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Hand in Graph Theory-9 (Dyxtra)

We learned in the beginning how to get from one place to another by the shortest path. That’s why we learned a searching algorithm called Breadth First Search. The algorithm is nice but the problem is that it assumes that every path takes the same time to traverse, meaning that all edges have the same cost. Movie Website Template Free – Movie Review Website Template – Marvel Website Template Free Download

In practical life most of the time it becomes stagnant, then we need Dextrose. When I first heard the name, I thought diextra was a terrible thing, but actually writing diextra is as easy as writing BFS, I’ll try to show you how to change BFS a bit and add a priority queue to make it diextra.

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Before we start Dextrous we are introduced to a little thing called path relaxation. Let the distance of each node from the source be d[]

in the array. For example, d[3] means that it takes a total distance of d[3] from the source to reach 3 by crossing different edges. If the distance is not known, then I will leave a value of infinity, that is, a very large value. And cost[u][v] contains u–v

Edge’s cost.

Suppose you visit different places and go from Farmgate to TSC in 10 minutes and again from Farmgate to Curzon Hall in 25 minutes. Then you have the information:

  d[TSC] = 10, d[Karjanhal] = 25

Now you can go to Curzon in 7 minutes from TSC.

  cost[tsc][cost] = 7

So instead of 25 minutes you can go to Karjanhal in just 10 + 7 = 17 minutes. As you have seen:

  d[tsc] + cost[tsc][carjon] < d[carjonhall]

So you can go to Curzon Hall by this new road and make d[Curzon Hall] = d[TSC] + cost[TSC][Curzon]!!

That’s what the picture above says. We go from u to v if d[u] + cost[u][v] < d[v]. And update d[v] to d[v]=d[u]+cost[u][v]

I will make In the future, if I can go to Karjanhal by another road in less time, then I will compare and update that road like this. It goes something like this:

if d[u] + cost[u][v] < d[v]:

d[v] = d[u] + cost[u][v]

if d[u] + cost[u][v] < d[v]:

d[v] = d[u] + cost[u][v]

If you have understood the above part, then 60% of the job of understanding Dijkstra is done. If you don’t understand, read again.

You must understand BFS well. In BFS we never need to visit a node twice, we check each time if a node is visited, if not we push that node to the queue and increase the distance by 1. In Dxtra we will keep the node in the queue in the same way but instead of updating with visited we will “relax” or update the new edge as above.

It is not assumed that the source is node number 1. then

Infinity is because we don’t know the distance of 2,3,4 yet, and the distance to the source is zero. Now you try to update as many nodes as you can from the source, like the previous BFS, if you can update it, push it to the queue. For example 1 – 2 we will advance along the edge because d[1]+2<d[2]

This condition is fulfilled. Then d[2] will become 2, similarly going from 1 to 3 d[3]

It will be 5.

But 5 is not the shortest distance to the 3rd node! We saw in BFS that a node is not updated more than once, that property does not work here. 2-3 edges forward from node 2 and update again then d[3]

in d[2]+1=3

will get So we see that in this case a node can be updated many times. (Question: What is the maximum number of times?)

We then modify the update part of the previous BFS sudocode so that a node can be updated repeatedly:

1 procedure BFSModified(G,source):

2 Q = queue(), distance[] = infinity

3 Q.enqueue(source)

4 distance[source] = 0

5 while Q is not empty

6 u ← Q.pop()

7 for all edges from u to v in G.adjacentEdges(v) do

8 if distance[u] + cost[u][v] < distance[v]

9 distance[v] = distance[u] + cost[u][v]

10 Q.enqueue(v)

11 end if

12 end for

13 end while

14 Return distance

We are constantly updating the code of the previous BFS just costing! This code will get you the shortest path to each node from the source but it’s very bad in terms of complexity! For that we need a priority queue.

In BFS we are working on a “first come first serve” basis when moving from 1 node to many nodes and again from those nodes. For example, in the above graph from 1 first to node 3 and then to node 2, I am working on 3 first, then I am working on 2.

See what the problem is here. Working with 3 earlier, we update the distance of 4 as distance 5 + 3 = 8. Later when 3 is updated again with 2 then the distance of 3 becomes 3, now the distance of 4 is updated again as 3 + 3 = 6. 4 was updated twice in total.

Scientist Dijkstra reasoned that if the nearest nodes were processed first, instead of operating on a “first come, first served” basis, updates would be needed much less often. If we had worked on 2 earlier, 3 would have been updated earlier and we would have got the shortest distance by updating 4 once! Simulate it with a pen.

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The idea is to work with the node closest to the source at any given time in the queue. That’s why we will replace the queue with a priority queue that pushes the node to the queue and brings the nearest node to the front. The difference is that earlier I pushed the empty node number, now the current distance i.e. d[u]

Its value must also be pushed.

Below I give a complete diaxtra pseudocode which is 1 to n

Finds the shortest path to the th node and prints the path as well:

1 procedure dijkstra(G, source):

2 Q = priority_queue(), distance[] = infinity

3 Q.enqueue({distance[source], source})

4 distance[source]=0

5 while Q is not empty

6 u = nodes in Q with minimum distance[]

7 remove u from the Q

9 if distance[u] + cost[u][v] < distance[v]

10 distance[v] = distance[u] + cost[u][v]

11 Q.enqueue({distance[v], v})

12 end if

13 end for

14 end while

15 Return distance

Here are some differences from the previous sudocode. Here priority queue is used instead of normal queue. So while popping from the queue, the closest node found so far from the source is popped and worked on that first.

The above sudocode calculates the distance of all remaining nodes from the source. If you just want to get the distance of a node, you can return it when it is popped from the queue.

Does Dijkstra’s algorithm work if there are negative edges? If there is a negative cycle then it will end up in an infinite loop, repeatedly updating to reduce costs. If there is a negative edge but no cycle then do not work. But if you don’t return the target as soon as it pops, it will work, but it won’t be the original Dijkstra algorithm anymore.


The complexity of BFS was O(V+E).

Where V is the number of nodes and E is the number of edges. Here too it will work as before but it will take O(logV) complexity to sort each time on the priority queue. Total: O(VlogV+E)


To work with negative cycles we need to know Bellman-Ford Algorithm. There too the edge is updated by relaxing, to a node at most n–1

May need to be updated every time that property is invoked.

Solve the following problems quickly to learn Dxtra better:


Not the best

New Traffic System

Happy Coding!

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