I have a set of 128bit number and the size of set < 2^32 ...so theoretically I can have a mapping function that maps all the 128bit numbers to 32 bit number ....how can I construct the mapping function ???
Seems like you are looking for a minimal perfect hash which maps n keys to n consecutive integers.
The wiki page link in the above sentence mentions two libraries which implement this.
Also see this for more detail: http://burtleburtle.net/bob/hash/perfect.html
Without knowing the nature of the input data, it's impossible to give the optimal hashing algorithm. But if the input is evenly distributed then you could use the lower 32 bits of the input. This means the possibility of collisions, so you have to deal with that.
The generic construction is to keep all your 128-bit values in a big array, sorted in ascending order. Then, each value is "mapped" to its index in the array. To "compute" the map, you do a binary search in the array, to get the precise index of the value in the array. With 232 values, the array has size 64 GB, and the binary search entails 35-or-so lookups in the array.
In all generality you cannot do really better than that. However, if your 128-bit values have a reasonably uniform spread (it depends from where they come), then the big array structure can be compressed by a large margin, especially if you can guarantee that all inputs to your map will always be part of the set of 128-bit values; my bet is that you can trim it down to a couple of gigabytes -- but the lookup will be more expensive.
For a more practical solution, you will have to work with the structure of your 128-bit values: where they come from, what they represent...
Set a position of your number as division of it's value on 2^32.
Related
I have a bunch of objects with unique 8 digit hexadecimal identifiers ex[fd4786ac] that I need to construct and look up quickly. Deletion is not a priority. These hexadecimal values are currently being stored as strings.
Considered a trie(or some variation of a trie), a skip list, and a some variation of a hash table. Using a skip list over a AVL tree would be preferable since it is likely these strings will be sequential but not guaranteed and tree re-balancing would be often. How ever I'm open to other data structures if they better suit my need.
A good choice would be to convert your keys into 32-bit integers, and then use a hash table.
If you want to write your own just for this use case, then:
Instead of hashing keys all the time or storing hash values, use a bijective hash function and use the hashes instead of the keys.
Since your keys are very small you should probably use open addressing -- it will save space and it's a little faster. Wikipedia will give you lots of choices for probing schemes. I currently like robin hood hashing: https://www.sebastiansylvan.com/post/robin-hood-hashing-should-be-your-default-hash-table-implementation/
Your 8 digit hexadecimal identifiers represent a 4byte (32bit) integer so you can use that as an index for a (quite large) array with 2^32 entries.
If the array contains pointers this would cost 64GB.
Most likely too much to keep it in RAM.
So if the number of elements is orders of magnitudes below 2^32, use a Hash-Map or a sorted ist (access O(logn) ).
In cases where I have a key for each element and I don't know the index of the element into an array, hashtables perform better than arrays (O(1) vs O(n)).
Why is that? I mean: I have a key, I hash it.. I have the hash.. shouldn't the algorithm compare this hash against every element's hash? I think there's some trick behind the memory disposition, isn't it?
In cases where I have a key for each element and I don't know the
index of the element into an array, hashtables perform better than
arrays (O(1) vs O(n)).
The hash table search performs O(1) in the average case. In the worst case, the hash table search performs O(n): when you have collisions and the hash function always returns the same slot. One may think "this is a remote situation," but a good analysis should consider it. In this case you should iterate through all the elements like in an array or linked lists (O(n)).
Why is that? I mean: I have a key, I hash it.. I have the hash..
shouldn't the algorithm compare this hash against every element's
hash? I think there's some trick behind the memory disposition, isn't
it?
You have a key, You hash it.. you have the hash: the index of the hash table where the element is present (if it has been located before). At this point you can access the hash table record in O(1). If the load factor is small, it's unlikely to see more than one element there. So, the first element you see should be the element you are looking for. Otherwise, if you have more than one element you must compare the elements you will find in the position with the element you are looking for. In this case you have O(1) + O(number_of_elements).
In the average case, the hash table search complexity is O(1) + O(load_factor) = O(1 + load_factor).
Remember, load_factor = n in the worst case. So, the search complexity is O(n) in the worst case.
I don't know what you mean with "trick behind the memory disposition". Under some points of view, the hash table (with its structure and collisions resolution by chaining) can be considered a "smart trick".
Of course, the hash table analysis results can be proven by math.
With arrays: if you know the value, you have to search on average half the values (unless sorted) to find its location.
With hashes: the location is generated based on the value. So, given that value again, you can calculate the same hash you calculated when inserting. Sometimes, more than 1 value results in the same hash, so in practice each "location" is itself an array (or linked list) of all the values that hash to that location. In this case, only this much smaller (unless it's a bad hash) array needs to be searched.
Hash tables are a bit more complex. They put elements in different buckets based on their hash % some value. In an ideal situation, each bucket holds very few items and there aren't many empty buckets.
Once you know the key, you compute the hash. Based on the hash, you know which bucket to look for. And as stated above, the number of items in each bucket should be relatively small.
Hash tables are doing a lot of magic internally to make sure buckets are as small as possible while not consuming too much memory for empty buckets. Also, much depends on the quality of the key -> hash function.
Wikipedia provides very comprehensive description of hash table.
A Hash Table will not have to compare every element in the Hash. It will calculate the hashcode according to the key. For example, if the key is 4, then hashcode may be - 4*x*y. Now the pointer knows exactly which element to pick.
Whereas if it has been an array, it will have to traverse through the whole array to search for this element.
Why is [it] that [hashtables perform lookups by key better than arrays (O(1) vs O(n))]? I mean: I have a key, I hash it.. I have the hash.. shouldn't the algorithm compare this hash against every element's hash? I think there's some trick behind the memory disposition, isn't it?
Once you have the hash, it lets you calculate an "ideal" or expected location in the array of buckets: commonly:
ideal bucket = hash % num_buckets
The problem is then that another value may have already hashed to that bucket, in which case the hash table implementation has two main choice:
1) try another bucket
2) let several distinct values "belong" to one bucket, perhaps by making the bucket hold a pointer into a linked list of values
For implementation 1, known as open addressing or closed hashing, you jump around other buckets: if you find your value, great; if you find a never-used bucket, then you can store your value in there if inserting, or you know you'll never find your value when searching. There's a potential for the searching to be even worse than O(n) if the way you traverse alternative buckets ends up searching the same bucket multiple times; for example, if you use quadratic probing you try the ideal bucket index +1, then +4, then +9, then +16 and so on - but you must avoid out-of-bounds bucket access using e.g. % num_buckets, so if there are say 12 buckets then ideal+4 and ideal+16 search the same bucket. It can be expensive to track which buckets have been searched, so it can be hard to know when to give up too: the implementation can be optimistic and assume it will always find either the value or an unused bucket (risking spinning forever), it can have a counter and after a threshold of tries either give up or start a linear bucket-by-bucket search.
For implementation 2, known as closed addressing or separate chaining, you have to search inside the container/data-structure of values that all hashed to the ideal bucket. How efficient this is depends on the type of container used. It's generally expected that the number of elements colliding at one bucket will be small, which is true of a good hash function with non-adversarial inputs, and typically true enough of even a mediocre hash function especially with a prime number of buckets. So, a linked list or contiguous array is often used, despite the O(n) search properties: linked lists are simple to implement and operate on, and arrays pack the data together for better memory cache locality and access speed. The worst possible case though is that every value in your table hashed to the same bucket, and the container at that bucket now holds all the values: your entire hash table is then only as efficient as the bucket's container. Some Java hash table implementations have started using binary trees if the number of elements hashing to the same buckets passes a threshold, to make sure complexity is never worse than O(log2n).
Python hashes are an example of 1 = open addressing = closed hashing. C++ std::unordered_set is an example of closed addressing = separate chaining.
The purpose of hashing is to produce an index into the underlying array, which enables you to jump straight to the element in question. This is usually accomplished by dividing the hash by the size of the array and taking the remainder index = hash%capacity.
The type/size of the hash is typically that of the smallest integer large enough to index all of RAM. On a 32 bit system this is a 32 bit integer. On a 64 bit system this is a 64 bit integer. In C++ this corresponds to unsigned int and unsigned long long respectively. To be pedantic C++ technically specifies minimum sizes for its primitives i.e. at least 32 bits and at least 64 bits, but that's beside the point. For the sake of making code portable C++ also provides a size_t primative which corresponds to the appropriate unsigned integer. You'll see that type a lot in for loops which index into arrays, in well written code. In the case of a language like Python the integer primitive grows to whatever size it needs to be. This is typically implemented in the standard libraries of other languages under the name "Big Integer". To deal with this the Python programming language simply truncates whatever value you return from the __hash__() method down to the appropriate size.
On this score I think it's worth giving a word to the wise. The result of arithmetic is the same regardless of whether you compute the remainder at the end or at each step along the way. Truncation is equivalent to computing the remainder modulo 2^n where n is the number of bits you leave intact. Now you might think that computing the remainder at each step would be foolish due to the fact that you're incurring an extra computation at every step along the way. However this is not the case for two reasons. First, computationally speaking, truncation is extraordinarily cheap, far cheaper than generalized division. Second, and this is the real reason as the first is insufficient, and the claim would generally hold even in its absence, taking the remainder at each step keeps the number (relatively) small. So instead of something like product = 31*product + hash(array[index]), you'll want something like product = hash(31*product + hash(array[index])). The primary purpose of the inner hash() call is to take something which might not be a number and turn it into one, where as the primary purpose of the outer hash() call is to take a potentially oversized number and truncate it. Lastly I'll note that in languages like C++ where integer primitives have a fixed size this truncation step is automatically performed after every operation.
Now for the elephant in the room. You've probably realized that hash codes being generally speaking smaller than the objects they correspond to, not to mention that the indices derived from them are again generally speaking even smaller still, it's entirely possible for two objects to hash to the same index. This is called a hash collision. Data structures backed by a hash table like Python's set or dict or C++'s std::unordered_set or std::unordered_map primarily handle this in one of two ways. The first is called separate chaining, and the second is called open addressing. In separate chaining the array functioning as the hash table is itself an array of lists (or in some cases where the developer feels like getting fancy, some other data structure like a binary search tree), and every time an element hashes to a given index it gets added to the corresponding list. In open addressing if an element hashes to an index which is already occupied the data structure probes over to the next index (or in some cases where the developer feels like getting fancy, an index defined by some other function as is the case in quadratic probing) and so on until it finds an empty slot, of course wrapping around when it reaches the end of the array.
Next a word about load factor. There is of course an inherent space/time trade off when it comes to increasing or decreasing the load factor. The higher the load factor the less wasted space the table consumes; however this comes at the expense of increasing the likelihood of performance degrading collisions. Generally speaking hash tables implemented with separate chaining are less sensitive to load factor than those implemented with open addressing. This is due to the phenomenon known as clustering where by clusters in an open addressed hash table tend to become larger and larger in a positive feed back loop as a result of the fact that the larger they become the more likely they are to contain the preferred index of a newly added element. This is actually the reason why the afore mentioned quadratic probing scheme, which progressively increases the jump distance, is often preferred. In the extreme case of load factors greater than 1, open addressing can't work at all as the number of elements exceeds the available space. That being said load factors greater than 1 are exceedingly rare in general. At time of writing Python's set and dict classes employ a max load factor of 2/3 where as Java's java.util.HashSet and java.util.HashMap use 3/4 with C++'s std::unordered_set and std::unordered_map taking the cake with a max load factor of 1. Unsurprisingly Python's hash table backed data structures handle collisions with open addressing where as their Java and C++ counterparts do it with separate chaining.
Last a comment about table size. When the max load factor is exceeded, the size of the hash table must of course be grown. Due to the fact that this requires that every element there in be reindexed, it's highly inefficient to grow the table by a fixed amount. To do so would incur order size operations every time a new element is added. The standard fix for this problem is the same as that employed by most dynamic array implementations. At every point where we need to grow the table we simply increase its size by its current size. This unsurprisingly is known as table doubling.
I think you answered your own question there. "shouldn't the algorithm compare this hash against every element's hash". That's kind of what it does when it doesn't know the index location of what you're searching for. It compares each element to find the one you're looking for:
E.g. Let's say you're looking for an item called "Car" inside an array of strings. You need to go through every item and check item.Hash() == "Car".Hash() to find out that that is the item you're looking for. Obviously it doesn't use the hash when searching always, but the example stands. Then you have a hash table. What a hash table does is it creates a sparse array, or sometimes array of buckets as the guy above mentioned. Then it uses the "Car".Hash() to deduce where in the sparse array your "Car" item is actually. This means that it doesn't have to search through the entire array to find your item.
Given an ordered sequence of around a few thousand 32 bit integers, I would like to know how measures of their disorder or entropy are calculated.
What I would like is to be able to calculate a single value of the entropy for each of two such sequences and be able to compare their entropy values to determine which is more (dis)ordered.
I am asking here, as I think I may not be the first with this problem and would like to know of prior work.
Thanks in advance.
UPDATE #1
I have just found this answer that looks great, but would give the same entropy if the integers were sorted. It only gives a measure of the entropy of the individual ints in the list and disregards their (dis)order.
Entropy calculation generally:
http://en.wikipedia.org/wiki/Entropy_%28information_theory%29
Furthermore, you have to sort your integers, then iterate over the sorted integer list to find out the frequency of your integers. Afterwards, you can use the formula.
I think I'll have to code a shannon entropy in 2D. Arrange the list of 32 bit ints as a series of 8 bit bytes and do a Shannons on that, then to cover how ordered they may be, take the bytes eight at a time and form a new list of bytes composed of bits 0 of the eight followed by bits 1 of the eight ... bits 7 of the 8; then the next 8 original bytes ..., ...
I'll see how it goes/codes...
Entropy is a function on probabilities, not data (arrays of ints, or files). Entropy is a measure of disorder, but when the function is modified to take data as input it loses this meaning.
The only true way one can generate a measure of disorder for data is to use Kolmogorov Complexity. Though this has problems too, in particular it's uncomputable and is not yet strictly well defined as one must arbitrarily pick a base language. This well-definedness can be solved if the disorder one is measuring is relative to something that is going to process the data. So when considering compression on a particular computer, the base language would be Assembly for that computer.
So you could define the disorder of an array of integers as follows:
The length of the shortest program written in Assembly that outputs the array.
This problem is a little similar to that solved by reservoir sampling, but not the same. I think its also a rather interesting problem.
I have a large dataset (typically hundreds of millions of elements), and I want to estimate the number of unique elements in this dataset. There may be anywhere from a few, to millions of unique elements in a typical dataset.
Of course the obvious solution is to maintain a running hashset of the elements you encounter, and count them at the end, this would yield an exact result, but would require me to carry a potentially large amount of state with me as I scan through the dataset (ie. all unique elements encountered so far).
Unfortunately in my situation this would require more RAM than is available to me (nothing that the dataset may be far larger than available RAM).
I'm wondering if there would be a statistical approach to this that would allow me to do a single pass through the dataset and come up with an estimated unique element count at the end, while maintaining a relatively small amount of state while I scan the dataset.
The input to the algorithm would be the dataset (an Iterator in Java parlance), and it would return an estimated unique object count (probably a floating point number). It is assumed that these objects can be hashed (ie. you can put them in a HashSet if you want to). Typically they will be strings, or numbers.
You could use a Bloom Filter for a reasonable lower bound. You just do a pass over the data, counting and inserting items which were definitely not already in the set.
This problem is well-addressed in the literature; a good review of various approaches is http://www.edbt.org/Proceedings/2008-Nantes/papers/p618-Metwally.pdf. The simplest approach (and most compact for very high accuracy requirements) is called Linear Counting. You hash elements to positions in a bitvector just like you would a Bloom filter (except only one hash function is required), but at the end you estimate the number of distinct elements by the formula D = -total_bits * ln(unset_bits/total_bits). Details are in the paper.
If you have a hash function that you trust, then you could maintain a hashset just like you would for the exact solution, but throw out any item whose hash value is outside of some small range. E.g., use a 32-bit hash, but only keep items where the first two bits of the hash are 0. Then multiply by the appropriate factor at the end to approximate the total number of unique elements.
Nobody has mentioned approximate algorithm designed specifically for this problem, Hyperloglog.
I was reading about this person's interview "at a well-known search company".
http://asserttrue.blogspot.com/2009/05/one-of-toughest-job-interview-questions.html
He was asked a question which led to him implementing a hash table. He said the following:
HASH = INITIAL_VALUE;
FOR EACH ( CHAR IN WORD ) {
HASH *= MAGIC_NUMBER
HASH ^= CHAR
HASH %= BOUNDS
}
RETURN HASH
I explained that the hash table array
length should be prime, and the BOUNDS
number is less than the table length,
but coprime to the table length.
Why should the BOUNDS number be less than the number of buckets? What does being coprime to the table length do? Isn't it supposed to be coprime to the BOUNDS?
I would hazard that he is completely wrong. BOUNDS should be the number of buckets or the last few buckets are going to be underused.
Further, the bounding of the output to the number of buckets should be OUTSIDE the hash function. This is an implementation detail of that particular hash table. You might have a very large table using lots of buckets and another using few. Both should share the same string->hash function
Further, if you read the page that you linked to it is quite interesting. I would have implemented his hash table as something like 10,000 buckets - For those who haven't read it, the article suggests ~ 4,000,000,000 buckets to store 1,000,000 or so possible words. For collisions, each bucket has a vector of word structures, each of those containing a count, a plaintext string and a hash (unique within the bucket). This would use far less memory and work better with modern caches since your working set would be much smaller.
To further reduce memory usage you could experiment with culling words from the hash during the input phase that look like they are below the top 100,000 based on the current count.
I once interviewed for a job at a well-known search company. I got the exact same question. I tried to tackle it by using hash table.
One thing that I learnt from that interview was that at a well-known search company, you do not propose hashes as solutions. You use any tree-like structure you like but you always use ordered structure, not hash table.
A simple explicit suffix tree would only use worst case maybe 500k memory (with a moderately efficient implementation, 4 byte character encodings, and relatively long English words that have minimal overlap) to do the same thing.
I think the guy in the article outsmarted himself.