I've faved a question here, and the most promising answer to-date implies "graph carvings". Problem is, I have no clue what it is (neither does the OP, apparently), and it sounds very promising and interesting for several uses. My Googlefu failed me on this topic, as I found no useful/free resource talking about them.
Can someone please tell me what is a 'graph carving', how I can make one for a graph, and how I can determine what makes a certain carving better suited for a task than another?
Please don't go too mathematical on me (or be ready to answer more questions): I understand what's a graph, what's a node and what's a vertex, I manage with big O notation, but I have no real maths background.
I think the answer given in the linked question is a little loose with terminology. I think it is describing a tree carving of a graph G. This is still not particularly google-friendly, I admit, but perhaps it will get you going on your way. The main application of this structure appears to be in one particular DFS algorithm, described in these two papers.
A possibly more clear description of the same algorithm may appear in this book.
I'm not sure stepping through this algorithm would be particularly helpful. It is a reasonably complex algorithm and the explanation would probably just parrot those given in the papers I linked. I can't claim to understand it very well myself. Perhaps the most fruitful approach would be to look at the common elements of those three links, and post specific questions about parts you don't understand.
Q1:
what is a 'graph carving'
There are two types of graph carving: Tree-Carving and Carving.
A tree-carving of a graph is a partition of the vertex set V into subsets V1,V2,...,Vk with the following properties. Each subset constitutes a node of a tree T. For every vertex v in Vj, all the neighbors of v in G belong either to Vj itself, or to Vi where Vi is adjacent to Vj in the tree T.
A carving of a graph is a partitioning of the vertex set V into a collection of subsets V1,V2,...Vk with the following properties. Each subset constitutes a node of a rooted tree T. Each non-leaf node Vj of T has a special vertex denoted by g(Vj) that belongs to p(Vj). For every vertex v in Vi, all the neighbors of v that are in ancestor sets of Vi belong to either
Vi or
Vj, where Vj is the parent of Vi in the tree T, or
Vl, where Vl is the grandparent in the tree T. In this case, however the neighbor of v can only be g(p(Vi))
Those defination referred from chapter 6 of book "Approximation Algorithms for NP-Hard problems" and paper1. (paper1 is picked from Gain's answer, thanks Gain.)
According to my understanding. Tree-Carving or Carving are a kind of representation (or a simplification) of an original graph G. So that the resulting new graph still preserve 'connection properties' of G, but with much smaller size(less vertex, less nodes). These two methods both somehow try to delete 'local' 'similar' information but to keep 'structure' 'vital' information. By merging some 'closed' vertices into one vertex and deleting some edges.
And It seems that Tree Carving is a little bit simpler and easier to understand Since in **Carving**, edges are allowed to go to a single vertex in the grapdhparenet node as well. It would preserve more information.
Q2:
how I can make one for a graph
I only know how to get a tree-carving.
You can refer the algorithm from paper1.
It's a Depth-First-Search based algorithm.
Do DFS, before return from an iteration, check whether this edges is 'bridge' edge or not. If yes, you need remove this 'bridge' and adding some 'back edge'.
You would get a DFS-partition which yields a tree-carving of G.
Q3:
how I can determine what makes a certain carving better suited for a task than another?
Sorry I don't know. I am also a new guy in graph theory.
If you have more question:
What's g function of g(Vj)?
a special node called gray node. go to paper1
What's p function of p(Vj)?
I am not sure. maybe p represent 'parent'. go to paper1
What's the back edge of node t?
some edge(u,v) s.t. u is a decent of t and v is a precedent of t. goto to paper1
What's bridge?
bridge wiki
Related
I'm currently writing my master thesis about clusterings in graphs. My prof said he wants the graph to be represented as a hash table. Because it needs less space than the adjency matrix and it is faster in checking if a edge exists between two vertices than adjency lists.
Anyway, I have a lot of problems understanding how a graph can be built with (perfect) hash functions. I know there should be two tables inside each other. The first includes every node and the second contains all the adjacent vertices. But how do I find a hash function that makes this correctly?
After I built the graph I have to assign a weight to each edge. Is it better to build a new graph or keep the old one? How can I assign the weights correctly to each edge and how do I save it?
And the last question: How fast can I do a degree query for one vertex? O(1)?
Sorry for all these questions but I read so many papers and I'm still confused.
Thank you in advance for any help!!!
Lisa
You have to ask your professor, but I would assume it is something simple.
E.g. let us say you have a triangle A,B,C then in the hash you just represent it as
A {B,C}
B {A,C}
C {A,B}
So the entry to the link A,B could be both from A and B.
We are given a undirected graph without loops.We have to check if it is possible to delete edges such that the degree of each vertex is one.
What should I try to this question? Should I use adjacency matrix or list.
Please suggest me the efficient way.
If the graph needs to be fully connected, it's possible if and only if
there are exactly two vertices and they have an edge between them.
If it does not need to be fully connected you must search for a
similar constellation. That is, you need to see if it is possible to
partition the graph into pairs of vertices with one edge in each
pair. That is what we are after. What do we know?
The graph has no loops. This means it must be a tree! (Not necessarily
binary, though). Maybe we can solve this eagerly by starting at
the bottom of the tree? We don't know what the bottom is; how do we
solve that? We can decide this ourselves, so I decide to pick all the
leaves as "bottom".
I now propose the following algorithm (whose efficiency you may
evaluate yourself, it isn't necessarily the best algorithm):
For each leaf L:
Let P be the parent of L, and Q the parent of P (Q may be NULL).
Is the degree of P <= 2? That is, does it have only one edge
connecting it to L, and possibly one connecting it to Q?
If no: Pick another leaf L and go to 1.1.
If yes: L and P can form a pair by removing the edge between P and Q. So
remove L and P from your memory (in some way; let's come back to
data structures later).
Do we have any vertices left in the graph?
No: The answer is "Yes, we can partition the graph by removing edges".
Only one: The answer is "No, we cannot partition the graph".
More:
Did we remove any nodes?
If yes: Go back to 1 and check all the current leaves.
If no: The answer is "No, we cannot partition the graph".
So what data structure do you use for this? I think it's easiest to
use a priority queue, perhaps based on a min-heap, with degree as
priority. You can use a linked list as a temporary storage for leaves
you have already visited but not removed yet. Don't forget to decrease
the priority of Q in step 1.2.2.
But you can do even better! Add all leaves (vertices with degree 1) to
a linked list. Loop through that list in step 1. In step 1.2.2, check
the degree of Q after removing L and P. If the degree is 1, add it to
the list of leaves!
I believe you can implement the entire algorithm recursively as well, but I let you
think about that.
I'm reading a book about algorithms ("Data Structures and Algorithms in C++") and have come across the following exercise:
Ex. 20. Modify cycleDetectionDFS() so that it could determine whether a particular edge is part of a cycle in an undirected graph.
In the chapter about graphs, the book reads:
Let us recall from a preceding section that depth-first search
guaranteed generating a spanning tree in which no elements of edges
used by depthFirstSearch() led to a cycle with other element of edges.
This was due to the fact that if vertices v and u belonged to edges,
then the edge(vu) was disregarded by depthFirstSearch(). A problem
arises when depthFirstSearch() is modified so that it can detect
whether a specific edge(vu) is part of a cycle (see Exercise 20).
Should such a modified depth-first search be applied to each edge
separately, then the total run would be O(E(E+V)), which could turn
into O(V^4) for dense graphs. Hence, a better method needs to be
found.
The task is to determine if two vertices are in the same set. Two
operations are needed to implement this task: finding the set to which
a vertex v belongs and uniting two sets into one if vertex v belongs
to one of them and w to another. This is known as the union-find
problem.
Later on, author describes how to merge two sets into one in case an edge passed to the function union(edge e) connects vertices in distinct sets.
However, still I don't know how to quickly check whether an edge is part of a cycle. Could someone give me a rough explanation of such algorithm which is related to the aforementioned union-find problem?
a rough explanation could be checking if a link is a backlink, whenever you have a backlink you have a loop, and whenever you have a loop you have a backlink (that is true for directed and undirected graphs).
A backlink is an edge that points from a descendant to a parent, you should know that when traversing a graph with a DFS algorithm you build a forest, and a parent is a node that is marked finished later in the traversal.
I gave you some pointers to where to look, let me know if that helps you clarify your problems.
Given a directed graph, I need to find all vertices v, such that, if u is reachable from v, then v is also reachable from u. I know that, the vertex can be find using BFS or DFS, but it seems to be inefficient. I was wondering whether there is a better solution for this problem. Any help would be appreciated.
Fundamentally, you're not going to do any better than some kind of search (as you alluded to). I wouldn't worry too much about efficiency: these algorithms are generally linear in the number of nodes + edges.
The problem is a bit underspecified, so I'll make some assumptions about your data structure:
You know vertex u (because you didn't ask to find it)
You can iterate both the inbound and outbound edges of a node (efficiently)
You can iterate all nodes in the graph
You can (efficiently) associate a couple bits of data along with each node
In this case, use a convenient search starting from vertex u (depth/breadth, doesn't matter) twice: once following the outbound edges (marking nodes as "reachable from u") and once following the inbound edges (marking nodes as "reaching u"). Finally, iterate through all nodes and compare the two bits according to your purpose.
Note: as worded, your result set includes all nodes that do not reach vertex u. If you intended the conjunction instead of the implication, then you can save a little time by incorporating the test in the second search, rather than scanning all nodes in the graph. This also relieves assumption 3.
a little out of my depth here and need to phone a friend. I've got a directed acyclical graph I need to traverse and I'm stumbling into to graph theory for the first time. I've been reading a lot about it lately but unfortunately I don't have time to figure this out academically. Can someone give me a kick with some help as to how to process this tree?
Here are the rules:
there are n root nodes (I call them "sources")
there are n end nodes
source nodes carry a numeric value
downstream nodes (I call them "worker" nodes) preform various operations on the incoming values like Add, Mult, etc.
As you can see from the graph below, nodes a, b, and c need to be processed before d, e, or f.
What's the proper order to walk this tree?
I would look into linearization of DAGs which should be achievable through Topological sorts.
Linearization, from what I remember, basically sorts in an order which holds to the invariant that for all nodes (Node_X) that have an outdegree to any other given node NodeA, NodeX appears before NodeA.
This would mean that, from your example, nodes a,b, and d would be processed first. Node c second. Nodes e and f, last.
http://en.wikipedia.org/wiki/Topological_sorting
You need to process the nodes via a Topological sort. The sort is not necessarily unique so there might be more than one available order (not that this should matter anyway).
The linked wikipedia page should have concrete algorithms to help you.