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)).
Related
Let G(V,E) be a directed weighted graph with edge lengths, where all of the edge lengths are positive except two of the edges have negative lengths. Given a fixed vertex s, give an algorithm computing shortest paths from s to any other vertex in O(e + v log(v)) time.
My work:
I am thinking about using the reweighting technique of Johnson's algorithm. And then, run Belford Algo once and apply Dijkstra v times. This will give me the time complexity as O(v^2 log v + ve).
This is the standard all pair shortest problem, As I only need one vertex (s) - my time complexity will be O(v log v + e) right?
For this kind of problem, changing the graph is often a lot easier than changing the algorithm. Let's call the two negative-weight edges N1 and N2; a path by definition cannot use the same edge more than once, so there are four kinds of path:
A. Those which use neither N1 nor N2,
B. Those which use N1 but not N2,
C. Those which use N2 but not N1,
D. Those which use both N1 and N2.
So we can construct a new graph with four copies of each node from the original graph, such that for each node u in the original graph, (u, A), (u, B), (u, C) and (u, D) are nodes in the new graph. The edges in the new graph are as follows:
For each positive weight edge u-v in the original graph, there are four copies of this edge in the new graph, (u, A)-(v, A) ... (u, D)-(v, D). Each edge in the new graph has the same weight as the corresponding edge in the original graph.
For the first negative-weight edge (N1), there are two copies of this edge in the new graph; one from layer A to layer B, and one from layer C to layer D. These new edges have weight 0.
For the second negative-weight edge (N2), there are two copies of this edge in the new graph; one from layer A to layer C, and one from layer B to layer D. These new edges have weight 0.
Now we can run any standard single-source shortest-path problem, e.g. Dijkstra's algorithm, just once on the new graph. The shortest path from the source to a node u in the original graph will be one of the following four paths in the new graph, whichever corresponds to a path of the lowest weight in the original graph:
(source, A) to (u, A) with the same weight.
(source, A) to (u, B) with the weight in the new graph minus the weight of N1.
(source, A) to (u, C) with the weight in the new graph minus the weight of N2.
(source, A) to (u, D) with the weight in the new graph minus the weights of N1 and N2.
Since the new graph has 4V vertices and 4E - 2 edges, the worst-case performance of Dijkstra's algorithm is O((4E - 2) + 4V log 4V), which simplifies to O(E + V log V) as required.
To ensure that a shortest path in the new graph corresponds to a genuine path in the original graph, it remains to be proved that a path from e.g. (source, A) to (u, B) will not use two copies of the same edge from the original graph. That is quite easy to show, but I'll leave it to you as something to think about.
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];
I'm trying to find out an efficient way of finding the shortest path between 2 nodes in a graph with positive edge costs that goes trough a subset of nodes.
More formally:
Given a graph G = (U, V) where U is the set of all nodes in the graph and V is the set of all edges in the graph, a subset of U called U' and a cost function say:
f : UxU -> R+
f(x, y) = cost to travel from node x to node y if there exists an edge
between node x and node y or 0 otherwise,
I have to find the shortest path between a source node and a target node that goes trough all the nodes in U'.
The order in which I visit the nodes in U' doesn't matter and I am allowed to visit a node more than once.
My original idea was to make use of Roy-Floyd algorithm to generate the cost matrix.
Then, for each permutation of the nodes in U' I would be computing the cost between the source and the target like this: f(source_node, P1) + f(P1, P2) + ... + f(Pk, target) saving the configuration for the lowest cost and then reconstructing the path.
The complexity for this approach is O(n3 + k!) O(n3 + k*k!), where n is the number of nodes in the graph and k the number of nodes in the subset U', which is off limits since I'll have to deal with graphs with maximum n = 2000 nodes out of which maximum n - 2 nodes will be part of the U' subset.
This is a generalization of travelling salesman. If U' == U, you get exactly TSP.
You can use the O(n^2 * 2^n) TSP algorithm, where the exponential factor for full scale TSP (2^n) will reduce to k = |U'|, so you'd get O(n^2 * 2^k).
This has the DP solution to TSP.
http://www.lsi.upc.edu/~mjserna/docencia/algofib/P07/dynprog.pdf
Combine the source node s and target node t into a single node z, and define the node set U'' := U' union {z}, with the distances to and from z defined by f(z,x) := f(s,x) and f(x,z) := f(x,t) for all x in U \ {s, t}. Compute shortest paths between all nodes of U'' and let f'(x,y) be the shortest distances, or infinity when there is no appropriate path. Voila, you now have a travelling salesman problem on the complete directed graph with vertices U'' and edge weights f'.