I want to check if my unidirected graph is a tree. Tree is an acyclic and connected graph. I have a function that checks if graph is connected. So it is enough to be a tree if graph is connected and |E|=|V|-1?
You are correct, E = V - 1 is sufficient to check that your graph is a tree.
The logic is that every tree begins with just a root note (V=1, E=0, so E=V-1), and from there, any time we add one node (V=V+1), we must also add exactly one edge (E=E+1). This makes the equation E=V-1 remain true for all trees.
A cycle occurs when we connect two existing nodes with a new edge (E=E+1 but V stays the same), rendering the equation E=V-1 false.
If it interests you, you may want to read about the more general formula v - e + f = 2, where f is the number of regions inside a graph, including the exterior region. (A tree only has an exterior region so f=1). This rule is called Euler's Formula, which you can read about on Wikipedia.
Connected: It means that for every pair of vertices you choose, there will always be a path between them.
|E|=|V|-1: if your graph has |V| vertices and you are given |E|=|V|-1 edges to connect them, then if you form a cycle, you won't be able to form a connected graph (some vertices will remain without edges). We can conclude that these conditions are enough.
Related
I'm working on path double cover problem. I have undirected connected graph G and and I change every edge to 2 directed edges and each of them is in opposite direction. Then the goal is to find set of paths(no loops) in this directed graph so that every vertex is used once as start of path and once as end of another path. Each of directed edges are used exactly once.
undirected graph G
directed graph G
For this example there is set of paths P={(1,2,4),(4,3,1),(2,1,3),(3,4,2)}.
There are currently known 2 graphs K3 and K5 (fully connected graphs with 3 and 5 vertices) which cannot be covered in this way.
I want to make script which will find me this covering or tell me if there isn't one. I tried to generate all possible paths and then search in them but for bigger graph this approach isn't usable (n! complexity). I don't know how to set up the recursion so I can keep track of what I've used. I don't care about time complexity but it would be awesome if you had any tip for doing it more quickly. :D
Thanks for any suggestions. :D
Your definition is a bit confusing- you say that you need to find a set of paths (no loops) in the directed graph, with 1 outgoing edge per vertex. There is no way for these edges not to form a loop (at most n - 1 edges can be tree edges).
I'm going to assume that you instead mean "only one cycle; no subcycles".
In that case, your task becomes that of determining whether your graph has a Hamiltonian Cycle or not.
We can use Ore's Theorem as a quick check:
If deg v + deg w ≥ n for every pair of distinct non-adjacent vertices v and w of G then G is Hamiltonian.
Note that this says "if" and not "iif" / "if and only if", so a graph be Hamiltonian, and not satisfy this check.
To take things one step further, we can use the Bondy–Chvátal theorem:
A graph is Hamiltonian if and only if its closure is Hamiltonian.
And we obtain its closure in a similar method to what we did for Ore's Theorem check- we repeatedly add a new edge connecting a nonadjacent pair of vertices u and v with deg(v) + deg(u) ≥ n until no more pairs with this property can be found.
Once this is done, we check whether the closure is Hamiltonian. If the closure is a complete graph, then it is Hamiltonian. I was unable to find any proof that the closure will be complete iif the graph g is Hamiltonian, however it does seem to happen with every example graph I can conjure up, so at least it may be a stronger correlation than Ore's Theorem.
In the end, you just need to determine if the graph has Hamiltonian Cycle. I've listed above two ways you can perform quadratic-time checks to positively identify some of such graph (maybe all, again- not sure of the completeness of the closure bit).
I have a type of directed acyclic graph, with some constraints.
There is only one "entry" vertex
There can be multiple leaf vertices
Once a path splits, anything under that path cannot reach into the other path (this will become clearer with some examples below)
There can be any number of "split" vertices. They can be nested.
A "split" vertex can split into any number of paths. The examples below only show 2 paths for each, but it could be more.
My challenge is the following: for each "split" vertex (any vertex that has at least 2 outgoing edges), find the vertices where its paths reconnect - if such a vertex exists. The solution should be as efficient as possible.
Example A:
example a
In this example, vertex A is a "split" vertex, and its "reconnect vertex" is F.
Example B:
example b
Here, there are two split vertices: A and E. For both of them vertex G is the reconnect vertex.
Example C:
example c
Now there are three split vertices: A, D and E. The corresponding reconnect vertices are:
A -> K
D -> K
E -> J
Example D:
example d
Here we have three split vertices again: A, D and E. But this time, vertex E doesn't have a reconnect vertex because one of the paths terminates early.
Sounds like what you want is:
Connect each vertex with out-degree 0 to a single terminal vertex
Construct the dominator tree of the edge-reversed graph. The linked wikipedia article points to a couple algorithms for doing this.
The "reconnect vertex" for a split vertex is its immediate dominator in the edge-reversed graph, i.e., its parent in that dominator tree. This is called its "postdominator" in your original graph. If it's the terminal vertex that you added, then it doesn't have a reconnect vertex in your original graph.
This is the problem of identifying post-dominators in compilers and program analysis. This is often used in the context of calculating control dependences in control flow graphs. "Advanced Compiler Design and Implementation" is a good reference on these topics.
If the graph does not have cycles, then the solution (a) suggested by #matt-timmermans will work.
If the graph has cycles, then solution (a) can report spurious post-dominators. In such cases, a network-flow based approach works better. The algorithm to calculate non-termination sensitive control dependence in this paper using this approach. The basic idea is
at every split node, inject a unique token into the graph along each outgoing edge and
propagate the tokens thru the graph subject to this constraint: if node n is reachable from split node m, then tokens arriving at node m pass thru node n only if all tokens of node m have arrived at node n.
At the end, node n post-dominates node m if all tokens of node m have arrived at node n.
First of all, I have to admit I'm not good at graph theory.
I have a weakly connected directed graph G=(V,E) where V is about 16 millions and E is about 180 millions.
For a given set S, which is a subset of V (size of S will be around 30), is it possible to find a weakly connected sub-graph G'=(V',E') where S is a subset of V' but try to keep the number of V' and E' as small as possible?
The graph G may change and I hope there's a way to find the sub-graph in real time. (When a process is writing into G, G will be locked, so don't worry about G get changed when your sub-graph calculation is still running.)
My current solution is find the shortest path for each pair of vertex in S and merge those paths to get the sub-graph. The result is OK but the running time is pretty expensive.
Is there a better way to solve this problem?
If you're happy with the results from your current approach, then it's certainly possible to do at least as well a lot faster:
Assign each vertex in S to a set in a disjoint set data structure: https://en.wikipedia.org/wiki/Disjoint-set_data_structure. Then:
Do a breadth-first-search of the graph, starting with S as the root set.
When you the search discovers a new vertex, remember its predecessor and assign it to the same set as its predecessor.
When you discover an edge that connects two sets, merge the sets and follow the predecessor links to add the connecting path to G'
Another way to think about doing exactly the same thing:
Sort all the edges in E according to their distance from S. You can use BFS discovery order for this
Use Kruskal's algorithm to generate a spanning tree for G, processing the edges in that order (https://en.wikipedia.org/wiki/Kruskal%27s_algorithm)
Pick a root in S, and remove any subtrees that don't contain a member of S. When you're done, every leaf will be in S.
This will not necessarily find the smallest possible subgraph, but it will minimize its maximum distance from S.
I have this question from Robert Sedgewick's book on algorithms.
Critical edges. An MST edge whose deletion from the graph would cause the
MST weight to increase is called a critical edge. Show how to find all critical edges in a
graph in time proportional to E log E. Note: This question assumes that edge weights
are not necessarily distinct (otherwise all edges in the MST are critical).
Please suggest an algorithm that solves this problem.
One approach I can think of does the job in time E.V.
My approach is to run the kruskal's algorithm.
But whenever we encounter an edge whose insertion in the MST creates a cycle and if that
cycle already contains an edge with the same edge weight, then, the edge already inserted will not be a critical edge (otherwise all other MST edges are critical edges).
Is this algorithm correct? How can I extend this algorithm to do the job in time E log E.
The condition you suggest for when an edge is critical is correct I think. But it's not necessary to actually find a cycle and test each of its edges.
The Kruskal algorithm adds edges in increasing weight order, so the sequence of edge additions can be broken into blocks of equal-weight edge additions. Within each equal-weight block, if there is more than one edge that joins the same two components, then all of these edges are non-critical, because any one of the other edges could be chosen instead. (I say they are all non-critical because we are not actually given a specific MST as part of the input -- if we were then this would identify a particular edge to call non-critical. The edge that Kruskal actually chooses is just an artefact of initial edge ordering or how sorting was implemented.)
But this is not quite sufficient: it might be that after adding all edges of weight 4 or less to the MST, we find that there are 3 weight-5 edges, connecting component pairs (1, 2), (2, 3) and (1, 3). Although no component pair is joined by more than 1 of these 3 edges, we only need (any) 2 of them -- using all 3 would create a cycle.
For each equal-weight block, having weight say w, what we actually need to do is (conceptually) create a new graph in which each component of the MST so far (i.e. using edges having weight < w) is a vertex, and there is an edge between 2 vertices whenever there is a weight-w edge between these components. (This may result in multi-edges.) We then run DFS on each component of this graph to find any cycles, and mark every edge belonging to such a cycle as non-critical. DFS takes O(nEdges) time, so the sum of the DFS times for each block (whose sizes sum to E) will be O(E).
Note that Kruskal's algorithm takes time O(Elog E), not O(E) as you seem to imply -- although people like Bernard Chazelle have gotten close to linear-time MST construction, TTBOMK no one has got there yet! :)
Yes, your algorithm is correct. We can prove that by comparing the execution of Kruskal's algorithm to a similar execution where the cost of some MST edge e is changed to infinity. Until the first execution considers e, both executions are identical. After e, the first execution has one fewer connected component than the second. This condition persists until an edge e' is considered that, in the second execution, joins the components that e would have. Since edge e is the only difference between the forests constructed so far, it must belong to the cycle created by e'. After e', the executions make identical decisions, and the difference in the forests is that the first execution has e, and the second, e'.
One way to implement this algorithm is using a dynamic tree, a data structure that represents a labelled forest. One configuration of this ADT supports the following methods in logarithmic time.
MakeVertex() - constructs and returns a fresh vertex.
Link(u, c, v) - vertices u and v must not be connected. Creates an unmarked edge from vertex u to vertex v with cost c.
Mark(u, v) - vertices u and v must be endpoints of an edge e. Marks e.
Connected(u, v) - indicates whether vertices u and v are connected.
FindMax(u, v) - vertices u and v must be connected. Returns the endpoints of an unmarked edge on the unique path from u to v with maximum cost, together with that cost. The endpoints of this edge are given in the order that they appear on the path.
I make no claim that this is a good algorithm in practice. Dynamic trees, like Swiss Army knives, are versatile but complicated and often not the best tool for the job. I encourage you to think about how to take advantage of the fact that we can wait until all of the edges are processed to figure out what the critical edges are.
Here is the full question:
Assume we have a directed graph G = (V,E), we want to find a graph G' = (V,E') that has the following properties:
G' has same connected components as G
G' has same component graph as G
E' is minimized. That is, E' is as small as possible.
Here is what I got:
First, run the strongly connected components algorithm. Now we have the strongly connected components. Now go to each strong connected component and within that SCC make a simple cycle; that is, a cycle where the only nodes that are repeated are the start/finish nodes. This will minimize the edges within each SCC.
Now, we need to minimize the edges between the SCCs. Alas, I can't think of a way of doing this.
My 2 questions are: (1) Does the algorithm prior to the part about minimizing edges between SCCs sound right? (2) How does one go about minimizing the edges between SCCs.
For (2), I know that this is equivalent to minimizing the number of edges in a DAG. (Think of the SCCs as the vertices). But this doesn't seem to help me.
The algorithm seems right, as long as you allow for closed walks (i.e. repeating vertices.) Proper cycles might not exist (e.g. in an "8" shaped component) and finding them is NP-hard.
It seems that it is sufficient to group the inter-component edges by ordered pairs of components they connect and leave only one edge in each group.
Regarding the step 2,minimize the edges between the SCCs, you could randomly select a vertex, and run DFS, only keeping the longest path for each pair of (root, end), while removing other paths. Store all the vertices searched in a list L.
Choose another vertex, if it exists in L, skip to the next vertex; if not, repeat the procedure above.