# Searching & Sorting Algorithms for Humans

05 Dec 2018

Searching and sorting are atomic concepts in Computer Science, and something I didn’t realise was going to be such a big thing before I learned to code. Efficient methods for rifling through data, or sorting it into order are gold dust. Thinking about the number of times your implementation is completely reliant on accessing data, you quickly realise how critical doing these basic operations quickly and accurately can be in any algorithm.

As a beginner to CS, I’m going to attempt to summarise and analyse the major algorithms one should be aware of in this post. I think they are generally fairly easy to understand and implement, and critical for those tech interviews where you have to reason between your quick and merge sorts.

I’m going to go through them one by one. In general optimal data structure choice is critical, however wherever I can I’m going to try and implement the above algorithms on a Python list. This is largely for simplicity.

## Selection Sort

### Motivation

This is amongst the simplest sorting algorithms, however it’s simplicity comes with a cost, namely it’s very expensive in terms of time to compute in comparison with other more nuanced algorithms.

We’d probably never want to implement this algorithm on it’s own, as it’s a bad choice compared to other sorting algorithms, but it could be useful as a component in more complex sorts.

Let’s motivate the algorithm with an example. Consider sorting a deck of numbered cards. As a human, the easiest way to do this is by finding the card with the lowest value and placing it at the beginning of the deck, the proceeding to the card with the second lowest value, etc, until all cards have been placed.

### Implementation

def smallest(l):
"""
Find the smallest element of a list, by iterating through the
whole list.
"""
idx_smallest = 0
smallest = l[idx_smallest]
found = False
i = 0
if l[i] <= smallest:
smallest = l[i]
idx_smallest = i

i += 1

if i == len(l):
found = True

return idx_smallest, smallest

def selection_sort(l):
"""
Sort by continuously finding the next smallest element and putting
that at the appropriate index.
"""

ordered = False
i = 0

while not ordered:

idx_smallest, val_smallest = smallest(l[i:])
if val_smallest <= l[i]:
l[i], l[i+idx_smallest] = l[i+idx_smallest], l[i]

i += 1

if i == len(l):
ordered = True

return l


### Analysis

As a minimum this algorithm will operate in $O(N^2)$ in the worst and average case. We can see this as being due to the need to iterate through the whole list $(N-i)$ times at the $i^{th}$ timestep in the sort in order to find the next smallest element. Giving a leading order complexity of $O(N^2)$ We can perform the entire sort in-place so we only take up $O(1)$ memory.

## Merge Sort

This is an interview classic. Importantly merge-sort follows the ‘divide and conquer’ paradigm for solving problems. This boils down to ‘dividing’ your problem into approachable sub-problems, solving these, and combining’ the sub-problem solutions together into a final solution.

Merge sort closely follows this paradigm. Intuitively, it operates by dividing an $n$ element list into two subsequences of $\frac{n}{2}$ elements each, sorting the subsequences, and finally merging the results. The recursion has a base case of when you have a list of only a single element which is already sorted.

Our major motivation here is to save on time-complexity, and the recursive division of our problem space gives us a clue to it being significantly less than $O(N^2)$ like for selection sort. However, an implementation example should clarify this.

To understand the merge operation a didactic example is sorting a deck of cards. Imagine two piles of cards, picking one up off one pile and one off the the other pile, comparing the two values and placing face down in a third pile the smallest card. We do this until one of the first two piles are empty, at which point we place the remaining cards from the non-empty pile down on the third pile.

We can sort by dividing the array to be sorted into two, and recursively calling sort on the two portions, finally merging the results.

### Implementation

def merge(left, right):
"""
Iterative merge operation.
"""
i, j = 0, 0
result = []

while i < len(left) and j < len(right):

if left[i] < right[j]:
result.append(left[i])
i+=1

else:
result.append(right[j])
j+=1

result.extend(left[i:])
result.extend(right[j:])

return result

def merge_sort(l):
"""
Recursively defined sort.
"""
if len(l) == 1:
return l

mid = len(l) // 2
left = merge_sort(l[:mid])
right = merge_sort(l[mid:])

return merge(left, right)


### Analysis

The implementation above is very neat, and pretty much directly translated from the pseudocode in CLRS (a book you should check out beyond this post).

What kind of complexity can we expect with this algorithm? Well, to get an idea we can estimate the number of elements in the recursion tree. This will give us an indication of the number of operations taking place, and therefore a time complexity. Inductively, we can infer that there are $\log(N) + 1$ levels in the recursion tree. For example, a list of length 1 would have $\log(1) + 1 = 0$ levels, as we only call the merge_sort function once. Furthermore, the complexity of the merge operation is $O(N)$ as we may have to iterate through the whole list in order to merge. So total complexity can be seen to be $O(N(\log(N) + 1)) = O(N\log(N))$. Space complexity is $O(N)$ due to the need to store the result of the merge operation.

## Quick Sort

Merge sort has a wasteful bit of memory usage in that we need to keep a record of the merged results at each step. Quick sort lets us avoid this by sorting in place. It’s worst case runtime is $O(N^2)$, however it has an average case runtime of $O(N\log(N))$ - same as merge sort, and the worst case can be avoided by a clever strategy on behalf of the programmer, however this will be much easier to explain in code, so let’s start with the implementation.

Similar to merge sort, quick sort applies the ‘divide-and-conquer’ paradigm. However, in order to understand exactly how, we need to understand the ‘partition’ procedure first, which is critical to the implementation of quick sort.

Partitioning is easier to grasp through an example. Consider the following array:

# unpartitioned array
array = [2, 1, 10, 3, 4, 9, 5]


Let’s choose the last element 5, as the ‘pivot’. We’re going to try and arrange the array such that when 5 is placed in it’s correct index all elements to it’s left are smaller than or equal to it, and all elements to it’s right are greater than or equal to it:

# partitioned array
array = [2, 1, 3, 4, 5, 9, 10]


Note that the array is still unsorted at this point.

The following function implements this partitioning:

def partition(l, low, high):
"""Partition function capable of operating on sub-arrays"""
pivot = l[high]
i = low - 1
for j in range(low, high):
if l[j] <= pivot:
i += 1
l[i], l[j] = l[j], l[i]

l[i + 1], l[high] = l[high], l[i + 1]

return i + 1


I like to think of this as moving a dividing line up the array from left to right (essentially the $i$ index above), and checking if everything to the left of this line is smaller than the pivot element, which we’ve chosen to be the last element of the array. Once the for loop has completed $i$ will equal the number of swaps, hence elements, that are smaller than the pivot - therefore will tell us where to place the pivot element, which happens in the last line. Furthermore, this all happens in place on the array, we just return the index of the pivot element.

Using this functionality, we can implement quick sort by recursively calling quicksort on partitioned arrays. The base case of partitioning an array with a single/no elements just returning itself.

### Implementation

def quick_sort(l, low, high):
"""Recursive quick sort implementation."""
if low < high:
p = partition(l, low, high)
quick_sort(l, low, p-1)
quick_sort(l, p+1, high)


### Analysis

We can see that we’ve again used the divide-and-conquer paradigm, where we divide with the partition function rather than naively down the middle like with merge sort. But what’s the runtime for quick sort? It’s doing quite a lot of complicated stuff, especially the partition function, so we should be a little more careful in our analysis.

Starting with the partitioning, in the worst case. This occurs when we the procedure produces a sub-problem to two with n elements and 0 elements respectively for every partition i.e. our array is anti-sorted [5, 4, 3, 2, 1] Then the partitioning will be of $O(N^2)$. Therefore in the worst case quick sort is as rubbish as insertion sort. However, this can be helped with clever choices of pivot - for example just picking a random pivot each time, or something that reflects the data you are actually trying to sort just so that you aren’t faced with this problem.

In the best case, we’d get two partitioned sub-problems of equal size. This is now of the same order of complexity as merge sort, as something approximately the same is happening, just differently! This results in a quick sort complexity of $O(N\log(N))$.

In fact, as detailed in CLRS sub-problems have to be extremely unbalanced to not achieve something approaching the best case performance, i.e. even if we get two sub-problems of size $\frac{9N}{10}$ and $\frac{N}{10}$ we’d still get $O(N\log(N))$ performance, however this is definitely something to detail in another post. For now I hope you’re happy to use this as a fact.

Binary search is very easy to understand and implement, and follows common sense. Assuming our array is sorted (which we are now armed to implement!). We first check the middle item of an array if our element is there we return, otherwise if our element is greater than this we check the top half using the same procedure, or vice versa the bottom half, until we find our element. This is quite naturally defined recursively.

### Implementation

def binary_search(l, element):
"""Recursive binary search implementation."""
mid = len(l) // 2

if l[mid] == element:
return mid

elif l[mid] > element:
binary_search(l[:mid-1], element)

else:
binary_search(l[mid+1:], element)


### Analysis

We are cutting our problem space in two every time as we head down the recursion tree. Therefore, the number of operations till we find our element is the sum of $N + N/2 + N/4 + … + 1$, where for $N$ elements we do $2^k$ operations, where k is the step. Hence we’d have the relationship $2^k = N$ between the number of elements in our array and the number of steps, so the runtime is $O(\log(N))$.

A good take away from this is that if your problem space is reducing in size with each step it’s complexity is likely to be log-ish.

It is possible, probably, to implement these algorithms using Python lists but to be honest it’s not worth the effort, as it might just obfuscate the logic of the algorithm - which is what I’m trying to get across. So for this section I’m going to mix things up a little, and implement the following algorithms defined against the following Graph data structure I’ve just made up. It’s important to note that this is a convenience object and doesn’t really serve any purpose other purpose than didactic.

from collections import defaultdict

class Graph:

def __init__(self):
self._graph = defaultdict(list)

def __iter__(self):
return iter((k, v) for k, v in self._graph.items())

def __getitem__(self, item):
return self._graph[item]

def __setitem__(self, u, v):
self._graph[u].append(v)

def __repr__(self):
return str(self._graph)

g = Graph()  # create graph object
g['a'] = 'b' # add nodes


If you’re interested in how graphs are implemented check out CLRS , I can’t sell this book enough. Here I’ve chosen to represent it with a adjacency list, using Python’s default dict object, which basically allows one to have a dict with keys which are lists. Here the keys represent nodes and the items in their lists represent adjacent nodes.

The defining feature of depth first search is that it exhausts a given route through a graph until it tries the next one. I.e. it will look for the children of a given node, and then one child’s children and so on until it’s reached the end, before moving onto another branch. In terms of a useful example, this would be a useful algorithm if we wanted to know when two distant nodes are connected, in contrast to breadth-first search as we’ll see, as it would prioritise getting to the target node in any way possible ASAP.

### Implementation

def visit(v, visited):
visited[v] = True

def dfs(root, graph, visited=None):
"""Operates on Graph objects"""
if not isinstance(visited, dict):
visited = {k: False for k, v in graph}

visit(root, visited)

for node in graph[root]:
if not visited[node]:
dfs(node, graph, visited)


### Analysis

Depth-first search visits every vertex once and checks every edge in the graph once. Therefore, DFS complexity is $O(V+E)$ This assumes that the graph is represented as an adjacency list.

Consider the motivating example for depth first search, finding whether two distant nodes are linked, similarly we might want to solve the converse problem - are two nearby nodes linked - for which depth first search would be rubbish. It might search most of the graph before realising that they are in fact next to each other! Breadth first search solves this problem by checking all of the child nodes of a given node first.

### Implementation

We can implement the algorithm using a queue, a very common error for candidates is apparently to assume that this can be implemented recursively. However if you think about it this can’t make sense, because you would need to step side-ways in your recursion tree rather than downwards, introducing another variable to keep a track of where you are! This could get very complicated, and it’s much much simpler to go iteratively.

from queue import Queue

def visit(v, visited):
visited[v] = True

def bfs(root, graph, visited=None, queue=None):
"""Operates on Graph objects"""

if not isinstance(visited, dict):
visited = {k: False for k, v in graph}

if not isinstance(queue, dict):
q = Queue()

visit(root, visited)

while q:
node_to_visit = q.get()

# visit child nodes, then put them in queue
for node in graph[node_to_visit]:
if not visited[node]:
visit(node, visited)
q.put(node)


### Analysis

This will actually have the same average runtime as breadth first search as the same operation is taking place. The worst case will depend on your application though, as discussed above.

## Conclusion

I hope you’ve enjoyed this whirlwind tour of some concepts in Computer Science, in all honesty this has served as a revision exericise for me. However I had fun writing it, and explaining something really makes you question your own understanding. I will be doing another post in the near future, doing the same thing but for atomic data structures.