How does a hash table work?

Question

I\'m looking for an explanation of how a hash table works - in plain English for a simpleton like me!

For example, I know it takes the key, calculates the hash (I a

Answer 1

You guys are very close to explaining this fully, but missing a couple things. The hashtable is just an array. The array itself will contain something in each slot. At a minimum you will store the hashvalue or the value itself in this slot. In addition to this you could also store a linked/chained list of values that have collided on this slot, or you could use the open addressing method. You can also store a pointer or pointers to other data you want to retrieve out of this slot.

It's important to note that the hashvalue itself generally does not indicate the slot into which to place the value. For example, a hashvalue might be a negative integer value. Obviously a negative number cannot point to an array location. Additionally, hash values will tend to many times be larger numbers than the slots available. Thus another calculation needs to be performed by the hashtable itself to figure out which slot the value should go into. This is done with a modulus math operation like:

uint slotIndex = hashValue % hashTableSize;

This value is the slot the value will go into. In open addressing, if the slot is already filled with another hashvalue and/or other data, the modulus operation will be run once again to find the next slot:

slotIndex = (remainder + 1) % hashTableSize;

I suppose there may be other more advanced methods for determining slot index, but this is the common one I've seen... would be interested in any others that perform better.

With the modulus method, if you have a table of say size 1000, any hashvalue that is between 1 and 1000 will go into the corresponding slot. Any Negative values, and any values greater than 1000 will be potentially colliding slot values. The chances of that happening depend both on your hashing method, as well as how many total items you add to the hash table. Generally, it's best practice to make the size of the hashtable such that the total number of values added to it is only equal to about 70% of its size. If your hash function does a good job of even distribution, you will generally encounter very few to no bucket/slot collisions and it will perform very quickly for both lookup and write operations. If the total number of values to add is not known in advance, make a good guesstimate using whatever means, and then resize your hashtable once the number of elements added to it reaches 70% of capacity.

I hope this has helped.

PS - In C# the GetHashCode() method is pretty slow and results in actual value collisions under a lot of conditions I've tested. For some real fun, build your own hashfunction and try to get it to NEVER collide on the specific data you are hashing, run faster than GetHashCode, and have a fairly even distribution. I've done this using long instead of int size hashcode values and it's worked quite well on up to 32 million entires hashvalues in the hashtable with 0 collisions. Unfortunately I can't share the code as it belongs to my employer... but I can reveal it is possible for certain data domains. When you can achieve this, the hashtable is VERY fast. :)

Answer 2

Short and sweet:

A hash table wraps up an array, lets call it internalArray. Items are inserted into the array in this way:

let insert key value =
    internalArray[hash(key) % internalArray.Length] <- (key, value)
    //oversimplified for educational purposes

Sometimes two keys will hash to the same index in the array, and you want to keep both values. I like to store both values in the same index, which is simple to code by making internalArray an array of linked lists:

let insert key value =
    internalArray[hash(key) % internalArray.Length].AddLast(key, value)

So, if I wanted to retrieve an item out of my hash table, I could write:

let get key =
    let linkedList = internalArray[hash(key) % internalArray.Length]
    for (testKey, value) in linkedList
        if (testKey = key) then return value
    return null

Delete operations are just as simple to write. As you can tell, inserts, lookups, and removal from our array of linked lists is nearly O(1).

When our internalArray gets too full, maybe at around 85% capacity, we can resize the internal array and move all of the items from the old array into the new array.

Answer 3

All of the answers so far are good, and get at different aspects of how a hashtable works. Here is a simple example that might be helpful. Lets say we want to store some items with lower case alphabetic strings as a keys.

As simon explained, the hash function is used to map from a large space to a small space. A simple, naive implementation of a hash function for our example could take the first letter of the string, and map it to an integer, so "alligator" has a hash code of 0, "bee" has a hash code of 1, "zebra" would be 25, etc.

Next we have an array of 26 buckets (could be ArrayLists in Java), and we put the item in the bucket that matches the hash code of our key. If we have more than one item that has a key that begins with the same letter, they will have the same hash code, so would all go in the bucket for that hash code so a linear search would have to be made in the bucket to find a particular item.

In our example, if we just had a few dozen items with keys spanning the alphabet, it would work very well. However, if we had a million items or all the keys all started with 'a' or 'b', then our hash table would not be ideal. To get better performance, we would need a different hash function and/or more buckets.

Answer 4

Usage and Lingo:

1. Hash tables are used to quickly store and retrieve data (or records).
2. Records are stored in buckets using hash keys
3. Hash keys are calculated by applying a hashing algorithm to a chosen value (the key value) contained within the record. This chosen value must be a common value to all the records.
4. Each bucket can have multiple records which are organized in a particular order.

Real World Example:

Hash & Co., founded in 1803 and lacking any computer technology had a total of 300 filing cabinets to keep the detailed information (the records) for their approximately 30,000 clients. Each file folder were clearly identified with its client number, a unique number from 0 to 29,999.

The filing clerks of that time had to quickly fetch and store client records for the working staff. The staff had decided that it would be more efficient to use a hashing methodology to store and retrieve their records.

To file a client record, filing clerks would use the unique client number written on the folder. Using this client number, they would modulate the hash key by 300 in order to identify the filing cabinet it is contained in. When they opened the filing cabinet they would discover that it contained many folders ordered by client number. After identifying the correct location, they would simply slip it in.

To retrieve a client record, filing clerks would be given a client number on a slip of paper. Using this unique client number (the hash key), they would modulate it by 300 in order to determine which filing cabinet had the clients folder. When they opened the filing cabinet they would discover that it contained many folders ordered by client number. Searching through the records they would quickly find the client folder and retrieve it.

In our real-world example, our buckets are filing cabinets and our records are file folders.

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An important thing to remember is that computers (and their algorithms) deal with numbers better than with strings. So accessing a large array using an index is significantly much faster than accessing sequentially.

As Simon has mentioned which I believe to be very important is that the hashing part is to transform a large space (of arbitrary length, usually strings, etc) and mapping it to a small space (of known size, usually numbers) for indexing. This if very important to remember!

So in the example above, the 30,000 possible clients or so are mapped to a smaller space.

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The main idea in this is to divide your entire data set into segments as to speed up the actual searching which is usually time consuming. In our example above, each of the 300 filing cabinet would (statistically) contain about 100 records. Searching (regardless the order) through 100 records is much faster than having to deal with 30,000.

You may have noticed that some actually already do this. But instead of devising a hashing methodology to generate a hash key, they will in most cases simply use the first letter of the last name. So if you have 26 filing cabinets each containing a letter from A to Z, you in theory have just segmented your data and enhanced the filing and retrieval process.

Hope this helps,

Jeach!

Answer 5

Here's an explanation in layman's terms.

Let's assume you want to fill up a library with books and not just stuff them in there, but you want to be able to easily find them again when you need them.

So, you decide that if the person that wants to read a book knows the title of the book and the exact title to boot, then that's all it should take. With the title, the person, with the aid of the librarian, should be able to find the book easily and quickly.

So, how can you do that? Well, obviously you can keep some kind of list of where you put each book, but then you have the same problem as searching the library, you need to search the list. Granted, the list would be smaller and easier to search, but still you don't want to search sequentially from one end of the library (or list) to the other.

You want something that, with the title of the book, can give you the right spot at once, so all you have to do is just stroll over to the right shelf, and pick up the book.

But how can that be done? Well, with a bit of forethought when you fill up the library and a lot of work when you fill up the library.

Instead of just starting to fill up the library from one end to the other, you devise a clever little method. You take the title of the book, run it through a small computer program, which spits out a shelf number and a slot number on that shelf. This is where you place the book.

The beauty of this program is that later on, when a person comes back in to read the book, you feed the title through the program once more, and get back the same shelf number and slot number that you were originally given, and this is where the book is located.

The program, as others have already mentioned, is called a hash algorithm or hash computation and usually works by taking the data fed into it (the title of the book in this case) and calculates a number from it.

For simplicity, let's say that it just converts each letter and symbol into a number and sums them all up. In reality, it's a lot more complicated than that, but let's leave it at that for now.

The beauty of such an algorithm is that if you feed the same input into it again and again, it will keep spitting out the same number each time.

Ok, so that's basically how a hash table works.

Technical stuff follows.

First, there's the size of the number. Usually, the output of such a hash algorithm is inside a range of some large number, typically much larger than the space you have in your table. For instance, let's say that we have room for exactly one million books in the library. The output of the hash calculation could be in the range of 0 to one billion which is a lot higher.

So, what do we do? We use something called modulus calculation, which basically says that if you counted to the number you wanted (i.e. the one billion number) but wanted to stay inside a much smaller range, each time you hit the limit of that smaller range you started back at 0, but you have to keep track of how far in the big sequence you've come.

Say that the output of the hash algorithm is in the range of 0 to 20 and you get the value 17 from a particular title. If the size of the library is only 7 books, you count 1, 2, 3, 4, 5, 6, and when you get to 7, you start back at 0. Since we need to count 17 times, we have 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, and the final number is 3.

Of course modulus calculation isn't done like that, it's done with division and a remainder. The remainder of dividing 17 by 7 is 3 (7 goes 2 times into 17 at 14 and the difference between 17 and 14 is 3).

Thus, you put the book in slot number 3.

This leads to the next problem. Collisions. Since the algorithm has no way to space out the books so that they fill the library exactly (or the hash table if you will), it will invariably end up calculating a number that has been used before. In the library sense, when you get to the shelf and the slot number you wish to put a book in, there's already a book there.

Various collision handling methods exist, including running the data into yet another calculation to get another spot in the table (double hashing), or simply to find a space close to the one you were given (i.e. right next to the previous book assuming the slot was available also known as linear probing). This would mean that you have some digging to do when you try to find the book later, but it's still better than simply starting at one end of the library.

Finally, at some point, you might want to put more books into the library than the library allows. In other words, you need to build a bigger library. Since the exact spot in the library was calculated using the exact and current size of the library, it goes to follow that if you resize the library you might end up having to find new spots for all the books since the calculation done to find their spots has changed.

I hope this explanation was a bit more down to earth than buckets and functions :)

Answer 6

This is how it works in my understanding:

Here's an example: picture the entire table as a series of buckets. Suppose you have an implementation with alpha-numeric hash-codes and have one bucket for each letter of the alphabet. This implementation puts each item whose hash code begins with a particular letter in the corresponding bucket.

Let's say you have 200 objects, but only 15 of them have hash codes that begin with the letter 'B.' The hash table would only need to look up and search through the 15 objects in the 'B' bucket, rather than all 200 objects.

As far as calculating the hash code, there is nothing magical about it. The goal is just to have different objects return different codes and for equal objects to return equal codes. You could write a class that always returns the same integer as a hash-code for all instances, but you would essentially destroy the usefulness of a hash-table, as it would just become one giant bucket.