Implementation of Dijkstra with adjacency list and priority queue in Ruby

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require 'pqueue'
def dijkstra(source, edges, num_vertices)
path_weights = Array.new(num_vertices, Float::INFINITY)
previous_shortest_path = Array.new(num_vertices, nil)
pq = PQueue.new([source])x,y
path_weights[source] = 0

while pq.size != 0
min_node = pq.pop
# Note: If we are interested only in shortest distance from source to a single target, we can break the for loop when the picked minimum distance vertex is equal to target

(edges[min_node] || ).each do |neighbor, weight|
if path_weights[neighbor] > path_weights[min_node] + weight
path_weights[neighbor] = path_weights[min_node] + weight
previous_shortest_path[neighbor] = min_node
pq.push(neighbor)
end
end
end

return [path_weights, previous_shortest_path]
end


Assuming that the priority queue uses a binary heap, this should be O(E log V)



Would this work for all graphs without negative cycles? Any obvious improvements or edge cases missing?







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    require 'pqueue'
    def dijkstra(source, edges, num_vertices)
    path_weights = Array.new(num_vertices, Float::INFINITY)
    previous_shortest_path = Array.new(num_vertices, nil)
    pq = PQueue.new([source])x,y
    path_weights[source] = 0

    while pq.size != 0
    min_node = pq.pop
    # Note: If we are interested only in shortest distance from source to a single target, we can break the for loop when the picked minimum distance vertex is equal to target

    (edges[min_node] || ).each do |neighbor, weight|
    if path_weights[neighbor] > path_weights[min_node] + weight
    path_weights[neighbor] = path_weights[min_node] + weight
    previous_shortest_path[neighbor] = min_node
    pq.push(neighbor)
    end
    end
    end

    return [path_weights, previous_shortest_path]
    end


    Assuming that the priority queue uses a binary heap, this should be O(E log V)



    Would this work for all graphs without negative cycles? Any obvious improvements or edge cases missing?







    share|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      require 'pqueue'
      def dijkstra(source, edges, num_vertices)
      path_weights = Array.new(num_vertices, Float::INFINITY)
      previous_shortest_path = Array.new(num_vertices, nil)
      pq = PQueue.new([source])x,y
      path_weights[source] = 0

      while pq.size != 0
      min_node = pq.pop
      # Note: If we are interested only in shortest distance from source to a single target, we can break the for loop when the picked minimum distance vertex is equal to target

      (edges[min_node] || ).each do |neighbor, weight|
      if path_weights[neighbor] > path_weights[min_node] + weight
      path_weights[neighbor] = path_weights[min_node] + weight
      previous_shortest_path[neighbor] = min_node
      pq.push(neighbor)
      end
      end
      end

      return [path_weights, previous_shortest_path]
      end


      Assuming that the priority queue uses a binary heap, this should be O(E log V)



      Would this work for all graphs without negative cycles? Any obvious improvements or edge cases missing?







      share|improve this question













      require 'pqueue'
      def dijkstra(source, edges, num_vertices)
      path_weights = Array.new(num_vertices, Float::INFINITY)
      previous_shortest_path = Array.new(num_vertices, nil)
      pq = PQueue.new([source])x,y
      path_weights[source] = 0

      while pq.size != 0
      min_node = pq.pop
      # Note: If we are interested only in shortest distance from source to a single target, we can break the for loop when the picked minimum distance vertex is equal to target

      (edges[min_node] || ).each do |neighbor, weight|
      if path_weights[neighbor] > path_weights[min_node] + weight
      path_weights[neighbor] = path_weights[min_node] + weight
      previous_shortest_path[neighbor] = min_node
      pq.push(neighbor)
      end
      end
      end

      return [path_weights, previous_shortest_path]
      end


      Assuming that the priority queue uses a binary heap, this should be O(E log V)



      Would this work for all graphs without negative cycles? Any obvious improvements or edge cases missing?









      share|improve this question












      share|improve this question




      share|improve this question








      edited May 14 at 4:24









      Mast

      7,32563484




      7,32563484









      asked May 13 at 22:03









      bgcode

      1012




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