Let G = (V, E) be a weighted, directed graph with weight function w : E → R. Give an O(V E)-time algorithm to find, for each vertex v ∈ V , the value δ*(v) = min{u∈V} {δ(u, v)}.
I don't understand the question. Could someone give me some ideas?
This basically means:
G = (V, E) having a graph with V vertices and E edges
weighted, directed graph with weight function w : E → R the graph is directed and weighted, where each edge has real value weight
O(V E)-time algorithm find the algorithm that runs in number of operations proportional to number of vertices multiplied by number of edges
for each vertex v ∈ V , the value δ*(v) = min{u∈V} {δ(u, v)} here they have not described what δ(u, v) means, but most probably this is the sum of weights of edges from vertex u to v. This basically asks you to find the minimum distance from vertex u to all vertices v.
And the answer to your question Bellman Ford.
Related
I have the following algorithm to find the minimum vertex cover of a tree. That is a minimal sized set S of vertices such that for every edge (v,u) in G either v is in S or u is in S.
I have been told the algorithm has linear time complexity, however I don't understand how this is the case, since isn't the number of edges incident to u of the order O(n) and so the complexity would be O(n^2)?
Let T = <V, E> be a Tree. That is, the vertex set is V, the edge set is E. Also suppose the cover set = C. The algorithm can be described as follows:
while V != [] do
Identify a leaf vertex v
Locate u = parent(v), the parent vertex of v.
Add u to C
Remove all the edges incident to u
return C.
In a tree, |E| = |V| - 1, so there are O(n) edges to deal with in total.
I am trying to design an algorithm where, given a connected weighted graph G = (V, E) and a subset of vertices U that is in V, will construct a minimum spanning tree such that all vertices in U are leaves (other vertices may also be leaves), or returns that no such tree exists (False).
This is all I got, adapting Prim's algorithm (fair warning, its really bad; don't even know if it works/is efficient or what data structures to use, I will accept literally any other correct algorithm instead):
Let x be an arbitrary node in G
Set S = {x}
While S != V:
Let (u,v) be the cheapest edge with u in S and v not in S
Add (u,v) to tree T if u is not in U, add v to S
If all u in U is in the tree T:
return T
Else:
return False
I also have a picture of what I think it would do to this graph I drew:
pic here
A proof that the algorithm is correct would also give me some peace of mind.
If all vertices u ∈ U are to be leaves in a solution, no u can be used in that solution to connect two other vertices. All vertices not in U must be connected by edges not incident to any u.
Remove U and all edges incident to U. Find the minimum spanning tree, then connect each u to the tree by the smallest-weighted edge available from those we removed.
Suppose we are given a directed graph G = (V, E) with potentially positive and negative edge lengths, but no negative cycles. Let s ∈ V be a given source
vertex. How to design an algorithm for the single-source shortest path problem that runs in time O(k(|V | + |E|)) if the shortest paths from s to any other vertex takes at most k edges?
Here`s O(k(|V | + |E|)) approach:
We can use Bellman-Ford algorithm with some modifications
Create array D[] to store shortest path from node s to some node u
initially D[s]=0, and all other D[i]=+oo (infinity)
Now after we iterate throught all edges k times and relax them, D[u] holds shortest path value from node s to u after <=k edges
Because any s-u shortest path is atmost k edges, we can end algorithm after k iterations over edges
Pseudocode:
for each vertex v in vertices:
D[v] := +oo
D[s] = 0
repeat k times:
for each edge (u, v) with weight w in edges:
if D[u] + w < D[v]:
D[v] = D[u] + w
Given a graph G = (V, E) and a subset v of V how do you compute a "minimal spanning tree" m that connects all the nodes in v together? That is to say, it can have paths that go through vertices not in v.
My first thought was that m must be a subset of M where M is the MST of G but here is a counter-example through this diagram: Find the MST of {B, C}. It is clearly the shortest path, the edge with weight 17, which is not a subset of M.
I'm having trouble reducing this problem/defining exactly what to run a classic MST algorithm on.
Thanks in advance!
Yes, this will be a homework (I am self-learning not for university) question but I am not asking for solution. Instead, I am hoping to clarify the question itself.
In CLRS 3rd edition, page 593, excise 22.1-5,
The square of a directed graph G = (V, E) is the graph G2 = (V, E2) such that (u,v) ∈ E2 if and only if G contains a path with at most two edges between u and v. Describe efficient algorithms for computing G2 from G for both the adjacency-list and adjacency-matrix representations of G. Analyze the running times of your algorithms.
However, in CLRS 2nd edition (I can't find the book link any more), page 530, the same exercise but with slightly different description:
The square of a directed graph G = (V, E) is the graph G2 = (V, E2) such that (u,w) ∈ E2 if and only if for some v ∈ V, both (u,v) ∈ E and (v,w) ∈ E. That is, G2 contains an edge between u and w whenever G contains a path with exactly two edges between u and w. Describe efficient algorithms for computing G2 from G for both the adjacency-list and adjacency-matrix representations of G. Analyze the running times of your algorithms.
For the old exercise with "exactly two edges", I can understand and can solve it. For example, for adjacency-list, I just do v->neighbour->neighbour.neighbour, then add (v, neighbour.neighbour) to the new E2.
But for the new exercise with "at most two edges", I am confused.
What does "if and only if G contains a path with at most two edges between u and v" mean?
Since one edge can meet the condition "at most two edges", if u and v has only one path which contains only one edge, should I add (u, v) to E2?
What if u and v has a path with 2 edges, but also has another path with 3 edges, can I add (u, v) to E2?
Yes, that's exactly what it means. E^2 should contain (u,v) iff E contains (u,v) or there is w in V, such that E contains both (u,w) and (w,v).
In other words, E^2 according to the new definition is the union of E and E^2 according to the old definition.
Regarding to your last question: it doesn't matter what other paths between u and v exist (if they do). So, if there are two paths between u and v, one with 2 edges and one with 3 edges, then (u,v) should be in E^2 (according to both definitions).
The square of a graph G, G^2 defined by those vertices V' for which d(u,v)<=2 and the eges G' of G^2 is all those edges of G which have both the end vertices From V'