Given a DAG and multiple sources S1 .. Sn, we define a valid node as a node that is reachable from all sources. Find out all the valid nodes which don't have a valid parent.
Effectively, find the direct-multi-source children of S1 .. Sn
Example: In the following DAG, S1, S2, S3 are the sources. Nodes a, b, c are reachable via all sources. However, node c has a parent which is also reachable via all sources.
So the top view is a, b
The naive algorithm for this would be to
For each source, find the set of reachable nodes
Take the intersection of these sets to find all valid nodes
Perform a multi-source BFS from all sources. When moving from parent -> child keep track of whether the parent was a valid node or not. Any valid node which was reached via a parent which is also a valid node gets discarded.
This algorithm takes O((V+E) * S) time because of step 1.
V = number of vertices
E = number of edges
S = number of sources
Is there a faster algorithm? Can this be done in O(V+E)?
Well, you can do this in O(V+E) time if you can union sets of size S in constant time.
In the easy case, you would use bitmasks to represent sets, so if S <= 64, for example, then you're good with using a 64-bit word to represent each set, and bitwise-OR for union.
Even if you have a lot more than 64 sources, a compressed set representation with an algorithm like this one is probably going to be your best bet.
The algorithm isn't difficult:
Topologically sort the graph, using Kahn's algorithm of DFS;
Initialize a set for each vertex. Sources get singleton sets containing only themselves, and other vertexes get empty sets. Also initialize each vertex as "not marked" and "not crossed out";
Process the vertexes in topological order. For each vertex, if it's "crossed out", then cross out all of its children. Otherwise replace its set with the union of its set and the sets of all of its parents (which have already been processed). If its set then contains all sources, then mark the vertex and cross out all of its children (which have not yet been processed).
Return all the vertexes that are marked.
Step 3 can be optimized.
SET T = valid nodes from step 2
LOOP V over valid nodes
LOOP P over valid node's parents
IF P is valid
DELETE V from T
T is the set of "top nodes"
You can "distribute" the data structures by searching from each source in order to label each vertex with the set of sources that reach it. Nodes where the set's cardinality is equal the number of sources are valid. Make a set V of these. Then make one more search of the whole graph removing vertices reached via a valid parent from V. At the end, V contains the answer.
The run time will I think be be O(V * E) assuming constant time set insert and remove. This algorithm takes set intersection out of the picture at the possible cost of more memory.
I have a directed acyclic graph created by users, where each node (vertex) of the graph represents an operation to perform on some data. The outputs of a node depend on its inputs (obviously), and that input is provided by its parents. The outputs are then passed on to its children. Cycles are guaranteed to not be present, so can be ignored.
This graph works on the same principle as the Shader Editor in Blender. Each node performs some operation on its input, and this operation can be arbitrarily expensive. For this reason, I only want to evaluate these operations when strictly required.
When a node is updated, via user input or otherwise, I need to reevaluate every node which depends on the output of the updated node. However, given that I can't justify evaluating the same node multiple times, I need a way to determine the correct order to update the nodes. A basic breadth-first traversal doesn't solve the problem. To see why, consider this graph:
A traditional breadth-first traversal would result in D being evaluated prior to B, despite D depending on B.
I've tried doing a breadth-first traversal in reverse (that is, starting with the O1 and O2 nodes, and traversing up the graph), but I seem to run into the same problem. A reversed breadth-first traversal will visit D before B, thus I2 before A, resulting in I2 being ordered after A, despite A depending on I2.
I'm sure I'm missing something relatively simple here, and I feel as though the reverse traversal is key, but I can't seem to wrap my head around it and get all the pieces to fit. I suppose one potential solution is to use the reverse traversal as intended, but rather than avoiding visiting each node more than once, just visiting each node each time it comes up, ensuring that it has a definitely correct ordering. But visiting each node multiple times and the exponential scaling that comes with that is a very unattractive solution.
Is there a well-known efficient algorithm for this type of problem?
Yes, there is a well known efficient algorithm. It's topological sorting.
Create a dictionary with all nodes and their corresponding in-degree, let's call it indegree_dic. in-degree is the number of parents/or incoming edges to that node. Have a set S of the nodes with in-degree equal to zero.
Taken from the Wikipedia page with some modification:
L ← Empty list that will contain the nodes sorted topologically
S ← Set of all nodes with no incoming edge that haven't been added to L yet
while S is not empty do
remove a node n from S
add n to L
for each child node m of n do
decrement m's indegree
if indegree_dic[m] equals zero then
delete m from indegree_dic
insert m into S
if indegree_dic has length > 0 then
return error (graph is not a DAG)
else
return L (a topologically sorted order)
This sort is not unique. I mention that because it has some impact on your algorithm.
Now, whenever a change happens to any of the nodes, you can safely avoid recalculation of any nodes that come before the changed node in your topologically sorted list, but need to nodes that come after it. You can be sure that all the parents are processed before their children if you follow the sorted list in your calculation.
This algorithm is not optimal, as there could be nodes after the changed node, that are not children of that node. Like in the following scenario:
A
/ \
B C
One correct topological sort would be [A, B, C]. Now, suppose B changes. You skip A because nothing has changed for it, but recalculate C because it comes after B. But you actually don't need to, because B has no effect on C whatsoever.
If the impact of this isn't big, you could use this algorithm and keep the implementation easier and less prone to bugs. But if efficiency is key, here are some ideas that may help:
You can do a topological sort each time and include the which node has change as a factor. When choosing nodes from S in the above algorithm, choose every other node that you can before you choose the changed node. In other words, you choose the changed node from S only when S has length 1. This guarantees that you process every node that isn't below the hierarchy of the changed node before it. This approach helps when the sorting is much cheaper then processing the nodes.
Another approach, which I'm not entirely sure is correct, is to look after the changed node in the topological sorted list and start processing only when you reach the first child of the changed node.
Another way relies on idea 1 but is helpful if you can do some pre-processing. You can create topological sorts for each case of one node being changed. When a node is changed, you try to put it in the ordering as late as possible. You save all these ordering in a node to ordering dictionary and based on which node has changed you choose that ordering.
I'd like to construct a binary tree from a quite unusual input. The input contains:
Total number of nodes.
The integer label of the root.
A list of all edges (vertices/nodes that are connected to each
other). The edges in the list are UNSORTED, there is only one rule for
determining left/right children - the child in the edge that appears
first in list is always on the left. The order of child/parent in the vertices pair is also random.
I've come up with some straighforward solutions but they require multiple searches through the list of all edges (I'd basically find the 2 edges that have the labeled root in them and repeat this process for all the subtrees.)
I imagine this straightforward approach would be VERY inefficient for trees with a big amount of nodes, but I can't come up with anything else.
Any ideas for more efficient algorithms to solve this?
Here's an example for better visualization:
INPUT: 5 NODES, ROOT LABELED 2, LIST OF EDGES: [(1,0),(1,2),(2,3),(1,4)]
The tree would look like this:
2
1 3
0 4
It is important to clarify whether the given edge list is stated to be directed or not.
If edges are given in a directed fashion (i.e. it is stated that any given edge A-B also includes the information that A is a parent of B) storing the edges in an adjacency list while recording number of incoming edges for each vertex in an array should be sufficient. Once you go through the array for the incoming edges, the vertex with 0 incoming edges(i.e. parents) should be the root. Then you can run a DFS in linear time complexity to traverse the graph and put it in any data structure that is best for your needs.
If the edges given are stated to be undirected, the scheme changes a bit. In that case, you don't have the concept of incoming and outgoing edge. In that case, as no structure for the array is specified(e.g. BST, etc.) you can basically consider any node with less than 3 edges as root and run DFS as mentioned above. (all the leaves and intermediary nodes with single child nodes)
A simple solution is: "Link all the edges in the tree that it!"
Start preparing a dictionary. If nodes don't exist by the start and end point, create them nodes. As it is random in nature, you can set their left and right pointers to NULL initially.
You have rule - " the child in the edge that appears first in list is always on the left.". So create child accordingly.
Also, you already know the root of the tree so you can iterate across the nodes you have constructed so far.
Through this you can generate tree in one shot.
Hope this helps!
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