I understand the "theory" behind it. Is some type or array of linked list where their position in the array is the result of doing "hashFuction(element) mod array.length" and you used the list to manage collisions.
My question is, what is actually the best length for the array? We are working with graph with max 20,000 nodes. But I think an array of 20,000 elements is already too inefficient.
I was thinking about creating an array with X length and then if it reaches that many elements do something like copy all element to an array of 2X, but the problem is that they would not have the same index for the elements and I can actually "copy" all the array, I would need to do for each element apply the hash function to find their new place and would be very very slow if I am talking about a 10,000 elements array.
Sorry for my grammar mistakes, English is not my native language.
What you've described is essentially a chained hash table, and your question boils down to how to size that table so that it's space efficient. I think you're overengineering your solution. Instead, just use a standard, off-the-shelf implementation of a hash table in your Programming Language of Choice. It's likely to be much more optimized than whatever you'd come up with and would probably have many fewer bugs.
Hope this helps!
Related
I was reading: https://en.wikipedia.org/wiki/Counting_sort and https://www.geeksforgeeks.org/counting-sort/
There is one little detail which I don't get at all, why to complicate things where they can be so much easier? What's the problem of allocating an array of size k where the field of numbers is [1...k] and count how many times each number appeared and lastly walking down the array and printing according to the counter in each cell.
What's the problem of allocating an array of size k where the field of numbers is [1...k] and count how many times each number appeared and lastly walking down the array and printing according to the counter in each cell.
From your phrase "how many times each number appeared", it sounds like you're picturing an array of positive integers, where you want to sort them in increasing order, and where you can use those integers directly as indices in your helper array?
But that's not what the Wikipedia article describes. The algorithm in the Wikipedia article is for an array whose elements can have whatever data-type we choose, provided there's a function key that maps from that data-type to the set of indices in the helper array, with the property that we want to stably sort elements according to the result of key (so, if key(x) < key(y) then we want to sort x before y, and if key(x) = key(y) then we want to keep x and y in the same order they originally had).
In particular, the counting-sort algorithm in the Wikipedia article is useful as a component of radix sort: first you sort by the last digit (using a key function that gives the last digit of a number), then by the second-to-last digit, and so on, until an array of numbers is sorted.
There is one little detail which I don't get at all, why to complicate things where they can be so much easier?
A pro tip: we all usually think that our own code is "easier" and that other people are "complicating things", because code is easier to write than to read, so the code that we understand best is the code that we've come up with ourselves.
As it happens, in this case the Wikipedia code really is more complicated, because it serves a much more general use-case than you were picturing; but in general, it's not a good idea to just assume that everyone will agree that your code is the easy version and that others' is unnecessarily complicated.
I am implementing a table in which each entry consists of two integers. The entries must be ordered in ascending order by key (according to the first integer of each set). All elements will be added to the table as the program is running and must be put in the appropriate slot. Time complexity is of utmost importance and I will only use the insert, remove, and iterate functions.
Which Java data structure is ideal for this implementation?
I was thinking LinkedHashMap, as it maps keys to values (each entry in my table is two values). It also provides O(1) insert/remove functionality. However, it is not sorted. If entries can be efficiently inserted in appropriate order as they come in, this is not a bad idea as the data structure would be sorted. But I have not read or thought of an efficient way to do this. (Maybe like a comparator)?
TreeMap has a time complexity of log(n) for both add and remove. It maintains sorted order and has an iterator. But can we do better than than log(n)?
LinkedList has O(1) add/remove. I could insert with a loop, but this seems inefficient as well.
It seems like TreeMap is the way to go. But I am not sure.
Any thoughts on the ideal data structure for this program are much appreciated. If I have missed an obvious answer, please let me know.
(It can be a data structure with a Set interface, as there will not be duplicates.)
A key-value pair suggests for a Map. As you need key based ordering it narrows down to a SortedMap, in your case a TreeMap. As far as keeping sorting elements in a data structure, it can't get better than O(logn). Look no further.
The basic idea is that you need to insert the key at a proper place. For that your code needs to search for that "proper place". Now, for searching like that, you cannot perform better than a binary search, which is log(n), which is why I don't think you can perform an insert better than log(n).
Hence, again, a TreeMap would be that I would advise you to use.
Moreover, if the hash values, that you state, (specially because there are no duplicates) can be enumerated (as in integer number, serial numbers or so), you could try using statically allocated arrays for doing that. Then you might get a complexity of O(1) perhaps!
I need a data structure that stores tuples and would allow me to do a query like: given tuple (x,y,z) of integers, find the next one (an upped bound for it). By that I mean considering the natural ordering (a,b,c)<=(d,e,f) <=> a<=d and b<=e and c<=f. I have tried MSD radix sort, which splits items into buckets and sorts them (and does this recursively for all positions in the tuples). Does anybody have any other suggestion? Ideally I would like the abouve query to happen within O(log n) where n is the number of tuples.
Two options.
Use binary search on a sorted array. If you build the keys ( assuming 32bit int)' with (a<<64)|(b<<32)|c and hold them in a simple array, packed one beside the other, you can use binary search to locate the value you are searching for ( if using C, there is even a library function to do this), and the next one is simply one position along. Worst case Performance is O(logN), and if you can do http://en.wikipedia.org/wiki/Interpolation_search then you might even approach O(log log N)
Problem with binary keys is might be tricky to add new values, might need gyrations if you will exceed available memory. But it is fast, only a few random memory accesses on average.
Alternatively, you could build a hash table by generating a key with a|b|c in some form, and then have the hash data pointing to a structure that contains the next value, whatever that might be. Possibly a little harder to create in the first place as when generating the table you need to know the next value already.
Problems with hash approach are it will likely use more memory than binary search method, performance is great if you don't get hash collisions, but then starts to drop off, although there a variations around this algorithm to help in some cases. Hash approach is possibly much easier to insert new values.
I also see you had a similar question along these lines, so I guess the guts of what I am saying is combine A,b,c to produce a single long key, and use that with binary search, hash or even b-tree. If the length of the key is your problem (what language), could you treat it as a string?
If this answer is completely off base, let me know and I will see if I can delete this answer, so you questions remains unanswered rather than a useless answer.
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.
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.