I am studying algorithms, and I have seen an exercise like this
I can overcome this problem with exponential time but. I don't know how to prove this linear time O(E+V)
I will appreciate any help.
Let G be the graph where the minimum spanning tree T is embedded; let A and B be the two trees remaining after (u,v) is removed from T.
Premise P: Select minimum weight edge (x,y) from G - (u,v) that reconnects A and B. Then T' = A + B + (x,y) is a MST of G - (u,v).
Proof of P: It's obvious that T' is a tree. Suppose it were not minimum. Then there would be a MST - call it M - of smaller weight. And either M contains (x,y), or it doesn't.
If M contains (x,y), then it must have the form A' + B' + (x,y) where A' and B' are minimum weight trees that span the same vertices as A and B. These can't have weight smaller than A and B, otherwise T would not have been an MST. So M is not smaller than T' after all, a contradiction; M can't exist.
If M does not contain (x,y), then there is some other path P from x to y in M. One or more edges of P pass from a vertex in A to another in B. Call such an edge c. Now, c has weight at least that of (x,y), else we would have picked it instead of (x,y) to form T'. Note P+(x,y) is a cycle. Consequently, M - c + (x,y) is also a spanning tree. If c were of greater weight than (x,y) then this new tree would have smaller weight than M. This contradicts the assumption that M is a MST. Again M can't exist.
Since in either case, M can't exist, T' must be a MST. QED
Algorithm
Traverse A and color all its vertices Red. Similarly label B's vertices Blue. Now traverse the edge list of G - (u,v) to find a minimum weight edge connecting a Red vertex with a Blue. The new MST is this edge plus A and B.
When you remove one of the edges then the MST breaks into two parts, lets call them a and b, so what you can do is iterate over all vertices from the part a and look for all adjacent edges, if any of the edges forms a link between the part a and part b you have found the new MST.
Pseudocode :
for(all vertices in part a){
u = current vertex;
for(all adjacent edges of u){
v = adjacent vertex of u for the current edge
if(u and v belong to different part of the MST) found new MST;
}
}
Complexity is O(V + E)
Note : You can keep a simple array to check if vertex is in part a of the MST or part b.
Also note that in order to get the O(V + E) complexity, you need to have an adjacency list representation of the graph.
Let's say you have graph G' after removing the edge. G' consists have two connected components.
Let each node in the graph have a componentID. Set the componentID for all the nodes based on which component they belong to. This can be done with a simple BFS for example on G'. This is an O(V) operation as G' only has V nodes and V-2 edges.
Once all the nodes have been flagged, iterate over all unused edges and find the one with the least weight that connects the two components (componentIDs of the two nodes will be different). This is an O(E) operation.
Thus the total runtime is O(V+E).
Related
I have a bipartite graph. I'll refer to red-nodes and black-nodes of the respective disjoint sets.
I would like to know how to find a connected induced subgraph that maximizes the number of red-nodes while ensuring that all black nodes in the subgraph have new valences less than or equal to 2. Where "induced" means that if two nodes are connected in the original graph and both exist in the subgraph then the edge between them is automatically included. Eventually I'd like to introduce non-negative edge-weights.
Can this be reduced to a standard graph algorithm? Hopefully one with known complexity and simple implementation.
It's clearly possible to grow a subgraph greedily. But is this best?
I'm sure that this problem belongs to NP-complete class, so there is no easy way to solve it. I would suggest you using constraint satisfaction approach. There are quite a few ways to formulate your problem, for example mixed-integer programming, MaxSAT or even pseudo-boolean constraints.
For the first try, I would recommend MiniZinc solver. For example, consider this example of defining and solving graph problems in MiniZinc.
Unfortunately this is NP-hard, so there are probably no polynomial-time algorithms to solve it. Here is a reduction from the NP-hard problem Independent Set, where we are given a graph G = (V, E) (with n = |V| and m = |E|) and an integer k, and the task is to determine whether it is possible to find a set of k or more vertices such that no two vertices in the set are linked by an edge:
For every vertex v_i in G, create a red vertex r_i in H.
For every edge (v_i, v_j) in G, create the following in H:
a black vertex b_ij,
n+1 red vertices t_ijk (1 <= k <= n+1),
n black vertices u_ijk (1 <= k <= n),
n edges (t_ijk, u_ijk) (1 <= k <= n)
n edges (t_ijk, u_ij{k-1}) (2 <= k <= n+1)
the three edges (r_i, b_ij), (r_j, b_ij), and (t_ij1, b_ij).
For every pair of vertices v_i, v_j, create the following:
a black vertex c_ij,
the two edges (r_i, c_ij) and (r_j, c_ij).
Set the threshold to m(n+1)+k.
Call the set of all r_i R, the set of all b_ij B, the set of all c_ij C, the set of all t_ij T, and the set of all u_ij U.
The general idea here is that we force each black vertex b_ij to choose at most 1 of the 2 red vertices r_i and r_j that correspond to the endpoints of the edge (i, j) in G. We do this by giving each of these b_ij vertices 3 outgoing edges, of which one (the one to t_ij1) is a "must-have" -- that is, any solution in which a t_ij1 vertex is not selected can be improved by selecting it, as well as the n other red vertices it connects to (via a "wiggling path" that alternates between vertices in t_ijk and vertices in u_ijk), getting rid of either r_i or r_j to restore the property that no black vertex has 3 or more neighbours in the solution if necessary, and then finally restoring connectedness by choosing vertices from C as necessary. (The c_ij vertices are "connectors": they exist only to ensure that whatever subset of R we include can be made into a single connected component.)
Suppose first that there is an IS of size k in G. We will show that there is a connected induced subgraph X with at least m(n+1)+k red nodes in H, in which every black vertex has at most 2 neighbours in X.
First, include in X the k vertices from R that correspond to the vertices in the IS (such a set must exist by assumption). Because these vertices form an IS, no vertex in B is adjacent to more than 1 of them, so for each vertex b_ij, we may safely add it, and the "wiggling path" of 2n+1 vertices beginning at t_ij1, into X as well. Each of these wiggling paths contains n+1 red vertices, and there are m such paths (one for each edge in G), so there are now m(n+1)+k red vertices in X. Finally, to ensure that X is connected, add to it every vertex c_ij such that r_i and r_j are both in X already: notice that this does not change the total number of red vertices in X.
Now suppose that there is a connected induced subgraph X with at least m(n+1)+k red nodes in H, in which every black vertex has at most 2 neighbours in X. We will show that there is an IS in G of size k.
The only red vertices in H are those in R and those in T. There are only n vertices in R, so if X does not contain all m wiggly paths, it must have at most (m-1)(n+1)+n = m(n+1)-1 red vertices, contradicting the assumption that it has at least m(n+1)+k red vertices. Thus X must contain all m wiggly paths. This leaves k other red vertices in X, which must be from R. No two of these vertices can be adjacent to the same vertex in B, since that B-vertex would then be adjacent to 3 vertices: thus, these k vertices correspond to an IS in G.
Since a YES-instance of IS implies a YES-instance to the constructed instance of your problem and vice versa, the solution to the constructed instance of your problem corresponds exactly to the solution to the IS instance; and since the construction is clearly polynomial-time, this establishes that your problem is NP-hard.
Hello this is my first question. I met a homework in algorithm and probability that I can't find a clue to calculate.
Question:
Computing Number of Triangles in a Graph: Given an undirected graph G = (V, E), a triangle in G is a clique of size 3 (formally, a set of nodes {u, v, w} is a triangle in G if (u, v), (v, w), (u, w) are all edges of G). Consider the following algorithm for approximating the number of triangles in a graph. First construct a sampled graph G' = (V, E') as follows. The vertex set of G' is same as that of G. For every e ∈ E, put e in E' with probability p (think of p as, say, 0.1). In this new sampled graph G', count the number of triangles and let T' be the number of triangles in G' (assume that you have given a black box subroutine to count the number of triangles in G' ). Then output T̃= T'/p.
Show that the expected value of T̃=T ,T is the triangle number of original graph G.
I am confusing that the edge in G or G' to form a triangle is not independent since two adjacent triangles in G might share the edge. And not the all the pair of vertices in G can form a edge in G', only those edges are in G will be present in G' with p. It's hard for me to think of the relationship of number of edges and number of triangles in G or G'.
Hope someone can give me some hints, even not the whole solution is OK.
the edge in G or G' to form a triangle is not independent since two adjacent triangles in G might share the edge
Doesn't matter. The sum of expectations is the expectation of the sum regardless of correlation, so you can reason about the triangles individually. (Higher moments, were you concerned about analyzing the estimation quality of this algorithm, would be trickier.)
I have a connected graph G=(V,E) V={1,2,...,n} and a cost function c:E->R
and a second partial graph G'=(V,T) where T={ for every vertex v∈ V find the neighbor with the minimum cost and add the new edge to T}
If G' graph has at least 2 connected components with the set of vertices we consider the graph H where
iff the set of edges (from the initial graph G) is not null.We define over the edges of H a cost function.
Let's say I choose V(H)={a,e,f} and E(H)={ae,af,fe} and
E12={ab,bc,bd,ed}
E23={eg,ef} E31={fc,fd}
c'(ae)=min{c(ab),c(bc),c(bd),c(ed)}=4
c'(af)=min{c(fc),c(fd)}=9
c'(fe)=min{c(eg),c(ef)}=8
Now for every edge e ∈ E(H) we note with e' the edge (from the original graph G)
for which this minimum is attained.
So e'={bc,df,eg} because bc=4 , df=9 and eg=8 and are the min edges that connect my components.
And I have a minimum spanning tree in H relative to the cost function c' and A' is the set of edges for this tree.
So A'={ae,fe} (I deleted the edge with the maximum cost=af from my graph H to create a min spanning tree)
and I have another set of edges A'={e'|e∈A'} and
is a min spanning tree in G relative to the function cost c.
But none of my edges from A' are the same with the ones from e'.
What I'm I doing wrong?
Looks like you're implementing Boruvka's algorithm. If you look at the notation, it says there's an edge from one new node vC1 to a new node vC2 if there are a pair of nodes x ∈ C1 and y ∈ C2 that are adjacentnin the original graph G. In other words, there's an edge between two new nodes if the connected components they correspond to in G' are adjacent in G. The cost of the edge running between them is then the lowest of the costs of any of the edges running between those CC's in the original graph G.
Love some guidance on this problem:
G is a directed acyclic graph. You want to move from vertex c to vertex z. Some edges reduce your profit and some increase your profit. How do you get from c to z while maximizing your profit. What is the time complexity?
Thanks!
The problem has an optimal substructure. To find the longest path from vertex c to vertex z, we first need to find the longest path from c to all the predecessors of z. Each problem of these is another smaller subproblem (longest path from c to a specific predecessor).
Lets denote the predecessors of z as u1,u2,...,uk and dist[z] to be the longest path from c to z then dist[z]=max(dist[ui]+w(ui,z))..
Here is an illustration with 3 predecessors omitting the edge set weights:
So to find the longest path to z we first need to find the longest path to its predecessors and take the maximum over (their values plus their edges weights to z).
This requires whenever we visit a vertex u, all of u's predecessors must have been analyzed and computed.
So the question is: for any vertex u, how to make sure that once we set dist[u], dist[u] will never be changed later on? Put it in another way: how to make sure that we have considered all paths from c to u before considering any edge originating at u?
Since the graph is acyclic, we can guarantee this condition by finding a topological sort over the graph. topological sort is like a chain of vertices where all edges point left to right. So if we are at vertex vi then we have considered all paths leading to vi and have the final value of dist[vi].
The time complexity: topological sort takes O(V+E). In the worst case where z is a leaf and all other vertices point to it, we will visit all the graph edges which gives O(V+E).
Let f(u) be the maximum profit you can get going from c to u in your DAG. Then you want to compute f(z). This can be easily computed in linear time using dynamic programming/topological sorting.
Initialize f(u) = -infinity for every u other than c, and f(c) = 0. Then, proceed computing the values of f in some topological order of your DAG. Thus, as the order is topological, for every incoming edge of the node being computed, the other endpoints are calculated, so just pick the maximum possible value for this node, i.e. f(u) = max(f(v) + cost(v, u)) for each incoming edge (v, u).
Its better to use Topological Sorting instead of Bellman Ford since its DAG.
Source:- http://www.utdallas.edu/~sizheng/CS4349.d/l-notes.d/L17.pdf
EDIT:-
G is a DAG with negative edges.
Some edges reduce your profit and some increase your profit
Edges - increase profit - positive value
Edges - decrease profit -
negative value
After TS, for each vertex U in TS order - relax each outgoing edge.
dist[] = {-INF, -INF, ….}
dist[c] = 0 // source
for every vertex u in topological order
if (u == z) break; // dest vertex
for every adjacent vertex v of u
if (dist[v] < (dist[u] + weight(u, v))) // < for longest path = max profit
dist[v] = dist[u] + weight(u, v)
ans = dist[z];
Instance of the problem:
Undirected and unweighted graph G=(V,E).
two source nodes a and b, two destination nodes c and d and a constant D(complete positive number).(we can assume that lambda(c,d),lambda(a,b)>D, when lambda(x,y) is the shortest path between x and y in G).
we have two peoples standing on the nodes a and b.
Definition:scheduler set-
A scheduler set is a set of orders such that in each step only one of the peoples make a move from his node v to one of v neighbors, when the starting position of them is in the nodes a,b and the ending position is in the nodes c,d.A "scheduler set" is missing-disorders if in each step the distance between the two peoples is > D.
I need to find an algorithm that decides whether there is a "missing-disorders scheduler set" or not.
any suggestions?
One simple solution would be to first solve all-pairs shortest paths using n breadth-first searches from every node in O(n * (n + m)).
Then create the graph of valid node pairs (x,y) with lambda(x, y) > D, with edges indicating the possible moves. There is an edge {(v,w), (x,y)} if v = x and there is an edge {w, y} in the original graph or if w = y and there is an edge {v, x} in the original graph. This new graph has O(n^2) nodes and O(nm) edges.
Now you just need to check whether (c, d) is reachable from (a, b) in the new graph. This can be achieved using DFS or BFS.
The total runtime be O(n * (n + m)).