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 Re: dijskra: is there something missing in the article? (response to post by mafoko) I've always thought Dijkstra does the following:set all L[i] to infinityset L[start] to zero.add all vertices to the priority queue.while the queue is not empty { top = the item from queue with lowest L[] remove top from queue mark top as visited update all L[]'s for neighbours of top (if L[i]+D[i,j]
 Re: dijskra: is there something missing in the article? (response to post by mafoko) I've always thought Dijkstra does the following:set all L[i] to infinityset L[start] to zero.add all vertices to the priority queue.while the queue is not empty { top = the item from queue with lowest L[] remove top from queue mark top as visited update all L[]'s for neighbours of top (if L[i]+D[i,j]
 Re: dijskra: is there something missing in the article? (response to post by mafoko) I've always thought Dijkstra does the following:set all L[i] to infinityset L[start] to zero.add all vertices to the priority queue.while the queue is not empty { top = the item from queue with lowest L[] remove top from queue mark top as visited update all L[]'s for neighbours of top (if L[i]+D[i,j]
 Re: dijskra: is there something missing in the article? (response to post by mafoko) I've always thought Dijkstra does the following:set all L[i] to infinityset L[start] to zero.add all vertices to the priority queue.while the queue is not empty { top = the item from queue with lowest L[] remove top from queue mark top as visited update all L[]'s for neighbours of top (if L[i]+D[i,j]