I am attempting to write a key-value store in rust, with a b+ tree for indexing. I am creating an index file to store the b+ tree, and a data file to store the values of the leaf nodes. I am splitting the files into constant size blocks on the data file and the index file. Where I am having a problem is in the index file.
What is the best format to store the tree? I am currently storing three item for internal nodes: the index value, the offset (the number of bytes) to its children in the index file and a bit to say whether the current node is a leaf node (which would mean the offset number points to the disk file), with the entire tree serialized as a preorder traversal.
But during insertion it will be very costly when a new level is added to the tree for example, and I will need to rewrite the entire tree (index) to the disk.
Is there any way to optimize / keep track of the new nodes created and only replace those instead of writing the entire tree?
Related
I'm reading an article about why we need B-Tree. It's telling me that B-Tree can decrease the number of IO whereas other trees, such as Red-Black tree, can't. And the number of IO equals to the height of B-Tree.
Here is an example.
We are looking for the value 9. With the B-Tree, there are three times of IO, but with the binary tree, there are maybe four times of IO.
Now I'm confused. Why can the B-Tree ensure that there are at most only three times of IO? In other words, who can ensure that the node 3 and the node 7 must be located at the same disk block? I've thought that the data structure of each node in a B-Tree may be an array so that they were sequential, and sequential data is normally located at the same disk block (seriously, I'm not sure...), but it seems that the data structure of the node in a B-Tree is a List, which means that they are not sequential. So as my understanding, it's also possible to generate two times of IO while accessing 3 and 7. In this case, can't we say that accessing 9 may also need 4 times of IO?
On disk, every B-tree node is in a single contiguous block, and every node contains thousands of keys.
For each key there is a pointer to the corresponding node in the next level. On disk, this "pointer" is the address of the contiguous block that contains the target node.
So, for example, if there are 10^9 leaf-level keys, there could be 1000000 leaf-level nodes. On the parent level, there are 1000000 keys that point to those nodes, distributed among 1000 parent nodes. On the root level, there are 1000 keys in a single node.
Is there general pseudocode or related data structure to get the nth value of a b-tree? For example, the eighth value of this tree is 13 [1,4,9,9,11,11,12,13].
If I have some values sorted in a b-tree, I would like to find the nth value without having to go through the entire tree. Is there a better structure for this problem? The data order could update anytime.
You are looking for order statistics tree. The idea of it, is in addition to any data stored in nodes - also store the size of the subtree in the node, and keep them updated in insertions and deletions.
Since you are "touching" O(logn) nodes for each insert/delete operation - keeping it up to date still keeps the O(logn) behavior of these.
FindKth() is then done by eliminating subtrees that their bigger index is still smaller than k, and checking the next one. Since you don't need to go to the depth of each subtree, only directly to the required one (and checking the nodes in the path to this element) - you need to "touch" O(logn) nodes, which makes this operation O(logn) as well.
The heap property says:
If A is a parent node of B then the key of node A is ordered with
respect to the key of node B with the same ordering applying across
the heap. Either the keys of parent nodes are always greater than or
equal to those of the children and the highest key is in the root node
(this kind of heap is called max heap) or the keys of parent nodes are
less than or equal to those of the children and the lowest key is in
the root node (min heap).
But why in this wiki, the Binary Heap has to be a Complete Binary Tree? The Heap Property doesn't imply that in my impression.
According to the wikipedia article you provided, a binary heap must conform to both the heap property (as you discussed) and the shape property (which mandates that it is a complete binary tree). Without the shape property, one would lose the runtime advantage that the data structure provides (i.e. the completeness ensures that there is a well defined way to determine the new root when an element is removed, etc.)
Every item in the array has a position in the binary tree, and this position is calculated from the array index. The positioning formula ensures that the tree is 'tightly packed'.
For example, this binary tree here:
is represented by the array
[1, 2, 3, 17, 19, 36, 7, 25, 100].
Notice that the array is ordered as if you're starting at the top of the tree, then reading each row from left-to-right.
If you add another item to this array, it will represent the slot below the 19 and to the right of the 100. If this new number is less than 19, then values will have to be swapped around, but nonetheless, that is the slot that will be filled by the 10th item of the array.
Another way to look at it: try constructing a binary heap which isn't a complete binary tree. You literally cannot.
You can only guarantee O(log(n)) insertion and (root) deletion if the tree is complete. Here's why:
If the tree is not complete, then it may be unbalanced and in the worst case, simply a linked list, requiring O(n) to find a leaf, and O(n) for insertion and deletion. With the shape requirement of completeness, you are guaranteed O(log(n)) operations since it takes constant time to find a leaf (last in array), and you are guaranteed that the tree is no deeper than log2(N), meaning the "bubble up" (used in insertion) and "sink down" (used in deletion) will require at most log2(N) modifications (swaps) of data in the heap.
This being said, you don't absolutely have to have a complete binary tree, but you just loose these runtime guarantees. In addition, as others have mentioned, having a complete binary tree makes it easy to store the tree in array format forgoing object reference representation.
The point that 'complete' makes is that in a heap all interior (not leaf) nodes have two children, except where there are no children left -- all the interior nodes are 'complete'. As you add to the heap, the lowest level of nodes is filled (with childless leaf nodes), from the left, before a new level is started. As you remove nodes from the heap, the right-most leaf at the lowest level is removed (and pushed back in at the top). The heap is also perfectly balanced (hurrah!).
A binary heap can be looked at as a binary tree, but the nodes do not have child pointers, and insertion (push) and deletion (pop or from inside the heap) are quite different to those procedures for an actual binary tree.
This is a direct consequence of the way in which the heap is organised. The heap is held as a vector with no gaps between the nodes. The parent of the i'th item in the heap is item (i - 1) / 2 (assuming a binary heap, and assuming the top of the heap is item 0). The left child of the i'th item is (i * 2) + 1, and the right child one greater than that. When there are n nodes in the heap, a node has no left child if (i * 2) + 1 exceeds n, and no right child if (i * 2) + 2 does.
The heap is a beautiful thing. It's one flaw is that you do need a vector large enough for all entries... unlike a real binary tree, you cannot allocate a node at a time. So if you have a heap for an indefinite number of items, you have to be ready to extend the underlying vector as and when needed -- or run some fragmented structure which can be addressed as if it was a vector.
FWIW: when stepping down the heap, I find it convenient to step to the right child -- (i + 1) * 2 -- if that is < n then both children are present, if it is == n only the left child is present, otherwise there are no children.
By maintaining binary heap as a complete binary gives multiple advantages such as
1.heap is complete binary tree so height of heap is minimum possible i.e log(size of tree). And insertion, build heap operation depends on height. So if height is minimum then their time complexity will be reduced.
2.All the items of complete binary tree stored in contiguous manner in array so random access is possible and it also provide cache friendliness.
In order for a Binary Tree to be considered a heap two it must meet two criteria. 1) It must have the heap property. 2) it must be a complete tree.
It is possible for a structure to have either of these properties and not have the other, but we would not call such a data structure a heap. You are right that the heap property does not entail the shape property. They are separate constraints.
The underlying structure of a heap is an array where every node is an index in an array so if the tree is not complete that means that one of the index is kept empty which is not possible beause it is coded in such a way that each node is an index .I have given a link below so that u can see how the heap structure is built
http://www.sanfoundry.com/java-program-implement-min-heap/
Hope it helps
I find that all answers so far either do not address the question or are, essentially, saying "because the definition says so" or use a similar circular argument. They are surely true but (to me) not very informative.
To me it became immediately obvious that the heap must be a complete tree when I remembered that you insert a new element not at the root (as you do in a binary search tree) but, rather, at the bottom right.
Thus, in a heap, a new element propagates from the bottom up - it is "moved up" within the tree till it finds a suitable place.
In a binary search tree a newly inserted element moves the other way round - it is inserted at the root and it "moves down" till it finds its place.
The fact that each new element in a heap starts as the bottom right node means that the heap is going to be a complete tree at all times.
We need to maintain mobileNumber and its location in memory.
The challenge is that we have more than 5 million of users
and storing the location for each user will be like hash map of 5 million records.
To resolve this problem, we have to work on ranges
We are given ranges of phone numbers like
range1 start="9899123446" end="9912345678" location="a"
range2 start="9912345679" end="9999999999" location="b"
A number can belong to single location only.
We need a data structure to store these ranges in the memory.
It has to support two functions
findLocation(Integer number) it should return the location name to
which number belongs
changeLocation( Integer Number , String range). It changes location of Number from old location to new location
This is completely in memory design.
I am planning to use tree structure with each node contains ( startofrange , endofrange ,location).
I will keep the nodes in sorted order. I have not finalized anything yet.
The main problem is-- when 2nd function to change location is called say 9899123448 location to b
The range1 node should split to 3 nodes 1st node (9899123446,9899123447,a)
2nd node (9899123448,9899123448,b) 3rd node (9899123449,9912345678,a).
Please suggest the suitable approach
Thanks in advance
Normally you can use specialized data structures to store ranges and implement the queries, e.g. Interval Tree.
However, since phone number ranges do not overlap, you can just store the ranges in a standard tree based data structure (Binary Search Tree, AVL Tree, Red-Black Tree, B Tree, would all work) sorted only by [begin].
For findLocation(number), use corresponding tree search algorithm to find the first element that has [begin] value smaller than the number, check its [end] value and verify if the number is in that range. If a match if found, return the location, otherwise the number is not in any range.
For changeLocation() operation:
Find the old node containing the number
If an existing node is found in step 1, delete it and insert new nodes
If no existing node is found, insert a new node and try to merge it with adjacent nodes.
I am assuming you are using the same operation for simply adding new nodes.
More practically, you can store all the entries in a database, build an index on [begin].
First of all range = [begin;end;location]
Use two structures:
Sorted array to store ranges begins
Hash-table to access ends and locations by begins
Apply following algo:
Use binary search to find "nearest less" value ob begin
Use hash-table to find end and location for begin
I'm been trying to implement and understand the split/merge operations on a treap. Every node has two keys: a heap key and a tree key. Looking at the heap keys you should see a valid heap and same with the tree keys.
Splitting a treap is easier than normal because you can just insert a dummy node with the maxmimum or minimum priority (depends on if it's a max-heap or min-heap). However, this link just says to assume that the splitting key isn't in the tree. However, what if I always want existing key inside the right tree, or the left tree? What do I do?
Find the node with the key in question.
Move it up to become the new root (by giving it a very high -- or very low -- priority).
Split off the left (or right) subtree.