Let be G a given undirected simple graph, with edge-weights w. Does there exist an algorithm with time complexity O((n+m)log^*(n+m)) that counts the number of node pairs (u,v) for which there exists a path from u to v with total weight under some given constant W? Looking for either an algorithm or a proof that no such algorithm exists.
I’ve tried union find + DFS yet it doesn’t seem that will only use up to n+m calls to Find/Union... I’ve also tried dis-proving the existence on an algorithm by solving APSP with time complexity lower than the lower bound, yet to no avail.
Successfully shown no such algorithm exists:
Assume by contradiction that such an algorithm exists.
Let G be a given undirected un-weighted graph, we’ll compute the diameter of the graph as follows:
Assign weight 1 to every edge on the graph. Thus the weighted diameter is equal to un-weighted diameter.
We find that the diameter of the graph is bounded by m
Perform a binary search using the algorithm to find the diameter of the graph (check for each value whether or not the count returns zero).
Overall, we find the diameter of G in O((n+m)log(n)log*(n+m)). The current best algorithms for finding a graph diameter are O(min(nm, ~n^2.4)). Thus, at the very least, if this algorithm succeeded then the time complexity of computing a diameter would decrease dramatically. Not entirely a proof, but good enough for the purpose I needed it for.
Longest Path
We have a graph G=(V,E), lengths l(e) in Z^(+) for each e in E, a positive integer K and two nodes s,t in V.
The question is if there is a simple path in G from s to t of length at least K ?
Show that the problem Longest Path belongs to NP.
Show that the problem Longest Path is NP-complete, reducing Hamiltonian Path to it.
Show that if the graph is directed and acyclic then the problem can be solved in time O(|V|+|E|).
Could you give me a hint how we could show that the problem belongs to NP?
Also, how can we reduce a problem to an other, in order to show that the latter is NP-complete?
EDIT:
So in order to show that the problem belongs to NP, do we have to draw a simple and count the sum of the lengths of the edges?
Do we say for example the following?
We see that the length of the path from the node s to the node t is equal to l((s,w))+l((w,t))=3+12=15, so there is no simple path in G from s to t of length at least K.
Or does it suffice the following?
"Given a a simple path , we can easily check if its length is at least K(by simply computing the sum of lengths of all edges in it). Thus, it is in NP."
EDIT 2: Also could you explain me further why we reduce the Hamiltonian path problem to this one in polynomial time by setting all edges' lengths equal to one and set K = |V| - 1 ?
EDIT 3: Suppose that we have a problem A and a problem B and it is known that B is NP-complete. If we want to show that A is also NP-complete, do we change the data of A in that way so that we have the same problem as the problem B and so we deduce that A is also NP-complete? Or have I understood it wrong?
EDIT 4: Also how can we show that if the graph is directed and acyclic then the problem can be solved in time O(|V|+|E|)?
EDIT 5: All edges'lengths of a Hamiltonian path are equal to 1, right? And if we have V vertices, the length of the longest path is at V-1, yes? But in our problems, the lengths of the edges aren't specific and K is also not a fixed number. So if we set all edges' lengths equal to one and set K = |V| - 1, don't we reduce our problem to the Hamiltonian path problem? Or have I understood it wrong?
To show that a problem is in NP, we need to show that it can be verified in polynomial time. Given a certificate(a simple path in this case), we can easily check that it length is at least K(by simply computing the sum of lengths of all edges in it). Thus, it is in NP.
Reduction from A to B means: given an instance of A, create an instance of B(to be more precise, we are interested in polynomial time reduction here) and solve it in order to solve the original problem. So how can we reduce the Hamiltonian path problem to this one in polynomial time? It is pretty straightforward: we can set all edges' lengths equal to one and set K = |V| - 1. Then we should try all pairs of vertices in the graph (s, t), s != t and if the solution for this problem returns true for at least one pair, return true. Otherwise, we should return false(checking that we have a path of length |V| - 1 in a graph where all edges have unit length is exactly the same thing as checking that a Hamiltonian path exists by its definition).
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 was wondering, that if there is a weighted graph G(V,E), and I need to find a single shortest path between any two vertices S and T in it then I could have used the Dijkstras algorithm. but I am not sure how this can be done when we need to find all the distinct shortest paths from S to T. Is it solvable on O(n) time? I had one more question like if we assume that the weights of the edges in the graph can assume values only in certain range lets say 1 <=w(e)<=2 will this effect the time complexity?
You can do it using Dijkstra. In Dijkstra's algorithm, you assign labels to each node based on the distance it has from your source and settle nodes according to this distance (smallest first). Once the target node is settled, you have the length of the shortest path. To retrieve the path (=the sequence of edges), you can keep track of the parent of each node you settle. To retrieve all possible paths, you have to account for multiple parents of each node.
A small example:
Suppose you have a graph which looks like this (all edges weight 1 for simplicity):
B
/ \
A C - D
\ /
E
When you do dijkstra to find the distance A->D, you would get 3. You would first settle node A with distance 0, then nodes B and E with distance 1, node C with distance 2 and finally node D with distance 3. To get the paths, you would remember the parents of the nodes when you settle them. In this case you would first set the parents of B and C =node A. You then need to set the parents of node C to B and E as they both supply the same length. Node D would get node C as a parent.
When you need the paths, you could just follow the parent pointers, starting at D and branching each time a node has multiple parents. This would give you a branch at node C.
The post mentioned in the comment above uses the same principles, although it does not allow you to actually extract the paths, it only gives you the count.
One final remark: Dijkstra is not a linear time algorithm as you need to do a lot of operations to maintain you queue and node sets.
(a) Using BFS from s to mark nodes visited as usual, in the meantime, changing following things:
(b) for each node v, it has a record L(v) representing the layer that v is on and a record f(v)
representing the number of incoming paths:
(c) everytime a node v1 is found in another node's v2 neighbors:
(d) if v1 is not in the queue, add it into queue, L(v1) = L(v2) + 1,f(v1)+ = f(v2)
(d) if v1 is already in the queue and L(v1) equals L(v2), do nothing.
(e) if v1 is already in the queue but L(v1) doe not equal to L(v2), f(v1)+ = f(v2)
(f) until in this modi ed BFS, t pop from the queue for the rst time, we have f(t)
(g) and this f(t) represents the number of di erent shortest paths from s to t
This might solve your problem.
I am looking for a Dijkstra's algorithm implementation, that also takes into consideration the number of nodes traversed.
What I mean is, a typical Dijkstra's algorithm, takes into consideration the weight of the edges connecting the nodes, while calculating the shortest route from node A to node B. I want to insert another parameter into this. I want the algorithm to give some weightage to the number of nodes traversed, as well.
So that the shortest route computed from A to B, under certain values, may not necessarily be the Shortest Route, but the route with the least number of nodes traversed.
Any thoughts on this?
Cheers,
RD
Edit :
My apologies. I should have explained better. So, lets say, the shortest route from
(A, B) is A -> C -> D -> E -> F -> B covering a total of 10 units
But I want the algorithm to come up with the route A -> M -> N -> B covering a total of 12 units.
So, what I want, is to be able to give some weightage to the number of nodes as well, not just the distance of the connected nodes.
Let me demonstrate that adding a constant value to all edges can change which route is "shortest" (least total weight of edges).
Here's the original graph (a triangle):
A-------B
\ 5 /
2 \ / 2
\ /
C
Shortest path from A to B is via C. Now add constant 2 to all edges. The shortest path becomes instead the single step from A directly to B (owing to "penalty" we've introduced for using additional edges).
Note that the number of edges used is (excluding the node you start from) the same as the number of nodes visited.
The way you can do that is adapt the weights of the edges to always be 1, so that you traverse 5 nodes, and you've gone a distance of "5". The algorithm would be the same at that point, optimizing for number of nodes traversed rather than distance traveled.
If you want some sort of hybrid, you need to determine how much importance to give to traversing a node and the distance. The weight used in calculations should look something like:
weight = node_importance * 1 + (1 - node_importance) * distance
Where node_importance would be a percentage which gauges how much distance is a factor and how much minimum node traversal is important. Though you may have to normalize the distances to be an average of 1.
I'm going to go out on a limb here, but have you tried the A* algorithm? I may have understood your question wrong, but it seems like A* would be a good starting point for what you want to do.
Check out: http://en.wikipedia.org/wiki/A*_search_algorithm
Some pseudo code there to help you out too :)
If i understood the question correctly then its best analogy would be that used to find the best network path.
In network communication a path may not only be selected because it is shortest but has many hop counts(node), thus may lead to distortion, interference and noise due to node connection.
So the best path calculation contains the minimizing the function of variables as in your case Distance and Hop Count(nodes).
You have to derive a functional equation that could relate the distance and node counts with quality.
so something as suppose
1 hop count change = 5 unit distance (which means the impact is same for 5unit distace or 1 node change)
so to minimize the loss you can use it in the linear equation.
minimize(distance + hopcount);
where hopcount can be expressed as distance.