Dijkstra的算法问题和python中的关键错误

Question

Trying to code Dijkstras algorithm from this psuedocode:

1: procedure ShortestPath
2:   for 0 ≤ i ≤ n do
3:     L(vi) = ∞
4:   end for
5:   L(a) = 0
6:   S = ∅
7:   while z /∈ S do
8:     u = vertex not in S with L(u) minimal
9:     S = S ∪ {u}
10:    for v /∈ S do
11:      if L(u) + µ(uv) < L(v) then
12:        L(v) = L(u) + µ(uv)
13:      end if
14:    end for
15:  end while
16:  return L(z)
17: end procedure

The code I have written is this:

from math import inf
def dijkstras(G,start,stop):
    L = {}
    for i in G[0]:
        L[i] = inf
    L[start] = 0
    visited = []
    print(L)
    while stop not in  visited:
        u = inf
        for i in L:
            if L[i] < u and L[i] not in visited:
                u = L[i]
                break
        visited.append(u)
        for v in G[0]:
            if v in visited:
                continue

            if {u,v} or {v,u} in G[1]:
                for i in G[1]:
                    if {u,v} or {v,u} == i[0]:
                        weight = i[1]
                        break
                    else:
                        weight = inf
                if L[u] + weight < L[v]:
                    L[v] = L[u] + weight
    return stop

It gives me KeyError = 0, I Presume this is to do with line L[v] = L[u] + weight. Other than that I think the code is correct. If anyone spots the issue please let me know, cheers.

Answer 1

In your code I see u = inf or u = L[...] , yet in original pseudocode u is a graph vertice, not weight. In other words you confused distances with vertices. Perhaps use strings to name vertices?

Provide example of your graph format. Is G[1] a dictionary with transition keys? list of pairs of a transition with its weight? Looks like in different place it is interpreted differently.

{u,v} or {v,u} == i[0] is obviously error, you likely meant {u,v} == i[0] or {v,u} == i[0]