After reading about how to implement a graph it seems I have basically two options:
Matrix
Adjacency list
In order to decide which implementation to use this post can be useful.
When an adjacency list is used to implement a graph the cost to know if there is an edge between two nodes may take linear time (for those nodes connected to all nodes).
That make me wonder: Why not to use a HashSet instead of a linked list in order to keep the neighbors of a node?
This will give us constant time to know if there is an edge between two nodes.
I'm sure must be a disadvantage using a Set instead of Linked list but I can't see it.
I think "list" is just a generic name. I've used a Set and it works perfectly well.
There is no specific reason to use a list instead of a set, here go through this link - Graph using set
Hope this helps!
Related
Isn't it always better when searching for shortest path to use for connected nodes lists instead of grid?
When using grid, you have to iterate over the grid every time, whereas using lists saves lots of time.
With adjacency matrix usually each check costs you O(n) time. It may be a bit slower than a list of connected nodes. However, you can do some fancy stuff with it. For example, if you want to delete a lot of edges, you can do it in O(1) using adjacency matrix (it may take a lot longer using a list of nodes depending on what data structure you use for it). Adjacency matrix is also a matrix. What do I mean by that? If you want to check in how many ways you can get from node A to node B in k steps, you can raise this matrix to the power of k, which is impossible to do with a list.
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 that can check, in a directed graph, if a vertex, let's say V2, is reachable from a vertex V1, without traversing all the vertices?
You might find a route to that node without traversing all the edges, and if so you can give a yes answer as soon as you do. Nothing short of traversing all the edges can confirm that the node isn't reachable (unless there's some other constraint you haven't stated that could be used to eliminate the possibility earlier).
Edit: I should add that it depends on how often you need to do queries versus how large (and dense) your graph is. If you need to do a huge number of queries on a relatively small graph, it may make sense to pre-process the data in the graph to produce a matrix with a bit at the intersection of any V1 and V2 to indicate whether there's a connection from V1 to V2. This doesn't avoid traversing the graph, but it can avoid traversing the graph at the time of the query. I.e., it's basically a greedy algorithm that assumes you're going to eventually use enough of the combinations that it's easiest to just traverse them all and store the result. Depending on the size of the graph, the pre-processing step may be slow, but once it's done executing a query becomes quite fast (constant time, and usually a pretty small constant at that).
Depth first search or breadth first search. Stop when you find one. But there's no way to tell there's none without going through every one, no. You can improve the performance sometimes with some heuristics, like if you have additional information about the graph. For example, if the graph represents a coordinate space like a real map, and most of the time you know that there's going to be a mostly direct path, then you can attempt to have the depth-first search look along lines that "aim towards the target". However, imagine the case where the start and end points are right next to each other, but with no vector inbetween, and to find it, you have to go way out of the way. You have to check every case in order to be exhaustive.
I doubt it has a name, but a breadth-first search might go like this:
Add V1 to a queue of nodes to be visited
While there are nodes in the queue:
If the node is V2, return true
Mark the node as visited
For every node at the end of an outgoing edge which is not yet visited:
Add this node to the queue
End for
End while
Return false
Create an adjacency matrix when the graph is created. At the same time you do this, create matrices consisting of the powers of the adjacency matrix up to the number of nodes in the graph. To find if there is a path from node u to node v, check the matrices (starting from M^1 and going to M^n) and examine the value at (u, v) in each matrix. If, for any of the matrices checked, that value is greater than zero, you can stop the check because there is indeed a connection. (This gives you even more information as well: the power tells you the number of steps between nodes, and the value tells you how many paths there are between nodes for that step number.)
(Note that if you know the number of steps in the longest path in your graph, for whatever reason, you only need to create a number of matrices up to that power. As well, if you want to save memory, you could just store the base adjacency matrix and create the others as you go along, but for large matrices that may take a fair amount of time if you aren't using an efficient method of doing the multiplications, whether from a library or written on your own.)
It would probably be easiest to just do a depth- or breadth-first search, though, as others have suggested, not only because they're comparatively easy to implement but also because you can generate the path between nodes as you go along. (Technically you'd be generating multiple paths and discarding loops/dead-end ones along the way, but whatever.)
In principle, you can't determine that a path exists without traversing some part of the graph, because the failure case (a path does not exist) cannot be determined without traversing the entire graph.
You MAY be able to improve your performance by searching backwards (search from destination to starting point), or by alternating between forward and backward search steps.
Any good AI textbook will talk at length about search techniques. Elaine Rich's book was good in this area. Amazon is your FRIEND.
You mentioned here that the graph represents a road network. If the graph is planar, you could use Thorup's Algorithm which creates an O(nlogn) space data structure that takes O(nlogn) time to build and answers queries in O(1) time.
Another approach to this problem would allow you to ignore all of the vertices. If you were to only look at the edges, you can produce a transitive closure array that will show you each vertex that is reachable from any other vertex.
Start with your list of edges:
Va -> Vc
Va -> Vd
....
Create an array with start location as the rows and end location as the columns. Fill the arrays with 0. For each edge in the list of edges, place a one in the start,end coordinate of the edge.
Now you iterate a few times until either V1,V2 is 1 or there are no changes.
For each row:
NextRowN = RowN
For each column that is true for RowN
Use boolean OR to OR in the results of that row of that number with the current NextRowN.
Set RowN to NextRowN
If you run this algorithm until the end, you will quickly have a complete list of all reachable vertices without looking at any of them. The runtime is proportional to the number of edges. This would work well with a reasonable implementation and a reasonable number of edges.
A slightly more complex version of this algorithm would be to only calculate the vertices reachable by V1. To do this, you would focus your scope on the ones that are currently reachable at any given time. You can also limit adding rows to only one time, since the other rows are never changing.
In order to be sure, you either have to find a path, or traverse all vertices that are reachable from V1 once.
I would recommend an implementation of depth first or breadth first search that stops when it encounters a vertex that it has already seen. The vertex will be processed on the first occurrence only. You need to make sure that the search starts at V1 and stops when it runs out of vertices or encounters V2.
I have a list of items (blue nodes below) which are categorized by the users of my application. The categories themselves can be grouped and categorized themselves.
The resulting structure can be represented as a Directed Acyclic Graph (DAG) where the items are sinks at the bottom of the graph's topology and the top categories are sources. Note that while some of the categories might be well defined, a lot is going to be user defined and might be very messy.
Example:
(source: theuprightape.net)
On that structure, I want to perform the following operations:
find all items (sinks) below a particular node (all items in Europe)
find all paths (if any) that pass through all of a set of n nodes (all items sent via SMTP from example.com)
find all nodes that lie below all of a set of nodes (intersection: goyish brown foods)
The first seems quite straightforward: start at the node, follow all possible paths to the bottom and collect the items there. However, is there a faster approach? Remembering the nodes I already passed through probably helps avoiding unnecessary repetition, but are there more optimizations?
How do I go about the second one? It seems that the first step would be to determine the height of each node in the set, as to determine at which one(s) to start and then find all paths below that which include the rest of the set. But is this the best (or even a good) approach?
The graph traversal algorithms listed at Wikipedia all seem to be concerned with either finding a particular node or the shortest or otherwise most effective route between two nodes. I think both is not what I want, or did I just fail to see how this applies to my problem? Where else should I read?
It seems to me that its essentially the same operation for all 3 questions. You're always asking "Find all X below node(s) Y, where X is of type Z". All you need is a generic mechanism for 'locate all nodes below node', (solves Q3) and then you can filter the results for 'nodetype=sink' (solves Q1). For Q2, you have the starting-point (your node set) and your ending point (any sink below the starting point) so your solution set is all paths from starting node specified to the sink. So I would suggest that what you basically have a is a tree, and basic tree-traversal algorithms would be the way to go.
Despite the fact that your graph is acyclic, the operations you cite remind me of similar aspects of control flow graph analysis. There is a rich set of algorithms based on dominance that may be applicable. For example, your third operation reminds me od computing dominance frontiers; I believe that algorithm would work directly if you temporarily introduce "entry" and "exit" nodes. The entry node connects the "given set of nodes" and the exit nodes connects the sinks.
Also see Robert Tarjan's basic algorithms.