I've looked at various other StackOverflow answer's and they all are different to what my lecturer has written in his slides.
Depth First Search has a time complexity of O(b^m), where b is the
maximum branching factor of the search tree and m is the maximum depth
of the state space. Terrible if m is much larger than d, but if search
tree is "bushy", may be much faster than Breadth First Search.
He goes on to say..
The space complexity is O(bm), i.e. space linear in length of action
sequence! Need only store a single path from the root to the leaf
node, along with remaining unexpanded sibling nodes for each node on
path.
Another answer on StackOverflow states that it is O(n + m).
Time Complexity: If you can access each node in O(1) time, then with branching factor of b and max depth of m, the total number of nodes in this tree would be worst case = 1 + b + b2 + β¦ + bm-1. Using the formula for summing a geometric sequence (or even solving it ourselves) tells that this sums to = (bm - 1)/(b - 1), resulting in total time to visit each node proportional to bm. Hence the complexity = O(bm).
On the other hand, if instead of using the branching factor and max depth you have the number of nodes n, then you can directly say that the complexity will be proportional to n or equal to O(n).
The other answers that you have linked in your question are similarly using different terminologies. The idea is same everywhere. Some solutions have added the edge count too to make the answer more precise, but in general, node count is sufficient to describe the complexity.
Space Complexity: The length of longest path = m. For each node, you have to store its siblings so that when you have visited all the children, and you come back to a parent node, you can know which sibling to explore next. For m nodes down the path, you will have to store b nodes extra for each of the m nodes. Thatβs how you get an O(bm) space complexity.
The complexity is O(n + m) where n is the number of nodes in your tree, and m is the number of edges.
The reason why your teacher represents the complexity as O(b ^ m), is probably because he wants to stress the difference between Depth First Search and Breadth First Search.
When using BFS, if your tree has a very large amount of spread compared to it's depth, and you're expecting results to be found at the leaves, then clearly DFS would make much more sense here as it reaches leaves faster than BFS, even though they both reach the last node in the same amount of time (work).
When a tree is very deep, and non-leaves can give information about deeper nodes, BFS can detect ways to prune the search tree in order to reduce the amount of nodes necessary to find your goal. Clearly, the higher up the tree you discover you can prune a sub tree, the more nodes you can skip.
This is harder when you're using DFS, because you're prioritize reaching a leaf over exploring nodes that are closer to the root.
I suppose this DFS time/space complexity is taught on an AI class but not on Algorithm class.
The DFS Search Tree here has slightly different meaning:
A node is a bookkeeping data structure used to represent the search
tree. A state corresponds to a configuration of the world. ...
Furthermore, two different nodes can contain the same world state if
that state is generated via two different search paths.
Quoted from book 'Artificial Intelligence - A Modern Approach'
So the time/space complexity here is focused on you visit nodes and check whether this is the goal state. #displayName already give a very clear explanation.
While O(m+n) is in algorithm class, the focus is the algorithm itself, when we store the graph as adjacency list and how we discover nodes.
Related
Ignoring space complexity, assuming each node in the tree is touched exactly once and considering DFS and BFS traversal time equivalent, what is the time complexity of traversing an n-ary tree?
Given that Big O notation is an asymptotic measure, meaning that we are looking a function that gives us a line or curve that best fits the problem as it is extended to more levels or branches.
My intuition tells me that for tree structures in general we would want a function of the sort b^l where b is the branching factor and l is the number of levels in the tree (for a full and complete tree).
However for a partial tree, it would make sense to take some sort of average of b and l, perhaps AVG(b)^AVG(l).
In looking for this answer I find many people are saying it is O(n) where n is the number of verticies in the tree (nodes -1). See:
What is the time complexity of tree traversal? and
Complexity of BFS in n-ary tree
But a linear solution in my mind does not model the cost (in time) that the algorithm will take as the tree adds additional levels (on average). Which is what I understand Big O notation is intended to do.
The height or branching factor of a tree are not the determining factors for the complexity of a complete traversal (whether it be BFS or DFS). It is only the number of vertices that is the determining factor.
If you want to express the complexity in terms of branching factor and height of the tree, then you are really asking what the relation is between these measures and the number of nodes in a tree.
As you already indicate, if the branching factor is 100 but it is only rarely that a node has more than 3 children, then the branching factor is not really telling us much. The same can be said about the height of the tree. A tree of height 4 and branching factor 2 can have between 5 and 31 nodes.
What to do then? Often, the worst case will be taken (maximising the number of nodes). This means that you'll translate branching factor and height to a tree that is perfect, i.e. where each node has the maximum number of children, except for the leaves of the tree, which are all on the same level.
The number of nodes π is then πβ+1-1, where π is the branching factor, and β the height (number of edges on the longest path from root to leaf).
That means the worst case time complexity for a given π and β is O(πβ+1)
Working with average is not really practical. The distribution of the branching factor may not be linear, and the distribution of leaf-depths might not be linear either, so working with averages is not going to give much insight.
As to the cost of adding a node: Once the insertion point is determined, the time complexity for adding a node is constant. It does not matter which that insertion increases the branching factor or increases the height of the tree.
There is some variation when it comes to finding a node if the tree is a search tree (like a BST or B-tree). In that case the height of the tree becomes important, and a search would cost O(β). If however the tree is self-balancing (like AVL or B-tree), this variation is limited and this complexity would be O(logπ)
apologies first, english is not my first language.
So here's my understanding on graph that's represented as adjancey list: It's usually used for sparse graph, which is the case for most of graphs, and it uses V (number of vertex) lists. so, V head pointers + 2e (# of edges) nodes for undirected graph. Therefore, space complexity = O(E+V)
Since any node can have upto V-1 edges (excluding itself) it has time complexity of O(V) to check a node's adjacency.
As to check all the edges, it takes O(2e + V) so O(v + e)
Now, since it's mostly used for sparse graph, it's rarely O(v) to check adjacency, but simply the number of edges a given vertex has (which is O(V) at worst since V-1 is the possible maximum)
What I'm wondering is, is it possible to make the list (the edge nodes) binary tree? So to find out whether node A is adjacent to node B, time complexity would be O(logn) and not linear O(n).
If it is possible, is it actually done quite often? Also, what is that kind of data structure called? I've been googling if such combinations are possible but couldn't find anything. I would be very grateful if anyone could explain this to me in detail as i'm new to data structure. Thank you.
Edit: I know binary search can be performed on arrays. I'm talking about linked list representation, I thought I made it obvious when I said heads to the lists but wow
There's no reason the adjacency list for each vertex couldn't be stored as a binary tree, but there are tradoffs.
As you say, this adjacency list representation is often used for sparse graphs. Often, "sparse graph" means that a particular vertex is adjacent to few others. So your "adjacency list" for a particular vertex would be very small. Whereas it's true that binary search is O(log n) and sequential search is O(n), when n is very small sequential search is faster. I've seen cases where sequential search beats binary search when n is smaller than 16. It depends on the implementation, of course, but don't count on binary search being faster for small lists.
Another thing to think about is memory. Linked list overhead is one pointer per node. Unless, of course, you're using a doubly linked list. Binary tree overhead is two pointers per node. Perhaps not a big deal, but if you're trying to represent a very large graph, that extra pointer will become important.
If the graph will be updated frequently at run time, you have to take that into account, too. Adding a new edge to a linked list of edges is an O(1) operation. But adding an edge to a binary tree will require O(log n). And you want to make sure you keep that tree balanced. An unbalanced tree starts to act like a linked list.
So, yes, you could make your adjacency lists binary trees. You have to decide whether it's worth the extra effort, based on your application's speed requirements and the nature of your data.
The run time of BFS is O(b^d)
b is the branching factor
d is the depth(# of level) of the graph from starting node.
I googled for awhile, but I still dont see anyone mention how they figure out this "b"
So I know branching factor means the "# of child that each node has"
Eg, branching factor for a binary Tree is 2.
so for a BFS graph , is that b= average all the branching factor of each node in our graph.
or b = MAX( among all branch factor of each node) ?
Also, no matter which way we pick the b, still seeming ambiguous to approach our run time.
For example , if our graph has 30000 nodes, only 5 nodes has 10000 branching, and all the rest 29955 nodes just have 10 branching. and we have the depth setup to be 100.
Seems O(b^d) is not making sense at this case.
Can someone explain a little bit. Thankyou!
The runtime that is more often quoted is that BFS is O(m + n) where m is the number of edges and n the number of nodes. This is because each vertex is processed once and each edge at most twice.
I think O(b^d) is used when using BFS on, say, brute-forcing a game of chess, where each position had a relatively constant branching factor and your engine needs to search a certain number of positions deep. For example, b is about 35 for chess and Deep Blue had a search depth of 6-8 (going up to 20).
In such cases, because the graph is relatively acyclic, b^d is roughly the same as m + n (they are equal for trees). O(b^d) is more useful as b is fixed and d is something you control.
in graphs O(b^d), the b = MAX. Since it is the worst case. check this link from princeton http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Breadth-first_search.html - go to time complexity portion
To quote from Artificial Intelligence - A modern approach by Stuart Russel and Peter Norvig:
Time and space complexity are always considered with respect to some measure of the prob- lem difficulty. In theoretical computer science, the typical measure is the size of the state space graph, |V | + |E|, where V is the set of vertices (nodes) of the graph and E is the set of edges (links). This is appropriate when the graph is an explicit data structure that is input to the search program. (The map of Romania is an example of this.) In AI, the graph is often represented implicitly by the initial state, actions, and transition model and is frequently infi- nite. For these reasons, complexity is expressed in terms of three quantities: b, the branching factor or maximum number of successors of any node; d, the depth of the shallowest goal node (i.e., the number of steps along the path from the root); and m, the maximum length of any path in the state space. Time is often measured in terms of the number of nodes generated during the search, and space in terms of the maximum number of nodes stored in memory. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how βredundantβ the paths in the state space are.
This should give you a clear insight about the difference between O(|V|+|E|) and b^d
My book here (Artificial intelligence A modern approach) says that the worst-case time and space complexity of a uniform-cost search algorithm would be O(b[C*/e]) , where b is the branching factor, C* is the cost of the optimal solution, and every action costs atleast e. But why is this so?
First, the complexity is O(B^(C/e)) [exponential in C/e].
To understand it, think of a simple example case first:
Let G=(V,E) be a graph, with branch factor B. The graph is unweighted (w(e) = 1 for each e).
Consider finding the shortest path from S to T.
In this case, the algorithm is actually a BFS, and it will discover all nodes in the path up to length SOL, where SOL is the length of the shortest path, which is O(B^|SOL|)
For the general case - the same idea holds, you need to discover all nodes up to cost C. So you discover nodes up to depth C/e, giving you O(B^(C/e)) total nodes needed to be explored.
The exponential factor is because: First level (root) has B^0=1 nodes, second level has B nodes. from each of these you discover B nodes, giving you B^2, ....
EDIT:
Missed it so far, but the title asks for space complexity and not time complexity. However, the answer remains the same, since a uniform cost search holds a visited set, for already visited nodes. Since each node you discover is also added to it - the answer remains O(B^(C/e))
C*/e means average number of nodes which should be visited during the search, and for visiting each of this nodes you should look at all possible b branches (at least root nodes), so you should check b[C*/e] node in your search. which is your search time complexity, this is by assuming process on each node takes O(1).
P.S: It's Ξ©(b[C*/e])in worst case
I am looking for an algorithm to split a tree with N nodes (where the maximum degree of each node is 3) by removing one edge from it, so that the two trees that come as the result have as close as possible to N/2. How do I find the edge that is "the most centered"?
The tree comes as an input from a previous stage of the algorithm and is input as a graph - so it's not balanced nor is it clear which node is the root.
My idea is to find the longest path in the tree and then select the edge in the middle of the longest path. Does it work?
Optimally, I am looking for a solution that can ensure that neither of the trees has more than 2N / 3 nodes.
Thanks for your answers.
I don't believe that your initial algorithm works for the reason I mentioned in the comments. However, I think that you can solve this in O(n) time and space using a modified DFS.
Begin by walking the graph to count how many total nodes there are; call this n. Now, choose an arbitrary node and root the tree at it. We will now recursively explore the tree starting from the root and will compute for each subtree how many nodes are in each subtree. This can be done using a simple recursion:
If the current node is null, return 0.
Otherwise:
For each child, compute the number of nodes in the subtree rooted at that child.
Return 1 + the total number of nodes in all child subtrees
At this point, we know for each edge what split we will get by removing that edge, since if the subtree below that edge has k nodes in it, the spilt will be (k, n - k). You can thus find the best cut to make by iterating across all nodes and looking for the one that balances (k, n - k) most evenly.
Counting the nodes takes O(n) time, and running the recursion visits each node and edge at most O(1) times, so that takes O(n) time as well. Finding the best cut takes an additional O(n) time, for a net runtime of O(n). Since we need to store the subtree node counts, we need O(n) memory as well.
Hope this helps!
If you see my answer to Divide-And-Conquer Algorithm for Trees, you can see I'll find a node that partitions tree into 2 nearly equal size trees (bottom up algorithm), now you just need to choose one of the edges of this node to do what you want.
Your current approach is not working assume you have a complete binary tree, now add a path of length 3*log n to one of leafs (name it bad leaf), your longest path will be within one of a other leafs to the end of path connected to this bad leaf, and your middle edge will be within this path (in fact after you passed bad leaf) and if you partition base on this edge you have a part of O(log n) and another part of size O(n) .