This is a homework question, but I think there's something missing from it. It asks:
Provide a sequence of m keys to fill a hash table implemented with linear probing, such that the time to fill it is minimum.
And then
Provide another sequence of m keys, but such that the time fill it is maximum. Repeat these two questions if the hash table implements quadratic probing
I can only assume that the hash table has size m, both because it's the only number given and because we have been using that letter to address a hash table size before when describing the load factor. But I can't think of any sequence to do the first without knowing the hash function that hashes the sequence into the table.
If it is a bad hash function, such that, for instance, it hashes every entry to the same index, then both the minimum and maximum time to fill it will take O(n) time, regardless of what the sequence looks like. And in the average case, where I assume the hash function is OK, how am I supposed to know how long it will take for that hash function to fill the table?
Aren't these questions linked to the hash function stronger than they are to the sequence that is hashed?
As for the second question, I can assume that, regardless of the hash function, a sequence of size m with the same key repeated m-times will provide the maximum time, because it will cause linear probing from the second entry on. I think that will take O(n) time. Is that correct?
Well, the idea behind these questions is to test your understanding of probing styles. For linear probing, if a collision occurs, you simply test the next cell. And it goes on like this until you find an available cell to store your data.
Your hash table doesn't need to be size m but it needs to be at least size m.
First question is asking that if you have a perfect hash function, what is the complexity of populating the table. Perfect hashing function addresses each element without collision. So for each element in m, you need O(1) time. Total complexity is O(m).
Second question is asking for the case that hash(X)=cell(0), which all of the elements will search till the first empty cell(just rear of the currently populated table).
For the first element, you probe once -> O(1)
For the second element, you probe twice -> O(2)
for the nth element, you probe n times -> O(n)
overall you have m elements, so -> O(n*(n+1)/2)
For quadratic probing, you have the same strategy. The minimum case is the same, but the maximum case will have O(nlogn). ( I didn't solve it, just it's my educated guess.)
This questions doesn't sound terribly concerned with the hash function, but it would be nice to have. You seem to pretty much get it, though. It sounds to me like the question is more concerned with "do you know what a worst-case list of keys would be?" than "do you know how to exploit bad hash functions?"
Obviously, if you come up with a sequence where all the entries hash to different locations, then you have O(1) insertions for O(m) time in total.
For what you are saying about hashing all the keys to the same location, each insertion should take O(n) if that's what you are suggesting. However, that's not the total time for inserting all the elements. Also, you might want to consider not literally using the same key over and over but rather using keys that would produce the same location in the table. I think, by convention, inserting the same key should cause a replacement, though I'm not 100% sure.
I'll apologize in advance if I gave too much information or left anything unclear. This question seems pretty cut-and-dried save the part about not actually knowing the hash function, and it was kind of hard to really say much without answering the whole question.
Related
I am creating a hashtable and inserting n elements in it from an unsorted array. As I am calling the hashfunction n times. Wouldn't the time complexity to create/insert the hashtable be O(n) ?
I tried searching everywhere, but they mention complexity in case of collisions, but don't cover how can I create a hashtable in O(1) in a perfect case as I will have to traverse the array in order to pick element one by one and put it in the hashtable?
When inserting a new record into a hash table, using a perfect hash function, an unused index entry (used to point to the record) will be found immediately giving O(1). Later, the record will be found immediately when searched for.
Of course, hash functions are seldom perfect. As the hash index starts to become populated the function will at times require two or more attempts to find an unused index entry to insert a new record and every later attempt to search for that record will also require two or more attempts before the correct index entry is found. So the actual search complexity of a hash table may end up as O(1.5) or more but that value is made up of searches where the record is most often found in the first attempt while others may require two or more.
I guess the trick is to find a hashing algorithm that is "good enough" which means a compromise between an index that isn't too big, an average complexity that is reasonably low and a worst case that is acceptable.
I posted on another search question here and showed how hashing could be used and a good hashing function be determined. The question required looking up a 64-bit value in a table containing 16000 records and replacing it with another value from the record. Computing the second value from the first was not possible. My algorithm sported an average search comnplexity of <1.5 with worst case of 14. The index size was a bit large for my taste but I think a more exhaustive search could have found a better one. In comparison, binary searches required, at best, about twice as many clock cycles to perform a lookup as the hash function did, probably because their time complexity was greater.
I am confused about the time complexity of hash table many articles state that they are "amortized O(1)" not true order O(1) what does this mean in real applications. What is the average time complexity of the operations in a hash table, in actual implementation not in theory, and why are the operations not true O(1)?
It's impossible to know in advance how many collisions you will get with your hash function, as well as things like needing to resize. This can add an element of unpredictability to the performance of a hash table, making it not true O(1). However, virtually all hash table implementations offer O(1) on the vast, vast, vast majority of inserts. This is the same as array inserting - it's O(1) unless you need to resize, in which case it's O(n), plus the collision uncertainty.
In reality, hash collisions are very rare and the only condition in which you'd need to worry about these details is when your specific code has a very tight time window in which it must run. For virtually every use case, hash tables are O(1). More impressive than O(1) insertion is O(1) lookup.
For some uses of hash tables, it's impossible to create them of the "right" size in advance, because it is not known how many elements will need to be held simultaneously during the lifetime of the table. If you want to keep fast access, you need to resize the table from time to time as the number of element grows. This resizing takes linear time with respect to the number of elements already in the table, and is usually done on an insertion, when the number elements passes a threshold.
These resizing operations can be made seldom enough that the amortized cost of insertion is still constant (by following a geometric progression for the size of the table, for instance doubling the size each time it is resized). But one insertion from time to time takes O(n) time because it triggers a resize.
In practice, this is not a problem unless you are building hard real-time applications.
Inserting a value into a Hash table takes, on the average case, O(1) time. The hash function is
computed, the bucked is chosen from the hash table, and then item is inserted. In the worst case scenario,
all of the elements will have hashed to the same value, which means either the entire bucket list must be
traversed or, in the case of open addressing, the entire table must be probed until an empty spot is found.
Therefore, in the worst case, insertion takes O(n) time
refer: http://www.cs.unc.edu/~plaisted/comp550/Neyer%20paper.pdf (Hash Table Section)
I am curious to know what is the reasoning that could overweighs towards using a self-balancing tree technique to store items than using a hash table.
I see that hash tables cannot maintain the insertion-order, but I could always use a linked list on top to store the insertion-order sequence.
I see that for small number of values, there is an added cost of of the hash-function, but I could always save the hash-function together with the key for faster lookups.
I understand that hash tables are difficult to implement than the straight-forward implementation of a red-black tree, but in a practical implementation wouldn't one be willing to go an extra mile for the trouble?
I see that with hash tables it is normal for collisions to occur, but with open-addressing techniques like double hashing that allow to save the keys in the hash table itself, hasn't the problem been reduced to the effect of not tipping the favor towards red black trees for such implementations?
I am curious if I am strictly missing a disadvantage of hash table that still makes red black trees quite viable data structure in practical applications (like filesystems, etc.).
Here is what I can think of:
There are kinds of data which cannot be hashed (or is too expensive to hash), therefore cannot be stored in hash tables.
Trees keep data in the order you need (sorted), not insertion order. You can't (effectively) do that with hash table, even if you run a linked list through it.
Trees have better worst-case performace
Storage allocation is another consideration. Every time you fill all of the buckets in a hash-table, you need to allocate new storage and re-hash everything. This can be avoided if you know the size of the data ahead of time. On the other hand, balanced trees don't suffer from this issue at all.
Just wanted to add :
Balanced binary trees have a predictable time of fetching a data [log n] independent of the type of data. Many times that may be important for your application to estimate the response times for your application. [hash tables may have unpredictable response times]. Remember for smaller n's as in most common use cases the difference in performance in an in-memory look up is hardly going to matter and the bottle neck of the system is going to be elsewhere and sometimes you just want to make the system much simpler to debug and analyze.
Trees are generally more memory efficient compared to hash tables and much simpler to implement without any analysis on the distribution of input keys and possible collisions etc.
In my humble opinion, self-balancing trees work pretty well as Academic topics. And I
do not know anything that can be qualified as a "straight-forward implementation of a
red-black tree".
In the real world, the memory wall makes them far less efficient than they are on paper.
With this in mind, hash tables are decent alternatives, especially if you don't practice
them the Academic style (forget about the table size constraint and you magically resolve
the table resize issue and almost all collision issues).
In a word: keep it simple. If that's simple for you then that's simple for your computer.
I think if you want to query for a range of keys instead of one key, self balanced tree structure will perform better than a hash table structure.
A few reasons I can think of:
Trees are dynamic (the space complexity is N), whereas hash tables are often implemented as arrays which are fixed size, which means they will often be initialized with K size, where K > N, so even if you only have 1 element in a hashmap, you might still have 100 empty slots that take up memory. Another effect of this is:
Increasing the size of an array-based hash table is costly (O(N) average time, O(N log N) worst case), whereas trees can grow in constant time (O(1)) + (time to locate insertion point (O(log N))
Elements in a tree can be gathered in sorted order (using ex: in-order-traversal). Thereby you often get a sorted list as a free perk with trees.
Trees can have a better worst-case performance vs a hashmap depending on how the hashmap is implemented (ex: hashmap with chaining will have O(N) worst case, whereas self-balanced trees can guarantee O(log N) worst case for all operations).
Both self-balanced trees and hashmaps have a worst-case efficiency of O(log N) in the best worst-case (assuming that the hashmap does handle colissions), but Hashmaps can have a better average-case performance (often close to O(1)), whereas Trees will have a constant O(log N). This is because even thou a hashmap can locate the insertion index in O(1), it has to account for hash colissions (more than one element hashing to the same array index), and thus in the best case degrades to a self-balanced tree (such as the Java implementation of hashmap), that is, each element in the hashmap can be implemented as a self-balanced tree, storing all elements which has hashed to the given array cell.
I've been reading a lot about Hash Tables and how to implement on in C and I think I have almost all the concepts in my head so I can start to code my own, I just have a couple of questions that I have yet to properly understand.
As a reference, I've been reading this:
http://eternallyconfuzzled.com/jsw_home.aspx
1) As I've read on the site above, a power of two or a prime number is recommended for the Hash Table size. This is basically an array and an array has a fixed size so I can quickly look up for the value I'm looking for. I can't declare a small array if I have a large input as it won't fit and I can't declare a very large array if my input data is not that large cause it's wasted memory.
What is the optimum size for the Hash Table? What should I base my decision on?
2) Also, on that site, there's a couple of hashing functions which I have yet to read them all. It also states that it's always best to use a good known algorithm and to roll my own. And I might do just that, I'll pick one from that site and test it out on my code and see if it minimizes collisions based on my input data.
What's bugging me is how I control the hash range? The hash can't return and integer larger than the Hash Table size or we'll have a serious problem. How do I deal with this?
1) What you are referring to is the load factor of the hash table - the percentage of buckets that are expected to be filled. Wikipedia has this to say:
With a good hash function, the average
lookup cost is nearly constant as the
load factor increases from 0 up to 0.7
or so. Beyond that point, the
probability of collisions and the cost
of handling them increases.
I believe the Java implementation (and probably others) resizes periodically to keep the load factor within an acceptable range.
2) Just use the modulo operator (%) to keep the bucket index legal. The second operator should be the size of your bucket array.
Pick a small size for your hash table. As you add stuff to your table, check to see what percentage of the table is being used; when it is greater than 70% full, make the table bigger. This also holds true as you remove elements-- make the table smaller when it is less than 60% full, for instance. Wikipedia has a good description of some strategies for dynamic resizing, but that's the general idea.
I only say this because you seem to have known input data:
If you know the rough order of magnitude of the amount of data you will be storing in the hash table, it's generally good enough to just create a table about that big. (You shouldn't worry about whether everything will fit. Instead, the right thing to think about is how many collisions you will have and how you will handle them.)
As for the right hash function, it's possible that the structure of your input will suggest which one will be correct. For instance, what aspects of your input are likely to be evenly distributed?
Which is faster to find an item in a hashtable or in a sorted list?
Algorithm complexity is a good thing to know, and hashtables are known to be O(1) while a sorted vector (in your case I guess it is better to use a sorted array than a list) will provide O(log n) access time.
But you should know that complexity notation gives you the access time for N going to the infinite. That means that if you know that your data will keep growing, complexity notation gives you some hint on the algorithm to chose.
When you know that your data will keep a rather low length: for instance having only a few entries in your array/hashtable, you must go with your watch and measure. So have a test.
For instance, in another problem: sorting an array. For a few entries bubble sort while O(N^2) may be quicker than .. the quick sort, while it is O(n log n).
Also, accordingly to other answers, and depending on your item, you must try to find the best hash function for your hashtable instance. Otherwise it may lead to dramatic bad performance for lookup in your hashtable (as pointed out in Hank Gay's answer).
Edit: Have a look to this article to understand the meaning of Big O notation .
Assuming that by 'sorted list' you mean 'random-accessible, sorted collection'. A list has the property that you can only traverse it element by element, which will result in a O(N) complexity.
The fastest way to find an element in a sorted indexable collection is by N-ary search, O(logN), while a hashtable without collissions has a find complexity of O(1).
Unless the hashing algorithm is extremely slow (and/or bad), the hashtable will be faster.
UPDATE: As commenters have pointed out, you could also be getting degraded performance from too many collisions not because your hash algorithm is bad but simply because the hashtable isn't big enough. Most library implementations (at least in high-level languages) will automatically grow your hashtable behind the scenes—which will cause slower-than-expected performance on the insert that triggers the growth—but if you're rolling your own, it's definitely something to consider.
The get operation in a SortedList is O(log n) while the same operation e a HashTable is O(1). So, normally, the HashTable would be much faster. But this depends on a number of factors:
The size of the list
Performance of the hashing algorithm
Number of collisions / quality of the hashing algorithm
It depends entirely on the amount of data you have stored.
Assuming you have enough memory to throw at it (so the hash table is big enough), the hash table will locate the target data in a fixed amount of time, but the need to calculate the hash will add some (also fixed) overhead.
Searching a sorted list won't have that hashing overhead, but the time required to do the work of actually locating the target data will increase as the list grows.
So, in general, a sorted list will generally be faster for small data sets. (For extremely small data sets which are frequently changed and/or infrequently searched, an unsorted list may be even faster, since it avoids the overhead of doing the sort.) As the data set becomes large, the growth of the list's search time overshadows the fixed overhead of hashing, and the hash table becomes faster.
Where that breakpoint is will vary depending on your specific hash table and sorted-list-search implementations. Run tests and benchmark performance on a number of typically-sized data sets to see which will actually perform better in your particular case. (Or, if the code already runs "fast enough", don't. Just use whichever you're more comfortable with and don't worry about optimizing something which doesn't need to be optimized.)
In some cases, it depends on the size of the collection (and to a lesser degree, implementation details). If your list is very small, 5-10 items maybe, I'd guess the list would be faster. Otherwise xtofl has it right.
HashTable would be more efficient for list containing more than 10 items. If the list has fewer than 10 items, the overhead due to hashing algo will be more.
In case you need a fast dictionary but also need to keep the items in an ordered fashion use the OrderedDictionary. (.Net 2.0 onwards)