When we are insert/lookup an key in a hash table, textbook said it is O(1) time. Yet, how is possible to have an O(1) lookup time? If the hash table store the key in a vector, it will cost O(N), if in a binary tree, it will be O(logN). I just can't image some data structure with O(1) accessing time.
Thanks!
The hashtable hashes your key and put it in array.
For example, hash(x) = 3, where x is your key. The table then puts it into array[3]. Accessing from array is O(1).
At a minimum, hash tables consist of an array and hash function. When an object is added to the table, the hash function is computed on the object and it is stored in the array at the index of the corresponding value that was computed. e.g., if hash(obj) = 2 then arr[2] = obj.
The average insert/lookup on a hash table is O(1).
However it is possible to have collisions when objects compute the same hash value.
In the general case there are "buckets" at each index of the array to handle these collisions. Meaning, all three objects are stored in some other data structure (maybe a linked list or another array) at the index of the hash table.
Therefore, the worst case for lookup on a hash table is O(n) because it is possible that all objects stored in the hash table have collided and are stored in the same bucket.
Technically speaking, hash table lookup, if there is no collision, is O(logn). This is because hashing time is linear with respect to the size (in bytes) of the identifier, and the smallest that a new identifier added to the hash table can be, for that identifier to be unique, is O(logn).
However, the log of all the computer memory in the world is just such a small number, which means that we have very good upper bounds on hash table identifier size. Case in point, log10 of the number of particles in the observable universe is estimated at slightly over 80; in log2 it's about 3.3 times as much. Logarithms grow very slowly.
As a result, most log terms can be treated as constant terms. It's just that traditionally we only apply this fact to hash tables, but not to search trees, in order to teach recurrence relations to students.
Related
I'm scanning a huge table (> 1B docs) so I'm using a parallel scan (using one segment per worker).
The table has a hash key and a sort key.
Intuitively a segment should contain a set of hash keys (including all their sort keys), so one hash key shouldn't appear in more than one segment, but I haven't found any documentation indicating this.
Does anyone know how does DynamoDB behave in this scenario?
Thanks
This is an interesting question. I thought it would be easy to find a document stating that each segment contains a disjoint range of hash keys, and the same hash key cannot appear in more than one segment - but I too failed to find any such document. I am curious if anyone else can find such a document. In the meantime, I can try to offer additional intuitions on why your conjecture is likely correct - but also might be wrong:
My first intuition would be that you are right:
DynamoDB uses the hash key, also known as a partition key to decide on which of the many storage nodes to store copy of this data. All of the items sharing the same partition key (with different sort key values) are stored together, in sort-key order, so they can be Queryed together in order. DynamoDB uses a hash function on the partition key to decide the placement of each item (hence the name "hash key").
Now, if DynamoDB needs to divide the task of scanning all the data into "segments", the most sensible thing for it to do is to divide the space of hash values (i.e., hash function of the hash keys) to different equal-sized pieces. This division is easy to do (just a numeric division by TotalSegments), it ensures roughly the same amount of items in each segment (assuming there are many different partitions), and it ensures that the scanning of each segment involves a different storage node, so the parallel scan can proceed faster than what a single storage node is capable of.
However, there is one indication that this might not be the entire story.
The DynamoDB documentation claims that
In general, there is no practical limit on the number of distinct sort key values per partition key value.
This means that in theory at least, your entire database, perhaps one petabyte of it, may be in a single partition with billions of different sort keys. Since Amazon's single storage node do have a size limit, it means DynamoDB must (unless the above statement is false) support splitting of a single huge partition into multiple storage nodes. This means that when GetItem is looking for a particular item, DynamoDB needs to know which sort key is on which storage node. It also means that a parallel scan might - possibly - divide this huge partition into pieces, all scanning the same partition but different sort-key ranges in it. I am not sure we can completely rule out this possibility. I am guessing it will never happen when you only have smallish partitions.
Every DynamoDB table has a "hashspace" and data is partitioned as per the hash value of the partition key. When a ParallelScan is intended and the TotalSegments and Segment values are provided, the table's complete hashspace is logically divided into these "Segments" such that TotalSegments cover the complete hash space, without overlapping. It is quite possible some segments here do not actually have any data corresponding to them, since there may not be any data in the hashspace allocated to the segment. This can be observed if the TotalSegments value chosen is very high for instance.
And for each Segment value passed in the Scan request (with TotalSegments value being constant), each Segment would return distinct items without any overlap.
FAQs
Q. Ideal Number for TotalSegments ?
-> You might need to experiment with values, find the sweet spot for your table, and the number of workers you use, until your application achieves its best performance.
Q. One or more segments do not return any records. Why?
-> This is possible if the hash range that is allocated as per the TotalSegments value does not have any items. In this case, the TotalSegments value can be decreased, for better performance.
Q. Scan for a segment failed midway. Can a Scan for that segment alone be retried now ?
-> As long as the TotalSegments value remains the same, a Scan for one of the segments can be re-run, since it would have the same hash range allocated at any given time.
Q. Can I perform a Scan for a single segment, without performing the Scan for other segments as per TotalSegments value?
-> Yes. Multiple Scan operations for different Segments are not linked/do not depend on previous/other Segment Scans.
My application requires a key value store. Following are some of the details regarding key values:
1) Number of keys (data type: string) can either be 256, 1024 or 4096.
2) Data type of values against each key is a list of integers.
3) The list of integers (value) against each key can vary in size
4) The largest size of the value can be around 10,000,000 integers
5) Some keys might contain very small list of integers
The application needs fast access to the list of integers against a specified key . However, this step is not frequent in the working of the application.
I need suggestions for best Key value stores for my case. I need fast retrieval of values against key and value size can be around 512 MB or more.
I checked Redis but it requires the store to be stored in memory. However, in the given scenario I think I should look for disk based key value stores.
LevelDB can fit your use case very well, as you have limited number of keys (given you have enough disk space for your requirements), and might not need a distributed solution.
One thing you need to specify is if (and how) you wish to modify the lists once in the db, as levelDB and many other general key-val stores do not have such atomic transactions.
If you are looking for a distributed db, cassandra is good, as it will also let you insert/remove individual list elements.
Other than collision detection and throwing a LinkedList in a hashtable, what are some other ways that a Hash Table can be implemented? Is collision detection the only way to achieve an efficient hash table?
Ultimately a finite sized hash table is going to have collisions, at least any generally programmed one. If your key is type string then the hash table has an infinite number of possible keys, but with a hash table, you have just a finite number of buckets. So fundamentally there has to be collisions. If you were to implement a hash table where it ignores collisions, then you would have a very strange, indeterministic data structure that would appear to remove elements at random.
Now, the data structure used on the backend doesn't have to be a linked list. You could implement it as a red-black tree and get log(n) performance out of a collision. You should checkout the article 5 Myths About Hash Tables and also this Stack Overflow question about HashMaps vs Maps.
Now, if you know something about you key type, say the key is a 2 character long string, then there are only a finite number of possible keys, you can then proceed to create a "hash" function that converts the key to a relatively small integer, you could create a look-up table that is guaranteed to not have collisions.
It is important to note that a well-implemented hash table will not suffer very much from collisions. There are bigger problems in the world like world hunger (or even how to implement an efficient hash function) than the computer having to traverse three nodes in a linked list once every 5 days.
Other than collision detection and throwing a LinkedList in a hashtable, what are some other ways that a Hash Table can be implemented?
Other ways include:
having another container type linked from the nodes where elements have collided, such as a balanced binary tree or vector/array
GCC's hash table underpinning std::unordered_X uses a single singly-linked list of values, and a contiguous array of buckets container iterators into the list; that's got some great characteristics including optimal iteration speed regardless of the current load_factor()
using open addressing / closed hashing, which - when an insert/find/erase finds another key in the bucket it has hashed to, uses some algorithm to find another bucket to look in instead (and so on until it finds the key, a deleted element it can insert over, or an unused bucket); there are a number of options for this kind of "probing", the simplest being a try-the-next-bucket approach, another being quadratic 1, 4, 9, 16..., another the use of alternative hash functions.
perfect hash functions (below)
Is collision detection the only way to achieve an efficient hash table?
sometimes it's possible to find a perfect hash function that won't have collisions, but that's generally only true for very limited input sets, whether due to the nature of the inputs (e.g. month and year of birth of living people only has order-of a thousand possible values), or because a small number are known at compile time (e.g. a set of 200 keywords for a compiler).
I would like to have a little clarification on the definiton of a bucket in SAS hashtable. The question is exactly about the hashexp parameter.
According to the SAS DOCs, hashexp is:
The hash object's internal table size, where the size of the hash table is 2n.
The value of HASHEXP is used as a power-of-two exponent to create the hash table size. For example, a value of 4 for HASHEXP equates to a hash table size of 24, or 16. The maximum value for HASHEXP is 20.
The hash table size is not equal to the number of items that can be stored. Imagine the hash table as an array of 'buckets.' A hash table size of 16 would have 16 'buckets.' Each bucket can hold an infinite number of items. The efficiency of the hash table lies in the ability of the hashing function to map items to and retrieve items from the buckets.
You should set the hash table size relative to the amount of data in the hash object in order to maximize the efficiency of the hash object lookup routines. Try different HASHEXP values until you get the best result. For example, if the hash object contains one million items, a hash table size of 16 (HASHEXP = 4) would work, but not very efficiently. A hash table size of 512 or 1024 (HASHEXP = 9 or 10) would result in the best performance.
The question is what exactly is a hash table size, while it is not a amount of data in the hash object?
Should it be understood as if we wanted to allocate as much memory as it may be neccessary but not less, no more. It is a power of two to get things work fast. But it does not limit the amount of data possibly used, it only indicates about how much is going to be used, right?
Paul Dorfman (the master of hashing) goes into a fair bit of detail on page 10 of this whitepaper:
http://www2.sas.com/proceedings/forum2008/037-2008.pdf
As I understand it, hashtables store their data in binary trees. Each bucket created by hashexp represents the number of binary trees that will be used to store the data. A hashexp of 0 would use a single tree, while a hashexp of 8 would use 256 trees. When a lookup is performed against the hash object, an internal algorithm determines which tree the key should exist in (based on the hashed value). It then checks that tree for the value. By automatically knowing which of the 256 trees to look in (for example) it would have saved itself 8 comparisons (2^8) when compared to a single binary tree.
The whole thing seems a lot more complex than that but that's my interpretation of why it works out faster.
As Rob Penridge pointed out, Paul Dorfman is indeed the SAS Hash Object Guru. Hashexp is not related to the size of the hash table, again as mentioned in Rob's answer.
If you have a table with 100obs and 10 numeric variables which is loaded into a hash table, then size of the hash table is simply 100obs*10vars*8bytes(assuming all numeric vars are stored as 8byte fields) 7.8KB give or take a 10%.
Remember that SAS dynamically allocates RAM space as records are added to the Hash table in memory, so you do not need to specify in advance what size it should be.[I've been using hash tables regularly, but cant think of any place where one can specify the size in advance].
General tip: if you want to know how big your hash table is going to be, run a PROC CONTENTS on the dataset you want to load into Hash table and multiply "Observation Length" & "No. of obs in dataset", this will give the memory size needed in bytes. If you have that much memory then you can load it into memory.
I don't have experience with hash tables outside of arrays/dictionaries in dynamic languages, so I recently found out that internally they're implemented by making a hash of the key and using that to store the value. What I don't understand is why aren't the values stored with the key (string, number, whatever) as the, well, key, instead of making a hash of it and storing that.
This is a near duplicate: Why do we use a hashcode in a hashtable instead of an index?
Long story short, you can check if a key is already stored VERY quickly, and equally rapidly store a new mapping. Otherwise you'd have to keep a sorted list of keys, which is much slower to store and retrieve mappings from.
what is hash table?
It is also known as hash map is a data structure used to implement an associative array.It is a structure that can map keys to values.
How it works?
A hash table uses a hash function to compute an index into an array of buckets or slots, from which the correct value can be found.
See the below diagram it clearly explains.
Advantages:
In a well-dimensioned hash table, the average cost for each lookup is independent of the number of elements stored in the table.
Many hash table designs also allow arbitrary insertions and deletions of key-value pairs.
In many situations, hash tables turn out to be more efficient than search trees or any other table lookup structure.
Disadvantages:
The hash tables are not effective when the number of entries is very small. (However, in some cases the high cost of computing the hash function can be mitigated by saving the hash value together with the key.)
Uses:
They are widely used in many kinds of computer software, particularly for associative arrays, database indexing, caches and sets.
What I don't understand is why aren't the values stored with the key (string, number, whatever) as the, well, key
And how do you implement that?
Computers know only numbers. A hash table is a table, i.e. an array and when we get right down to it, an array can only addressed via an integral nonnegative index. Everything else is trickery. Dynamic languages that let you use string keys – they use trickery.
And one such trickery, and often the most elegant, is just computing a numerical, reproducible “hash” number of the key and using that as the index.
(There are other considerations such as compaction of the key range but that’s the foremost issue.)
In a nutshell: Hashing allows O(1) queries/inserts/deletes to the table. OTOH, a sorted structure (usually implemented as a balanced BST) makes the same operations take O(logn) time.
Why take a hash, you ask? How do you propose to store the key "as the key"? Ask yourself this, if you plan to store simply (key,value) pairs, how fast will your lookups/insertions/deletions be? Will you be running a O(n) loop over the entire array/list?
The whole point of having a hash value is that it allows all keys to be transformed into a finite set of hash values. This allows us to store keys in slots of a finite array (enabling fast operations - instead of searching the whole list you only search those keys that have the same hash value) even though the set of possible keys may be extremely large or infinite (e.g. keys can be strings, very large numbers, etc.) With a good hash function, very few keys will ever have the same hash values, and all operations are effectively O(1).
This will probably not make much sense if you are not familiar with hashing and how hashtables work. The best thing to do in that case is to consult the relevant chapter of a good algorithms/data structures book (I recommend CLRS).
The idea of a hash table is to provide a direct access to its items. So that is why the it calculates the "hash code" of the key and uses it to store the item, insted of the key itself.
The idea is to have only one hash code per key. Many times the hash function that generates the hash code is to divide a prime number and uses its remainer as the hash code.
For example, suppose you have a table with 13 positions, and an integer as the key, so you can use the following hash function
f(x) = x % 13
What I don't understand is why aren't
the values stored with the key
(string, number, whatever) as the,
well, key, instead of making a hash of
it and storing that.
Well, how do you propose to do that, with O(1) lookup?
The point of hashtables is basically to provide O(1) lookup by turning the key into an array index and then returning the content of the array at that index. To make that possible for arbitrary keys you need
A way to turn the key into an array index (this is the hash's purpose)
A way to deal with collisions (keys that have the same hash code)
A way to adjust the array size when it's too small (causing too many collisions) or too big (wasting space)
Generally the point of a hash table is to store some sparse value -- i.e. there is a large space of keys and a small number of things to store. Think about strings. There are an uncountable number of possible strings. If you are storing the variable names used in a program then there is a relatively small number of those possible strings that you are actually using, even though you don't know in advance what they are.
In some cases, it's possible that the key is very long or large, making it impractical to keep copies of these keys. Hashing them first allows for less memory usage as well as quicker lookup times.
A hashtable is used to store a set of values and their keys in a (for some amount of time) constant number of spots. In a simple case, let's say you wanted to save every integer from 0 to 10000 using the hash function of i % 10.
This would make a hashtable of 1000 blocks (often an array), each having a list 10 elements deep. So if you were to search for 1234, it would immediately know to search in the table entry for 123, then start comparing to find the exact match. Granted, this isn't much better than just using an array of 10000 elements, but it's just to demonstrate.
Hashtables are very useful for when you don't know exactly how many elements you'll have, but there will be a good number fewer collisions on the hash function than your total number of elements. (Which makes the hash function "hash(x) = 0" very, very bad.) You may have empty spots in your table, but ideally a majority of them will have some data.
The main advantage of using a hash for the purpose of finding items in the table, as opposed to using the original key of the key-value pair (which BTW, it typically stored in the table as well, since the hash is not reversible), is that..
...it allows mapping the whole namespace of the [original] keys to the relatively small namespace of the hash values, allowing the hash-table to provide O(1) performance for retrieving items.
This O(1) performance gets a bit eroded when considering the extra time to dealing with collisions and such, but on the whole the hash table is very fast for storing and retrieving items, as opposed to a system based solely on the [original] key value, which would then typically be O(log N), with for example a binary tree (although such tree is more efficient, space-wise)
Also consider speed. If your key is a string and your values are stored in an array, your hash can access any element in 'near' constant time. Compare that to searching for the string and its value.