Skip to main content

Dijkstra's algorithm - Part 1 - Tutorial


This will be a 3 Part series of posts where I will be implementing the Dijkstra's Shortest Path algorithm in Python. The three parts will be


1) Representing the Graph
2) Priority Queue
3) The Algorithm

To represent a graph we'll be using an Adjacency List. Another alternative is using an Adjacency Matrix, but for a sparse graph an Adjacency List is more efficient.

Adjacency List

An Adjacency List is usually represented as a HashTable (or an Array) where an entry at `u` contains a Linked List. The Linked List contains `v` and optionally another parameter `w`. Here `u` and `v` are node(or vertex) labels and `w` is the weight of the edge. By Traversing the linked list we obtain the immediate neighbours of `u`. Visually, it looks like this.


For implementing this in Python, we'll be using the dict() for the main HashTable. For the Linked List we can use a list of 2 sized tuples (v,w). 

Sidenote: Instead of a list of tuples, you can use a dict(), with the key as `v` and value as weight (or a tuple of edge properties). The  Advantage being you can check for adjacency faster. In Dijkstra, we don't need that check, so I'll stick with a list for now.
So here's the python code

The above program simply reads 3 integers per line from the data piped to it (or a file passed as an argument)

I'll be using the undirected graph from Wikipedia for reference. Here's the graph followed by its textual representation.



(Sorry about the overlapping edge labels, I'm new to Graphviz.)

1 2 7
1 3 9
1 6 14
2 3 10
2 4 15
3 4 11
3 6 2
4 5 6
5 6 9

So what have we accomplished so far? We essentially have stored the entire graph as a data structure, where we can iterate over the neighbours of any vertex using something like
That's it for this part. In the next part, we'll be building our very own Priority Queue with a little help from Python's Standard modules. Until next time...

Comments

Popular posts from this blog

Find Increasing Triplet Subsequence - Medium

Problem - Given an integer array A[1..n], find an instance of i,j,k where 0 < i < j < k <= n and A[i] < A[j] < A[k]. Let's start with the obvious solution, bruteforce every three element combination until we find a solution. Let's try to reduce this by searching left and right for each element, we search left for an element smaller than the current one, towards the right an element greater than the current one. When we find an element that has both such elements, we found the solution. The complexity of this solution is O(n^2). To reduce this even further, One can simply apply the longest increasing subsequence algorithm and break when we get an LIS of length 3. But the best algorithm that can find an LIS is O(nlogn) with O( n ) space . An O(nlogn) algorithm seems like an overkill! Can this be done in linear time? The Algorithm: We iterate over the array and keep track of two things. The minimum value iterated over (min) The minimum increa...

Merge k-sorted lists - Medium

Problem - Given k-sorted lists, merge them into a single sorted list. A daft way of doing this would be to copy all the list into a new array and sorting the new array. ie O(n log(n)) The naive method would be to simply perform k-way merge similar to the auxiliary method in Merge Sort. But that is reduces the problem to a minimum selection from a list of k-elements. The Complexity of this algorithm is an abysmal O(nk). Here's how it looks in Python. We maintain an additional array called Index[1..k] to maintain the head of each list. We improve upon this by optimizing the minimum selection process by using a familiar data structure, the Heap! Using a MinHeap, we extract the minimum element from a list and then push the next element from the same list into the heap, until all the list get exhausted. This reduces the Time complexity to O(nlogk) since for each element we perform O(logk) operations on the heap. An important implementation detail is we need to keep track ...

2SUM - Medium

Problem : Given an array with distinct elements. Find out if there exists an instance where sum of two distinct elements is equal to a given constant K. Say, A = [1,2,3,4], K = 7. Output should be 3,4 The naive solution for this problem is pretty trivial. Brute force every pair-sum and check if the sum is K. You could be a little smart about this. You initially sort the array O(nlogn) and iterate the array pairwise and quit when the sum exceeds K, you can quit the inner loop since the array is sorted the sum can only increase. For the rest of the article, I'll be assuming K = 0, this doesn't affect the solutions in any way and it's trivial to introduce a new variable into the code without major changes. Here's the code, should be easy to follow... I'll be using Java today. Now that the naive solution is out of the way, let's try and do some clever stuff. So notice we can afford to sort the array since the brute solution was already O(n^2) (which ...