I am trying to understand the time complexity of some efficient methods of detecting cycles in a graph.
Two approaches for doing this are explained here. I would assume that time complexity is provided in terms of worst-case.
The first is union-find, which is said to have a time complexity of O(Vlog E).
The second uses a DFS based approach and is said to have a time complexity of O(V+E). If I am correct this is a more efficient complexity asymptotically than O(Vlog E). It is also convenient that the DFS-based approach can be used for directed and undirected graphs.
My issue is that I fail to see how the second approach can be considered to run in O(V+E) time because DFS runs in O(V+E) time and the algorithm checks the nodes adjacent to any discovered nodes for the starting node. Surely this would mean that the algorithm runs in O(V2) time because up to V-1 adjacent nodes might have to be traversed over for each discovered node? It is obviously impossible for more than one node to require the traversal of n-1 adjacent nodes but from my understanding this would still be the upper bound of the runtime.
Hopefully someone understands why I think this and can help me to understand why the complexity is O(V+E).
The algorithm, based on DFS, typically maintains a "visited" boolean variable for each vertex, which contains one bit of information - this vertex was already visited or not. So, none of vertices can be visited more than once.
If the graph is connected, then starting the DFS from any vertex will give you an answer right away. If the graph is a tree, then all the vertices will be visited in a single call to the DFS. If the graph is not a tree, then a single call to the DFS will find a cycle - and in this case not all the vertices might be visited. In both cases the subgraph, induced by all the already visited vertices, will be a tree at each step of the DFS lookup - so the total number of traversed edges will be O(V). Because of that we can reduce the time complexity estimate O(V+E) of the cycle detection algorithm to O(V).
Starting the DFS from all vertices of the graph is necessary in the case when the graph consists of a number of connected components - the "visited" boolean variable guarantees that the DFS won't traverse the same component again and again.
Given a directed graph, starting point, ending point and a time limit.
At each vertices, there's a "value" that represents how much attractive this location is. There's also a cost of traveling from vertices to another.
What i need is to find a path that is within the time limit and has the maximum "value".
To answer the question in part, note that the considered problem is NP-hard as it contains the Knapsack problem as a subproblem using the following reduction.
Given an instance of the Knapsack problem given by n items defined by profits p_1,...,p_n and weights w_1,...,w_n and a target capacity C, define a graph as follows.
To the left there is a source node s and a terminal node t to the right. For the i-th item define two nodes yes_i and no_i, which correspond to selecting and not selecting the respective item. The attractiveness of the yes-nodes is p_i and the attractiveness of the no-nodes is zero.
The pairs of nodes can be imagined as arranged in columns between the source and the terminal. Each node has in two in-edges from the previous column (except for the first colum, which is connected to the source). The weight of each of these edges is w_i if and only if they are the in-edges of a yes-node and zero if they are in-edges of a no-node. The last column is connected to the terminal.
Each path from s to t must column-wise decide whether or not to select the item of the respective colum; likewise, any selection of items corresponds to a path from s to t. By definition of the edge weights, the total weight of the path is equal to the total weight of the selection of items, while the total weight of the selected yes-nodes is equal to the the total weight of the selected items.
In total, we obtain a bijection of the feasible solutions of the Knapsack instance and the constructed instance of the described path problem. In total, this means that the problem in the question is NP-hard as it contains the Knapsack problem (which is known to be NP-hard) as a subproblem.
Alternatively the NP-hardness can be seen by setting the attractiveness of all locations in a complete graph to one; this special case of the problem is the decision version of the Hamiltonian Path problem, which is known to be NP-complete.
I´ve been trying to find an algorithm to search if a graph is connected. The graph is undirected and I only want to find a solution (there can be multiple) or if there is none. I was looking for a alg. that performs near linear time, maybe O(logN) or O(NlogN).
Can DFS be up to the task or is there another alternative for this specific problem?
It's going to depend on how you define N, if N is number of vertices, the input itself can be of size O(N^2), and you will be needing to read all of it (unless you have some specific ordering of the input, and than that might change).
DFS runs in O(|V|+|E|) (number of nodes + number of edges), and can find if the graph is connected by simply counting the number of new vertices you discover, and when done, checking if this number is |V|.
I have a collection of 15M (Million) DAGs (directed acyclic graphs - directed hypercubes actually) that I would like to remove isomorphisms from. What is the common algorithm for this? Each graph is fairly small, a hybercube of dimension N where N is 3 to 6 (for now) resulting in graphs of 64 nodes each for N=6 case.
Using networkx and python, I implemented it like this which works for small sets like 300k (Thousand) just fine (runs in a few days time).
def isIsomorphicDuplicate(hcL, hc):
"""checks if hc is an isomorphism of any of the hc's in hcL
Returns True if hcL contains an isomorphism of hc
Returns False if it is not found"""
#for each cube in hcL, check if hc could be isomorphic
#if it could be isomorphic, then check if it is
#if it is isomorphic, then return True
#if all comparisons have been made already, then it is not an isomorphism and return False
for saved_hc in hcL:
if nx.faster_could_be_isomorphic(saved_hc, hc):
if nx.fast_could_be_isomorphic(saved_hc, hc):
if nx.is_isomorphic(saved_hc, hc):
return True
return False
One better way to do it would be to convert each graph to its canonical ordering, sort the collection, then remove the duplicates. This bypasses checking each of the 15M graphs in a binary is_isomophic() test, I believe the above implementation is something like O(N!N) (not taking isomorphic time into account) whereas a clean convert all to canonical ordering and sort should take O(N) for the conversion + O(log(N)N) for the search + O(N) for the removal of duplicates. O(N!N) >> O(log(N)N)
I found this paper on Canonical graph labeling, but it is very tersely described with mathematical equations, no pseudocode: "McKay's Canonical Graph Labeling Algorithm" - http://www.math.unl.edu/~aradcliffe1/Papers/Canonical.pdf
tldr: I have an impossibly large number of graphs to check via binary isomorphism checking. I believe the common way this is done is via canonical ordering. Do any packaged algorithms or published straightforward to implement algorithms (i.e. have pseudocode) exist?
Here is a breakdown of McKay ’ s Canonical Graph Labeling Algorithm, as presented in the paper by Hartke and Radcliffe [link to paper].
I should start by pointing out that an open source implementation is available here: nauty and Traces source code.
Ok, let's do this! Unfortunately this algorithm is heavy in graph theory, so we need some terms. First I will start by defining isomorphic and automorphic.
Isomorphism:
Two graphs are isomorphic if they are the same, except that the vertices are labelled differently. The following two graphs are isomorphic.
Automorphic:
Two graphs are automorphic if they are completely the same, including the vertex labeling. The following two graphs are automorphic. This seems trivial, but turns out to be important for technical reasons.
Graph Hashing:
The core idea of this whole thing is to have a way to hash a graph into a string, then for a given graph you compute the hash strings for all graphs which are isomorphic to it. The isomorphic hash string which is alphabetically (technically lexicographically) largest is called the "Canonical Hash", and the graph which produced it is called the "Canonical Isomorph", or "Canonical Labelling".
With this, to check if any two graphs are isomorphic you just need to check if their canonical isomporphs (or canonical labellings) are equal (ie are automorphs of each other). Wow jargon! Unfortuntately this is even more confusing without the jargon :-(
The hash function we are going to use is called i(G) for a graph G: build a binary string by looking at every pair of vertices in G (in order of vertex label) and put a "1" if there is an edge between those two vertices, a "0" if not. This way the j-th bit in i(G) represents the presense of absence of that edge in the graph.
McKay ’ s Canonical Graph Labeling Algorithm
The problem is that for a graph on n vertices, there are O( n! ) possible isomorphic hash strings based on how you label the vertices, and many many more if we have to compute the same string multiple times (ie automorphs). In general we have to compute every isomorph hash string in order to find the biggest one, there's no magic sort-cut. McKay's algorithm is a search algorithm to find this canonical isomoprh faster by pruning all the automorphs out of the search tree, forcing the vertices in the canonical isomoprh to be labelled in increasing degree order, and a few other tricks that reduce the number of isomorphs we have to hash.
(1) Sect 4: the first step of McKay's is to sort vertices according to degree, which prunes out the majority of isomoprhs to search, but is not guaranteed to be a unique ordering since there may be more than one vertex of a given degree. For example, the following graph has 6 vertices; verts {1,2,3} have degree 1, verts {4,5} have degree 2 and vert {6} has degree 3. It's partial ordering according to vertex degree is {1,2,3|4,5|6}.
(2) Sect 5: Impose artificial symmetry on the vertices which were not distinguished by vertex degree; basically we take one of the groups of vertices with the same degree, and in turn pick one at a time to come first in the total ordering (fig. 2 in the paper), so in our example above, the node {1,2,3|4,5|6} would have children { {1|2,3|4,5|6}, {2|1,3|4,5|6}}, {3|1,2|4,5|6}} } by expanding the group {1,2,3} and also children { {1,2,3|4|5|6}, {1,2,3|5|4|6} } by expanding the group {4,5}. This splitting can be done all the way down to the leaf nodes which are total orderings like {1|2|3|4|5|6} which describe a full isomorph of G. This allows us to to take the partial ordering by vertex degree from (1), {1,2,3|4,5|6}, and build a tree listing all candidates for the canonical isomorph -- which is already a WAY fewer than n! combinations since, for example, vertex 6 will never come first. Note that McKay evaluates the children in a depth-first way, starting with the smallest group first, this leads to a deeper but narrower tree which is better for online pruning in the next step. Also note that each total ordering leaf node may appear in more than one subtree, there's where the pruning comes in!
(3) Sect. 6: While searching the tree, look for automorphisms and use that to prune the tree. The math here is a bit above me, but I think the idea is that if you discover that two nodes in the tree are automorphisms of each other then you can safely prune one of their subtrees because you know that they will both yield the same leaf nodes.
I have only given a high-level description of McKay's, the paper goes into a lot more depth in the math, and building an implementation will require an understanding of this math. Hopefully I've given you enough context to either go back and re-read the paper, or read the source code of the implementation.
This is indeed an interesting problem.
I would approach it from the adjacency matrix angle. Two isomorphic graphs will have adjacency matrices where the rows / columns are in a different order. So my idea is to compute for each graph several matrix properties which are invariant to row/column swaps, off the top of my head:
numVerts, min, max, sum/mean, trace (probably not useful if there are no reflexive edges), norm, rank, min/max/mean column/row sums, min/max/mean column/row norm
and any pair of isomorphic graphs will be the same on all properties.
You could make a hash function which takes in a graph and spits out a hash string like
string hashstr = str(numVerts)+str(min)+str(max)+str(sum)+...
then sort all graphs by hash string and you only need to do full isomorphism checks for graphs which hash the same.
Given that you have 15 million graphs on 36 nodes, I'm assuming that you're dealing with weighted graphs, for unweighted undirected graphs this technique will be way less effective.
This is an interesting question which I do not have an answer for! Here is my two cents:
By 15M do you mean 15 MILLION undirected graphs? How big is each one? Any properties known about them (trees, planar, k-trees)?
Have you tried minimizing the number of checks by detecting false positives in advance? Something includes computing and comparing numbers such as vertices, edges degrees and degree sequences? In addition to other heuristics to test whether a given two graphs are NOT isomorphic. Also, check nauty. It may be your way to check them (and generate canonical ordering).
If all your graphs are hypercubes (like you said), then this is trivial: All hypercubes with the same dimension are isomorphic, hypercubes with different dimension aren't. So run through your collection in linear time and throw each graph in a bucket according to its number of nodes (for hypercubes: different dimension <=> different number of nodes) and be done with it.
since you mentioned that testing smaller groups of ~300k graphs can be checked for isomorphy I would try to split the 15M graphs into groups of ~300k nodes and run the test for isomorphy on each group
say: each graph Gi := VixEi (Vertices x Edges)
(1) create buckets of graphs such that the n-th bucket contains only graphs with |V|=n
(2) for each bucket created in (1) create subbuckets such that the (n,m)-th subbucket contains only graphs such that |V|=n and |E|=m
(3) if the groups are still too large, sort the nodes within each graph by their degrees (meaning the nr of edges connected to the node), create a vector from it and distribute the graphs by this vector
example for (3):
assume 4 nodes V = {v1, v2, v3, v4}. Let d(v) be v's degree with d(v1)=3, d(v2)=1, d(v3)=5, d(v4)=4, then find < := transitive hull ( { (v2,v1), (v1,v4), (v4,v3) } ) and create a vector depening on the degrees and the order which leaves you with
(1,3,4,5) = (d(v2), d(v1), d(v4), d(v3)) = d( {v2, v1, v4, v3} ) = d(<)
now you have divided the 15M graphs into buckets where each bucket has the following characteristics:
n nodes
m edges
each graph in the group has the same 'out-degree-vector'
I assume this to be fine grained enough if you are expecting not to find too many isomorphisms
cost so far: O(n) + O(n) + O(n*log(n))
(4) now, you can assume that members inside each bucket are likely to be isomophic. you can run your isomorphic-check on the bucket and only need to compare the currently tested graph against all representants you have already found within this bucket. by assumption there shouldn't be too many, so I assume this to be quite cheap.
at step 4 you also can happily distribute the computation to several compute nodes, which should really speed up the process
Maybe you can just use McKay's implementation? It is found here now: http://pallini.di.uniroma1.it/
You can convert your 15M graphs to the compact graph6 format (or sparse6) which nauty uses and then run the nauty tool labelg to generate the canonical labels (also in graph6 format).
For example - removing isomorphic graphs from a set of random graphs:
#gnp.py
import networkx as nx
for i in range(100000):
graph = nx.gnp_random_graph(10,0.1)
print nx.generate_graph6(graph,header=False)
[nauty25r9]$ python gnp.py > gnp.g6
[nauty25r9]$ cat gnp.g6 |./labelg |sort |uniq -c |wc -l
>A labelg
>Z 10000 graphs labelled from stdin to stdout in 0.05 sec.
710
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.