Given undirected, connected graph G={V,E}, a vertex in V(G), label him v, and a weight function f:E->R+(Positive real numbers), I need to find a MST such that v's degree is minimal. I've already noticed that if all the edges has unique weight, the MST is unique, so I believe it has something to do with repetitive weights on edges. I though about running Kruskal's algorithm, but when sorting the edges, I'll always consider edges that occur on v last. For example, if (a,b),(c,d),(v,e) are the only edges of weight k, so the possible permutations of these edges in the sorted edges array are: {(a,b),(c,d),(v,e)} or {(c,d),(a,b),(v,e)}. I've ran this variation over several graphs and it seems to work, but I couldn't prove it. Does anyone know how to prove the algorithm's correct (Meaning proving v's degree is minimal), or give a contrary example of the algorithm failing?
First note that Kruskal's algorithm can be applied to any weighted graph, whether or not it is connected. In general it results in a minimum-weight spanning forest (MSF), with one MST for each connected component. To prove that your modification of Kruskal's algorithm succeeds in finding the MST for which v has minimal degree, it helps to prove the slightly stronger result that if you apply your algorithm to a possibly disconnected graph then it succeeds in finding the MSF where the degree of v is minimized.
The proof is by induction on the number, k, of distinct weights.
Basis Case (k = 1). In this case weights can be ignored and we are trying to find a spanning forest in which the degree of v is minimized. In this case, your algorithm can be described as follows: pick edges for as long as possible according to the following two rules:
1) No selected edge forms a cycle with previously selected edges
2) An edge involving v isn't selected unless any edge which doesn't
involve v violates rule 1.
Let G' denote the graph from which v and all incident edges have been removed from G. It is easy to see that the algorithm in this special case works as follows. It starts by creating a spanning forest for G'. Then it takes those trees in the forest that are contained in v's connected component in the original graph G and connects each component to v by a single edge. Since the components connected to v in the second stage can be connected to each other in no other way (since if any connecting edge not involving v exists it would have been selected by rule 2) it is easy to see that the degree of v is minimal.
Inductive Case: Suppose that the result is true for k and G is a weighted graph with k+1 distinct weights and v is a specified vertex in G. Sort the distinct weights in increasing order (so that weight k+1 is the longest of the distinct weights -- say w_{k+1}). Let G' be the sub-graph of G with the same vertex set but with all edges of weight w_{k+1} removed. Since the edges are sorted in the order of increasing weight, note that the modified Kruskal's algorithm in effect starts by applying itself to G'. Thus -- by the induction hypothesis prior to considering edges of weight w_{k+1}, the algorithm has succeeded in constructing an MSF F' of G' for which the degree, d' of v in G' is minimized.
As a final step, modified Kruskal's applied to the overall graph G will merge certain of the trees in F' together by adding edges of weight w_{k+1}. One way to conceptualize the final step is the think of F' as a graph where two trees are connected exactly when there is an edge of weight w_{k+1} from some node in the first tree to some node in the second tree. We have (almost) the basis case with F'. Modified Kruskal's will add edged of weight w_{k+1} until it can't do so anymore -- and won't add an edge connecting to v unless there is no other way to connect to trees in F' that need to be connected to get a spanning forest for the original graph G.
The final degree of v in the resulting MSF is d = d'+d" where d" is the number of edges of weight w_{k+1} added at the final step. Neither d' nor d" can be made any smaller, hence it follows that d can't be made any smaller (since the degree of v in any spanning forest can be written as the sum of the number of edges whose weight is less than w_{k+1} coming into v and the number off edges of weight w_{k+1} coming into v).
QED.
There is still an element of hand-waving in this, especially with the final step -- but Stack Overflow isn't a peer-reviewed journal. Anyway, the overall logic should be clear enough.
One final remark -- it seems fairly clear that Prim's algorithm can be similarly modified for this problem. Have you looked into that?
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.
I am having problem with shortest path in directed weighted graph. I know Dijkstra, BFS, DFS. However, I have a set of vertices S for starting points and a set of vertices E to end. S and E doesn't overlap. So how can I find the set of edges with minimal sum of edge weight? The edge set doesn't have to include all vertices in S, but have to reach all vertices in E. Should I start with Dijkstra on all permutation of {Si, Ei} and optimize or I miss any important algorithm I should know? Or even I am over-thinking....
If I understand you correctly, you want to find the tree of minimal weight in the graph that contains all the vertices of E and at least one vertex from S.
The problem is called general Steiner tree, and it is NP-hard. So the best you can probably hope for is an exponential-time algorithm or some kind of approximation (the minimum spanning tree of the whole graph comes to mind, maybe after removing some unneeded subtrees).
There is a simple DP solution that works in O(2^n * (n + m)): Let f(S) be the cost of the minimum tree in the graph that spans all the nodes in S. It can be shown that there is such a tree T such that the weight of T \ {x} is f(S \ {x}) for some x, so the transition can be done in O(n + m).
I'm trying to find an efficient method of detecting whether a given graph G has two different minimal spanning trees. I'm also trying to find a method to check whether it has 3 different minimal spanning trees. The naive solution that I've though about is running Kruskal's algorithm once and finding the total weight of the minimal spanning tree. Later , removing an edge from the graph and running Kruskal's algorithm again and checking if the weight of the new tree is the weight of the original minimal spanning tree , and so for each edge in the graph. The runtime is O(|V||E|log|V|) which is not good at all, and I think there's a better way to do it.
Any suggestion would be helpful,
thanks in advance
You can modify Kruskal's algorithm to do this.
First, sort the edges by weight. Then, for each weight in ascending order, filter out all irrelevant edges. The relevant edges form a graph on the connected components of the minimum-spanning-forest-so-far. You can count the number of spanning trees in this graph. Take the product over all weights and you've counted the total number of minimum spanning trees in the graph.
You recover the same running time as Kruskal's algorithm if you only care about the one-tree, two-trees, and three-or-more-trees cases. I think you wind up doing a determinant calculation or something to enumerate spanning trees in general, so you likely wind up with an O(MM(n)) worst-case in general.
Suppose you have a MST T0 of a graph. Now, if we can get another MST T1, it must have at least one edge E different from the original MST. Throw away E from T1, now the graph is separated into two components. However, in T0, these two components must be connected, so there will be another edge across this two components that has exactly the same weight as E (or we could substitute the one with more weight with the other one and get a smaller ST). This means substitute this other edge with E will give you another MST.
What this implies is if there are more than one MSTs, we can always change just a single edge from a MST and get another MST. So if you are checking for each edge, try to substitute the edge with the ones with the same weight and if you get another ST it is a MST, you will get a faster algorithm.
Suppose G is a graph with n vertices and m edges; that the weight of any edge e is W(e); and that P is a minimal-weight spanning tree on G, weighing Cost(W,P).
Let δ = minimal positive difference between any two edge weights. (If all the edge weights are the same, then δ is indeterminate; but in this case, any ST is an MST so it doesn't matter.) Take ε such that δ > n·ε > 0.
Create a new weight function U() with U(e)=W(e)+ε when e is in P, else U(e)=W(e). Compute Q, an MST of G under U. If Cost(U,Q) < Cost(U,P) then Q≠P. But Cost(W,Q) = Cost(W,P) by construction of δ and ε. Hence P and Q are distinct MSTs of G under W. If Cost(U,Q) ≥ Cost(U,P) then Q=P and distinct MSTs of G under W do not exist.
The method above determines if there are at least two distinct MSTs, in time O(h(n,m)) if O(h(n,m)) bounds the time to find an MST of G.
I don't know if a similar method can treat whether three (or more) distinct MSTs exist; simple extensions of it fall to simple counterexamples.
Here is an excise
Suppose we are given the minimum spanning tree T of a given graph G
(with n vertices and m edges) and a new edge e = (u, v) of weight w
that we will add to G. Give an efficient algorithm to find the minimum
spanning tree of the graph G + e. Your algorithm should run in O(n)
time to receive full credit.
I have this idea:
In the MST, just find out the path between u and v. Then find the edge (along the path) with maximum weight; if the maximum weight is bigger than w, then remove that edge from the MST and add the new edge to the MST.
The tricky part is how to do this in O(n) time and it is also I get stuck.
The question is that how the MST is stored. In normal Prim's algorithm, the MST is stored as a parent array, i.e., each element is the parent of the according vertex.
So suppose the excise give me a parent array indicating the MST, how can I release the above algorithm in O(n)?
First, how can I identify the path between u and v from the parent array? I can have two ancestor arrays for u and v, then check on the common ancestor, then I can get the path, although in backwards. I think for this part, to find the common ancestor, at least I have to do it in O(n^2), right?
Then, we have the path. But we still need to find the weight of each edge along the path. Since I suppose the graph will use adjacency-list for Prim's algorithm, we have to do O(m) (m is the number of edges) to locate each weight of the edge.
...
So I don't see it is possible to do the algorithm in O(n). Am I wrong?
The idea you have is right. Note that, finding the path between u and v is O(n). I'll assume you have a parent array identifying the MST. tracking the path (for max edge) from u to v or u to root vertex should take only O(n). If you reach root vertex, just track the path from v to u or root vertex.
Now that you have the path from u -> u1 ... -> max_path_vert1 -> max_path_vert2 -> ... -> v, remove the edge max_path_vert1->max_path_vert2 (assuming this is greater than the added edge) and reverse the parents for u->...->max_path_vert1 and mark parent[u] = v.
Edit: More explanation for clarity
Note that, in MST there will be exactly one path between any pair of vertices. So, if you can trace from u->y and v->y, you have only traced through atmost n vertices. If you traced more than n vertices that means you visited a vertex twice, which will not happen in an MST. Ok, now hopefully you're convinced it's O(n) to track from u->y and v->y. Once you have these paths, you have established a path from u->v. Do you see how? I'm assuming this is an undirected graph, since finding MST for directed graph is a different concept in itself. For undirected graph, when you have a path from x->y you have a path from y-x. So, u->y->v exist. You don't even need to trace back from y->v, since weights for v->y will be same as that of y->v. Just find the edge with the maximum weight when you trace from u->y and v->y.
Now for finding edge weights in O(1); how are you storing your current weights? Adjacency list or adjacency matrix? For O(1) access, store it the way parent vertex array is stored. So, weight[v] = weight(v, parent[v]). So, you'll have O(1) access. Hope this helps.
Well - your solution is correct.
But regarding implementation, I dont see why you are using G instead of T to find the path between u and v. Using any search traversal in T for the path between u and v, will give you O(n). - That is, you can assume that v is the root and performs a Depth-First Search algorithm [in this case, you will have to assume all neighbors of v as children] - and stop the DFS once you find u - then, the nodes in the stack corresponds to the path between u and v.
It is easy afterward to find the cost of each edge in the path (O(n)), and it is easy as well to delete/add edges. In total O(n).
Does that help somehow ?
Or maybe you are getting O(n^2) - according to my understanding - because you access the children of a vertex v in T in O(n) -- Here, you have to present your data structure as a mapped array so that the cost is reduced to O(1). [for instace, {a,b,c,u,w}(vertices) -> {0,1,2,3,4}(indices of vertices).