I can't figure out how to proceed with the following problem.
Say I have an unoriented graph with an end node and a start node, I need to find the shortest path between these two nodes, but the path must include all mustpass node types.
There can be up to 10 of these types. This means that I should visit at least one node of each type (marked with a letter in the image) and then go to the end. Once I visit one of the nodes of type B, I may, but need not, visit other nodes of type B. The nodes that are marked with a number simply form a path and do not need to be visited.
This question is very similar to this. There it was suggested to find the shortest path between all the crucial nodes and then use the DFS algorithm. Can I apply the same algorithm to this problem?
Let N be the number of vertices and M be the number of edges.
Break down the solution into two phases.
Phase 1:
Compute the distance between each pair of edges. If the edges are not weighted, this can be easily done by starting a BFS from each node, spending a total of O(N(N+M)) time. If the edges are weighted, you can you the Dijkstra's algorithm on each node, spending a total of O(N(NlogN+M)) time.
After this phase, we have computed dist(x,y) for any pair of nodes x,y.
Phase 2:
Now that we can query the distance between any pair of nodes in O(1) using the precomputed values in phase 1, it is time to put everything together. I can propose two possibilities here
Possibility 1:
Us a similar approach as in the thread you linked. There are 1! factorial orders in which you could visit each node. Let's say we have fixed one possible order [s,t1,t2,...,t10,e] (where s and e are the start / end nodes, and ti represents a type) and we are trying to find out what would be the most optimal way to visit the nodes from start to finish in that order. Since each type can have multiple nodes belonging to it, it is not as simple as querying distances for every consecutive types t_i and t_{i+1}.
What we should do instead if, for every node x, compute the fastest way to reach the end node from node x while respecting the order of types [s,t1,...,t10,e]. So if x is of type t_i, then we are looking for a way to reach e from x while visiting nodes of types t_{i+1}, t_{i+2}, ... t_{10} in that order. Call this value dp[x] (where dp stands for dynamic-programming).
We are looking to find dp[s]. How do we compute dp[x] for a given node x? Simple - iterate through all nodes y of type t_{i+1} and consider d(x,y) + dp[y]. Then we have dp[x] = min{dist(x,y) + dp[y] for all y of type t_{i+1}}. Note that we need to compute dp[x] starting from the nodes of type t_10 all the way back to the nodes of type t_1.
The complexity here is O(10! * N^2)
Possibility 2:
There is actually a much faster way to find the answer and reduce the complexity to O(2^10 * N^3) (which can give massive gains for large N, and especially for larger number of node types (like 20 instead of 10)).
To accomplish this we do the following. For each subset S of the set of types {1,2,...10}, and for each pair of nodes x, y with types in S define
dp[S][x][y], which represents the fastest way to traverse the graph starting from node x, ending in node y and visiting all at least one node for every type in S. Note that we don't care about the actual order. To compute dp[S][x][y] for a given (S,x,y), all we need to do is go over all the possibilities for the second type to visit (node z with type t3). then we update dp[S][x][y] according dist(x,z) + dp[S-t1][z][y] (where t1 is the type of the node x). The number of all the possible subsets along with start and end nodes is 2^10 * N^2. To compute each dp, we consider N possibilities for the second node to visit. So overall we get O(2^10 * N^3)
Note: in all of my analysis above, you can replace the value 10, with a more general K, representing the number of different types possible.
I have a directed, weighted routing graph (ca. 10^5 edges, 4 edges per node, lots of circles).
Each edge has a cost associated with it. How can I rate the "connectedness" of each node? It should be a measure of how cheap it is to reach other nodes from this one.
How does everything change if every node gets a reliability factor (the probability that the chosen path containing that node will fail and a new one must be found)?
Thanks for your help
I believe that the problem you've put forth in many terms matches the use case of the PageRank algorithm.
I won't discuss how the algorithm works in general since there are a lot many blogs/videos available online which already explains it in great detail. One of my personal favourite short video on the same is this.
Now lets see how does the algorithm fits for your use case. Let's define connectedness of a node x as C(x). We can rephrase your given statement "how cheap it is to reach other nodes from this node" to "how likely are we to end up on the given node in a random walk across the graph such that we are biased to take edges whose costs are less".
The statement to a large extent relates to the ideology behind the PageRank algorithm. We just need to consider how to include the edge cost for our working.
The original PageRank algorithm uniformly divides the page rank of a given node to all it's adjacent node (denoted as PR(y) / OUT(y) in formula). We on the other hand needs to be more biased towards edges with lower cost for which I'll recommend modifying the formula to,
(SUM-EDGES-COST(y) - EDGE-COST(x, y)) * (C(y) / SUM-EDGES-COST(y))
instead of the traditional C(x) / OUT(x). We take the difference (SUM-EDGES-COST(y) - EDGE-COST(x, y)) since in our scenario lower edge cost means more connectedness. Another possibility is to apply softmax function to the edge cost for each node as a normalisation strategy.
As to answer the part about having a reliability factor, given by R(x) for a node x, we can just multiply it directly to C(x) in the formula.
To wrap things up,
should match your given scenario.
What I've presented here is just one possibility which I can think from the top of my mind and it's highly likely that it just might not work out. All I can hope is that it helps you out in some or the other way. Cheers! :)
I have the following complete weighted graph, where each weight represents the probability of a vertice belonging to the same category as the next. I know a priori the category for which some of the vertices belong to; how would I be able to classify every other vertice?
In a more detailed manner I can describe the problem as following; From all the vertices N and clusters C, we have a set where we know for sure the specific cluster which a node belongs: P(v_n|C_n)=1. From the graph given we also know for each node, the probability of every other belonging to the same cluster as it: P(v_n1∩C_n2). From this, how can we estimate the cluster for every other node?
Let w_i be a vector where w_i[j] is the probability of node j, being in the cluser, at iteration i.
We define w_i:
w_0[j] = 1 j is given node in the class
0 otherwise
w_{i}[j] = P(j | w_{i-1})
Where: P(j | w_{i-1}) is the probability j being in the cluster, assuming we know the probabilities for each other node k to be in it, as w_{i-1}[k].
We can calculate the above probability:
P(j | w_{i-1}) = 1- (1- w_{i-1}[0]*c(0,j))*(1- w_{i-1}[1]*c(1,j))*...*(1- w_{i-1}[n-1]*c(n-1,j))
in here:
w_{i-1} is the output of last iteration.
c(x,y) is the weight of edge (x,y)
c(x,x) = 1
Repeat until convergence, and in the converged vector (let it be w), the probability of j being in the cluster is w[j]
Explanation for the probability function:
In order for a node NOT to be in the set, it needs all the other nodes will "decide" not to share it.
So, the probability for that happening is:
(1- w_{i-1}[0]*c(0,j))*(1- w_{i-1}[1]*c(1,j))*...*(1- w_{i-1}[n-1]*c(n-1,j))
^ ^ ^
node 0 doesn't share node 1 doesn't share node n-1 doesn't share
In order to be in the class, at least one node need to "share", so the probability for that happening is the complemantory, which is the formula we derived for P(j | w_{i-1})
You should start from the definition of the result. How should you show the probabilities of belonging?
The result, IMHO, should be a set of categories and a table: rows for vertices and columns for categories, and in the cells there will be possibilities of belonging of that vertice to this category.
Your graph can set some probabilities of belonging only if you already have some start known probabilities. I.e, that table would be already partly filled.
While filling the table according to the start values and weights of edges we would surely come to the situation, when we are getting different probabilities in the cells, coming into it by different ways. One more point should be set: can we change the start values in the table or they are set hardly? The same question for the weights of the edges.
Now the task is partly defined, and the part is very, very small. You even don't know the number of categories!
After you set all these rules and numbers, all is quite trivial - use The Gauss Method of Lesser Squares. As for iterative way, be careful - you don't know beforehand if the solution is stable or if it exists. If not, the iteration won't converge, and the whole that piece of code you wrote for it is for nothing. And by Gauss method you are getting a set of linear equations, and the standard algorithms are written to solve it for all cases. And at the end you have not only the solution, but the possible mistake for every final value.
I'm searching for an algorithm to find pairs of adjacent nodes on a hexagonal (honeycomb) graph that minimizes a cost function.
each node is connected to three adjacent nodes
each node "i" should be paired with exactly one neighbor node "j".
each pair of nodes defines a cost function
c = pairCost( i, j )
The total cost is then computed as
totalCost = 1/2 sum_{i=1:N} ( pairCost(i, pair(i) ) )
Where pair(i) returns the index of the node that "i" is paired with. (The sum is divided by two because the sum counts each node twice). My question is, how do I find node pairs that minimize the totalCost?
The linked image should make it clearer what a solution would look like (thick red line indicates a pairing):
Some further notes:
I don't really care about the outmost nodes
My cost function is something like || v(i) - v(j) || (distance between vectors associated with the nodes)
I'm guessing the problem might be NP-hard, but I don't really need the truly optimal solution, a good one would suffice.
Naive algos tend to get nodes that are "locked in", i.e. all their neighbors are taken.
Note: I'm not familiar with the usual nomenclature in this field (is it graph theory?). If you could help with that, then maybe that could enable me to search for a solution in the literature.
This is an instance of the maximum weight matching problem in a general graph - of course you'll have to negate your weights to make it a minimum weight matching problem. Edmonds's paths, trees and flowers algorithm (Wikipedia link) solves this for you (there is also a public Python implementation). The naive implementation is O(n4) for n vertices, but it can be pushed down to O(n1/2m) for n vertices and m edges using the algorithm of Micali and Vazirani (sorry, couldn't find a PDF for that).
This seems related to the minimum edge cover problem, with the additional constraint that there can only be one edge per node, and that you're trying to minimize the cost rather than the number of edges. Maybe you can find some answers by searching for that phrase.
Failing that, your problem can be phrased as an integer linear programming problem, which is NP-complete, which means that you might get dreadful performance for even medium-sized problems. (This does not necessarily mean that the problem itself is NP-complete, though.)
Given an undirected graph, I want to generate all subgraphs which are trees of size N, where size refers to the number of edges in the tree.
I am aware that there are a lot of them (exponentially many at least for graphs with constant connectivity) - but that's fine, as I believe the number of nodes and edges makes this tractable for at least smallish values of N (say 10 or less).
The algorithm should be memory-efficient - that is, it shouldn't need to have all graphs or some large subset of them in memory at once, since this is likely to exceed available memory even for relatively small graphs. So something like DFS is desirable.
Here's what I'm thinking, in pseudo-code, given the starting graph graph and desired length N:
Pick any arbitrary node, root as a starting point and call alltrees(graph, N, root)
alltrees(graph, N, root)
given that node root has degree M, find all M-tuples with integer, non-negative values whose values sum to N (for example, for 3 children and N=2, you have (0,0,2), (0,2,0), (2,0,0), (0,1,1), (1,0,1), (1,1,0), I think)
for each tuple (X1, X2, ... XM) above
create a subgraph "current" initially empty
for each integer Xi in X1...XM (the current tuple)
if Xi is nonzero
add edge i incident on root to the current tree
add alltrees(graph with root removed, N-1, node adjacent to root along edge i)
add the current tree to the set of all trees
return the set of all trees
This finds only trees containing the chosen initial root, so now remove this node and call alltrees(graph with root removed, N, new arbitrarily chosen root), and repeat until the size of the remaining graph < N (since no trees of the required size will exist).
I forgot also that each visited node (each root for some call of alltrees) needs to be marked, and the set of children considered above should only be the adjacent unmarked children. I guess we need to account for the case where no unmarked children exist, yet depth > 0, this means that this "branch" failed to reach the required depth, and cannot form part of the solution set (so the whole inner loop associated with that tuple can be aborted).
So will this work? Any major flaws? Any simpler/known/canonical way to do this?
One issue with the algorithm outlined above is that it doesn't satisfy the memory-efficient requirement, as the recursion will hold large sets of trees in memory.
This needs an amount of memory that is proportional to what is required to store the graph. It will return every subgraph that is a tree of the desired size exactly once.
Keep in mind that I just typed it into here. There could be bugs. But the idea is that you walk the nodes one at a time, for each node searching for all trees that include that node, but none of the nodes that were searched previously. (Because those have already been exhausted.) That inner search is done recursively by listing edges to nodes in the tree, and for each edge deciding whether or not to include it in your tree. (If it would make a cycle, or add an exhausted node, then you can't include that edge.) If you include it your tree then the used nodes grow, and you have new possible edges to add to your search.
To reduce memory use, the edges that are left to look at is manipulated in place by all of the levels of the recursive call rather than the more obvious approach of duplicating that data at each level. If that list was copied, your total memory usage would get up to the size of the tree times the number of edges in the graph.
def find_all_trees(graph, tree_length):
exhausted_node = set([])
used_node = set([])
used_edge = set([])
current_edge_groups = []
def finish_all_trees(remaining_length, edge_group, edge_position):
while edge_group < len(current_edge_groups):
edges = current_edge_groups[edge_group]
while edge_position < len(edges):
edge = edges[edge_position]
edge_position += 1
(node1, node2) = nodes(edge)
if node1 in exhausted_node or node2 in exhausted_node:
continue
node = node1
if node1 in used_node:
if node2 in used_node:
continue
else:
node = node2
used_node.add(node)
used_edge.add(edge)
edge_groups.append(neighbors(graph, node))
if 1 == remaining_length:
yield build_tree(graph, used_node, used_edge)
else:
for tree in finish_all_trees(remaining_length -1
, edge_group, edge_position):
yield tree
edge_groups.pop()
used_edge.delete(edge)
used_node.delete(node)
edge_position = 0
edge_group += 1
for node in all_nodes(graph):
used_node.add(node)
edge_groups.append(neighbors(graph, node))
for tree in finish_all_trees(tree_length, 0, 0):
yield tree
edge_groups.pop()
used_node.delete(node)
exhausted_node.add(node)
Assuming you can destroy the original graph or make a destroyable copy I came up to something that could work but could be utter sadomaso because I did not calculate its O-Ntiness. It probably would work for small subtrees.
do it in steps, at each step:
sort the graph nodes so you get a list of nodes sorted by number of adjacent edges ASC
process all nodes with the same number of edges of the first one
remove those nodes
For an example for a graph of 6 nodes finding all size 2 subgraphs (sorry for my total lack of artistic expression):
Well the same would go for a bigger graph, but it should be done in more steps.
Assuming:
Z number of edges of most ramificated node
M desired subtree size
S number of steps
Ns number of nodes in step
assuming quicksort for sorting nodes
Worst case:
S*(Ns^2 + MNsZ)
Average case:
S*(NslogNs + MNs(Z/2))
Problem is: cannot calculate the real omicron because the nodes in each step will decrease depending how is the graph...
Solving the whole thing with this approach could be very time consuming on a graph with very connected nodes, however it could be paralelized, and you could do one or two steps, to remove dislocated nodes, extract all subgraphs, and then choose another approach on the remainder, but you would have removed a lot of nodes from the graph so it could decrease the remaining run time...
Unfortunately this approach would benefit the GPU not the CPU, since a LOT of nodes with the same number of edges would go in each step.... and if parallelization is not used this approach is probably bad...
Maybe an inverse would go better with the CPU, sort and proceed with nodes with the maximum number of edges... those will be probably less at start, but you will have more subgraphs to extract from each node...
Another possibility is to calculate the least occuring egde count in the graph and start with nodes that have it, that would alleviate the memory usage and iteration count for extracting subgraphs...
Unless I'm reading the question wrong people seem to be overcomplicating it.
This is just "all possible paths within N edges" and you're allowing cycles.
This, for two nodes: A, B and one edge your result would be:
AA, AB, BA, BB
For two nodes, two edges your result would be:
AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB
I would recurse into a for each and pass in a "template" tuple
N=edge count
TempTuple = Tuple_of_N_Items ' (01,02,03,...0n) (Could also be an ordered list!)
ListOfTuple_of_N_Items ' Paths (could also be an ordered list!)
edgeDepth = N
Method (Nodes, edgeDepth, TupleTemplate, ListOfTuples, EdgeTotal)
edgeDepth -=1
For Each Node In Nodes
if edgeDepth = 0 'Last Edge
ListOfTuples.Add New Tuple from TupleTemplate + Node ' (x,y,z,...,Node)
else
NewTupleTemplate = TupleTemplate + Node ' (x,y,z,Node,...,0n)
Method(Nodes, edgeDepth, NewTupleTemplate, ListOfTuples, EdgeTotal
next
This will create every possible combination of vertices for a given edge count
What's missing is the factory to generate tuples given an edge count.
You end up with a list of possible paths and the operation is Nodes^(N+1)
If you use ordered lists instead of tuples then you don't need to worry about a factory to create the objects.
If memory is the biggest problem you can use a NP-ish solution using tools from formal verification. I.e., guess a subset of nodes of size N and check whether it's a graph or not. To save space you can use a BDD (http://en.wikipedia.org/wiki/Binary_decision_diagram) to represent the original graph's nodes and edges. Plus you can use a symbolic algorithm to check if the graph you guessed is really a graph - so you don't need to construct the original graph (nor the N-sized graphs) at any point. Your memory consumption should be (in big-O) log(n) (where n is the size of the original graph) to store the original graph, and another log(N) to store every "small graph" you want.
Another tool (which is supposed to be even better) is to use a SAT solver. I.e., construct a SAT formula that is true iff the sub-graph is a graph and supply it to a SAT solver.
For a graph of Kn there are approximately n! paths between any two pairs of vertices. I haven't gone through your code but here is what I would do.
Select a pair of vertices.
Start from a vertex and try to reach the destination vertex recursively (something like dfs but not exactly). I think this would output all the paths between the chosen vertices.
You could do the above for all possible pairs of vertices to get all simple paths.
It seems that the following solution will work.
Go over all partitions into two parts of the set of all vertices. Then count the number of edges which endings lie in different parts (k); these edges correspond to the edge of the tree, they connect subtrees for the first and the second parts. Calculate the answer for both parts recursively (p1, p2). Then the answer for the entire graph can be calculated as sum over all such partitions of k*p1*p2. But all trees will be considered N times: once for each edge. So, the sum must be divided by N to get the answer.
Your solution as is doesn't work I think, although it can be made to work. The main problem is that the subproblems may produce overlapping trees so when you take the union of them you don't end up with a tree of size n. You can reject all solutions where there is an overlap, but you may end up doing a lot more work than needed.
Since you are ok with exponential runtime, and potentially writing 2^n trees out, having V.2^V algorithms is not not bad at all. So the simplest way of doing it would be to generate all possible subsets n nodes, and then test each one if it forms a tree. Since testing whether a subset of nodes form a tree can take O(E.V) time, we are potentially talking about V^2.V^n time, unless you have a graph with O(1) degree. This can be improved slightly by enumerating subsets in a way that two successive subsets differ in exactly one node being swapped. In that case, you just have to check if the new node is connected to any of the existing nodes, which can be done in time proportional to number of outgoing edges of new node by keeping a hash table of all existing nodes.
The next question is how do you enumerate all the subsets of a given size
such that no more than one element is swapped between succesive subsets. I'll leave that as an exercise for you to figure out :)
I think there is a good algorithm (with Perl implementation) at this site (look for TGE), but if you want to use it commercially you'll need to contact the author. The algorithm is similar to yours in the question but avoids the recursion explosion by making the procedure include a current working subtree as a parameter (rather than a single node). That way each edge emanating from the subtree can be selectively included/excluded, and recurse on the expanded tree (with the new edge) and/or reduced graph (without the edge).
This sort of approach is typical of graph enumeration algorithms -- you usually need to keep track of a handful of building blocks that are themselves graphs; if you try to only deal with nodes and edges it becomes intractable.
This algorithm is big and not easy one to post here. But here is link to reservation search algorithm using which you can do what you want. This pdf file contains both algorithms. Also if you understand russian you can take a look to this.
So you have a graph with with edges e_1, e_2, ..., e_E.
If I understand correctly, you are looking to enumerate all subgraphs which are trees and contain N edges.
A simple solution is to generate each of the E choose N subgraphs and check if they are trees.
Have you considered this approach? Of course if E is too large then this is not viable.
EDIT:
We can also use the fact that a tree is a combination of trees, i.e. that each tree of size N can be "grown" by adding an edge to a tree of size N-1. Let E be the set of edges in the graph. An algorithm could then go something like this.
T = E
n = 1
while n<N
newT = empty set
for each tree t in T
for each edge e in E
if t+e is a tree of size n+1 which is not yet in newT
add t+e to newT
T = newT
n = n+1
At the end of this algorithm, T is the set of all subtrees of size N. If space is an issue, don't keep a full list of the trees, but use a compact representation, for instance implement T as a decision tree using ID3.
I think problem is under-specified. You mentioned that graph is undirected and that subgraph you are trying to find is of size N. What is missing is number of edges and whenever trees you are looking for binary or you allowed to have multi-trees. Also - are you interested in mirrored reflections of same tree, or in other words does order in which siblings are listed matters at all?
If single node in a tree you trying to find allowed to have more than 2 siblings which should be allowed given that you don't specify any restriction on initial graph and you mentioned that resulting subgraph should contain all nodes.
You can enumerate all subgraphs that have form of tree by performing depth-first traversal. You need to repeat traversal of the graph for every sibling during traversal. When you'll need to repeat operation for every node as a root.
Discarding symmetric trees you will end up with
N^(N-2)
trees if your graph is fully connected mesh or you need to apply Kirchhoff's Matrix-tree theorem