Algorithms

These algorithms are coded in a python-like pseudocode, but in some cases the pseudocode is actually valid python!

Complexity

In the analysis of algorithms, many assumptions about hardware and basic operations must be made, e.g array access is constant time.

Sorting Algorithms

• Sorting is important because we can search sorted arrays very efficiently, in $O(\log n)$ time.
• In the worst case, $\Theta(n)$ exchanges will be needed.
• Comparison-based algorithms require at least $\Omega(n\lg n)$ comparisons.

Insertion Sort

Maintain a sorted section of the array, then insert new items into the correct position.

def insertion_sort(a):
for i in range(1, len(a)):
# assert first i positions sorted
j = i - 1
while j >= 0 and a[j] > a[j+1]:
swap(j, j+1)
j -= 1

• Inserting the last element needs at most $n-1$ comparisons and swaps. The second last element requires $n-2$…
• So insertion sort is $O(n^2)$. That said, it has a very small constant term so is often faster than $O(n \lg n)$ algorithms for small n.
• It is stable as long as we only swap if the element is larger than the key.

Selection sort

At each iteration, find the minimum of the remaining array and swap it to the current index.

def selection_sort(a):
for i in range(len(a)):
swap(i, argmin(a[i:end]))


$O(n^2)$ and unstable. Its only advantage is that it is easy to analyse.

Binary insertion sort

Same as insertion sort, except we find the correct position using binary partitioning.

def binary_insertion_sort(a):
for i in range(1, len(a)):
hi = i
lo = 0
while lo < hi:
j = (hi + lo) // 2
if a[i] > a[j]:
lo = j + 1
else:
hi = j

# Swap a[i] into the right place
tmp = a[i]
for j from i - 1 down to (hi - 1):
a[j+1] = a[j]
a[hi] = tmp


Binary insertion sort will be preferred to insertion sort when comparisons are expensive, but the swapping costs mean that it is still $O(n^2)$.

Bubblesort

In each pass, go through the list swapping adjacent elements as needed. If no swaps are done in a pass, the array is sorted.

def bubblesort(a):
while True:
didSwap = False
for i in range(len(a) - 1):
if a[i] > a[i+1]:
swap(i, i+1)
didSwap = True
if not didSwap:
break

• In the worst case, an element will be n positions away from its final position, so the complexity is $O(n^2)$.
• Stable

Mergesort

Divide and conquer algorithm that splits the list in two then recursively sorts each half, before merging sorted lists.

def mergesort(a, lo, hi):
if lo < hi:
mid = (lo + hi) // 2
mergesort(a, lo, mid)
mergesort(a, mid+1, hi)
merge(a, lo, mid, hi)

def merge(a, lo, mid, hi):
# both these subarrays are sorted
l = a[lo: mid]
r = a[mid+1 : hi]

aux = [] * (len(l) + len(r))

i = lo
j = mid + 1

for k in range(len(aux)):
if i > mid: # fill using right only
aux[k] = aux[j]
j += 1
else if j > hi: # fill using left only
aux[k] = a[i]
i += 1
else if a[i] <= a[j]: # otherwise compare
aux[k] = a[i]
i += 1
else:
aux[k] = a[j]
j++

• $\Theta (n \lg n)$ runtime, but requires $O(n)$ extra space.
• Mergesort is stable because there is no reordering of equal elements.
• Mergesort can instead be implemented bottom-up, merging pairs, then pairs of pairs, then pairs of fours, etc.
def mergesort(a):
step = 1
while (step < len(a)):
for lo in range(0, len(a), 2*step):
mid = lo + step - 1;
hi = min(lo + 2*step - 1, len(a) - 1);
merge(aux, lo, mid, hi);


Quicksort

Choose the last item as the pivot, then partition the array into items ≤ the pivot and items > pivot. Put the pivot in the middle then recursively sort left and right.

def quicksort(a):
pivot = a[len(a) - 1]
i = 0
j = len(a) - 2

while i <= j:
if a[i] > pivot and a[j] <= pivot:
swap(i, j)
i += 1
j -= 1
else if a[i] <= pivot:
i += 1
else: j -= 1

# ASSERT i == j + 1
# ASSERT all items to the left of i <= pivot
swap(j, len(a) - 1)
quicksort(a[0:j])
quicksort(a[j+1:end])

• $O(n \lg n)$ average case, $O(n^2)$ worst case.
• Requires $O(\lg n)$ additional space to store stack frames, but $O(n)$ in the worst case.
• Unstable.

Order statistics

A quicksort-like algorithm can be used to compute the median and, more generally, order statistics.

1. Select a pivot and partition the array into subarrays of size $p$ and $n-p$
2. If $k < p$, recursively look for the kth item in the lower partition.
3. Otherwise recurse into the upper partition to find rank $k-p$

This has recurrence:

However, the worst case is $O(n^2)$ as with quicksort. There exists a guaranteed linear time algorithm but it is much more complicated.

Heapsort

1. Turn the array into a max-heap in $O(n)$
2. Swap last item with max, reduce heapsize, then heapify down.
3. Repeat until heapsize is 0.
def heapsort(a):
for i in range(len(a) // 2, 0 included):
heapify(a[i], i, len(a))

for k in range(len(a), 1):
# a[0:k] is a max-heap
# a[k:end] is sorted
swap(0, k - 1)
heapify(a, 0, k-1)

def heapify(a, iRoot, iEnd):
if a[iRoot] satisfies max-heap:
return

j = largest child of iRoot
swap(iRoot, j)
heapify(a, j, iEnd)


Runtime $O(n \lg n)$

Counting sort

Counting sort does not require comparisons. Assuming that the inputs are positive integers within some range, it counts the number of each element, then finds the cumulative sum, from which we can identify exactly where a given element should go.

def counting_sort(a):
count = [0] * max(a)
for x in a:
count[x] += 1
# cumulate
for i in range(1, len(count)):
count[i] += count[i-1]

sorted = [0] * len(a)
for i in range(len(a) - 1, 0 included):
idx = count[a[i]] - 1
sorted[idx] = a[i]
count[a[i]] -= 1
return sorted

• It is a stable sorting algorithm, with $\Theta(n)$ cost.

Bucketsort

Bucketsort creates an array of buckets (linked lists), with the assumption that elements will fall into these buckets uniformly. We can then run insertion sort within each bucket before concatenating the buckets into a sorted array.

def bucket_sort(a):
# assuming that elements are drawn uniformly from [0,1]
n = len(a)
bucketWidth = 1/n
buckets = new [] of length n

for x in a:
idx = int(x / bucketWidth)
bucket[idx].next = x

sorted = []
for b in buckets:
insertion_sort(b)
sorted.append(b)
return b


We use insertion sort because each bucket should contain only one element on average. But the worst case is still $O(n^2)$ as a result.

Assuming all elements have the same number of digits, we use a stable sort each column starting from the least significant digit.

def radix_sort(a, d):
for i in range(1, d):
stable_sort(a on digit i)


$O(n)$ if counting sort is used for digits.

Dynamic programming TODO

Dynamic programming tends to be useful when problems have the following features:

1. There exist many choices each with some ‘score’
2. The optimal solution is composed of optimal solutions to subproblems
3. The subproblems overlap.

Graph algorithms

DFS

Used to traverse or search a graph.

def dfs(g, s):
for v in g.vertices:
v.visited = False
s.visited = True

stack = Stack()
stack.push(s)

while not stack.empty():
v = stack.pop()
for w in v.neighbours:
if not w.visited:
stack.push(w)
w.visited = True

• $O(V+E)$ runtime

BFS

Used to traverse or search a graph.

def bfs(g, s):
for v in g.vertices:
v.visited = False
s.visited = True

q = new Queue()
q.push(s)

while not q.empty():
v = q.pop()
for w in v.neighbours():
if not w.visited:
q.push(w)
w.visited = True

• To use DFS or BFS to find a path, we just have to update a previous field for each node, then walk back from the target to the start.
• $O(V+E)$ runtime

Dijkstra

• After running Dijkstra, the distance field contains the minimum distance from s to that vertex.
• Similar to BFS, except we use a priority queue to store vertices. If we visit a vertex that has already been seen, we update its distance and its position in the priority queue.
def dijkstra(g, s):
for v in g.vertices:
v.distance = infinity
s.distance = 0

pq = PriorityQueue(sortkey = lambda v: v.distance)
pq.push(s)

while not pq.empty():
v = pq.popmin()
for (w, edgecost) in v.neighbours:
dist = v.distance + edgecost
if dist < w.distance:
w.distance = dist

if w in pq:
pq.decreasekey(w)
else:
pq.push(w)

• $O(E + V \log V)$ runtime

Bellman equation

• Used to find minweight path (i.e same as Dijkstra but works for negative weights)
• $W_{ij}$ is the minweight action to go from state i to j:
• The minimal weight path from i to j in l steps is denoted by $M_{ij}^{(l)}$. We can compute it with dynamic programming.
• This can be formulated as a matrix multiplication, where $x \wedge y \equiv \min(x,y), n = |V|$.
• This is a brute force algorithm requiring $\log V$ matrix multiplications, so runtime is $O(V^3 \log V)$.

Bellman-Ford

• Used to find the minweight path (i.e same as Dijkstra but works for negative weights)