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?
Related
So the question came up about whether tombstones should be included when calculating the load factor of a hash table.
I thought that, given that the load factor is used to determine when to expand capacity, tombstones should not be included. An obvious example is if you almost fill and then remove every value in a hash table. Here insertions are super easy (no collisions) so I believe the load factor shouldn't include them.
But you could look at this and think that with all the tombstones lookups will be slow (potentially searching almost the entire space).
So I thought I'd ask the question. Should the load factor of a hashtable include tombstones in the calculation?
Load factor is not an essential part of hash table data structure -- it is the way to define rules of behaviour for the dymamic system (growing/shrinking hash table is a dynamic system).
Moreover, in my opinion, in 95% of modern hash table cases this way is over simplified, dynamic systems behave suboptimally. It's advantages:
Well, simplicity of understanding and implementation.
Hash table data structure shouldn't store many numbers with some thresholds -- likely only one number. This is meaningful when hash table is very small and the size of the header affects total data structure memory efficiency (in bytes to store an entry).
In certain (and common) case: append/update only hash table, more complex models of behaviour degenerate to the "just load factor" model, in other words, load factor model defines relatively optimal behaviour.
See also my answer on load factor model. I prefer [min load, target load, max load] + growth factor frame model.
If you develop general-purpose hash table with tombstones, I think you can just pick up my results (below). I spend maybe several weeks solely developing this model. Maybe you can make some improvements or further research, I would be glad.
Two main hash table dynamic behaviour patterns are targeted:
growing hash table (maybe in growing phase), with little or no removals
initial fill of hash table, when proper capacity was not specified (or unknown)
hash table that remains of the same or nearly the same size, number of removals is equal or nearly equal to number of insertions
caches with upper size bound, LRUs, tables with entry expires
Two thresholds are defined:
max size (i. e. number of alive entries), table size * max load
min number of free (i. e. empty, without alive entry nor tombstone) slots, computed by magic formula.
If hash table size exceeds max size, we assume we are in the "growing pattern", rehash to the table size to be able to store current size * growth factor entries, i. e. choose table size closest possible to current size * growth factor / target load.
If the number of free slots becomes below than min number of free slots, we are in "cache pattern", rehash "to the current size", i. e. to the table size closest possible to current size / target load.
Read the source where all the above logics are coded.
Also, article Tombstones purge from hashtable: theory and practice sheds some light.
If you develop specially purposed hash table, which dymanic properties are known (or could be studied), I recommend you to develop your own model, fitting your case. Don't rely on pure math and CS theory, evaluate your model in benchmarks.
I recently implemented an algorithm in Java that used a hash table. I compared it to a few other algorithms with rather large data input sizes such as 100000.
The thing that has struck me is that once my data input size exceeds 10000 the performance of the hash table drops dramatically. To emphasise this drop, what took 4000 ms with input size 1000 suddenly goes up to 172000 ms for input size 5000.
Can anyone please explain to me what the reason for this is? I'd really like to know.
Thanks!
This question is way too ambiguous for anyone to give a definitive answer, but if I had to guess I would say that you are encountering collisions. The stock implementation of java's HashMap uses linked lists to hold the entries whose keys' hashes collide, which will certainly happen if the hashCode method has been incorrectly defined; perhaps returning a constant value.
Having said that, if you're just measuring elapsed time, that doesn't tell you too much. Perhaps you crossed a threshold that caused a major garbage collection to occur. You should try to measure performance after your JVM and hash table are sufficiently warmed up, and take lots of measurements and consider their average, before coming to any conclusions.
If we had an assignment:
Given a block of binary data, count the frequency of the bytes within it.
And you were supposed to do this in C, the answer would be trivial and reasonably fast even for larger binary blocks. How would one go about implementing this in a purely functional language, without side effects?
For example, if you wrote a function that accepted freqency counts for each byte and the rest of the list of bytes, and returned modified frequency counts, it would have to do awful lot of work for data set of 100M bytes.
Also, if you sorted the data and then somehow counted the amount of subsequent same-valued bytes, the sort itself would take a lot of time.
Is there a reasonable way to implement this?
The straightforward way to do it is indeed to pass in and return data structures mapping bytes to counts. This would probably be implemented as some kind of tree (since that's what you get out of the standard library containers, as far as I know). In pure functional programming when you're passed in a tree and you need to return a new tree with a difference in only one node, the returned tree ends up sharing almost all of its structure and data with the original tree.
There is some overhead in traversing the tree to get to the count, but since you're counting bytes the tree is only ever smaller than 256 elements, so the overhead is log(255), which is a constant. It doesn't get larger for large data sets - it doesn't change the big-oh complexity of the algorithm. That's actually true even if you use the greatest possible overhead of copying around a full 256-entry array of counts with no sharing.
If you want to optimise this, you can take advantage of the fact that the "intermediate" frequency counts are never needed except as part of the computation of the next set of counts. That means you can use various techniques for getting the implementation to use destructive updates even while you're still semantically writing functional code. An STref in Haskell is basically letting you do this manually.
Theoretically the compiler could notice that you're replacing a never-needed-again value with a new one, so it could do the update in place for you. I don't know whether or not any actual production ready compilers are currently able to make this optimisation.
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.
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.