000006
2097152 0.1599999999999997
4194304 0.29000000000000004
8388608 0.5900000000000003
16777216 1.1099999999999999

fit takes the lists of ns and ts and fits it with a curve of the form a * n**exp, where exp is a given exponent and a is chosen so that the line goes through a particular point in the sequence, usually the last.

In [7]:

def fit(ns, ts, exp=1.0, index=-1):
    """Fits a curve with the given exponent.
    
    ns: sequence of problem sizes
    ts: sequence of times
    exp: exponent of the fitted curve
    index: index of the element the fitted line should go through
    
    returns: sequence of fitted times

    
    """
    # Use the element with the given index as a reference point, 
    # and scale all other points accordingly.
    nref = ns[index]
    tref = ts[index]

    tfit = []
    for n in ns:
        ratio = n / nref
        t = ratio**exp * tref
        tfit.append(t)

    return tfit

plot_timing_test plots the results.

In [8]:

def plot_timing_test(ns, ts, label='', color='blue', exp=1.0, scale='log'):
    """Plots data and a fitted curve.

    ns: sequence of n (problem size)
    ts: sequence of t (run time)
    label: string label for the data curve
    color: string color for the data curve
    exp: exponent (slope) for the fitted curve
    """
    tfit = fit(ns, ts, exp)
    fit_label = 'exp = %d' % exp
    plt.plot(ns, tfit, label=fit_label, color='0.7', linestyle='dashed')
    plt.plot(ns, ts, 'o-', label=label, color=color, alpha=0.7)
    plt.xlabel('Problem size (n)')
    plt.ylabel('Runtime (seconds)')
    plt.xscale(scale)
    plt.yscale(scale)
    plt.legend()

Here are the results from append_list. When we plot ts versus ns on a log-log scale, we should get a straight line. If the order of growth is linear, the slope of the line should be 1.

In [9]:

plot_timing_test(ns, ts, 'append_list', exp=1)

For small values of n, the runtime is so short that we're probably not getting an accurate measurement of just the operation we're interested in. But as n increases, runtime seems to converge to a line with slope 1.

That suggests that performing append n times is linear, which suggests that a single append is constant time.

list.pop¶

Now let's try that with pop, and specifically with pop(0), which pops from the left side of the list.

In [10]:

def pop_left_list(n):
    t= []
    for i in range(n):
        t.append(i)
    for _ in range(n):
        t.pop(0)
    return t

ns, ts = run_timing_test(pop_left_list)
plot_timing_test(ns, ts, 'pop_left_list', exp=1)

1024 0.0
2048 0.0
4096 0.009999999999999787
8192 0.009999999999999787
16384 0.040000000000000036
32768 0.1299999999999999
65536 0.6299999999999999
131072 2.7600000000000007

That doesn't look very good. The runtimes increase more steeply than the line with slope 1. Let's try slope 2.

In [11]:

plot_timing_test(ns, ts, 'pop_left_list', exp=2)

The last few points converge on the line with slope 2, which suggests that pop(0) is quadratic.

Exercise: What happens if you pop from the end of the list? Write a function called pop_right_list that pops the last element instead of the first. Use run_timing_test to characterize its performance. Then use plot_timing_test with a few different values of exp to find the slope that best matches the data. What conclusion can you draw about the order of growth for popping an element from the end of a list?

In [12]:

# Solution

def pop_right_list(n):
    t= []
    for i in range(n):
        t.append(i)
    for _ in range(n):
        t.pop(-1)
    return t

ns, ts = run_timing_test(pop_right_list)
plot_timing_test(ns, ts, 'pop_right_list', exp=1)

1024 0.0
2048 0.0
4096 0.0
8192 0.0
16384 0.0
32768 0.009999999999999787
65536 0.009999999999999787
131072 0.030000000000001137
262144 0.03999999999999915
524288 0.08000000000000185
1048576 0.16000000000000014
2097152 0.3099999999999987
4194304 0.6699999999999999
8388608 1.2800000000000011

Sorting¶

We expect sorting to be n log n. On a log-log scale, that doesn't look like a straight line, so there's no simple test whether it's really n log n. Nevertheless, we can plot results for sorting lists with different lengths, and see what it looks like.

In [13]:

def sort_list(n):
    t = np.random.random(n)
    t.sort()
    return t

ns, ts = run_timing_test(sort_list)
plot_timing_test(ns, ts, 'sort_list', exp=1)

1024 0.0
2048 0.0
4096 0.009999999999999787
8192 0.0
16384 0.0
32768 0.0
65536 0.009999999999999787
131072 0.009999999999999787
262144 0.019999999999999574
524288 0.040000000000000924
1048576 0.08000000000000007
2097152 0.16999999999999993
4194304 0.34999999999999964
8388608 0.75
16777216 1.549999999999999

It sure looks like sorting is linear, so that's surprising. But remember that log n changes much more slowly than n. Over a wide range of values, n log n can be hard to distinguish from an algorithm with linear growth. As n gets bigger, we would expect this curve to be steeper than slope 1. But often, for practical problem sizes, n log n is as good as linear.

Bisection search¶

We expect bisection search to be log n, which is so fast it is hard to measure the way we've been doing it.

To make it work, I create the sorted list ahead of time and use the parameter hi to specify which part of the list to search. Also, I have to run each search 100,000 times so it takes long enough to measure.

In [14]:

t = np.random.random(16777216)
t.sort()

from bisect import bisect

def search_sorted_list(n):
    for i in range(100000):
        index = bisect(t, 0.1, hi=n) 
    return index

ns, ts = run_timing_test(search_sorted_list, max_time=0.2)
plot_timing_test(ns, ts, 'search_sorted_list', exp=1)

1024 0.16999999999999815
2048 0.19000000000000128
4096 0.1999999999999993
8192 0.21000000000000085

It looks like the runtime increases slowly as n increases, but it's definitely not linear. To see if it's constant time, we can compare it to the line with slope 0.

In [15]:

plot_timing_test(ns, ts, 'search_sorted_list', exp=0)

Nope, looks like it's not constant time, either. We can't really conclude that it's log n based on this curve alone, but the results are certainly consistent with that theory.

Dictionary methods¶

Exercise: Write methods called add_dict and lookup_dict, based on append_list and pop_left_list. What is the order of growth for adding and looking up elements in a dictionary?

In [16]:

# Solution

def add_dict(n):
    d = {}
    for i in range(n):
        d[i] = 1
    return d

ns, ts = run_timing_test(add_dict)
plot_timing_test(ns, ts, 'add_dict', exp=1)

1024 0.0
2048 0.0
4096 0.0
8192 0.0
16384 0.010000000000001563
32768 0.0
65536 0.00999999999999801
131072 0.010000000000001563
262144 0.019999999999999574
524288 0.05000000000000071
1048576 0.13000000000000256
2097152 0.2099999999999973
4194304 0.4299999999999997
8388608 0.7699999999999996
16777216 1.6000000000000014

In [17]:

# Solution

def lookup_dict(n):
    d = {}
    for i in range(n):
        d[i] = 1
    total = 0
    for i in range(n):
        total += d[i]
    return d

ns, ts = run_timing_test(lookup_dict)
plot_timing_test(ns, ts, 'lookup_dict', exp=1)

1024 0.0
2048 0.0
4096 0.0
8192 0.010000000000001563
16384 0.0
32768 0.0
65536 0.019999999999999574
131072 0.019999999999999574
262144 0.05999999999999872
524288 0.07000000000000028
1048576 0.16999999999999815
2097152 0.33999999999999986
4194304 0.6799999999999997
8388608 1.3500000000000014

In [18]:

# Solution

# Adding `n` elements to a dictionary takes linear time, 
# so adding a single element is constant time.

# Same with looking up `n` elements.

Implementing a hash table¶

The reason Python dictionaries can add and look up elements in constant time is that they are based on hash tables. In this section, we'll implement a hash table in Python. Remember that this example is for educational purposes only. There is no practical reason to write a hash table like this in Python.

We'll start with a simple linear map, which is a list of key-value pairs.

In [19]:

class LinearMap(object):
    """A simple implementation of a map using a list of tuples
    where each tuple is a key-value pair."""

    def __init__(self):
        self.items = []

    def add(self, k, v):
        """Adds a new item that maps from key (k) to value (v).
        Assumes that they keys are unique."""
        self.items.append((k, v))

    def get(self, k):
        """Looks up the key (k) and returns the corresponding value,
        or raises KeyError if the key is not found."""
        for key, val in self.items:
            if key == k:
                return val
        raise KeyError

First let's make sure it works:

In [20]:

def test_map(m):
    s = string.ascii_lowercase

    for k, v in enumerate(s):
        m.add(k, v)

    for k in range(len(s)):
        print(k, m.get(k))

In [21]:

m = LinearMap()
test_map(m)

0 a
1 b
2 c
3 d
4 e
5 f
6 g
7 h
8 i
9 j
10 k
11 l
12 m
13 n
14 o
15 p
16 q
17 r
18 s
19 t
20 u
21 v
22 w
23 x
24 y
25 z

Now let's see how long it takes to add n elements.

In [22]:

def add_linear_map(n):
    d = LinearMap()
    for i in range(n):
        d.add(i, 1)
    return d

ns, ts = run_timing_test(add_linear_map)
plot_timing_test(ns, ts, 'add_linear_map', exp=1)

1024 0.010000000000005116
2048 0.0
4096 0.0
8192 0.0
16384 0.00999999999999801
32768 0.00999999999999801
65536 0.00999999999999801
131072 0.030000000000001137
262144 0.060000000000002274
524288 0.14000000000000057
1048576 0.25
2097152 0.5599999999999987
4194304 1.0800000000000018

Adding n elements is linear, so each add is constant time. How about lookup?

In [23]:

def lookup_linear_map(n):
    d = LinearMap()
    for i in range(n):
        d.add(i, 1)
    total = 0
    for i in range(n):
        total += d.get(i)
    return d

ns, ts = run_timing_test(lookup_linear_map)
plot_timing_test(ns, ts, 'lookup_linear_map', exp=2)

1024 0.020000000000003126
2048 0.06999999999999673
4096 0.25
8192 1.0400000000000027

Looking up n elements is $O(n^2)$ (notice that exp=2). So each lookup is linear.

Let's see what happens if we break the list of key-value pairs into 100 lists.

In [24]:

class BetterMap(object):
    """A faster implementation of a map using a list of LinearMaps
    and the built-in function hash() to determine which LinearMap
    to put each key into."""

    def __init__(self, n=100):
        """Appends (n) LinearMaps onto (self)."""
        self.maps = []
        for i in range(n):
            self.maps.append(LinearMap())

    def find_map(self, k):
        """Finds the right LinearMap for key (k)."""
        index = hash(k) % len(self.maps)
        return self.maps[index]

    def add(self, k, v):
        """Adds a new item to the appropriate LinearMap for key (k)."""
        m = self.find_map(k)
        m.add(k, v)

    def get(self, k):
        """Finds the right LinearMap for key (k) and looks up (k) in it."""
        m = self.find_map(k)
        return m.get(k)

In [25]:

m = BetterMap()
test_map(m)

0 a
1 b
2 c
3 d
4 e
5 f
6 g
7 h
8 i
9 j
10 k
11 l
12 m
13 n
14 o
15 p
16 q
17 r
18 s
19 t
20 u
21 v
22 w
23 x
24 y
25 z

The run time is better (we get to a larger value of n before we run out of time).

In [26]:

def lookup_better_map(n):
    d = BetterMap()
    for i in range(n):
        d.add(i, 1)
    total = 0
    for i in range(n):
        total += d.get(i)
    return d

ns, ts = run_timing_test(lookup_better_map)
plot_timing_test(ns, ts, 'lookup_better_map', exp=1)

1024 0.0
2048 0.010000000000001563
4096 0.00999999999999801
8192 0.030000000000001137
16384 0.08999999999999986
32768 0.3000000000000007
65536 1.1700000000000017

The order of growth is hard to characterize. It looks steeper than the line with slope 1. Let's try slope 2.

In [27]:

plot_timing_test(ns, ts, 'lookup_better_map', exp=2)

It might be converging to the line with slope 2, but it's hard to say anything conclusive without running larger problem sizes.

Exercise: Go back and run run_timing_test with a larger value of max_time and see if the run time converges to the line with slope 2. Just be careful not to make max_time to big.

Now we're ready for a complete implementation of a hash map.

In [28]:

class HashMap(object):
    """An implementation of a hashtable using a BetterMap
    that grows so that the number of items never exceeds the number
    of LinearMaps.

    The amortized cost of add should be O(1) provided that the
    implementation of sum in resize is linear."""

    def __init__(self):
        """Starts with 2 LinearMaps and 0 items."""
        self.maps = BetterMap(2)
        self.num = 0

    def get(self, k):
        """Looks up the key (k) and returns the corresponding value,
        or raises KeyError if the key is not found."""
        return self.maps.get(k)

    def add(self, k, v):
        """Resize the map if necessary and adds the new item."""
        if self.num == len(self.maps.maps):
            self.resize()

        self.maps.add(k, v)
        self.num += 1

    def resize(self):
        """Makes a new map, twice as big, and rehashes the items."""
        new_map = BetterMap(self.num * 2)

        for m in self.maps.maps:
            for k, v in m.items:
                new_map.add(k, v)

        self.maps = new_map

In [29]:

m = HashMap()
test_map(m)

0 a
1 b
2 c
3 d
4 e
5 f
6 g
7 h
8 i
9 j
10 k
11 l
12 m
13 n
14 o
15 p
16 q
17 r
18 s
19 t
20 u
21 v
22 w
23 x
24 y
25 z

Exercise: Write a function called lookup_hash_map, based on lookup_better_map, and characterize its run time.

If things go according to plan, the results should converge to a line with slope 1. Which means that n lookups is linear, which means that each lookup is constant time. Which is pretty much magic.

In [30]:

# Solution

def lookup_hash_map(n):
    d = HashMap()
    for i in range(n):
        d.add(i, 1)
    total = 0
    for i in range(n):
        total += d.get(i)
    return d

ns, ts = run_timing_test(lookup_hash_map)
plot_timing_test(ns, ts, 'lookup_hash_map', exp=1)

1024 0.0
2048 0.010000000000005116
4096 0.0799999999999983
8192 0.030000000000001137
16384 0.11999999999999744
32768 0.19000000000000483
65536 0.4399999999999977
131072 0.8599999999999994
262144 1.5999999999999943

In [ ]: