Here is an excise:
Let G be a weighted directed graph with n vertices and m edges, where all edges have positive weight. A directed cycle is a directed path that starts and ends at the same vertex and contains at least one edge. Give an O(n^3) algorithm to find a directed cycle in G of minimum total weight. Partial credit will be given for an O((n^2)*m) algorithm.
Here is my algorithm.
I do a DFS. Each time when I find a back edge, I know I've got a directed cycle.
Then I will temporarily go backwards along the parent array (until I travel through all vertices in the cycle) and calculate the total weights.
Then I compare the total weight of this cycle with min. min always takes the minimum total weights. After the DFS finishes, our minimum directed cycle is also found.
Ok, then about the time complexity.
To be honest, I don't know the time complexity of my algorithm.
For DFS, the traversal takes O(m+n) (if m is the number of edges, and n is the number of vertices). For each vertex, it might point back to one of its ancestors and thus forms a cycle. When a cycle is found, it takes O(n) to summarise the total weights.
So I think the total time is O(m+n*n). But obviously it is wrong, as stated in the excise the optimal time is O(n^3) and the normal time is O(m*n^2).
Can anyone help me with:
Is my algorithm correct?
What is the time complexity if my algorithm is correct?
Is there any better algorithm for this problem?
You can use Floyd-Warshall algorithm here.
The Floyd-Warshall algorithm finds shortest path between all pairs of vertices.
The algorithm is then very simple, go over all pairs (u,v), and find the pair that minimized dist(u,v)+dist(v,u), since this pair indicates on a cycle from u to u with weight dist(u,v)+dist(v,u). If the graph also allows self-loops (an edge (u,u)) , you will also need to check them alone, because those cycles (and only them) were not checked by the algorithm.
pseudo code:
run Floyd Warshall on the graph
min <- infinity
vertex <- None
for each pair of vertices u,v
if (dist(u,v) + dist(v,u) < min):
min <- dist(u,v) + dist(v,u)
pair <- (u,v)
return path(u,v) + path(v,u)
path(u,v) + path(v,u) is actually the path found from u to v and then from v to u, which is a cycle.
The algorithm run time is O(n^3), since floyd-warshall is the bottle neck, since the loop takes O(n^2) time.
I think correctness in here is trivial, but let me know if you disagree with me and I'll try to explain it better.
Is my algorithm correct?
No. Let me give a counter example. Imagine you start DFS from u, there are two paths p1 and p2 from u to v and 1 path p3 from v back to u, p1 is shorter than p2.
Assume you start by taking the p2 path to v, and walk back to u by path p3. One cycle found but apparently it's not minimum. Then you continue exploring u by taking the p1 path, but since v is fully explored, the DFS ends without finding the minimum cycle.
"For each vertex, it might point back to one of its ancestors and thus forms a cycle"
I think it might point back to any of its ancestors which means N
Also, how are u going to mark vertexes when you came out of its dfs, you may come there again from other vertex and its going to be another cycle. So this is not (n+m) dfs anymore.
So ur algo is incomplete
same here
3.
During one dfs, I think the vertex should be either unseen, or check, and for checked u can store the minimum weight for the path to the starting vertex. So if on some other stage u find an edge to that vertex u don't have to search for this path any more.
This dfs will find the minimum directed cycle containing first vertex. and it's O(n^2) (O(n+m) if u store the graph as list)
So if to do it from any other vertex its gonna be O(n^3) (O(n*(n+m))
Sorry, for my english and I'm not good at terminology
I did a similar kind of thing but i did not use any visited array for dfs (which was needed for my algorithm to work correctly) and hence i realised that my algorithm was of exponential complexity.
Since, you are finding all cycles it is not possible to find all cycles in less than exponential time since there can be 2^(e-v+1) cycles.
Related
I know that there are a ton of questions out there about the time complexity of BFS which is : O(V+E)
However I still struggle to understand why is the time complexity O(V+E) and not O(V*E)
I know that O(V+E) stands for O(max[V,E]) and my only guess is that it has something to do with the density of the graph and not with the algorithm itself unlike say Merge Sort where it's time complexity is always O(n*logn).
Examples I've thought of are :
A Directed Graph with |E| = |V|-1 and yeah the time complexity will be O(V)
A Directed Graph with |E| = |V|*|V-1| and the complexity would in fact be O(|E|) = O(|V|*|V|) as each vertex has an outgoing edge to every other vertex besides itself
Am I in the right direction? Any insight would be really helpful.
Your "examples of thought" illustrate that the complexity is not O(V*E), but O(E). True, E can be a large number in comparison with V, but it doesn't matter when you say the complexity is O(E).
When the graph is connected, then you can always say it is O(E). The reason to include V in the time complexity, is to cover for the graphs that have many more vertices than edges (and thus are disconnected): the BFS algorithm will not only have to visit all edges, but also all vertices, including those that have no edges, just to detect that they don't have edges. And so we must say O(V+E).
The complexity comes off easily if you walk through the algorithm. Let Q be the FIFO queue where initially it contains the source node. BFS basically does the following
while Q not empty
pop u from Q
for each adjacency v of u
if v is not marked
mark v
push v into Q
Since each node is added once and removed once then the while loop is done O(V) times. Also each time we pop u we perform |adj[u]| operations where |adj[u]| is the number of
adjacencies of u.
Therefore the total complexity is Sum (1+|adj[u]|) over all V which is O(V+E) since the sum of adjacencies is O(E) (2E for undirected graph and E for a directed one)
Consider a situation when you have a tree, maybe even with cycles, you start search from the root and your target is the last leaf of your tree. In this case you will traverse all the edges before you get into your destination.
E.g.
0 - 1
1 - 2
0 - 2
0 - 3
In this scenario you will check 4 edges before you actually find a node #3.
It depends on how the adjacency list is implemented. A properly implemented adjacency list is a list/array of vertices with a list of related edges attached to each vertex entry.
The key is that the edge entries point directly to their corresponding vertex array/list entry, they never have to search through the vertex array/list for a matching entry, they can just look it up directly. This insures that the total number of edge accesses is 2E and the total number of vertex accesses is V+2E. This makes the total time O(E+V).
In improperly implemented adjacency lists, the vertex array/list is not directly indexed, so to go from an edge entry to a vertex entry you have to search through the vertex list which is O(V), which means that the total time is O(E*V).
It is said that DFS can't be used to find the shortest path in the unweighted graph. I have read multiple post and blogs but not get satisfied as a little modification in DFS can make it possible.
I think if we use Modified DFS in this way, then we can find the shortest distances from the source.
Initialise a array of distances from root with infinity and distance of root from itself as 0.
While traversing, we keep track of no. of edges. On moving forward increment no. of edges and while back track decrement no. of edges. And each time check if(dist(v) > dist(u) + 1 ) then dist(v) = dist(u) + 1.
In this way we can find the shortest distances from the root using DFS. And in this way, we can find it in O(V+E) instead of O(ElogV) by Dijkstra.
If I am wrong at some point. Please tell me.
Yes, if the DFS algorithm is modified in the way you mentioned, it can be used to find the shortest paths from a root in an unweighted graph. The problem is that in modifying the algorithm you have fundamentally changed what it is.
It may seem like I am exaggerating as the change looks minor superficially but it changes it more than you might think.
Consider a graph with n nodes numbered 1 to n. Let there be an edge between each k and k + 1. Also, let 1 be connected to every node.
Since DFS can pick adjacent neighbors in any order, let's also assume that this algorithm always picks them in increasing numerical order.
Now try running algorithm in your head or your computer with root 1.
First the algorithm will reach n in n-1 steps using edges between 1-2, 2-3 and so on. Then after backtracking, the algorithm moves on to the second neighbor of 1, namely 3. This time there will be n-2 steps.
The same process will repeat until the algorithm finally sees 1-n.
The algorithm will need O(n ^ 2) rather than O(n) steps to finish. Remember that V = n & E = 2 * n - 3. So it is not O(V + E).
Actually, the algorithm you have described will always finish in O(V^2) on unweighted graphs. I will leave the proof of this claim as an exercise for the reader.
O(V^2) is not that bad. Especially if a graph is dense. But since BFS already provides an answer in O(V + E), nobody uses DFS for shortest distance calculation.
In an unweighted graph, you can use a breadth-first search (not DFS) to find shortest paths in O(E) time.
In fact, if all edges have the same weight, then Dijkstra's algorithm and breadth-first search are pretty much equivalent -- reduceKey() is never called, and the priority queue can be replaced with a FIFO queue, since newly added vertices never have smaller weight than previously-added ones.
Your modification to DFS does not work, because once you visit a vertex, you will not examine its children again, even if its weight changes. You will get the wrong answer for this graph if you follow S->A before S->B
S---->A---->C---->D---->E---->T
\ /
------->B-----/
The way Depth First Search on graphs is defined, it only visits each node once. When it encounters a node that was visited before, it backtracks.
So assume you have a triangle with nodes A, B, C and you want to find the shortest path from A to B. One possibility of a DFS traversal is A -> C -> B and you are done. This however is not the shortest path.
Hi I am struggling with this question. It is the following:
Devise an algorithm to find the lowest-weight cycle(i.e. of all cycles in the graph, the one with the smallest sum of edge weights), in a weighted, directed graph G = (V,E). Briefly justify the runtime and space complexity. Assume all edges are non-negative. It should run in O(|V||E|log|V|) time.
Hint: Use multiple calls to Dijkstra's algorithm.
I have seen solutions that use Floyd-Warshall but I was wondering how we would do this using Dijkstra's and how to do it within the time constraint given.
I have a few points of confusion:
Firstly, how do we even know how many cycles are in the graph and how
to check those?
Also, why is it |E||V|log|V|? By my understanding you should traverse
through all the vertices therefore making it |V|log|V|.
This is for my personal learning so if anyone has an example they could use, it would greatly help me! I am not really looking for pseudo-code - just a general algorithm to understand how using the shortest path from one node to all nodes is going to help us solve this problem.
Thank you!
Call Dijkstra's algorithm from each vertex to find the shortest path to itself, if one exists. The shortest path from any vertex to itself is the smallest cycle.
Dijkstra's algorithm takes O(|E| log |V|), so total time is O(|V||E| log |V|).
Note that this time can be worse than Floyd-Warshall, because there can be O(|V|^2) edges in the graph.
Call Dijkstra's algorithm |V| times, using each vertex in V as the start vertex. Store the results in a matrix, where dist(u,v) is the shortest path length from u to v.
For each pair of vertices (u,v), dist(u,v) is the shortest parth from u to v and dist(v,u) is the shortest path from v to u. Therefore, dist(u,v) + dist(v, u) is the length of the lowest weight cycle containing u and v. For each pair of vertices, computer this value and take the minimum. This is the lowest weigtht cycle.
What is the exact difference between Dijkstra's and Prim's algorithms? I know Prim's will give a MST but the tree generated by Dijkstra will also be a MST. Then what is the exact difference?
Prim's algorithm constructs a minimum spanning tree for the graph, which is a tree that connects all nodes in the graph and has the least total cost among all trees that connect all the nodes. However, the length of a path between any two nodes in the MST might not be the shortest path between those two nodes in the original graph. MSTs are useful, for example, if you wanted to physically wire up the nodes in the graph to provide electricity to them at the least total cost. It doesn't matter that the path length between two nodes might not be optimal, since all you care about is the fact that they're connected.
Dijkstra's algorithm constructs a shortest path tree starting from some source node. A shortest path tree is a tree that connects all nodes in the graph back to the source node and has the property that the length of any path from the source node to any other node in the graph is minimized. This is useful, for example, if you wanted to build a road network that made it as efficient as possible for everyone to get to some major important landmark. However, the shortest path tree is not guaranteed to be a minimum spanning tree, and the sum of the costs on the edges of a shortest-path tree can be much larger than the cost of an MST.
Another important difference concerns what types of graphs the algorithms work on. Prim's algorithm works on undirected graphs only, since the concept of an MST assumes that graphs are inherently undirected. (There is something called a "minimum spanning arborescence" for directed graphs, but algorithms to find them are much more complicated). Dijkstra's algorithm will work fine on directed graphs, since shortest path trees can indeed be directed. Additionally, Dijkstra's algorithm does not necessarily yield the correct solution in graphs containing negative edge weights, while Prim's algorithm can handle this.
Dijkstra's algorithm doesn't create a MST, it finds the shortest path.
Consider this graph
5 5
s *-----*-----* t
\ /
-------
9
The shortest path is 9, while the MST is a different 'path' at 10.
Prim and Dijkstra algorithms are almost the same, except for the "relax function".
Prim:
MST-PRIM (G, w, r) {
for each key ∈ G.V
u.key = ∞
u.parent = NIL
r.key = 0
Q = G.V
while (Q ≠ ø)
u = Extract-Min(Q)
for each v ∈ G.Adj[u]
if (v ∈ Q)
alt = w(u,v) <== relax function, Pay attention here
if alt < v.key
v.parent = u
v.key = alt
}
Dijkstra:
Dijkstra (G, w, r) {
for each key ∈ G.V
u.key = ∞
u.parent = NIL
r.key = 0
Q = G.V
while (Q ≠ ø)
u = Extract-Min(Q)
for each v ∈ G.Adj[u]
if (v ∈ Q)
alt = w(u,v) + u.key <== relax function, Pay attention here
if alt < v.key
v.parent = u
v.key = alt
}
The only difference is pointed out by the arrow, which is the relax function.
The Prim, which searches for the minimum spanning tree, only cares about the minimum of the total edges cover all the vertices. The relax function is alt = w(u,v)
The Dijkstra, which searches for the minimum path length, so it cares about the edge accumulation. The relax function is alt = w(u,v) + u.key
Dijsktra's algorithm finds the minimum distance from node i to all nodes (you specify i). So in return you get the minimum distance tree from node i.
Prims algorithm gets you the minimum spaning tree for a given graph. A tree that connects all nodes while the sum of all costs is the minimum possible.
So with Dijkstra you can go from the selected node to any other with the minimum cost, you don't get this with Prim's
The only difference I see is that Prim's algorithm stores a minimum cost edge whereas Dijkstra's algorithm stores the total cost from a source vertex to the current vertex.
Dijkstra gives you a way from the source node to the destination node such that the cost is minimum. However Prim's algorithm gives you a minimum spanning tree such that all nodes are connected and the total cost is minimum.
In simple words:
So, if you want to deploy a train to connecte several cities, you would use Prim's algo. But if you want to go from one city to other saving as much time as possible, you'd use Dijkstra's algo.
Both can be implemented using exactly same generic algorithm as follows:
Inputs:
G: Graph
s: Starting vertex (any for Prim, source for Dijkstra)
f: a function that takes vertices u and v, returns a number
Generic(G, s, f)
Q = Enqueue all V with key = infinity, parent = null
s.key = 0
While Q is not empty
u = dequeue Q
For each v in adj(u)
if v is in Q and v.key > f(u,v)
v.key = f(u,v)
v.parent = u
For Prim, pass f = w(u, v) and for Dijkstra pass f = u.key + w(u, v).
Another interesting thing is that above Generic can also implement Breadth First Search (BFS) although it would be overkill because expensive priority queue is not really required. To turn above Generic algorithm in to BFS, pass f = u.key + 1 which is same as enforcing all weights to 1 (i.e. BFS gives minimum number of edges required to traverse from point A to B).
Intuition
Here's one good way to think about above generic algorithm: We start with two buckets A and B. Initially, put all your vertices in B so the bucket A is empty. Then we move one vertex from B to A. Now look at all the edges from vertices in A that crosses over to the vertices in B. We chose the one edge using some criteria from these cross-over edges and move corresponding vertex from B to A. Repeat this process until B is empty.
A brute force way to implement this idea would be to maintain a priority queue of the edges for the vertices in A that crosses over to B. Obviously that would be troublesome if graph was not sparse. So question would be can we instead maintain priority queue of vertices? This in fact we can as our decision finally is which vertex to pick from B.
Historical Context
It's interesting that the generic version of the technique behind both algorithms is conceptually as old as 1930 even when electronic computers weren't around.
The story starts with Otakar Borůvka who needed an algorithm for a family friend trying to figure out how to connect cities in the country of Moravia (now part of the Czech Republic) with minimal cost electric lines. He published his algorithm in 1926 in a mathematics related journal, as Computer Science didn't existed then. This came to the attention to Vojtěch Jarník who thought of an improvement on Borůvka's algorithm and published it in 1930. He in fact discovered the same algorithm that we now know as Prim's algorithm who re-discovered it in 1957.
Independent of all these, in 1956 Dijkstra needed to write a program to demonstrate the capabilities of a new computer his institute had developed. He thought it would be cool to have computer find connections to travel between two cities of the Netherlands. He designed the algorithm in 20 minutes. He created a graph of 64 cities with some simplifications (because his computer was 6-bit) and wrote code for this 1956 computer. However he didn't published his algorithm because primarily there were no computer science journals and he thought this may not be very important. The next year he learned about the problem of connecting terminals of new computers such that the length of wires was minimized. He thought about this problem and re-discovered Jarník/Prim's algorithm which again uses the same technique as the shortest path algorithm he had discovered a year before. He mentioned that both of his algorithms were designed without using pen or paper. In 1959 he published both algorithms in a paper that is just 2 and a half page long.
Dijkstra finds the shortest path between it's beginning node
and every other node. So in return you get the minimum distance tree from beginning node i.e. you can reach every other node as efficiently as possible.
Prims algorithm gets you the MST for a given graph i.e. a tree that connects all nodes while the sum of all costs is the minimum possible.
To make a story short with a realistic example:
Dijkstra wants to know the shortest path to each destination point by saving traveling time and fuel.
Prim wants to know how to efficiently deploy a train rail system i.e. saving material costs.
Directly from Dijkstra's Algorithm's wikipedia article:
The process that underlies Dijkstra's algorithm is similar to the greedy process used in Prim's algorithm. Prim's purpose is to find a minimum spanning tree that connects all nodes in the graph; Dijkstra is concerned with only two nodes. Prim's does not evaluate the total weight of the path from the starting node, only the individual path.
Here's what clicked for me: think about which vertex the algorithm takes next:
Prim's algorithm takes next the vertex that's closest to the tree, i.e. closest to some vertex anywhere on the tree.
Dijkstra's algorithm takes next the vertex that is closest to the source.
Source: R. Sedgewick's lecture on Dijkstra's algorithm, Algorithms, Part II: https://coursera.org/share/a551af98e24292b6445c82a2a5f16b18
I was bothered with the same question lately, and I think I might share my understanding...
I think the key difference between these two algorithms (Dijkstra and Prim) roots in the problem they are designed to solve, namely, shortest path between two nodes and minimal spanning tree (MST). The formal is to find the shortest path between say, node s and t, and a rational requirement is to visit each edge of the graph at most once. However, it does NOT require us to visit all the node. The latter (MST) is to get us visit ALL the node (at most once), and with the same rational requirement of visiting each edge at most once too.
That being said, Dijkstra allows us to "take shortcut" so long I can get from s to t, without worrying the consequence - once I get to t, I am done! Although there is also a path from s to t in the MST, but this s-t path is created with considerations of all the rest nodes, therefore, this path can be longer than the s-t path found by the Dijstra's algorithm. Below is a quick example with 3 nodes:
2 2
(s) o ----- o ----- o (t)
| |
-----------------
3
Let's say each of the top edges has the cost of 2, and the bottom edge has cost of 3, then Dijktra will tell us to the take the bottom path, since we don't care about the middle node. On the other hand, Prim will return us a MST with the top 2 edges, discarding the bottom edge.
Such difference is also reflected from the subtle difference in the implementations: in Dijkstra's algorithm, one needs to have a book keeping step (for every node) to update the shortest path from s, after absorbing a new node, whereas in Prim's algorithm, there is no such need.
The simplest explanation is in Prims you don't specify the Starting Node, but in dijsktra you (Need to have a starting node) have to find shortest path from the given node to all other nodes.
The key difference between the basic algorithms lies in their different edge-selection criteria. Generally, they both use a priority queue for selecting next nodes, but have different criteria to select the adjacent nodes of current processing nodes: Prim's Algorithm requires the next adjacent nodes must be also kept in the queue, while Dijkstra's Algorithm does not:
def dijkstra(g, s):
q <- make_priority_queue(VERTEX.distance)
for each vertex v in g.vertex:
v.distance <- infinite
v.predecessor ~> nil
q.add(v)
s.distance <- 0
while not q.is_empty:
u <- q.extract_min()
for each adjacent vertex v of u:
...
def prim(g, s):
q <- make_priority_queue(VERTEX.distance)
for each vertex v in g.vertex:
v.distance <- infinite
v.predecessor ~> nil
q.add(v)
s.distance <- 0
while not q.is_empty:
u <- q.extract_min()
for each adjacent vertex v of u:
if v in q and weight(u, v) < v.distance:// <-------selection--------
...
The calculations of vertex.distance are the second different point.
Dijkstras algorithm is used only to find shortest path.
In Minimum Spanning tree(Prim's or Kruskal's algorithm) you get minimum egdes with minimum edge value.
For example:- Consider a situation where you wan't to create a huge network for which u will be requiring a large number of wires so these counting of wire can be done using Minimum Spanning Tree(Prim's or Kruskal's algorithm) (i.e it will give you minimum number of wires to create huge wired network connection with minimum cost).
Whereas "Dijkstras algorithm" will be used to get the shortest path between two nodes while connecting any nodes with each other.
Dijkstra's algorithm is a single source shortest path problem between node i and j, but Prim's algorithm a minimal spanning tree problem. These algorithm use programming concept named 'greedy algorithm'
If you check these notion, please visit
Greedy algorithm lecture note : http://jeffe.cs.illinois.edu/teaching/algorithms/notes/07-greedy.pdf
Minimum spanning tree : http://jeffe.cs.illinois.edu/teaching/algorithms/notes/20-mst.pdf
Single source shortest path : http://jeffe.cs.illinois.edu/teaching/algorithms/notes/21-sssp.pdf
#templatetypedef has covered difference between MST and shortest path. I've covered the algorithm difference in another So answer by demonstrating that both can be implemented using same generic algorithm that takes one more parameter as input: function f(u,v). The difference between Prim and Dijkstra's algorithm is simply which f(u,v) you use.
At the code level, the other difference is the API.
You initialize Prim with a source vertex, s, i.e., Prim.new(s); s can be any vertex, and regardless of s, the end result, which are the edges of the minimum spanning tree (MST) are the same. To get the MST edges, we call the method edges().
You initialize Dijkstra with a source vertex, s, i.e., Dijkstra.new(s) that you want to get shortest path/distance to all other vertices. The end results, which are the shortest path/distance from s to all other vertices; are different depending on the s. To get the shortest paths/distances from s to any vertex, v, we call the methods distanceTo(v) and pathTo(v) respectively.
They both create trees with the greedy method.
With Prim's algorithm we find minimum cost spanning tree. The goal is to find minimum cost to cover all nodes.
with Dijkstra we find Single Source Shortest Path. The goal is find the shortest path from the source to every other node
Prim’s algorithm works exactly as Dijkstra’s, except
It does not keep track of the distance from the source.
Storing the edge that connected the front of the visited vertices to the next closest vertex.
The vertex used as “source” for Prim’s algorithm is
going to be the root of the MST.
What is the exact difference between Dijkstra's and Prim's algorithms? I know Prim's will give a MST but the tree generated by Dijkstra will also be a MST. Then what is the exact difference?
Prim's algorithm constructs a minimum spanning tree for the graph, which is a tree that connects all nodes in the graph and has the least total cost among all trees that connect all the nodes. However, the length of a path between any two nodes in the MST might not be the shortest path between those two nodes in the original graph. MSTs are useful, for example, if you wanted to physically wire up the nodes in the graph to provide electricity to them at the least total cost. It doesn't matter that the path length between two nodes might not be optimal, since all you care about is the fact that they're connected.
Dijkstra's algorithm constructs a shortest path tree starting from some source node. A shortest path tree is a tree that connects all nodes in the graph back to the source node and has the property that the length of any path from the source node to any other node in the graph is minimized. This is useful, for example, if you wanted to build a road network that made it as efficient as possible for everyone to get to some major important landmark. However, the shortest path tree is not guaranteed to be a minimum spanning tree, and the sum of the costs on the edges of a shortest-path tree can be much larger than the cost of an MST.
Another important difference concerns what types of graphs the algorithms work on. Prim's algorithm works on undirected graphs only, since the concept of an MST assumes that graphs are inherently undirected. (There is something called a "minimum spanning arborescence" for directed graphs, but algorithms to find them are much more complicated). Dijkstra's algorithm will work fine on directed graphs, since shortest path trees can indeed be directed. Additionally, Dijkstra's algorithm does not necessarily yield the correct solution in graphs containing negative edge weights, while Prim's algorithm can handle this.
Dijkstra's algorithm doesn't create a MST, it finds the shortest path.
Consider this graph
5 5
s *-----*-----* t
\ /
-------
9
The shortest path is 9, while the MST is a different 'path' at 10.
Prim and Dijkstra algorithms are almost the same, except for the "relax function".
Prim:
MST-PRIM (G, w, r) {
for each key ∈ G.V
u.key = ∞
u.parent = NIL
r.key = 0
Q = G.V
while (Q ≠ ø)
u = Extract-Min(Q)
for each v ∈ G.Adj[u]
if (v ∈ Q)
alt = w(u,v) <== relax function, Pay attention here
if alt < v.key
v.parent = u
v.key = alt
}
Dijkstra:
Dijkstra (G, w, r) {
for each key ∈ G.V
u.key = ∞
u.parent = NIL
r.key = 0
Q = G.V
while (Q ≠ ø)
u = Extract-Min(Q)
for each v ∈ G.Adj[u]
if (v ∈ Q)
alt = w(u,v) + u.key <== relax function, Pay attention here
if alt < v.key
v.parent = u
v.key = alt
}
The only difference is pointed out by the arrow, which is the relax function.
The Prim, which searches for the minimum spanning tree, only cares about the minimum of the total edges cover all the vertices. The relax function is alt = w(u,v)
The Dijkstra, which searches for the minimum path length, so it cares about the edge accumulation. The relax function is alt = w(u,v) + u.key
Dijsktra's algorithm finds the minimum distance from node i to all nodes (you specify i). So in return you get the minimum distance tree from node i.
Prims algorithm gets you the minimum spaning tree for a given graph. A tree that connects all nodes while the sum of all costs is the minimum possible.
So with Dijkstra you can go from the selected node to any other with the minimum cost, you don't get this with Prim's
The only difference I see is that Prim's algorithm stores a minimum cost edge whereas Dijkstra's algorithm stores the total cost from a source vertex to the current vertex.
Dijkstra gives you a way from the source node to the destination node such that the cost is minimum. However Prim's algorithm gives you a minimum spanning tree such that all nodes are connected and the total cost is minimum.
In simple words:
So, if you want to deploy a train to connecte several cities, you would use Prim's algo. But if you want to go from one city to other saving as much time as possible, you'd use Dijkstra's algo.
Both can be implemented using exactly same generic algorithm as follows:
Inputs:
G: Graph
s: Starting vertex (any for Prim, source for Dijkstra)
f: a function that takes vertices u and v, returns a number
Generic(G, s, f)
Q = Enqueue all V with key = infinity, parent = null
s.key = 0
While Q is not empty
u = dequeue Q
For each v in adj(u)
if v is in Q and v.key > f(u,v)
v.key = f(u,v)
v.parent = u
For Prim, pass f = w(u, v) and for Dijkstra pass f = u.key + w(u, v).
Another interesting thing is that above Generic can also implement Breadth First Search (BFS) although it would be overkill because expensive priority queue is not really required. To turn above Generic algorithm in to BFS, pass f = u.key + 1 which is same as enforcing all weights to 1 (i.e. BFS gives minimum number of edges required to traverse from point A to B).
Intuition
Here's one good way to think about above generic algorithm: We start with two buckets A and B. Initially, put all your vertices in B so the bucket A is empty. Then we move one vertex from B to A. Now look at all the edges from vertices in A that crosses over to the vertices in B. We chose the one edge using some criteria from these cross-over edges and move corresponding vertex from B to A. Repeat this process until B is empty.
A brute force way to implement this idea would be to maintain a priority queue of the edges for the vertices in A that crosses over to B. Obviously that would be troublesome if graph was not sparse. So question would be can we instead maintain priority queue of vertices? This in fact we can as our decision finally is which vertex to pick from B.
Historical Context
It's interesting that the generic version of the technique behind both algorithms is conceptually as old as 1930 even when electronic computers weren't around.
The story starts with Otakar Borůvka who needed an algorithm for a family friend trying to figure out how to connect cities in the country of Moravia (now part of the Czech Republic) with minimal cost electric lines. He published his algorithm in 1926 in a mathematics related journal, as Computer Science didn't existed then. This came to the attention to Vojtěch Jarník who thought of an improvement on Borůvka's algorithm and published it in 1930. He in fact discovered the same algorithm that we now know as Prim's algorithm who re-discovered it in 1957.
Independent of all these, in 1956 Dijkstra needed to write a program to demonstrate the capabilities of a new computer his institute had developed. He thought it would be cool to have computer find connections to travel between two cities of the Netherlands. He designed the algorithm in 20 minutes. He created a graph of 64 cities with some simplifications (because his computer was 6-bit) and wrote code for this 1956 computer. However he didn't published his algorithm because primarily there were no computer science journals and he thought this may not be very important. The next year he learned about the problem of connecting terminals of new computers such that the length of wires was minimized. He thought about this problem and re-discovered Jarník/Prim's algorithm which again uses the same technique as the shortest path algorithm he had discovered a year before. He mentioned that both of his algorithms were designed without using pen or paper. In 1959 he published both algorithms in a paper that is just 2 and a half page long.
Dijkstra finds the shortest path between it's beginning node
and every other node. So in return you get the minimum distance tree from beginning node i.e. you can reach every other node as efficiently as possible.
Prims algorithm gets you the MST for a given graph i.e. a tree that connects all nodes while the sum of all costs is the minimum possible.
To make a story short with a realistic example:
Dijkstra wants to know the shortest path to each destination point by saving traveling time and fuel.
Prim wants to know how to efficiently deploy a train rail system i.e. saving material costs.
Directly from Dijkstra's Algorithm's wikipedia article:
The process that underlies Dijkstra's algorithm is similar to the greedy process used in Prim's algorithm. Prim's purpose is to find a minimum spanning tree that connects all nodes in the graph; Dijkstra is concerned with only two nodes. Prim's does not evaluate the total weight of the path from the starting node, only the individual path.
Here's what clicked for me: think about which vertex the algorithm takes next:
Prim's algorithm takes next the vertex that's closest to the tree, i.e. closest to some vertex anywhere on the tree.
Dijkstra's algorithm takes next the vertex that is closest to the source.
Source: R. Sedgewick's lecture on Dijkstra's algorithm, Algorithms, Part II: https://coursera.org/share/a551af98e24292b6445c82a2a5f16b18
I was bothered with the same question lately, and I think I might share my understanding...
I think the key difference between these two algorithms (Dijkstra and Prim) roots in the problem they are designed to solve, namely, shortest path between two nodes and minimal spanning tree (MST). The formal is to find the shortest path between say, node s and t, and a rational requirement is to visit each edge of the graph at most once. However, it does NOT require us to visit all the node. The latter (MST) is to get us visit ALL the node (at most once), and with the same rational requirement of visiting each edge at most once too.
That being said, Dijkstra allows us to "take shortcut" so long I can get from s to t, without worrying the consequence - once I get to t, I am done! Although there is also a path from s to t in the MST, but this s-t path is created with considerations of all the rest nodes, therefore, this path can be longer than the s-t path found by the Dijstra's algorithm. Below is a quick example with 3 nodes:
2 2
(s) o ----- o ----- o (t)
| |
-----------------
3
Let's say each of the top edges has the cost of 2, and the bottom edge has cost of 3, then Dijktra will tell us to the take the bottom path, since we don't care about the middle node. On the other hand, Prim will return us a MST with the top 2 edges, discarding the bottom edge.
Such difference is also reflected from the subtle difference in the implementations: in Dijkstra's algorithm, one needs to have a book keeping step (for every node) to update the shortest path from s, after absorbing a new node, whereas in Prim's algorithm, there is no such need.
The simplest explanation is in Prims you don't specify the Starting Node, but in dijsktra you (Need to have a starting node) have to find shortest path from the given node to all other nodes.
The key difference between the basic algorithms lies in their different edge-selection criteria. Generally, they both use a priority queue for selecting next nodes, but have different criteria to select the adjacent nodes of current processing nodes: Prim's Algorithm requires the next adjacent nodes must be also kept in the queue, while Dijkstra's Algorithm does not:
def dijkstra(g, s):
q <- make_priority_queue(VERTEX.distance)
for each vertex v in g.vertex:
v.distance <- infinite
v.predecessor ~> nil
q.add(v)
s.distance <- 0
while not q.is_empty:
u <- q.extract_min()
for each adjacent vertex v of u:
...
def prim(g, s):
q <- make_priority_queue(VERTEX.distance)
for each vertex v in g.vertex:
v.distance <- infinite
v.predecessor ~> nil
q.add(v)
s.distance <- 0
while not q.is_empty:
u <- q.extract_min()
for each adjacent vertex v of u:
if v in q and weight(u, v) < v.distance:// <-------selection--------
...
The calculations of vertex.distance are the second different point.
Dijkstras algorithm is used only to find shortest path.
In Minimum Spanning tree(Prim's or Kruskal's algorithm) you get minimum egdes with minimum edge value.
For example:- Consider a situation where you wan't to create a huge network for which u will be requiring a large number of wires so these counting of wire can be done using Minimum Spanning Tree(Prim's or Kruskal's algorithm) (i.e it will give you minimum number of wires to create huge wired network connection with minimum cost).
Whereas "Dijkstras algorithm" will be used to get the shortest path between two nodes while connecting any nodes with each other.
Dijkstra's algorithm is a single source shortest path problem between node i and j, but Prim's algorithm a minimal spanning tree problem. These algorithm use programming concept named 'greedy algorithm'
If you check these notion, please visit
Greedy algorithm lecture note : http://jeffe.cs.illinois.edu/teaching/algorithms/notes/07-greedy.pdf
Minimum spanning tree : http://jeffe.cs.illinois.edu/teaching/algorithms/notes/20-mst.pdf
Single source shortest path : http://jeffe.cs.illinois.edu/teaching/algorithms/notes/21-sssp.pdf
#templatetypedef has covered difference between MST and shortest path. I've covered the algorithm difference in another So answer by demonstrating that both can be implemented using same generic algorithm that takes one more parameter as input: function f(u,v). The difference between Prim and Dijkstra's algorithm is simply which f(u,v) you use.
At the code level, the other difference is the API.
You initialize Prim with a source vertex, s, i.e., Prim.new(s); s can be any vertex, and regardless of s, the end result, which are the edges of the minimum spanning tree (MST) are the same. To get the MST edges, we call the method edges().
You initialize Dijkstra with a source vertex, s, i.e., Dijkstra.new(s) that you want to get shortest path/distance to all other vertices. The end results, which are the shortest path/distance from s to all other vertices; are different depending on the s. To get the shortest paths/distances from s to any vertex, v, we call the methods distanceTo(v) and pathTo(v) respectively.
They both create trees with the greedy method.
With Prim's algorithm we find minimum cost spanning tree. The goal is to find minimum cost to cover all nodes.
with Dijkstra we find Single Source Shortest Path. The goal is find the shortest path from the source to every other node
Prim’s algorithm works exactly as Dijkstra’s, except
It does not keep track of the distance from the source.
Storing the edge that connected the front of the visited vertices to the next closest vertex.
The vertex used as “source” for Prim’s algorithm is
going to be the root of the MST.