We define ROMAN-SUBSET as the following problem:
INPUT: Directed graph G = ( V , E ) and a positive integer k
OUTPUT: If there is a subset R of V such that | R | <= k , and
such that every directed circuit in G includes at least one vertex
from R , then the output should be "TRUE", otherwise, it should be
"FALSE".
Assuming that the Vertex Cover (VC) problem is NP-complete, I must prove that ROMAN-SUBSET is also NP-complete. From what I understand, that means taking the VC input, modifying it, and then showing that plugging it into the ROMAN-SUBSET algorithm will yield the result of the VC problem.
I'm having a really tough time coming up with the transformation. I know that the VC input is a graph G and an integer k, and the problem is whether or not there exists a subset R of V that covers every edge in G, such that |R| <= k. So clearly, the R and k are similar from ROM to VC, but my difficulty is identifying how to transform the graph so that 1 vertex in every directed cycle (for ROM) corresponds to every edge (for VC). How can I go about modifying the graph to prove that VC can be reduced to ROM?
Thanks!
Here is the construction.
Take undirected graph G = (V, E) as in VC.
Now define the directed graph G1 = (V, E1), where for every edge (u,v) in E there are two edges (u,v) and (v,u) in E1.
In other words the new graph is the same as the old one, but every undirected edge is replaced with two directed edges that form a 2-cycle.
The claim is that from ROM on G1 follows VC on G.
Indeed, suppose that the answer for ROM on G1 is FALSE. Then for every choice of a set of less than k vertices there exists a cycle not in this set. So there exists an edge whose endpoints are not in the set. But this means that for the same choice of the set of less than k vertices in G there exists an edge whose endpoints are not in the set, so VC's answer is FALSE.
Conversely, suppose that the answer for ROM on G1 is TRUE. Then there exists a subset of V containing less than k vertices, so that given any cycle there exists at least one vertex in the cycle, which is in the set. But this means that for any edge in E one of its endpoints in in the set, because an edge in E corresponds to a 2-cycle in E1. Thus the answer for VC is TRUE.
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.
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).
I'm pretty sure this problem is P and not NP, but I'm having difficulty coming up with a polynomially bound algorithm to solve it.
You can :
check that number of edges in the graph is n(n-1)/2.
check that each vertice is connected to exaclty n-1 distinct vertices.
This will run in O(V²), which is polynomial.
Hope it helped.
Here's an O(|E|) algorithm that also has a small constant.
It's trivial to enumerate every edge in a complete graph. So all you need to do is scan your edge list and verify that every such edge exists.
For each edge (i, j), let f(i, j) = i*|V| + j. Assuming vertices are numbered 0 to |V|-1.
Let bitvec be a bit vector of length |V|2, initialized to 0.
For each edge (i, j), set bitvec[f(i, j)] = 1.
G is a complete graph if and only if every element of bitvec == 1.
This algorithm not only touches E once, but it's also completely vectorizable if you have a scatter instruction. That also means it's trivial to parallelize.
Here is an O(E) algorithm:
Use O(E) as it is input time, to scan the graph
Meanwhile, record each vertex p's degree, increase degree only if the neighbor is not p itself (self-connecting edge) and is not a vertex q where p and q has another edge counted already (multiple edge), these checking can be done in O(1)
Check if all vertex's degree is |V|-1, this step is O(V), if Yes then it is a complete graph
Total is O(E)
For a given graph G = (V,E), check for each pair u, v in the V, and see if edge (u,v) is in E.
The total number of u, v pairs are |V|*(|V|-1)/2. As a result, with a time complexity of O(|V|^2), you can check and see if a graph is complete or not.
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];
Dominating Set (DS) := given an undirected graph G = (V;E), a set of
vertices S V is a dominating set if for every vertex in V , there is a vertex in
S that is adjacent to v. Entire vertex set V is a trivial dominating set in
any graph.
Find minimum size dominating set for a tree.
I'll attempt to prove this in a more formal way.
OUTLINE
To prove your greedy algorithm is correct, you need to prove two things:
First, that your greedy choice is valid and can always be used in the formation of an optimal solution, and
second, that your problem has an optimal substructure property, that is, you can form an optimal solution from optimal solutions to subproblems of your own problem.
Greedy Choice: In your tree T = (V, E), find a vertex v in the tree with the highest number of leaves. Add it to your dominant set.
Optimal Substructure
T' = (V', E') such that:
V' = V \ ({a : a ϵ V, a is adjacent to v, and a's degree ≤ 2} ∪ {v})
E' = E - any edge involving any of the removed vertices
In other words
Look for a vertex with the highest number of leaves, remove any of its adjacent vertices with degree less than or equal to 2, then remove v itself, and add it to your dominant set. Repeat this until you have no vertices left.
PROOF
Greedy choice proof
For any leaf l, it must be that either itself or its parent is in the dominant set. In our case, the vertex v we would have chosen is in this situation.
Let A = {v1 , v2 , ... , vk} be a minimum dominant set of T. If A already has v as member, we are done. If it does not, we see two situations:
v has some neighbouring leaf l. Then, l must be part of the dominant set, otherwise our set is not dominating the entire tree. We can simply thus form A' = {A - {l} + {v}} and still be a dominant set. Since |A'| = |A|, A' is still optimal.
v does not have any neighbouring leaves l. Then, because v was chosen such that it has the highest number of leaves, then no vertex in T have any leaves. Then T is not a tree. Contradiction.
Thus, we will always be able to form an optimal solution with our greedy choice.
Optimal Substructure proof
Suppose that A is a minimum dominant set for T = (V, E), but that A' = A \ {v} is not a minimum dominant set for T' as defined above.
Make a minimum dominant set for T', call it B. As aforementioned, |B| < |A'|. It can be shown that B' = B ∪ {v} is a dominating set for T. Then, since |A'| = |A| - 1, |B'| = |B| + 1, we get |B'| < |A|. This is contradictory, since we assumed that A is an minimum independent set. Thus it must be that A' is also a minimum independent set of T'.
Proving B' = B ∪ {v} is a dominating set for T:
v may have had adjacent vertices adjacent not in T'. We will show that any vertices that were not considered in T' will be dominated by vertices in B' (This means that we picked our set optimally): Let y be some vertex adjacent to v and not in T'. By definition of T', y can only have degree 1 or 2. Now, y is dominated by v. If y is a leaf, then we are done. However, if y is of degree 2, then y is connected another node which is necessarily in the dominant set of B. This is because, when we removed v to make T', the degree of y became 1, meaning that y or its parent was necessarily added to the dominant set. Hence, B' is a dominant set for T.
1- Always start from leafs
2- Add their parent to DS and cut the children
3- Mark parent's of selected parent as already dominated
4- After completing process , check whether those marked nodes has a children that is not
dominated and add them to DS
Good luck