Which node data structure to use for a trie - algorithm

I am using trie for the first time.I wanted to know which is the best data structure to use for a trie while deciding which is the next branch that one is supposed to traverse. I was looking among an array,a hashmap and a linked list.

Each of these options has their advantages and disadvantages.
If you store the child nodes in an array, then you can look up which child to visit extremely efficiently by just indexing into the array. However, the space usage per node will be high: O(|Σ|), where Σ is the set of letters that your words can be formed from, even if most of those children are null.
If you store the child nodes in a linked list, then the time required to find a child will be O(|Σ|), since you may need to scan across all of the nodes of the linked list to find the child you want. On the other hand, the space efficiency will be quite good, because you only store the children that you're using. You could also consider using a fixed-sized array here, which has even better space usage but leads to very expensive insertions and deletions.
If you store the child nodes in a hash table, then the (expected) time to find a child will be O(1) and the memory usage will only be proportional (roughly) to the number of children you have. Interestingly, because you know in advance what values you're going to be hashing, you could consider using a dynamic perfect hash table to ensure worst-case O(1) lookups, at the expense of some precomputation.
Another option would be to store the child nodes in a binary search tree. This gives rise to the ternary search tree data structure. This choice is somewhere between the linked list and hash table options - the space usage is low and you can perform predecessor and successor queries efficiently, but there's a slight increase in the cost of performing a lookup due to the search cost in the BST. If you have a static trie where insertions never occur, you can consider using weight-balanced trees as the BSTs at each point; this gives excellent runtime for searches (O(n + log k), where n is the length of the string to search for and k is the total number of words in the trie).
In short, the array lookups are fastest but its space usage in the worst case is the worst. A statically-sized array has the best memory usage but expensive insertions and deletions. The hash table has decently fast lookups and good memory usage (on average). Binary search trees are somewhere in the middle. I would suggest using the hash table here, though if you put a premium on space and don't care about lookup times the linked list might be better. Also, if your alphabet is small (say, you're making a binary trie), the array overhead won't be too bad and you may want to use that.
Hope this helps!

If you are trying to build trie just for alphabets, I would suggest to use array and then use particia tree (space optimized trie).
http://en.wikipedia.org/wiki/Radix_tree
This will allow you to do fast lookup with array and doesn't waste too much of space if branching factor of certain node is low.

Related

Why don't we use AVL tree for hash table's item storage?

Recently, I was looking at the hash table which is using chaining as linked list. I came to the possibility of using the "chain" as AVL tree.
Therefore, each buckets in the hash table will have little AVL tree's root pointers. Wikipedia says hash table's worst case is O(n) (http://en.wikipedia.org/wiki/Hash_table). However, if we use each bucket's "chain" as AVL tree, we can bring it down to O(ln n).
Am I missing something?
As far as I know we can replace a linked list with an AVL tree.
Wouldn't such ADT be better than single AVL tree or hash table with linked list chaining?
I searched the internet and could not find such ADT.
This is discussed directly in the Wikipedia article your referenced:
Separate chaining with other structures
Instead of a list, one can use any other data structure that supports the required operations. For example, by using a self-balancing tree, the theoretical worst-case time of common hash table operations (insertion, deletion, lookup) can be brought down to O(log n) rather than O(n). However, this approach is only worth the trouble and extra memory cost if long delays must be avoided at all costs (e.g., in a real-time application), or if one must guard against many entries hashed to the same slot (e.g., if one expects extremely non-uniform distributions, or in the case of web sites or other publicly accessible services, which are vulnerable to malicious key distributions in requests).
In Java, standard HashMap use red-black trees in buckets, if buckets size exceeds constant 8; they are linearized back to the singly-linked list if bucket becomes smaller than 6 entries; apparently real world tests showed that for smaller buckets managing them as trees loses more due to general complexity of this data structure and extra memory footprint (because tree entries should hold at least 2 references to other entries, singly-linked entries hold only one reference), than gain from theoretically better asymptotic complexity.
I would also add, that for best performance hash table should be configured so that most buckets has only one entry (i. e. they are not even lists, just sole entries), marginally less should contain two entries and only exceptional buckets occasionally should have 3 or more entries. So holding 1-3 entries in tree makes absolutely no sense, compared to simple linked list.

Why implement a Hashtable with a Binary Search Tree?

When implementing a Hashtable using an array, we inherit the constant time indexing of the array. What are the reasons for implementing a Hashtable with a Binary Search Tree since it offers search with O(logn)? Why not just use a Binary Search Tree directly?
If the elements don't have a total order (i.e. the "greater than" and "less than" is not be defined for all pairs or it is not consistent between elements), you can't compare all pairs, thus you can't use a BST directly, but nothing's stopping you from indexing the BST by the hash value - since this is an integral value, it obviously has a total order (although you'd still need to resolve collision, that is have a way to handle elements with the same hash value).
However, one of the biggest advantages of a BST over a hash table is the fact that the elements are in order - if we order it by hash value, the elements will have an arbitrary order instead, and this advantage would no longer be applicable.
As for why one might consider implementing a hash table using a BST instead of an array, it would:
Not have the disadvantage of needing to resize the array - with an array, you typically mod the hash value with the array size and resize the array if it gets full, reinserting all elements, but with a BST, you can just directly insert the unchanging hash value into the BST.
This might be relevant if we want any individual operation to never take more than a certain amount of time (which could very well happen if we need to resize the array), with the overall performance being secondary, but there might be better ways to solve this problem.
Have a reduced risk of hash collisions since you don't mod with the array size and thus the number of possible hashes could be significantly bigger. This would reduce the risk of getting the worst-case performance of a hash table (which is when a significant portion of the elements hash to the same value).
What the actual worst-case performance is would depend on how you're resolving collisions. This is typically done with linked-lists for O(n) worst case performance. But we can also achieve O(log n) performance with BST's (as is done in Java's hash table implementation if the number of elements with some hash are above a threshold) - that is, have your hash table array where each element points to a BST where all elements have the same hash value.
Possibly use less memory - with an array you'd inevitably have some empty indices, but with a BST, these simply won't need to exist. Although this is not a clear-cut advantage, if it's an advantage at all.
If we assume we use the less common array-based BST implementation, this array will also have some empty indices and this would also require the occasional resizing, but this is a simply memory copy as opposed to needing to reinsert all elements with updated hashes.
If we use the typical pointer-based BST implementation, the added cost for the pointers would seemingly outweigh the cost of having a few empty indices in an array (unless the array is particularly sparse, which tends to be a bad sign for a hash table anyway).
But, since I haven't personally ever heard of this ever being done, presumably the benefits are not worth the increased cost of operations from expected O(1) to O(log n).
Typically the choice is indeed between using a BST directly (without hash values) and using a hash table (with an array).
Pros:
Potentially use less space b/c we don't allocate a large array
Can iterate through the keys in order, sometimes useful
Cons:
You'd have O(log N) lookup time, which is worse than the guaranteed O(1) for a chained hash table.
Since the requirements of a Hash Table are O(1) lookup, it's not a Hash Table if it has logarithmic lookup times. Granted, since collision is an issue with the array implementation (well, not likely an issue), using a BST could offer benefits in that regard. Generally, though, it's not worth the tradeoff - I can't think of a situation where you wouldn't want guaranteed O(1) lookup time when using a Hash Table.
Alternatively, there is the possibility of an underlying structure to guarantee logarithmic insertion and deletion via a BST variant, where each index in the array has a reference to the corresponding node in the BST. A structure like that could get sort of complex, but would guarantee O(1) lookup and O(logn) insertion/deletion.
I found this looking to see if anyone had done it. I guess maybe not.
I came up with an idea this morning of implementing a binary tree as an array consisting of rows stored by index. Row 1 has 1, row 2 has 2, row 3 has 4 (yes, powers of two). The advantage of this structure is a bit shift and addition or subtraction can be used to walk the tree instead of using extra memory to store bi- or uni-directional references.
This would allow you to rapidly search for a hash value based on some sort of hashable input, to discover if the value exists in some other store. Or for a hash collision (or partial collision) search. I can't think of many other uses for it but for these it would be phenomenally fast. Very likely a lot of the rotation operations would happen entirely in cpu cache and be written out in nice linear blobs to main memory.
Its main utility would be with sorting input values of a random nature. If the blobs in the array were two parts, like a hash, and an identifier for another store, you could do the comparisons very fast and insert very fast to discover where an item bearing a hash value is kept in another location (like the UUID of a filesystem node or maybe even the filename, or other short identifiable string).
I'll leave it to others to dream of other ways to use it but I'm using it for a graph theoretic proof of work search table for identifying partial collisions for a variant of Cuckoo Cycle.
I am just now working on the walk formula, and here it is:
i = index of array element
Walk Up (go to parent):
i>>1-(i+1)%2
(Obviously you probably need to test if i is zero)
Walk Left (down and left):
i<<1+2
(this and the next would also need to test against 2^depth of the structure, so it doesn't walk off the edge and fall back to the root)
Walk Right (down and right):
i<<1+1
As you can see, each walk is a short formula based on the index. A bit shift and addition for going left and right, and a bit shift, addition and modulus for ascending. Two instructions to move down, 4 to move up (in assembler, or as above in C and other HLL operator notation)
edit:
I can see from further commentary that the benefit of slashing the insert time definitely would be of benefit. But I don't think that a conventional vector based binary tree would provide nearly as much benefit as a dense version. A dense version, where all the nodes are in a contiguous array, when it is searched, naturally will travel in a linear fashion through the memory, which should help reduce cache misses and thus reduce the latency of the searches significantly, as well as the fact that there is a latency hit with memory in accessing randomly compared to streaming through blocks sequentially.
https://github.com/calibrae-project/bast/blob/master/pkg/bast/bast.go
This is my current state of a WiP to implement what I am calling a Bifurcation Array Search Tree. For the purpose of a fast insert/delete and not horribly slow search through a sorted collection of hashes, I think that this would be of quite large benefit for cases where there is a lot of data coming and going through the structure, or more to the point, beneficial for more realtime applications.

Datastructure for fast and efficient search

I have to store the sorted data in a data structure.
The data structure I want to use is heap or binary search tree.
But I am confused which one would better serve the requirement i.e. fast and efficient searching.
----MORE DETAILS---
I am designing an application that receive data from a source(say a data grid) and then store it into a data structure. The data that comes from data GRID station is in the form of sorted digits. The sorted data can be in ascending or descending order.
now I have to search the data. and the process should be efficient and fast.
A heap will only let you search quickly for the minimum element (find it in O(1) time, remove it in O(log n) time). If you design it the other way, it will let you find the maximum, but you don't get both. To search for arbitrary elements quickly (in O(log n) time), you'll want the binary search tree.
For efficient searching, one would definitely prefer a binary search tree.
To search for a value in a heap may require that you search the entire tree - you can't guarantee that some value may not appear on either the left or right subtree (unless one of the children is already greater than the target value, but this isn't guaranteed to happen).
So searching in a heap takes O(n), where-as it takes O(log n) in a (self-balancing) binary search tree.
A heap is only really preferred if you're primarily interested in finding and/or removing the minimum / maximum, along with insertions.
Either can be constructed in O(n) if you're given already-sorted data.
You mentioned a sorted data structure, but in the "more details" in your question I don't really see that a sorted data structure is required (it doesn't matter too much that that's the form in which your data is already in), but it really depends on exactly what type of queries you will do.
If you're only going to search for exact values, you don't really need a sorted data structure, and can use a hash table instead, which supports expected O(1) lookups.
Let me make a list of potential data structures and we'll elaborate:
Binary search tree - it contains sorted data so adding new elements is costly (O(log n) I think). When you search through it you can use the binary search which is O(log n). IT is memory efficient and it doesn't need much additional memory.
Hash table (http://en.wikipedia.org/wiki/Hash_table) - every element is stored with a Hash. You can get element by providing the hash. Your elements don't need to be sortable, they only need to provide hashing method. Accessing elements is O(1) which I suppose is pretty decent one :)
I myself usually use hashtables but it depends on what exactly you need to store and how often you add or delete elements.
Check this also: Advantages of Binary Search Trees over Hash Tables
So in my opinion out of heap and binary search list, use Hash table.
I would go with hash table with separate chaining with an AVLTree (I assume collision occurs). It will work better than O(logn) where n is number of items. After getting the index with hash function, m items will be in this index where m is less than or equal to n. (It is usually much smaller, but never more).
O(1) for hashing and O(logm) for searching in AVLTree. This is faster than binary search for sorted data.

What class of problem would one use a binary search tree to solve?

I've seen this data structure talked about a lot, but I am unclear as to what sort of problem would demand such a data structure (over alternative representations). I've never needed one, but perhaps that's because I don't quite grok it. Can you enlighten me?
One example of where you would use a binary search tree would be a sorted list of values where you want to be able to quickly add elements.
Consider using an array for this purpose. You have very fast access to read random values, but if you want to add a new value, you have to find the place in the array where it belongs, shift everything over, and then insert the new value.
With a binary search tree, you simply traverse the tree looking for where the value would be if it were in the tree already, and then add it there.
Also, consider if you want to find out if your sorted array contains a particular value. You have to start at one end of the array and compare the value you're looking for to each individual value until you either find the value in the array, or pass the point where it would have been. With a binary search tree, you greatly reduce the number of comparisons you are likely to have to make. Just a quick caveat, however, it is definitely possible to contrive situations where the binary search tree requires more comparisons, but these are the exception, not the rule.
One thing I've used it for in the past is Huffman decoding (or any variable-bit-length scheme).
If you maintain your binary tree with the characters at the leaves, each incoming bit decides whether you move to the left or right node.
When you reach a leaf node, you have your decoded character and you can start on the next one.
For example, consider the following tree:
.
/ \
. C
/ \
A B
This would be a tree for a file where the predominant letter was C (by having less bits used for common letters, the file is shorter than it would be for a fixed-bit-length scheme). The codes for the individual letters are:
A: 00 (left, left).
B: 01 (left, right).
C: 1 (right).
The class of problems you use then for are those where you want to be able to both insert and access elements reasonably efficiently. As well as unbalanced trees (such as the Huffman example above), you can also use balanced trees which make the insertions a little more costly (since you may have to rebalance on the fly) but make lookups a lot more efficient since you're traversing the minimum possible number of nodes.
from wiki
Self-balancing binary search trees can be used in a natural way to construct and maintain ordered lists, such as priority queues. They can also be used for associative arrays; key-value pairs are simply inserted with an ordering based on the key alone. In this capacity, self-balancing BSTs have a number of advantages and disadvantages over their main competitor, hash tables. One advantage of self-balancing BSTs is that they allow fast (indeed, asymptotically optimal) enumeration of the items in key order, which hash tables do not provide. One disadvantage is that their lookup algorithms get more complicated when there may be multiple items with the same key.
Self-balancing BSTs can be used to implement any algorithm that requires mutable ordered lists, to achieve optimal worst-case asymptotic performance. For example, if binary tree sort is implemented with a self-balanced BST, we have a very simple-to-describe yet asymptotically optimal O(n log n) sorting algorithm. Similarly, many algorithms in computational geometry exploit variations on self-balancing BSTs to solve problems such as the line segment intersection problem and the point location problem efficiently. (For average-case performance, however, self-balanced BSTs may be less efficient than other solutions. Binary tree sort, in particular, is likely to be slower than mergesort or quicksort, because of the tree-balancing overhead as well as cache access patterns.)
Self-balancing BSTs are flexible data structures, in that it's easy to extend them to efficiently record additional information or perform new operations. For example, one can record the number of nodes in each subtree having a certain property, allowing one to count the number of nodes in a certain key range with that property in O(log n) time. These extensions can be used, for example, to optimize database queries or other list-processing algorithms.

Advantages of Binary Search Trees over Hash Tables

What are the advantages of binary search trees over hash tables?
Hash tables can look up any element in Theta(1) time and it is just as easy to add an element....but I'm not sure of the advantages going the other way around.
One advantage that no one else has pointed out is that binary search tree allows you to do range searches efficiently.
In order to illustrate my idea, I want to make an extreme case. Say you want to get all the elements whose keys are between 0 to 5000. And actually there is only one such element and 10000 other elements whose keys are not in the range. BST can do range searches quite efficiently since it does not search a subtree which is impossible to have the answer.
While, how can you do range searches in a hash table? You either need to iterate every bucket space, which is O(n), or you have to look for whether each of 1,2,3,4... up to 5000 exists.
(what about the keys between 0 and 5000 are an infinite set? for example keys can be decimals)
Remember that Binary Search Trees (reference-based) are memory-efficient. They do not reserve more memory than they need to.
For instance, if a hash function has a range R(h) = 0...100, then you need to allocate an array of 100 (pointers-to) elements, even if you are just hashing 20 elements. If you were to use a binary search tree to store the same information, you would only allocate as much space as you needed, as well as some metadata about links.
One "advantage" of a binary tree is that it may be traversed to list off all elements in order. This is not impossible with a Hash table but is not a normal operation one design into a hashed structure.
In addition to all the other good comments:
Hash tables in general have better cache behavior requiring less memory reads compared to a binary tree. For a hash table you normally only incur a single read before you have access to a reference holding your data. The binary tree, if it is a balanced variant, requires something in the order of k * lg(n) memory reads for some constant k.
On the other hand, if an enemy knows your hash-function the enemy can enforce your hash table to make collisions, greatly hampering its performance. The workaround is to choose the hash-function randomly from a family, but a BST does not have this disadvantage. Also, when the hash table pressure grows too much, you often tend to enlargen and reallocate the hash table which may be an expensive operation. The BST has simpler behavior here and does not tend to suddenly allocate a lot of data and do a rehashing operation.
Trees tend to be the ultimate average data structure. They can act as lists, can easily be split for parallel operation, have fast removal, insertion and lookup on the order of O(lg n). They do nothing particularly well, but they don't have any excessively bad behavior either.
Finally, BSTs are much easier to implement in (pure) functional languages compared to hash-tables and they do not require destructive updates to be implemented (the persistence argument by Pascal above).
The main advantages of a binary tree over a hash table is that the binary tree gives you two additional operations you can't do (easily, quickly) with a hash table
find the element closest to (not necessarily equal to) some arbitrary key value (or closest above/below)
iterate through the contents of the tree in sorted order
The two are connected -- the binary tree keeps its contents in a sorted order, so things that require that sorted order are easy to do.
A (balanced) binary search tree also has the advantage that its asymptotic complexity is actually an upper bound, while the "constant" times for hash tables are amortized times: If you have a unsuitable hash function, you could end up degrading to linear time, rather than constant.
A binary tree is slower to search and insert into, but has the very nice feature of the infix traversal which essentially means that you can iterate through the nodes of the tree in a sorted order.
Iterating through the entries of a hash table just doesn't make a lot of sense because they are all scattered in memory.
A hashtable would take up more space when it is first created - it will have available slots for the elements that are yet to be inserted (whether or not they are ever inserted), a binary search tree will only be as big as it needs to be. Also, when a hash-table needs more room, expanding to another structure could be time-consuming, but that might depend on the implementation.
A binary search tree can be implemented with a persistent interface, where a new tree is returned but the old tree continues to exist. Implemented carefully, the old and new trees shares most of their nodes. You cannot do this with a standard hash table.
BSTs also provide the "findPredecessor" and "findSuccessor" operations (To find the next smallest and next largest elements) in O(logn) time, which might also be very handy operations. Hash Table can't provide in that time efficiency.
From Cracking the Coding Interview, 6th Edition
We can implement the hash table with a balanced binary search tree (BST) . This gives us an O(log n) lookup time. The advantage of this is potentially using less space, since we no longer allocate a large array. We can also iterate through the keys in order, which can be useful sometimes.
GCC C++ case study
Let's also get some insight from one of the most important implementations in the world. As we will see, it actually matches out theory perfectly!
As shown at What is the underlying data structure of a STL set in C++?, in GCC 6.4:
std::map uses BST
std::unordered_map uses hashmap
So this already points out to the fact that you can't transverse a hashmap efficiently, which is perhaps the main advantage of a BST.
And then, I also benchmarked insertion times in hash map vs BST vs heap at Heap vs Binary Search Tree (BST) which clearly highlights the key performance characteristics:
BST insertion is O(log), hashmap is O(1). And in this particular implementation, hashmap is almost always faster than BST, even for relatively small sizes
hashmap, although much faster in general, has some extremely slow insertions visible as single points in the zoomed out plot.
These happen when the implementation decides that it is time to increase its size, and it needs to be copied over to a larger one.
In more precise terms, this is because only its amortized complexity is O(1), not the worst case, which is actually O(n) during the array copy.
This might make hashmaps inadequate for certain real-time applications, where you need stronger time guarantees.
Related:
Binary Trees vs. Linked Lists vs. Hash Tables
https://cs.stackexchange.com/questions/270/hash-tables-versus-binary-trees
If you want to access the data in a sorted manner, then a sorted list has to be maintained in parallel to the hash table. A good example is Dictionary in .Net. (see http://msdn.microsoft.com/en-us/library/3fcwy8h6.aspx).
This has the side-effect of not only slowing inserts, but it consumes a larger amount of memory than a b-tree.
Further, since a b-tree is sorted, it is simple to find ranges of results, or to perform unions or merges.
It also depends on the use, Hash allows to locate exact match. If you want to query for a range then BST is the choice. Suppose you have a lots of data e1, e2, e3 ..... en.
With hash table you can locate any element in constant time.
If you want to find range values greater than e41 and less than e8, BST can quickly find that.
The key thing is the hash function used to avoid a collision. Of course, we cannot totally avoid a collision, in which case we resort to chaining or other methods. This makes retrieval no longer constant time in worst cases.
Once full, hash table has to increase its bucket size and copy over all the elements again. This is an additional cost not present over BST.
Binary search trees are good choice to implement dictionary if the keys have some total order (keys are comparable) defined on them and you want to preserve the order information.
As BST preserves the order information, it provides you with four additional dynamic set operations that cannot be performed (efficiently) using hash tables. These operations are:
Maximum
Minimum
Successor
Predecessor
All these operations like every BST operation have time complexity of O(H). Additionally all the stored keys remain sorted in the BST thus enabling you to get the sorted sequence of keys just by traversing the tree in in-order.
In summary if all you want is operations insert, delete and remove then hash table is unbeatable (most of the time) in performance. But if you want any or all the operations listed above you should use a BST, preferably a self-balancing BST.
A hashmap is a set associative array. So, your array of input values gets pooled into buckets. In an open addressing scheme, you have a pointer to a bucket, and each time you add a new value into a bucket, you find out where in the bucket there are free spaces. There are a few ways to do this- you start at the beginning of the bucket and increment the pointer each time and test whether its occupied. This is called linear probing. Then, you can do a binary search like add, where you double the difference between the beginning of the bucket and where you double up or back down each time you are searching for a free space. This is called quadratic probing.
OK. Now the problems in both these methods is that if the bucket overflows into the next buckets address, then you need to-
Double each buckets size- malloc(N buckets)/change the hash function-
Time required: depends on malloc implementation
Transfer/Copy each of the earlier buckets data into the new buckets data. This is an O(N) operation where N represents the whole data
OK. but if you use a linkedlist there shouldn't be such a problem right? Yes, In linked lists you don't have this problem. Considering each bucket to begin with a linked list, and if you have 100 elements in a bucket it requires you to traverse those 100 elements to reach the end of the linkedlist hence the List.add(Element E) will take time to-
Hash the element to a bucket- Normal as in all implementations
Take time to find the last element in said bucket- O(N) operation.
The advantage of the linkedlist implementation is that you don't need the memory allocation operation and O(N) transfer/copy of all buckets as in the case of the open addressing implementation.
So, the way to minimize the O(N) operation is to convert the implementation to that of a Binary Search Tree where find operations are O(log(N)) and you add the element in its position based on it's value. The added feature of a BST is that it comes sorted!
Hash Tables are not good for indexing. When you are searching for a range, BSTs are better. That's the reason why most database indexes use B+ trees instead of Hash Tables
Binary search trees can be faster when used with string keys. Especially when strings are long.
Binary search trees using comparisons for less/greater which are fast for strings (when they are not equal). So a BST can quickly answer when a string is not found.
When it's found it will need to do only one full comparison.
In a hash table. You need to calculate the hash of the string and this means you need to go through all bytes at least once to compute the hash. Then again, when a matching entry is found.

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