The time complexity for 1-NN search using a KD tree (i.e. balanced binary tree) is in what range?
Assume there are N points in the dataset
There's a hint here, but still can't figure it out: https://www.coursera.org/lecture/ml-clustering-and-retrieval/complexity-of-nn-search-with-kd-trees-BkZTg
A) O(N2) - O(N3)
B) O(log N) - O(N)
C) O(N logN) - O(N2)
D) None of the above
TLDR. The best and average case for finding the nearest neighbor with a kd-tree is going to be O(logβ(N)). But the worst case can be closer to O(N). Hence the answer is B) O(log N) - O(N)
Assuming the kd-tree is already built, each level of the tree normally bifurcates the entire range of points across one of the given dimensions. Just like a binary tree. So if the tree is perfectly balanced and each leaf node consists of just one point, there will be approximately N leaves across at the bottom of a tree that is logβ(N) levels high. Hence, to find the original point in the tree is always logβ(N).
But you aren't looking for the original point, you're looking for it's closest neighbor. So that's where it gets complicated. In practice, your leaf nodes don't contain just one point. They contain some reasonable number of points (~ logβ(N) or some small number like "10") that are considered to be in the same box or "cluster".
So when you find the initial point, you can immediately do a distance computation on the other points in the leaf node cluster to find an initial candidate for nearest neighbor. Hence, the highest probability is that the nearest neighbor is in the same cluster with a lesser probability its in one of the adjacent leaf node clusters. As you recurse up the tree, you have to decide if recursing the other child node is needed based on the dimension and mid-point value of each node. But if you've already found the nearest neighbor, you probably won't do too many more recursions down the tree.
But it's theoretically possible, that the initial bifurcation of the set of points split the original point with it closest neighbor. And in some crazy layout of points, you wind up having to do a distance computation with most of the points in the tree. Hence, O(N). Try making a 2-d graph of points and creating this scenario yourself.
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)
Assume we have n data points in a dataset. For a given point, we can order each of the n-1 other points based on its (metric) distance to that point.
What is the most efficient way to calculate this for every point in the dataset given we have a metric distance function such as an L-norm?
The naive approach seems to be to sort the list of distances for each point in turn, at a cost of O(n log n) per point, i.e. O(n^2 log n) for all points. Using a k-d tree seems no better given the full tree must be traversed each time.
Is there a better way that can take advantage of the triangle inequality, for example?
Since your output is O(n^2) you will not be able to get better than that.
I think this boils down to how fast you can rank all other points with respect to the distance to a point q. If you have an index structure (e.g., a KD-Tree or an R-Tree) you can use distance browsing to sort all other points w.r.t. q.
The basic idea of distance browsing is to have a priority queue pq whose entries are sorted by minimum distance to q. pq can contain points and entries of the index structure. You start with putting the root entry of the index structure into pq. Then you start poping elements from pq. When you encounter an entry (node) you resolve it and put the children back in to pq. When you encounter a point then you found your next nearest neighbour of q.
Overall an index structure has O(n) entries. Poping an element from pq is O(log |pq|). This makes the runtime O(n * log |pq|). The question is how many elements will be in pq in average.
I don't have a proof for this but a quick sketch let's me assume that the average number of elements in the queue should be around O(sqrt(n)) for L_1 and 2D space. Note, that the size of the queue heavily depends on the distance metric and the dimension of your points.
Putting all this together you can build the index structure (O(n log n)) and then for each point q rank all other points (O(n * log(sqrt(n))))
Overall this gives you a runtime of O(n * log(n) + n^2 * log(sqrt(n))).
However, to echo #MBo: this is a large hassle for little improvement over O(n^2 * log (n))
BLACK_PATH(T,x)
if x==NIL
then return TRUE
if COLOR(x)==BLACK
then return BLACK_PATH(T,left(x)) || BLACK_PATH(T,right(x))
return FALSE
The exercises asks to analyse the complexity of this procedure. I believe the reccurrence is the following
T(n)<=2T(2n/3)+O(1)
Using the recursion tree I obtain T(n)=O(n). Is this correct?
The complexity of this method is linear (O(n)) in the worst case with regards to the number of elements in the tree.
Using the master theorem in terms of the total number of nodes here is difficult because it does not take into account the properties of a red black tree. While it is true in general for heaps that every subtree of a tree with n nodes has max 2n/3 nodes, it is also true that for red black trees every subtree has at max n/2 black nodes. This is because red black trees are balanced with respect to black nodes (every path downwards to a leaf node from an arbitrary node has the same number of black nodes).
Most importantly: because the number of total nodes is not asymptotically higher than the number of black nodes you can, by analyzing the complexity purely with regards to the total number of black nodes, implicitly analyze the complexity with regards to the total number of nodes.
So rather than using T(n)<=2T(2n/3)+O(1) you should use T(m)<=T(m/2)+O(1) where m is the number of black nodes which gives you O(m) and because, as previously discussed, O(m)==O(n), we have O(n).
Another way to think about it: So long as you can understand that this algorithm is O(n) when all the nodes in the tree are black, you should be able to understand that it could only possibly require fewer operations if some of the nodes in the tree are red, since regardless of where the red node is every node in the subtree rooted at that red node will be ignored and not visited by this recursive algorithm. So it can only be O(n) or better, establishing O(n) as your worst case.
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.
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) .