Learning Python with Advent of Code Walkthroughs

Dazbo's Advent of Code solutions, written in Python

The Python Journey - Shortest Path Algorithms - BFS, Dijkstra, and A*

Graphs

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Useful Links

Graphs and Graph TheoryNetworkXA* AlgorithmHeapq

Page Contents

Overview

These algorithms help you find optimum paths through a graph.

Honestly, you won’t get far in Advent-of-Code if you don’t know them. Eric loves challenges like…

Here are my three top reasons why you should learn this:

  1. You’ll get the right answer!
  2. It’ll save you LOADS of time and effort.
  3. These algorithms are cool!!

Algorithm Summary

First, let me provide a quick summary and comparison. Then we’ll get into the details after.

Algorithm Summary When To Use It
Breadth First Search (BFS) Aka "flood fill". Explores equally in all directions. Uses a FIFO queue. A list will work, but a deque is best. For finding the shortest path where the _cost_ is constant for any move. Or, for finding every point in a graph. Or for finding ALL paths from a to b.
Dijkstra’s Algorithm Flood fill + a cost-based optimisation. I.e. prioritises paths to explore, favouring lower cost paths. Uses a _priority queue_, i.e. heapq. When movement costs may vary.
A* Optimised for a single destination. I.e. prioritises paths that _seem_ to be approaching the goal. Still uses a heapq, but uses some sort of cost heuristic. Depending on the value of the distance heuristic, it may not add much value over Dijkstra, and may actually be slower. When we have an idea of the direction we need to follow.

The Final Frontier!

All these algorithms utilise the concept of an expanding frontier. This expanding frontier is sometimes called a flood fill. It gives us a way to visit every allowed location on a map.

How the frontier works:

  1. Pick and remove a location from the frontier. Let’s call it current.
  2. Expand the frontier by finding all of the valid next moves from current. Let’s call them neighbours. Add the neighbours to the frontier.

Here’s some code that will perform a flood fill to explore every valid location in the graph:

frontier = Queue() # For BFS, we just use a FIFO queue.
frontier.append(start)
explored = set()  # To stop us exploring the same location more than once.
explored.add(start)

# keep going until the frontier is empty; i.e. when we've explored all the valid nodes
while frontier:   
   current = frontier.popleft()  # pop the first item off the FIFO queue
   for next in graph.neighbors(current):  # get all the valid neighbours
      if next not in explored:
         frontier.append(next) # add it to the frontier
         explored.add(next)  # mark it as explored

Breadth First Search (BFS)

BFS can be used to find a path from a to b. It can also be used to find paths from a to all locations in the graph.

It is guaranteed to find a solution if one exists. The frontier expands at an equal rate in all directions, and uses a FIFO (first-in, first-out) queue to implement the frontier.

How It Works

So the sequence of exploration looks like this:

Tree

The numbers represent the sequence of traversal. But note that it is possible for the same node to appear in the tree more than once.

When we find a node that matches our goal, we stop. Thus, a BFS is just a flood fill, with a goal.

The Queue

We can implement the queue in many ways:

Implementing the BFS

Here’s some pseudocode that shows how to implement the BFS:

function BFS(graph, root):
  queue = new Queue()
  label root as explored  # we know about this node, but not its children
  queue.enqueue(root)
  while queue not empty:
    current = queue.pop()
    if current is goal:
      return current

    for neighbour in graph.adjacent_nodes(current):
      if neighbour is not explored:
        label neighbour as explored
        queue.enqueue(neighbour)

There are a couple of convenient ways to label a node as explored:

  1. We can store them in a set.
  2. We can add them to a dictionary

If we use a set:

frontier = deque() # For BFS, we need a FIFO queue. A deque is perfect.
frontier.append(start)
explored = set()  # To stop us exploring the same location more than once.
explored.add(start)

# keep going until the frontier is empty; i.e. when we've explored all the valid nodes
while frontier:   
    current = frontier.popleft()  # pop the first item off the FIFO queue

    if current == goal:
        break

    for next in graph.neighbors(current):  # get all the valid neighbours
        if next not in explored:
            frontier.append(next) # add it to the frontier
            explored.add(next)  # mark it as explored

if current != goal:
    # there was no solution!

Keeping Track of Distance

If we’re only interested in the shortest distance to a point, but we don’t need to create the actual path, a cool trick is to keep track of the distance in our queue. E.g.

frontier = deque() # For BFS, we need a FIFO queue. A deque is perfect.
frontier.append(start, 0) # Queue always includes (point, steps)
explored = set()  # To stop us exploring the same location more than once.
explored.add(start)

# keep going until the frontier is empty; i.e. when we've explored all the valid nodes
while frontier:   
    current, steps = frontier.popleft()  # pop the first item off the FIFO queue

    if current == goal:
        break

    for next in graph.neighbors(current):  # get all the valid neighbours
        if next not in explored:
            frontier.append(next, steps+1) # add it to the frontier
            explored.add(next)  # mark it as explored

if current != goal:
    # there was no solution!

# distance from start to goal = steps

If we use a dictionary instead of a set, we can create a breadcrumb trail to each point we’ve explored.

Let’s explore every valid point in the graph:

frontier = deque() # For BFS, we just use a FIFO queue. A deque is perfect.
frontier.append(start)
came_from = {}  #  To stop us exploring the same location more than once; and for breadcrumbs
came_from[start] = None  # This marks the last breadcrumb

# keep going until the frontier is empty; i.e. when we've explored all the valid nodes
while frontier:   
    current = frontier.popleft()  # pop the first item off the FIFO queue

    for next in graph.neighbors(current):  # get all the valid neighbours
        if next not in came_from:
            frontier.append(next) # add it to the frontier
            came_from[next] = current

We can then create the breadcrumb path, like this:

current = goal  # start at the end
path = []       # to store every point in a specific path
while current != start: 
   path.append(current)
   current = came_from[current]  # get the next node in the trail
path.append(start) # optional - depends if we want the first node to be included or not
path.reverse() # optional - depends whether we want start->end or end->start

BFS Examples

Dijkstra’s Algorithm

This algorithm is like BFS. But instead of expanding the frontier in all directions with equal weight - called an unweighted pathfinder algorithm - we instead prioritise popping the node that has the lowest overall cost.

Dijkstra’s algorithm is perfect for:

How It Works

Check out the illustration below, that shows how the algorithm finds our goal, H:

Dijkstra Walkthrough

Dijkstra’s Algorithm Examples

A* Algorithm

The A* Algorithm is a lot like BFS. But instead of expanding frontier in all directions equally - called an unweighted pathfinder algorithm - we prioritise popping from the frontier in the direction that we believe takes us closer to our goal. Thus, a weighted pathfinder algorithm.

To do this, we need two key changes, compared to the BFS:

  1. We need to use a heapq. Recall that a heapq allows us to pop elements based on a priority.
  2. We need to provide a heuristic function, i.e. a function that computes some sort of cost. The heuristic estimates the cost to get from any given node to the goal. This cost is used to determine the priority. Typically, the heuristic will be a function that estimates distance, such as a Manhattan distance function.

Turning a BFS into an A*

Here’s a BFS that creates a breadcrumb path. The priority of the pop is simply the number of steps taken from the starting point.

def solve_with_bfs(start:Point, goal:Point) -> tuple[int, dict]:
    queue = []
    heapq.heappush(queue, (0, start)) # posn, steps
    came_from = {} # use a dict so we can build a breadcrumb trail
    came_from[start] = None

    while queue:
        step_count, curr_posn = heapq.heappop(queue)

        # Check if the target is reached
        if curr_posn == goal:
            return step_count, came_from

        # Explore adjacent hexagons
        for dx, dy in Hex.hex_vectors.values():
            adjacent = curr_posn + Point(dx, dy)
            if adjacent not in came_from:
                came_from[adjacent] = curr_posn
                heapq.heappush(queue, (step_count + 1, adjacent))

    assert False, "We should never get here."

Here I’ve changed the code to an A*. Now I’m popping from the heapq, using a priority that is sum of the number of steps taken, and the estimated distance to the goal.

def solve_with_astar(start: Point, goal: Point) -> tuple[int, dict]:
    """ Obtain the fastest path from origin to goal using A*.
    Returns: tuple[int, dict]: steps required, breadcrumbs
    """
    queue = []
    came_from = {} # use a dict so we can build a breadcrumb trail
    came_from[start] = None
    heapq.heappush(queue, (0, 0, start)) # distance heuristic, steps, posn

    while queue:
        _, step_count, curr_posn = heapq.heappop(queue)

        # Check if the target is reached
        if curr_posn == goal:
            return step_count, came_from

        # Explore adjacent hexagons
        for dx, dy in Hex.hex_vectors.values():
            adjacent = curr_posn + Point(dx, dy)
            if adjacent not in came_from:
                came_from[adjacent] = curr_posn
                heuristic_cost = Hex.distance(adjacent, goal)
                new_cost = step_count + 1 + heuristic_cost # heuristic is combination of distance to goal, and step count
                heapq.heappush(queue, (new_cost, step_count + 1, adjacent))

    assert False, "We should never get here."

I compared these two solutions when solving 2017 day 11. The BFS took 30 seconds, and the A* took just 20ms!!

A* Examples

Credit Where It’s Due

When I first learned these algorithms, the best guide I came across was this one. I wrote a lot of notes when I first read it, and in writing this page today, I’ve been referring to my previous notes. So, the content I’ve written here will likely have a lot in common with the amazing content from https://www.redblobgames.com.

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