I'm trying to represent a transitive relation (in a database) and having a hard time working out the best data structure.
Basically, the data structure is a series of pairs A → B such that if A → B and B → C, then implicitly A → C. It's important to me to be able to identify which entries are original input and which entries exist implicitly. Asking if A → C is equivalent to me having a digraph and asking if there exists a path from A to C in that digraph.
I could just represent the original entries, but if I do than then it takes a lot of time to determine if two items are related, since I need to search for all possible paths and this is rather slow.
Alternatively, I can store the original edges, as well as a listing of all paths. This makes adding a new edge easy, because when I add A → B I can just take the Cartesian product of paths ending in A and the paths ending in B and put them together. This has some significant space overhead of O(n2) in the worst case, but has the nice property that lookups, by far the most common operation, will be constant time. The issue is deleting, where I cannot think of anything really other than recalculating all paths that may or may not run through the edge deleted, and this can be really nasty.
Does anyone have any better ideas?
Technical notes: the digraph may be cyclic, but the relation is reflexive so I don't need to represent the reflexivity or store anything about it.
This is called the Reachability problem.
It would seem that you want an efficient online algorithm, which is an open problem, and an area of much research.
See my similar question on cs.SE: An incrementally-condensed transitive-reduction of a DAG, with efficient reachability queries, where I reference several related querstions across stackexchange:
Related:
What is the fastest deterministic algorithm for dynamic digraph reachability with no edge deletion?
What is the fastest deterministic algorithm for incremental DAG reachability?
Does an algorithm exist to efficiently maintain connectedness information for a DAG in presence of inserts/deletes?
Is there an online-algorithm to keep track of components in a changing undirected graph?
Dynamic shortest path data structure for DAG
Note that even though some algorithm might be for a DAG only, if it supports condensation (that is, collapsing strongly connected components into one node, since they are considered equal, ie. they relate back and forth), it is equivalent; after condensation, you can query the graph for the representative node in place of any of the condensed nodes (because they were both reachable from each-other, and thusly related to the rest of the graph in exactly the same way).
My conclusion is that as-of-yet there does not seem to be an efficient way to do this (on the order of O(log n) queries for a dynamic graph, with output-sensitive update times on the condensed graph). For less efficient ways, see the related links above.
The closest practical algorithm I found was here (source), which is an interesting read. I am not sure how easy/practical this data-structure or any data structure in any paper you will find, would be to adapt it to a database.
PS. Consider asking CS-related questions on cs.stackexchange.com in the future.
Related
Using a disjoint set forest, we can efficiently check incrementally whether a graph has become connected while adding edges to it. (Or rather, it allows us to check whether two vertices already belong to the same connected component.) This is useful, for example, in Kruskal's algorithm for computing the minimum spanning tree.
Is there a similarly efficient data structure and algorithm to check whether a graph has become disconnected while removing edges from it?
More formally, I have a connected graph G = (V, E) from which I'm iteratively removing edges one by one. Without loss of generality, let's say these are e_1, e_2, and so on. I want to stop right before G becomes disconnected, so I need a way to check whether the removal of e_i will make the graph disconnected. This can of course be done by removing e_i and then doing a breadth-first or depth-first search from an arbitrary vertex, but this is an O(|E|) operation, making the entire algorithm O(|E|²). I'm hoping there is a faster way by somehow leveraging results from the previous iteration.
I've looked at partition refinement but it doesn't quite solve this problem.
There’s a family of data structures that are designed to support queries of the form “add an edge between two nodes,” “delete an edge between two nodes,” and “is there a path from node x to node y?” This is called the dynamic connectivity problem and there are many data structures that you can use to solve it. Unfortunately, these data structures tend to be much more complicated than a disjoint-set forest, so you may want to search around to see if you can find a ready-made implementation somewhere.
The layered forest structure of Holm et al works by maintaining a hierarchy of spanning forests for the graph and doing some clever bookkeeping with edges to avoid rescanning them when seeing if two parts of the graph are still linked. It supports adding and deleting edges in amortized time O(log2 n) and queries of the form “are these two nodes connected?” in time O(log n / log log n).
A more recent randomized structure called the randomized cutset was developed by Kapron et al. It has worst-case O(log5 n) costs for all three operations (IIRC).
There might be a better way to solve your particular problem that doesn’t require these heavyweight approaches, but I figured I’d mention these in case they’re helpful.
Suppose I have a file with a one-liner (joke) on each line. I want to sort the jokes by how funny I find them. My first thought is to implement any sorting algorithm (preferably one that makes as few comparisons as possible) and having the comparison algorithm take my input; I'd just sit there and choose which of each pair of jokes it presented me was funnier.
There's a problem with that. My joke preference is not a total order. It lacks transitivity. For example, I might think that B is funnier than A when presented them, and that C is funnier than B, but when presented A and C somehow I find A to be funnier than C. If “>” means “is funnier than,” this means that C > B and B > A does not imply C > A. All sorting algorithms’ correctness depends on this.
But it still seems that there should be an algorithm that sorts the list of jokes so that the one at the top is most preferred over other jokes, and the one at the bottom is least preferred over other jokes, even if there are individual exceptions.
I don’t know how to Google this. Is there an algorithm for this kind of preference sorting? The answer here is not applicable because it forces the user’s preference to be transitive.
If you represent your decisions as a directed graph, where each joke is a node and each directed edge indicates one joke being better than the other, then you can retrieve an ordering by constructing the path which follows the edges and visits each node exactly once.
This type of graph is called a Tournament, and the path is a Hamiltonian path. I've got good news for you Bub, a Hamiltonian is proven to exist if the graph is strongly connected. Strongly connected just means that every node can be reached from every node, obeying the direction of the edges, so keep adding edges until this is true.
Tournament: https://en.wikipedia.org/wiki/Tournament_(graph_theory)
Hamiltonian Path: https://en.wikipedia.org/wiki/Hamiltonian_path
Here's my situation. I have a graph that has different sets of data being added at different times. For example, set1 might have a few thousand nodes and then set2 comes in later and we apply business logic to create edges from set1 to set2(and disgard any Vertices from set1 that do not have edges to set2). Then at a later point, we get set3, set4, and so on and the same process applies between each set and its previous set.
Question, what's the best way to organize this? What I did before was name the nodes set1-xx, set2-xx,etc.. The problem I faced was when I was trying to run analytics between the current set and the previous set I would have to run a loop through the entire graph and look for all the nodes that started with 'setx'. It took a long time as the graph grew, so I thought of another solution which was to create a node called 'set1' and have it connected to all nodes for that particular set. I am testing it but I was wondering if there way a more efficient way or a build in way of handling data structures like this? Is there a way to somehow segment data like this?
I think a general solution would be application but if it helps I'm using neo4j(so any specific solution to that database would be good as well).
You have a very special type of a directed graph, called a layered graph.
The choice of the data structure depends primarily on the expected graph density (how many nodes from a previous set/layer are typically connected to a node in the current set/layer) and on the operations that you need to perform on it most of the time. It is definitely a good idea to have each layer directly represented by a numeric index (that is, the outermost structure will be an array of sets/layers), and presumably you can also use one array of vertices per layer. However, the list of edges per vertex (out only, or in and out sets of edges depending on whether you ever traverse the layers backward) may be any of the following:
Linked list of vertex identifiers; this is good if the graph is very sparse and edges are often added/removed.
Sorted array of vertex identifiers; this is good if the graph is quite sparse and immutable.
Array of booleans, indexed by vertex identifiers, determining whether a given vertex is or is not linked by an edge from the current vertex; this is good if the graph is dense.
The "vertex identifier" can take many forms. For example, it can be an index into the array of vertices on the next layer.
Your second solution is what I would do- create a setX node and connect all nodes belonging to that set to setX. That way your data is partitioned and it is easier to query.
Here is the problem:
assuming two persons are registered in a social networking website, how to decide whether they are connected or not?
my analysis (after reading more): actually, the question is looking for - the shortest path from A to B in a graph. I think both BFS and Dijkstra's Algorithms works here and time complexity is exactly the same (O(V+E)) because it is an unweighted graph, so we can't take advantage of the priority queue. So, a simple queue could resolve the problem. But, both of them doesnt resolve the problem that: find the path between them.
Bidrectrol should be a better solution at this point.
To find a path between the two, you should begin with a breadth first search. First find all neighbors of A, then find all neighbors of all neighbors of A, etc. Once B is hit, not only do you have a path from A to B, but you also have a shortest such path.
Dijkstra's algorithm rocks, and you may be able to speed this up by working from both end, i.e. find neighbors of A and neighbors of B, and compare.
If you do a depth first search, then you're following one path at a time. This will be much much slower.
If you do dfs for finding whether two people are connected on a social network, then it will take too long!
You already know the two persons, so you should use Bidirectional Search.. But, simple bidirectional search won't be enough for a graph as big as a social networking site. You will have to use some heuristics. Wikipedia page has some links to it.
You may also be able to use A* search. From wikipedia : "A* uses a best-first search and finds the least-cost path from a given initial node to one goal node (out of one or more possible goals)."
Edit: I suggest A* because "The additional complexity of performing a bidirectional search means that the A* search algorithm is often a better choice if we have a reasonable heuristic." So, if you can't form a reasonable heuristic, then use Bidirectional search. (Forming a good heuristic is never easy ;).)
One way is to use Union Find, add all links union(from,to), and if find(A) is find(B) is True then A and B are connected. This avoids the recursive search but it actually computes the connectivity of all pairs and doesn't give you the paths that connects A and B.
I think that the true criteria is: there are at least N paths between A and B shorter then K, or A and B are connected diectly. I would go with K = 3 and N near 5, i.e. have 5 common friends.
Note: answer edited.
Any method might end up being very slow. If you need to do this repeatedly, it's best to find the connected components of the graph, after which the task becomes a trivial O(1) operation: if two people are in the same component, they are connected.
Note that finding connected components for the first time might be slow, but keeping them updated as new edges/nodes are added to the graph is fast.
There are several methods for finding connected components.
One method is to construct the Laplacian of the graph, and look at its eigenvalues / eigenvectors. The number of zero eigenvalues gives you the number of connected components. The non-zero elements of the corresponding eigenvectors gives the nodes belonging to the respective components.
Another way is along the following lines:
Create a transformation table of nodes. Element n of the array contains the index of the node that node n transforms to.
Loop through all edges (i,j) in the graph (denoting a connection between i and j):
Compute recursively which node do i and j transform to based on the current table. Let us denote the results by k and l. Update entry k to make it transform to l. Update entries i and j to point to l as well.
Loop through the table again, and update each entry to point directly to the node it recursively transforms to.
Now nodes in the same connected component will have the same entry in the transformation table. So to check if two nodes are connected, just check if they transform to the same value.
Every time a new node or edge is added to the graph, the transformation table needs to be updated, but this update will be much faster than the original calculation of the table.
Is there an algorithm or heuristics for graph isomorphism?
Corollary: A graph can be represented in different different drawings.
What s the best approach to find different drawing of a graph?
It is a hell of a problem.
In general, the basic idea is to simplify the graph into a canonical form, and then perform comparison of canonical forms. Spanning trees are generated with this objective, but spanning trees are not unique, so you need to have a canonical way to create them.
After you have canonical forms, you can perform isomorphism comparison (relatively) easy, but that's just the start, since non-isomorphic graphs can have the same spanning tree. (e.g. think about a spanning tree T and a single addition of an edge to it to create T'. These two graphs are not isomorph, but they have the same spanning tree).
Other techniques involve comparing descriptors (e.g. number of nodes, number of edges), which can produce false positive in general.
I suggest you to start with the wiki page about the graph isomorphism problem. I also have a book to suggest: "Graph Theory and its applications". It's a tome, but worth every page.
As from you corollary, every possible spatial distribution of a given graph's vertexes is an isomorph. So two isomorph graphs have the same topology and they are, in the end, the same graph, from the topological point of view. Another matter is, for example, to find those isomorph structures enjoying particular properties (e.g. with non crossing edges, if exists), and that depends on the properties you want.
One of the best algorithms out there for finding graph isomorphisms is VF2.
I've written a high-level overview of VF2 as applied to chemistry - where it is used extensively. The post touches on the differences between VF2 and Ullmann. There is also a from-scratch implementation of VF2 written in Java that might be helpful.
A very similar problem - graph automorphism - can be solved by saucy, which is available in source code. This finds all symmetries of a graph. If you have two graphs, join them into one and any isomorphism can be discovered as an automorphism of the join.
Disclaimer: I am one of co-authors of saucy.
There are algorithms to do this -- however, I have not had cause to seriously investigate them as of yet. I believe Donald Knuth is either writing or has written on this subject in his Art of Computing series during his second pass at (re)writing them.
As for a simple way to do something that might work in practice on small graphs, I would recommend counting degrees, then for each vertex, also note the set of degrees for those vertexes that are adjacent. This will then give you a set of potential vertex isomorphisms for each point. Then just try all those (via brute force, but choosing the vertexes in increasing order of potential vertex isomorphism sets) from this restricted set. Intuitively, most graph isomorphism can be practically computed this way, though clearly there would be degenerate cases that might take a long time.
I recently came across the following paper : http://arxiv.org/abs/0711.2010
This paper proposes "A Polynomial Time Algorithm for Graph Isomorphism"
My project - Griso - at sf.net: http://sourceforge.net/projects/griso/ with this description:
Griso is a graph isomorphism testing utility written in C++. It is based on my own POLYNOMIAL-TIME (in this point the salt of the project) algorithm. See Griso's sample input/output on http://funkybee.narod.ru/graphs.htm page.
nauty and Traces
nauty and Traces are programs for computing automorphism groups of graphs and digraphs [*]. They can also produce a canonical label. They are written in a portable subset of C, and run on a considerable number of different systems.
AutGroupGraph command in GRAPE's package of GAP.
bliss: another symmetry and canonical labeling program.
conauto: a graph ismorphism package.
As for heuristics: i've been fantasising about a modified Ullmann's algorithm, where you don't only use breadth first search but mix it with depth first search the way, that first you use breadth first search intensively, than you set a limit for breadth analysis and go deeper after checking a few neighbours, and you lower the breadh every step at some amount. This is practically how i find my way on a map: first locate myself with breadth first search, then search the route with depth first search - largely, and this is the best evolution of my brain has ever invented. :) On the long term some intelligence may be added for increasing breadth first search neighbour count at critical vertexes - for example where there are a large number of neighbouring vertexes with the same edge count. Like checking your actual route sometimes with the car (without a gps).
I've found out that the algorithm belongs in the category of k-dimension Weisfeiler-Lehman algorithms, and it fails with regular graphs. For more here:
http://dabacon.org/pontiff/?p=4148
Original post follows:
I've worked on the problem to find isomorphic graphs in a database of graphs (containing chemical compositions).
In brief, the algorithm creates a hash of a graph using the power iteration method. There might be false positive hash collisions but the probability of that is exceedingly small (i didn't had any such collisions with tens of thousands of graphs).
The way the algorithm works is this:
Do N (where N is the radius of the graph) iterations. On each iteration and for each node:
Sort the hashes (from the previous step) of the node's neighbors
Hash the concatenated sorted hashes
Replace node's hash with newly computed hash
On the first step, a node's hash is affected by the direct neighbors of it. On the second step, a node's hash is affected by the neighborhood 2-hops away from it. On the Nth step a node's hash will be affected by the neighborhood N-hops around it. So you only need to continue running the Powerhash for N = graph_radius steps. In the end, the graph center node's hash will have been affected by the whole graph.
To produce the final hash, sort the final step's node hashes and concatenate them together. After that, you can compare the final hashes to find if two graphs are isomorphic. If you have labels, then add them (on the first step) in the internal hashes that you calculate for each node.
There is more background here:
https://plus.google.com/114866592715069940152/posts/fmBFhjhQcZF
You can find the source code of it here:
https://github.com/madgik/madis/blob/master/src/functions/aggregate/graph.py