Assume we have a DAG with a single leaf Node leafN. Each node is an object containing list of parent nodes. Node class can be something like class Node(Object, List[Node].
How to construct a DAG from leafN?
For e.g., given leaf node leafD with two parents, leafD(Object#123, List(leafC(leafA(null)), leafB(leafA(null)))) it's DAG will be:
leafA
/ \
leafB leafC
\ /
leafD
i.e., leafA(leafB(leafD(null)), leafC(leafD(null))) (I am ignoring the objects from every node for clarity.)
In short we have a leaf node with parent pointers which themselves have parent pointers and finally, after applying an algorithm, we want a root node with pointers to children nodes which further have pointers to children nodes.
Code ain't required, algorithm or links to any will suffice.
This is pretty simple, once you realise that notionally only the direction of the edges in the DAG has changed.
Instead of the Parent ---> Child node relationship, you have a Child ---> Parent relationship.
As such, all you need to do is construct a DAG from the given relationships, assuming that the Child node is actually the parent, and the parent nodes are childs.
So you end up with
leafD
↙ ↘
leafB leafC
↘ ↙
leafA
Now, all you need to do is reverse the edges in the DAG.
This is pretty simple, One way to do it would be to walk through the DAG, and get all the edge relationships, and then reverse them one by one.
Another would be to do this recursively using level order traversal.
Check this question for more solutions around how to reverse a DAG.
All you need to do is to put your nodes into some data structure that allows you to find a node easily. If your integer values in your node are sequential (or close to it) and start at a low number then you can use an array and just put the node with integer i in position i of the array (This makes including the integer in the node somewhat redundant).
This is all explained well on this site (but note that in their implementation the list is of the edges to, not the edges from) http://algs4.cs.princeton.edu/42directed/
Related
Given an array of binary trees find whether any two trees share a node, not value wise, but "pointer" wise. At the bottom I provided an example.
My approach was to iterate through all the trees and store all the leaves (pointers) from each tree into a list, then check if list has any duplicates, but that's a rather slow approach. Is there perhaps a quicker way to solve this?
In the worst case you will have to traverse all nodes (all pointers) to find a shared node (pointer), as it might happen to be the last one visited. So the best time complexity we can expect to have is O(𝑚+𝑛) where 𝑚 and 𝑛 represent the number of nodes in either tree.
We can achieve this time complexity if we store the pointers from the first tree in a hash set and then traverse the pointers of the second tree to see if any of those is in the set. Assuming that get/set operations on a hash set have an amortized constant time complexity, the overal time complexity will be O(𝑚+𝑛).
If the same program is responsible for constructing the trees, then a reuse of the same node can be detected upon insertion. For instance, reuse of the same node in multiple trees can be completely avoided by having the insert method of your tree only take a value as argument, never a node instance. The method will then encapsulate the actual creation of the node, guaranteeing its uniqueness.
An idea for O(#nodes) time and O(1) space. It does more traversal work than simple traversals using a hash table, but it doesn't have the cost of using a hash table. I don't know what's better. Might depend on the language.
For two trees
Create one extra node. Do a Morris traversal of the first tree. It only modifies right child pointers, so we can use left child pointers for marking nodes as seen. For every tree node without left child, set our extra node as left child. Whenever checking a left child pointer, treat our extra node like a null pointer, i.e., don't visit it. After the traversal, the tree structure is restored, and all originally left-child-less tree nodes now point to our extra node as left child. That includes all leaf nodes.
Do a Morris traversal of the second tree. Again treat pointers to our extra node like null pointers. If we ever do encounter our extra node, we know the trees share a node. If not, then we know the trees don't share a node, since if they did share any, they'd also share a leaf node (just go down from any shared node to a leaf node, that's also shared), and all leafs nodes of the first tree are marked. After the traversal, the second tree is restored.
Do a Morris traversal of the first tree again, this time removing our extra node, restoring the original null pointers.
For an array of more than two trees
Mark the first tree as above. Check the second tree as above. Mark the second tree. Check the third. Mark the third. Check the fourth. Mark the fourth. Etc. When you found a shared node or there are no more trees, unmark the marked trees.
Every shared node must have two parents, or an ancestor with two parents.
LOOP over nodes
IF node has two parents
MARK node as shared
Mark all descendants as shared.
The definition of a B tree I have read in various of books all contains the following
Every node except the root node has to be at least half full
If the root node is an index node, it must have at least two children.
I presume that the second special case is to allow a B tree to have, say, only one key and still be valid. However, if the B tree has many nodes, is it still allowed for the root node to have only two subtrees? Won't this break the guarantee of B tree like easy splitting and joining operation?
However, if the B tree has many nodes, is it still allowed for the root node to have only two subtrees?
Yes, the root is special-cased because every other internal node has siblings that it can merge with.
Suppose that we delete a key and that, as a result, some internal node has too few children. We have two options in the usual B-tree algorithms: have this node take some children from its siblings or just merge siblings outright (possibly propagating the deficiency toward the root). Neither is an option for the root, so we just exempt it from the minimum children requirement. This increases the max height for a given number of keys by at most one, so the asymptotic running time of operations is unaffected.
Is it conceptually possible to have a tree where you traverse it by starting at a given leaf node (rather than the root node) and use parent pointers to get to the root?
I ask this since I saw someone implement a tree and they used an array to hold all of the leaf nodes/external nodes and each of the leaf/external nodes point only to their parent nodes and those parent point to their parent node etc. until you get to the root node which has no parents. Their implementation thus would require you to start at one of the leaves to get to anywhere in the tree and you would cannot go "down" the tree since her tree nodes do not have any children pointers, only parent pointers.
I found this implementation interesting since I haven't seen anything like it but I was curious if it could still be considered it a "tree". I have never seen a tree where you start traversal at the leaves, instead of root. I have also never seen a tree where the tree nodes only have parent pointers and no children pointers.
Yep, this structure exists. It's often called a spaghetti stack.
Spaghetti stacks are useful for representing the "is a part of" relation. For example, if you want to represent a class hierarchy in a way that makes upcasting efficient, then you might represent the class hierarchy as a spaghetti stack in which the node for each type stores a pointer to its parent type. That way, it's easy to find whether an upcast is valid by just walking upward from the node.
They're also often used in compilers to track scoping information. Each node represents one scope in the program, and each node has a pointer to the node representing the scope one level above it.
You can also think of a blockchain this way. Each block stores a backreference to its parent block. By starting at any block and tracing backwards to the root, you can recover the state encoded by that block.
Hope this helps!
If an array A of leaf nodes are given, traversal is possible. If only a single leaf node is given, I don't know how to traverse the tree. Pseudocode:
// initial step
add all nodes in A to a queue Q
//removeNode(Q) returns pointer to node at front of Q
while((node = removeNode(Q)) != NULL)
/* do your operation on node */
add node->parent to Q
How to find a loop in a binary tree? I am looking for a solution other than marking the visited nodes as visited or doing a address hashing. Any ideas?
Suppose you have a binary tree but you don't trust it and you think it might be a graph, the general case will dictate to remember the visited nodes. It is, somewhat, the same algorithm to construct a minimum spanning tree from a graph and this means the space and time complexity will be an issue.
Another approach would be to consider the data you save in the tree. Consider you have numbers of hashes so you can compare.
A pseudocode would test for this conditions:
Every node would have to have a maximum of 2 children and 1 parent (max 3 connections). More then 3 connections => not a binary tree.
The parent must not be a child.
If a node has two children, then the left child has a smaller value than the parent and the right child has a bigger value. So considering this, if a leaf, or inner node has as a child some node on a higher level (like parent's parent) you can determine a loop based on the values. If a child is a right node then it's value must be bigger then it's parent but if that child forms a loop, it means he is from the left part or the right part of the parent.
3.a. So if it is from the left part then it's value is smaller than it's sibling. So => not a binary tree. The idea is somewhat the same for the other part.
Testing aside, in what form is the tree that you want to test? Remeber that every node has a pointer to it's parent. An this pointer points to a single parent. So depending of the format you tree is in, you can take advantage from this.
As mentioned already: A tree does not (by definition) contain cycles (loops).
To test if your directed graph contains cycles (references to nodes already added to the tree) you can iterate trough the tree and add each node to a visited-list (or the hash of it if you rather prefer) and check each new node if it is in the list.
Plenty of algorithms for cycle-detection in graphs are just a google-search away.
I have a large (100,000+ nodes) Directed Acyclic Graph (DAG) and would like to run a "visitor" type function on each node in order, where order is defined by the arrows in the graph. i.e. all parents of a node are guaranteed to be visited before the node itself.
If two nodes do not refer to each other directly or indirectly, then I don't care which order they are visited in.
What's the most efficient algorithm to do this?
You would have to perform a topological sort on the nodes, and visit the nodes in the resulting order.
The complexity of such algorithm is O(|V|+|E|) which is quite good. You want to traverse all nodes, so if you would want a faster algorithm than that, you would have to solve it without even looking at all edges, which would be dangerous, because one single edge could havoc the order completely.
There are some answers here:
Good graph traversal algorithm
and here:
http://en.wikipedia.org/wiki/Topological_sorting
In general, after visiting a node, you should visit its related nodes, but only the nodes that are not already visited. In order to keep track of the visited nodes, you need to keep the IDs of the nodes in a set (or map), or you can mark the node as visited (somehow).
If you care about the topological order, you must first get hold of a collection of all the un-traversed links ("remaining links") to a node, sorted by the id of the referenced node (typically: map(node-ID -> link-count)). If you haven't got that, you might need to build it using an approach similar to the one above. Then, start by visiting a node whose remaining incoming link count is zero. For each link from that node, reduce the remaining link count for each related node, adding the related node to the set of nodes-to-visit (or just visiting the node) if the count reaches zero.
As mentioned in the other answers, this problem can be solved by Topological Sorting.
A very simple algorithm for that (not the most efficient):
Keep an array (or map) indegree[] where indegree[node]=number of incoming edges of node
while there is at least one node n with indegree[n]=0:
for each node n in nodes where indegree[n]>0:
visit(n)
indegree[n]=-1 # mark n as visited
for each node x adjacent to n:
indegree[x]=indegree[x]-1 # its parent has been visited, so one less edge coming into it
You can traverse a DAG in O(N) (without any topsort) by just running your dfs from every node with zero indegree, because those will be the valid "starting point". This will work because graph has no cycles, those zero indegree nodes must exist, and must traverse the whole graph.