I am trying to write Dijkstra’s Algorithm, however I am struggling on how to ‘say’ certain things in code.
To visualize, here are the columns I want represented using arrays:
max_nodes
A B C Length Predecessor Visited/Unvisited
A 0 1 2 -1 U
B 1 0 1 -1 U
C 2 1 0 -1 U
So, there will be several arrays, as seen in my code below:
def dijkstra (graph, start, end)
network[max_nodes][max_nodes]
state [max_nodes][length]
state2 [max_nodes][predecessor]
state3 [max_nodes][visited]
initialNode = 0
for nodes in graph:
D[max_nodes][length] = -1
P[max_nodes][predecessor] = ""
V[max_nodes][visited] = false
for l in graph:
length = lengthFromSource[node] + graph[node][l]
if length < lengthFromSourceNode[w]:
state[l][length] = x
state2[l][predecessor]
state3[l][visited] = true
x +=1
The part in bold is where I am stuck on – I am trying to implement this section of the algorithm:
3. For current node, consider all its unvisited neighbors and calculate their tentative distance. For example, if current node (A) has distance of 6, and an edge connecting it with another node (B) is 2, the distance to B through A will be 6+2=8. If this distance is less than the previously recorded distance, overwrite the distance
4. When we are done considering all neighbors of the current node, mark it as visited. A visited node will not be checked ever again; its distance recorded now is final and minimal
I think I am on the right track, i’m just stuck on how to say ‘start at a node, get the length from source to a node, if length is smaller, overwrite previous value, then move to next node
First, I assume this is a homework problem, as the best suggest is to not bother writing it yourself, but to find an existing implementation on the web. Here’s one that looks pretty good, for example.
Assuming you do need to reinvent the wheel, the code referenced there uses dictionaries to store the node data. So you feed it something like:
This seems a more intuitive way of presenting your graph information. Visited nodes and distances can be kept in dictionaries as well.