Isn't there a Dijkstra algorithm? It's hard to remember the name, and it's unlikely that you'll implement it in practice, such as the shortest path problem of weighted graphs, so you'll forget it soon. .. ..

What is Dijkstra's algorithm?

Dijkstra's algorithm is an algorithm that finds the shortest path of a graph. I think many people have heard only the name. It's easy and simple to do, but it's a little difficult to understand, but it's a relatively easy-to-understand algorithm if you understand each one. Breadth-first search can be used to solve unweighted, maze searches, etc., but if each side is weighted, you will have to calculate the entire route after all. Assuming that all vertices are passed only once, if there are E sides, it will be O (E!) And the amount of calculation will explode. It's a little tough to calculate this.

The Dijkstra algorithm is an algorithm that efficiently solves such problems.

By the way, the cost of each side must be a non-negative value (0 or more). If a negative number is included, the Bellman-Ford method etc. will be used.

procedure

The Dijkstra procedure is fairly simple.

Set the shortest distance 0 as the starting point
Move to the shortest distance known and the shortest distance from the points that have not been traced yet
Set the shortest distance between the vertices connected to that vertex. At this time, if the shortest distance of the vertex can be updated, update it.
Do this until you know the shortest distance of all vertices

Well, I think it's a little difficult to say, so let's see an example.

Example

It's an analog method, but I was able to express it the most. .. .. Consider the shortest path in the figure below.

Green is the top of the movement with the shortest path determined. Red is the starting point. The number at each vertex is the shortest path at that time dijkstra algo.gif

As you can see, you can see that the shortest path of the adjacent vertices is calculated from the starting point by moving to the shortest vertex at that time in order.

Implementation

Now let's get into the implementation. Let's implement the above procedure in a straightforward manner.

function main(nodes) {
    const start = nodes[0]
    //Record visited vertices
    const visited = new Set()
    const routesFromStart = new Map()
     //Record the distance from the starting point

    routesFromStart.set(start, {distance: 0})
    for(const n of nodes) {
        if(n != start) {
            //Substitute infinity for all vertices except the start
            routesFromStart.set(n, {distance: Number.MAX_VALUE})
        }
    }
    let current = start
    let routes = new Map()
    while(current != null) {
        visited.add(current)
        for(const edge of current.edges) {
             //Calculates the shortest distance from that vertex to adjacent vertices and updates if it is lower than the calculated value
            if(edge.cost + routesFromStart.get(current).distance < routesFromStart.get(edge.to).distance) {
                routesFromStart.set(edge.to, {distance: edge.cost + routesFromStart.get(current).distance})
                routes.set(current, edge.to)
            }
        }
        let cheapestNodeDistance = Number.MAX_VALUE
        current = null
        //Select the smallest vertex from the calculated vertices for the shortest unvisited distance

        for(const city of routesFromStart.keys()) {
            if(!visited.has(city) && cheapestNodeDistance > routesFromStart.get(city).distance){
                cheapestNodeDistance = routesFromStart.get(city).distance
                current = city
            }
        }
    }
    return routesFromStart.get(nodes[nodes.length - 1]).distance
}

Assuming that each vertex is V, this code goes around the loop for selecting the smallest vertex in the loop that goes around the number of each side at the maximum, so the amount of calculation is O (V ^ 2 + E) for each memory. It is O (V) because it is necessary to record the vertices.

Implementation using Priority Queue

Well, as you may have noticed, this code actually allows you to optimize the logic that determines the smallest vertices. That is the priority queue. Priority queues require O (logN) complexity to insert and retrieve, but the complexity of determining the smallest vertices is faster than a linear search.

Priority Queue is not implemented as standard in JavaScript, so it is implemented in Python.


def dijkstra(nodes):
    start_node = nodes[0]
    routes_from_start = {n: math.inf for n in nodes}

    #Set zero to first vertex
    routes_from_start[start_node] = 0
    
    minHeap = []

    #Add first vertex
    heappush(minHeap, (0, start_node))

    #Search until heap runs out
    while minHeap:
        (cost, current_node) = heappop(minHeap)

        #Since the priority key is duplicated, check here
        if cost > routes_from_start[current_node]:
            continue

        for node in current_node.routes:
            price_info = current_node.routes[node]
            if routes_from_start[node] > price_info + routes_from_start[current_node]:
                routes_from_start[node] = price_info + routes_from_start[current_node]
                #Add value to priority when updated
                heappush(minHeap, (price_info + routes_from_start[current_node], node))

    return routes_from_start[nodes[-1]]

Let's take another opportunity to explain Priority Queue. If the calculation efficiency is better and each vertex is V and each side is V, heap is operated for the number of times of O (V) and E that sets V to map, so O (ElogE). You can see that it can be calculated by the total O (V + ElogE) of. This is more efficient than the first algorithm.

Memorize the route

Now I know the shortest cost. But this problem is the "shortest path" problem. When you find the shortest cost, you usually want to know the route. Let's improve the above code.

def dijkstra(nodes):
    start_node = nodes[0]
    routes_from_start = {n: math.inf for n in nodes}

    #Set zero to first vertex
    routes_from_start[start_node] = 0
    
    minHeap = []


    #Add first vertex
    heappush(minHeap, (0, start_node))
    path = collections.defaultdict(Node)

    #Search until heap runs out
    while minHeap:
        (cost, current_node) = heappop(minHeap)

        #Since the priority key is duplicated, check here
        if cost > routes_from_start[current_node]:
            continue

        for node in current_node.routes:
            price_info = current_node.routes[node]
            if routes_from_start[node] > price_info + routes_from_start[current_node]:
                routes_from_start[node] = price_info + routes_from_start[current_node]
                #Record the node that updates the shortest distance
                path[node.id] = current_node.id
                #Add value to priority when updated
                heappush(minHeap, (price_info + routes_from_start[current_node], node))

    current_node = nodes[-1].id
    path_array = []

    #Follow the node that recorded the shortest distance from the goal
    while current_node:
        path_array.append(current_node)
        if current_node not in path:
            break
        current_node = path[current_node]

    return routes_from_start[nodes[-1]], path_array[::-1]

The Dijkstra algorithm knows which node updates the shortest distance, so you can record it and follow it last. The amount of calculation will increase by the number of nodes with the shortest distance.

By the way, why is this the shortest path?

By the way, I think many people have thought this way after looking at it so far. Sure, the algorithm is simple and it's not too difficult to implement. But why can you find the shortest distance? Let's check it lightly

Assuming that the vertex in L is the shortest distance from the start S, it would be nice to say that the shortest vertex connected from there is also the shortest distance from S.

Now, we will move to the shortest vertex included in T, so if the smallest point is i, then d [i] = min (T), isn't it? Now, assuming that each vertex is k, it is certain that the shortest distance d [k] is d [k]> = d [i]. Because d [i] is the smallest point and each vertex is non-negative. You can prove it inductively by doing it one after another.

Well, if you think about it, it's a recurrence formula.

The shortest distance between the vertices of L adjacent to d [i] = min (k ⊂ T) + i

Dynamic programming is used for recurrence formulas

The recurrence formula is DP, isn't it? This article is very helpful for DP (https://qiita.com/drken/items/a5e6fe22863b7992efdb)

Then, how will the value be updated in DP?

The value will be updated like this. The vertical axis is the number of trials. The horizontal axis is the apex.

What the Dijkstra algorithm was a kind of DP.

Summary

The Dijkstra algorithm I've tried, but once you understand it, it's fairly easy to understand. After that, when I tried to implement it and encountered a similar problem due to an algorithm problem, this was that time! !! I want to solve it like that.

As an aside, I also want to write a DP article

Click here for the explained youtube channel https://youtu.be/jz8b0q5R1Ss

reference

http://www.lab2.kuis.kyoto-u.ac.jp/~shuichi/algintro/alg-6s.pdf https://www.youtube.com/watch?v=X1AsMlJdiok

[JAVA] Dijkstra algorithm for beginners