I am having trouble calculating the time analysis of for the following algorithm on any arbitrary tree of size N.
Question is:
Consider the following algorithm,
which makes the following assumptions. x and y are the roots of two binary
trees, Tx and Ty. Left(z) is a pointer to the left child of node z in either
tree, and Right(z) points to the right child. If the node doesn't have a
left or right child, the pointer returns \NIL". Each node z also has a eld
Size(z) which returns the number of nodes in the sub-tree rooted at z.
Size(NIL) is defi ned to be 0. The algorithm SameTree(x; y) returns a
boolean answer that says whether or not the trees rooted at x and y are
the same if you ignore the difference between left and right pointers.
Program: SameTree(x,y: Nodes): Boolean;
IF Size(x) 6= Size(y) THEN return False; halt.
IF x = NIL THEN return T rue; halt.
IF (SameTree(Left(x); Left(y)) AND SameTree(Right(x); Right(y)))
OR (SameTree(Right(x); Left(y)) AND SameTree(Left(x); Right(y)))
THEN return T rue; halt.
Return False; halt
Give the time analysis to run the above algorithm on any arbitrary tree of size N. I got O(nlog2^3) for dense graphs and O(n) for less dense graphs. Am I right? Can someone help me determine the time costs please?
Well let's use the Master principle. We shell consider the worst case where line 4 checks the condition before the OR and then checks the condition after it on EACH recursive call.
We will also simplify it by assuming the binaries trees are less or more balanced (has almost the same amount of nodes in each son of each node in the tree).
You have:
T(n) = 4*T(n/2)+2.
Look at http://en.wikipedia.org/wiki/Master_theorem to understand what I will do next:
We have case 1 from the Master theorem.
log in base 2 of 4 is 2. so the correct answer is O(n^2). This is the analysis for the General Case. If you wish a more precise analysis, you need to tell us much more on the odds for your tree to be balanced, unbalanced and what is the chance of it built in such a way that line 4 will be activating both conditions in each recursive call.
Average cases are much more complicated.
Related
I'm trying to figure out this data structure, but I don't understand how can we
tell there are O(log(n)) subtrees that represents the answer to a query?
Here is a picture for illustration:
Thanks!
If we make the assumption that the above is a purely functional binary tree [wiki], so where the nodes are immutable, then we can make a "copy" of this tree such that only elements with a value larger than x1 and lower than x2 are in the tree.
Let us start with a very simple case to illustrate the point. Imagine that we simply do not have any bounds, than we can simply return the entire tree. So instead of constructing a new tree, we return a reference to the root of the tree. So we can, without any bounds return a tree in O(1), given that tree is not edited (at least not as long as we use the subtree).
The above case is of course quite simple. We simply make a "copy" (not really a copy since the data is immutable, we can just return the tree) of the entire tree. So let us aim to solve a more complex problem: we want to construct a tree that contains all elements larger than a threshold x1. Basically we can define a recursive algorithm for that:
the cutted version of None (or whatever represents a null reference, or a reference to an empty tree) is None;
if the node has a value is smaller than the threshold, we return a "cutted" version of the right subtree; and
if the node has a value greater than the threshold, we return an inode that has the same right subtree, and as left subchild the cutted version of the left subchild.
So in pseudo-code it looks like:
def treelarger(some_node, min):
if some_tree is None:
return None
if some_node.value > min:
return Node(treelarger(some_node.left, min), some_node.value, some_node.right)
else:
return treelarger(some_node.right, min)
This algorithm thus runs in O(h) with h the height of the tree, since for each case (except the first one), we recurse to one (not both) of the children, and it ends in case we have a node without children (or at least does not has a subtree in the direction we need to cut the subtree).
We thus do not make a complete copy of the tree. We reuse a lot of nodes in the old tree. We only construct a new "surface" but most of the "volume" is part of the old binary tree. Although the tree itself contains O(n) nodes, we construct, at most, O(h) new nodes. We can optimize the above such that, given the cutted version of one of the subtrees is the same, we do not create a new node. But that does not even matter much in terms of time complexity: we generate at most O(h) new nodes, and the total number of nodes is either less than the original number, or the same.
In case of a complete tree, the height of the tree h scales with O(log n), and thus this algorithm will run in O(log n).
Then how can we generate a tree with elements between two thresholds? We can easily rewrite the above into an algorithm treesmaller that generates a subtree that contains all elements that are smaller:
def treesmaller(some_node, max):
if some_tree is None:
return None
if some_node.value < min:
return Node(some_node.left, some_node.value, treesmaller(some_node.right, max))
else:
return treesmaller(some_node.left, max)
so roughly speaking there are two differences:
we change the condition from some_node.value > min to some_node.value < max; and
we recurse on the right subchild in case the condition holds, and on the left if it does not hold.
Now the conclusions we draw from the previous algorithm are also conclusions that can be applied to this algorithm, since again it only introduces O(h) new nodes, and the total number of nodes can only decrease.
Although we can construct an algorithm that takes the two thresholds concurrently into account, we can simply reuse the above algorithms to construct a subtree containing only elements within range: we first pass the tree to the treelarger function, and then that result through a treesmaller (or vice versa).
Since in both algorithms, we introduce O(h) new nodes, and the height of the tree can not increase, we thus construct at most O(2 h) and thus O(h) new nodes.
Given the original tree was a complete tree, then it thus holds that we create O(log n) new nodes.
Consider the search for the two endpoints of the range. This search will continue until finding the lowest common ancestor of the two leaf nodes that span your interval. At that point, the search branches with one part zigging left and one part zagging right. For now, let's just focus on the part of the query that branches to the left, since the logic is the same but reversed for the right branch.
In this search, it helps to think of each node as not representing a single point, but rather a range of points. The general procedure, then, is the following:
If the query range fully subsumes the range represented by this node, stop searching in x and begin searching the y-subtree of this node.
If the query range is purely in range represented by the right subtree of this node, continue the x search to the right and don't investigate the y-subtree.
If the query range overlaps the left subtree's range, then it must fully subsume the right subtree's range. So process the right subtree's y-subtree, then recursively explore the x-subtree to the left.
In all cases, we add at most one y-subtree in for consideration and then recursively continue exploring the x-subtree in only one direction. This means that we essentially trace out a path down the x-tree, adding in at most one y-subtree per step. Since the tree has height O(log n), the overall number of y-subtrees visited this way is O(log n). And then, including the number of y-subtrees visited in the case where we branched right at the top, we get another O(log n) subtrees for a total of O(log n) total subtrees to search.
Hope this helps!
I was reading this answer on how to check if a BST is height balanced, and really hooked by the bonus question:
Suppose the tree is massively unbalanced. Like, a million nodes deep on one side and three deep on the other. Is there a scenario in which this algorithm blows the stack? Can you fix the implementation so that it never blows the stack, even when given a massively unbalanced tree?
What would be a good strategy here?
I am thinking to do a level order traversal and track the depth, if a leaf is found and current node depth is bigger than the leaf node depth + 2, then it's not balanced. But how to combine this with height checking?
Edit: below is the implementation in the linked answer
IsHeightBalanced(tree)
return (tree is empty) or
(IsHeightBalanced(tree.left) and
IsHeightBalanced(tree.right) and
abs(Height(tree.left) - Height(tree.right)) <= 1)
To review briefly: a tree is defined as being either null or a root node with pointers .left to a left child and .right to a right child, where each child is in turn a tree, the root node appears in neither child, and no node appears in both children. The depth of a node is the number of pointers that must be followed to reach it from the root node. The height of a tree is -1 if it's null or else the maximum depth of a node that appears in it. A leaf is a node whose children are null.
First let me note the two distinct definitions of "balanced" proposed by answerers of the linked question.
EL-balanced A tree is EL-balanced if and only if, for every node v, |height(v.left) - height(v.right)| <= 1.
This is the balance condition for AVL trees.
DF-balanced A tree is DF-balanced if and only if, for every pair of leaves v, w, we have |depth(v) - depth(w)| <= 1. As DF points out, DF-balance for a node implies DF-balance for all of its descendants.
DF-balance is used for no algorithm known to me, though the balance condition for binary heaps is very similar, requiring additionally that the deeper leaves be as far left as possible.
I'm going to outline three approaches to testing balance.
Size bounds for balanced trees
Expand the recursive function to have an extra parameter, maxDepth. For each recursive call, pass maxDepth - 1, so that maxDepth roughly tracks how much stack space is left. If maxDepth reaches 0, report the tree as unbalanced (e.g., by returning "infinity" for the height), since no balanced tree that fits in main memory could possibly be that tall.
This approach relies on an a priori size bound on main memory, which is available in practice if not in all theoretical models, and the fact that no subtrees are shared. (PROTIP: unless you're very careful, your subtrees will be shared at some point during development.) We also need height bounds on balanced trees of at most a given size.
EL-balanced Via mutual induction, we prove a lower bound, L(h), on the number of nodes belonging to an EL-balanced tree of a given height h.
The base cases are
L(-1) = 0
L(0) = 1,
more or less by definition. The inductive case is trickier. An EL-balanced tree of height h > 0 is a node with an EL-balanced child of height h - 1 and another EL-balanced child of height either h - 1 or h - 2. This means that
L(h) = 1 + L(h - 1) + min(L(h - 2), L(h - 1)).
Add 1 to both sides and rearrange.
L(h) + 1 = L(h - 1) + 1 + min(L(h - 2) + 1, L(h - 1) + 1).
A little while later (spoiler), we find that
L(h) <= phi^(h + 2)/sqrt(5),
where phi = (1 + sqrt(5))/2 ~ 1.618.
maxDepth then should be set to the floor of the base-phi logarithm of the maximum number of nodes, plus a small constant that depends on fenceposty things.
DF-balanced Rather than write out an induction proof, I'm going to appeal to your intuition that the worst case is a complete binary tree with one extra leaf on the bottom. Then the proper setting for maxDepth is the base-2 logarithm of the maximum number of nodes, plus a small constant.
Iterative deepening depth-first search
This is the theoretician's version of the answer above. Because, for some reason, we don't know how much RAM our computer has (and with logarithmic space usage, it's not as though we need a tight bound), we again include the maxDepth parameter, but this time, we use it to truncate the tree implicitly below the specified depth. If the height of the tree comes back below the bound, then we know that the algorithm ran successfully. Alternatively, if the truncated tree is unbalanced, then so is the whole tree. The problem case is when the truncated tree is balanced but with height equal to maxDepth. Then we increase maxDepth and retry.
The simplest retry strategy is to increase maxDepth by 1 every time. Since balanced trees with n nodes have height O(log n), the running time is O(n log n). In fact, for DF-balanced trees, the running time is also O(n), since, except for the last couple traversals, the size of the truncated tree increases by a factor of 2 each time, leading to a geometric series.
Another strategy, doubling maxDepth each time, gives an O(n) running time for EL-balanced trees, since the largest tree of height h, with 2^(h + 1) - 1 nodes, is much smaller than the smallest tree of height 2h, with approximately (phi^2)^h nodes. The downside of doubling is that we may use twice as much stack space. With increase-by-1, however, in the family of minimum-size EL-balanced trees we constructed implicitly in defining L(h), the number of nodes at depth h - k in the tree of height h is polynomial of degree k. Accordingly, the last few scans will incur some superlinear term.
Temporarily mutating pointers
If there are parent pointers, then it's easy to traverse depth-first in place, because the parent pointers can be used to derive the relevant information on the stack in an efficient manner. If we don't have parent pointers but can mutate the tree temporarily, then, for descent into a child, we can cannibalize the pointer to that child to store temporarily the node's parent. The problem is determining on the way up whether we came from a left or a right child. If we can sneak a bit (say because pointers are 2-byte aligned, or because there's a spare bit in the balance factor, or because we're copying the tree for stop-and-copy garbage collection and can determine which arena we're in), then that's one way. Another test assumes that the tree is a binary search tree. It turns out that we don't need additional assumptions, however: Explain Morris inorder tree traversal without using stacks or recursion .
The one fly in the ointment is that this approach only works, as far as I know, on DF-balance, since there's no space on the stack to put the partial results for EL-balance.
Let's say I have binary trees A and B and I want to know if A is a "part" of B. I am not only talking about subtrees. What I want to know is if B has all the nodes and edges that A does.
My thoughts were that since tree is essentially a graph, and I could view this question as a subgraph isomorphism problem (i.e. checking to see if A is a subgraph of B). But according to wikipedia this is an NP-complete problem.
http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem
I know that you can check if A is a subtree of B or not with O(n) algorithms (e.g. using preorder and inorder traversals to flatten the trees to strings and checking for substrings). I was trying to modify this a little to see if I can also test for just "parts" as well, but to no avail. This is where I'm stuck.
Are there any other ways to view this problem other than using subgraph isomorphism? I'm thinking there must be faster methods since binary trees are much more restricted and simpler versions of graphs.
Thanks in advance!
EDIT: I realized that the worst case for even a brute force method for my question would only take O(m * n), which is polynomial. So I guess this isn't a NP-complete problem after all. Then my next question is, is there an algorithm that is faster than O(m*n)?
I would approach this problem in two steps:
Find the root of A in B (either BFS of DFS)
Verify that A is contained in B (giving that starting node), using a recursive algorithm, as below (I concocted same crazy pseudo-language, because you didn't specify the language. I think this should be understandable, no matter your background). Note that a is a node from A (initially the root) and b is a node from B (initially the node found in step 1)
function checkTrees(node a, node b) returns boolean
if a does not exist or b does not exist then
// base of the recursion
return false
else if a is different from b then
// compare the current nodes
return false
else
// check the children of a
boolean leftFound = true
boolean rightFound = true
if a.left exists then
// try to match the left child of a with
// every possible neighbor of b
leftFound = checkTrees(a.left, b.left)
or checkTrees(a.left, b.right)
or checkTrees(a.left, b.parent)
if a.right exists then
// try to match the right child of a with
// every possible neighbor of b
leftFound = checkTrees(a.right, b.left)
or checkTrees(a.right, b.right)
or checkTrees(a.right, b.parent)
return leftFound and rightFound
About the running time: let m be the number of nodes in A and n be the number of nodes in B. The search in the first step takes O(n) time. The running time of the second step depends on one crucial assumption I made, but that might be wrong: I assumed that every node of A is equal to at most one node of B. If that is the case, the running time of the second step is O(m) (because you can never search too far in the wrong direction). So the total running time would be O(m + n).
While writing down my assumption, I start to wonder whether that's not oversimplifying your case...
you could compare the trees in bottom-up as follows:
for each leaf in tree A, identify the corresponding node in tree B.
start a parallel traversal towards the root in both trees from the nodes just matched.
specifically, move to the parent of a node in A and subsequently move towards the root in B until you either encounter the corresponding node in B (proceed) or a marked node in A (see below, if a match in B is found proceed, else fail) or the root of B (fail)
mark all nodes visited in A.
you succeed, if you haven't failed ;-).
the main part of the algorithm runs in O(e_B) - in the worst case, all edges in B are visited a constant number of times. the leaf node matching will run in O(n_A * log n_B) if there the B vertices are sorted, O(n_A * log n_A + n_B * log n_B + n) = O(n_B * log n_B) (sort each node set, lienarly scan the results thereafter) otherwise.
EDIT:
re-reading your question, abovementioned step 2 is even easier, as for matching nodes in A, B, their parents must match too (otheriwse there would be a mismatch between the edge sets). no effect on worst-case run time, of course.
Heres the problem, and my attempted solution.
My solution:
1. Run a topological sort on the tree, which runs in linear time BigTheta(E+V) where E is the number of edges and V the number of vertices. This puts it in a linked list which also takes constant time.
2. A vertex u would be an ancestor if it has a higher finishing time than vertex v.
3. Look at the 2 vertice's in the linked list and compare their finishing time and return true or false depending on the result from step 2.
Does this sound correct or am i missing something?
I don't think your understanding of what "constant time" means is quite correct. "...time BigTheta(E+V) where E is the number of edges and V the number of vertices" is linear time, not constant time.
Granted, you are allowed to take linear time for the pre-processing, so that's ok, but how are you going to do your step 3 ("Look at the 2 vertice's in the linked list") in constant time?
Here is an approach that will work for any tree (not only binary). The pre-processing step is to perform an Euler Tour of the tree (this is just a DFS traversal) and create a list out of this tour. When you visit a node for the first time you append it to the list and when you visit it the last time you append it to the list.
Example:
x
/ \
y z
The list will look like: [b(x), b(y), e(y), b(z), e(z), e(x)]. Here b(x) means enter x and e(x) means leave x. Now once you have this list, you can answer the query is x an ancestor of y by performing the test b(x) is before b(y) and e(y) is before e(x) in the list.
The question is how can you do this in constant time?
For static trees (which is the case for you), you can use a lookup table (aka array) to store the b/e, now the test takes constant time. So this solves your problem.
A binary tree can be encoded using two functions l and r
such that for a node n, l(n) give the left child of n, r(n)
give the right child of n.
A branch of a tree is a path from the root to a leaf, the
length of a branch to a particular leaf is the number of
arcs on the path from the root to that leaf.
Let MinBranch(l,r,x) be a simple recursive algorithm for
taking a binary tree encoded by the l and r functions
together with the root node x for the binary tree and
returns the length of the shortest branch of the binary
tree.
Give the pseudocode for this algorithm.
OK, so basically this is what I've come up with so far:
MinBranch(l, r, x)
{
if x is None return 0
left_one = MinBranch(l, r, l(x))
right_one = MinBranch(l, r, r(x))
return {min (left_one),(right_one)}
}
Obviously this isn't great or perfect. I'd be greatful if
people can help me get this perfect and working - any help
will be appreciated.
I doubt anyone will solve homework for you straight-up. A clue: the return value must surely grow higher as the tree gets bigger, right? However I don't see any numeric literals in your function except 0, and no addition operators either. How will you ever return larger numbers?
Another angle on the same issue: anytime you write a recursive function, it helps to enumerate "what are all the conditions where I should stop calling myself? what I return in each circumstance?"
You're on the right approach, but you're not quite there; your recursive algorithm will always return 0. (the logic is almost right, though...)
note that the length of the sub-branches is one less than the length of the branch; so left_one and right_one should be 1 + MinBranch....
Steping through the algorithm with some sample trees will help uncover off-by-one errors like this one...
It looks like you almost have it, but consider this example:
4
3 5
When you trace through MinBranch, you'll see that in your
MinBranch(l,r,4) call:
left_one = MinBranch(l, r, l(x))
= MinBranch(l, r, l(4))
= MinBranch(l, r, 3)
= 0
That makes sense, after all, 3 is a leaf node, so of course the distance
to the closest leaf node is 0. The same happens for right_one.
But you then wind up here:
return {min (left_one),(right_one)}
= {min (0), (0) }
= 0
but that's clearly wrong, because this node (4) is not a leaf node. Your
code forgot to count the current node (oops!). I'm sure you can manage
to fix that.
Now, actually, they way you're doing this isn't the fastest, but I'm not
sure if that's relevant for this exercise. Consider this tree:
4
3 5
2
1
Your algorithm will count up the left branch recursively, even though it
could, hypothetically, bail out if you first counted the right branch
and noted that 3 has a left, so its clearly longer than 5 (which is a
leaf). But, of course, counting the right branch first doesn't always
work!
Instead, with more complicated code, and probably a tradeoff of greater
memory usage, you can check nodes left-to-right, top-to-bottom (just
like English reading order) and stop at the first leaf you find.
What you've created can be thought of as a depth-first search. However, given what you're after (shortest branch), this may not be the most efficent approach. Think about how your algorithm would perform on a tree that was very heavy on the left side (of the root node), but had only one node on the right side.
Hint: consider a breadth-first search approach.
What you have there looks like a depth first search algorithm which will have to search the entire tree before you come up with a solution. what you need is the breadth first search algorithm which can return as soon as it finds the solution without doing a complete search