In an algorithmics course our teacher covered "virtual initialization" where you allocate memory for an operation, but don't initialize all the values since the problem space might be too large compared to the values that actually need to be calculated (for example a dictionary or set), which wastes a lot of time for setting a default value. The basic principle is to have two arrays that point to each other (index each other) and keep track of how many variables have been assigned. let's say we have an array a and we want to find if a[i] contains a valid value, we can use array b as an index to a like so:
I had a look at the python time complexity table at https://wiki.python.org/moin/TimeComplexity and it mentions that a set membership test in the worst case can be of O(n). I'm not sure where to find the exact implementation of the set function but after a bit of googling, most people mention that it uses a hash table. My main questions are:
How can a hash table be used to check if a value is valid or not? We can hash every value and read the result, but that doesn't mean the output isn't garbage (we can be reading whatever was written to that address when malloc was called).
virtual initialization completely avoids the problem with collisions in a hash table, so why is it not a better solution than using a hash table?
When you implement a hash table using an array, you need a flag in each entry to indicate whether it's currently in use. This is needed to deal with hash function collisions -- if two values have the same hash code, you can't put them both in the same element of the array.
This means that when you allocate the array, you have to go through it, initializing all these flags. And you have to do this again whenever you grow the hash table.
"Virtual initialization" avoids this. The algorithm you pasted is used to tell whether an a[k] is in use. It uses a second array b that contains indexes into a. When inserting an element, a[k].p contains an index j in b, and b[j] contains k. If these two match, and also j is lower than the limit of all indexes, you know that the array element is in use.
To clear an entry, you simply set a[k].p = 0, since all valid entries have p between 1 and n.
This is an example of a time-space tradeoff in algorithm design. To avoid the time spent initializing the array, we allocate a second array. If you have lots of available memory, this can be an acceptable tradeoff.
Related
I know that pop the last element of the list takes O(1).
And after reading this post
What is the time complexity of popping elements from list in Python?
I notice that if we pop an arbitrary number from a list takes O(n) since all the pointers need to shift one position up.
But for the set, there is no order and no index. So I am not sure if there is still pointers in set?
If not, would the pop() for set is O(1)?
Thanks.
On modern CPython implementations, pop takes amortized constant-ish time (I'll explain further). On Python 2, it's usually the same, but performance can degrade heavily in certain cases.
A Python set is based on a hash table, and pop has to find an occupied entry in the table to remove and return. If it searched from the start of the table every time, this would take time proportional to the number of empty leading entries, and it would get slower with every pop.
To avoid this, the standard CPython implementation tries to remember the position of the last popped entry, to speed up sequences of pops. CPython 3.5+ has a dedicated finger member in the set memory layout to store this position, but earlier versions abuse the hash field of the first hash table entry to store this index.
On any Python version, removing all elements from a set with a sequence of pop operations will take time proportional to the size of the underlying hash table, which is usually within a small constant factor of the original number of elements (unless you'd already removed a bunch of elements). Mixing insertions and pop can interfere heavily with this on Python 2 if inserted elements land in hash table index 0, trashing the search finger. This is much less of an issue on Python 3.
I was wonder what if i had 1 million users, how much time will take it to loop every account to check for the username and email if the registration page to check if they are already used or not, or check if the username and password and correct in the log in page, won't it take ages if i do it in a traditional for loop?
Rather than give a detailed technical answer, I will try to give a theoretical illustration of how to address your concerns. What you seem to be saying is this:
Linear search might be too slow when logging users in.
I imagine what you have in mind is this: a user types in a username and password, clicks a button, and then you loop through a list of username/password combinations and see if there is a matching combination. This takes time that is linear in terms of the number of users in the system; if you have a million users in the system, the loop will take about a thousand times as long as when you just had a thousand users... and if you get a billion users, it will take a thousand times longer over again.
Whether this is a performance problem in practice can only be determined through testing and requirements analysis. However, if it is determined to be a problem, then there is a place for theory to come to the rescue.
Imagine one small improvement to our original scheme: rather than storing the username/password combinations in arbitrary order and looking through the whole list each time, imagine storing these combinations in alphabetic order by username. This enables us to use binary search, rather than linear search, to determine whether there exists a matching username:
check the middle element in the list
if the target element is equal to the middle element, you found a match
otherwise, if the target element comes before the middle element, repeat binary search on the left half of the list
otherwise, if the target element comes after the middle element, repeat binary search on the right half of the list
if you run out of list without finding the target, it's not in the list
The time complexity of this is logarithmic in terms of the number of users in the system: if you go from a thousand users to a million users, the time taken goes up by a factor of roughly ten, rather than one thousand as was the case for linear search. This is already a vast improvement over linear search and for any realistic number of users is probably going to be efficient enough. However, if additional performance testing and requirements analysis determine that it's still too slow, there are other possibilities.
Imagine now creating a large array of username/password pairs and whenever a pair is added to the collection, a function is used to transform the username into a numeric index. The pair is then inserted at that index in the array. Later, when you want to find whether that entry exists, you use the same function to calculate the index, and then check just that index to see if your element is there. If the function that maps the username to indices (called a hash function; the index is called a hash) is perfect - different strings don't map to the same index - then this unambiguously tells you whether your element exists. Notably, under somewhat reasonable assumptions, the time to make this determination is mostly independent from the number of users currently in the system: you can get (amortized) constant time behavior from this scheme, or something reasonably close to it. That means the performance hit from going from a thousand to a million users might be negligible.
This answer does not delve into the ugly real-world minutia of implementing these ideas in a production system. However, real world systems to implement these ideas (and many more) for precisely the kind of situation presented.
EDIT: comments asked for some pointers on actually implementing a hash table in Python. Here are some thoughts on that.
So there is a built-in hash() function that can be made to work if you disable the security feature that causes it to produce different hashes for different executions of the program. Otherwise, you can import hashlib and use some hash function there and convert the output to an integer using e.g. int.from_bytes. Once you get your number, you can take the modulus (or remainder after division, using the % operator) w.r.t. the capacity of your hash table. This gives you the index in the hash table where the item gets put. If you find there's already an item there - i.e. the assumption we made in theory that the hash function is perfect turns out to be incorrect - then you need a strategy. Two strategies for handling collisions like this are:
Instead of putting items at each index in the table, put a linked list of items. Add items to the linked list at the index computed by the hash function, and look for them there when doing the search.
Modify the index using some deterministic method (e.g., squaring and taking the modulus) up to some fixed number of times, to see if a backup spot can easily be found. Then, when searching, if you do not find the value you expected at the index computed by the hash method, check the next backup, and so on. Ultimately, you must fall back to something like method 1 in the worst case, though, since this process could fail indefinitely.
As for how large to make the capacity of the table: I'd recommend studying recommendations but intuitively it seems like creating it larger than necessary by some constant multiplicative factor is the best bet generally speaking. Once the hash table begins to fill up, this can be detected and the capacity expanded (all hashes will have to be recomputed and items re-inserted at their new positions - a costly operation, but if you increase capacity in a multiplicative fashion then I imagine this will not be too frequent an issue).
I'm teaching myself data structures through this python book and I'd appreciate if someone can correct me if I'm wrong since a hash set seems to be extremely similar to a hash map.
Implementation:
A Hashset is a list [] or array where each index points to the head of a linkedlist
So some hash(some_item) --> key, and then list[key] and then add to the head of a LinkedList. This occurs in O(1) time
When removing a value from the linkedlist, in python we replace it with a placeholder because hashsets are not allowed to have Null/None values, correct?
When the list[] gets over a certain % of load/fullness, we copy it over to another list
Regarding Time Complexity Confusion:
So one question is, why is Average search/access O(1) if there can be a list of N items at the linkedlist at a given index?
Wouldnt the average case be the searchitem is in the middle of its indexed linkedlist so it should be O(n/2) -> O(n)?
Also, when removing an item, if we are replacing it with a placeholder value, isn't this considered a waste of memory if the placeholder is never used?
And finally, what is the difference between this and a HashMap other than HashMaps can have nulls? And HashMaps are key/value while Hashsets are just value?
For your first question - why is the average time complexity of a lookup O(1)? - this statement is in general only true if you have a good hash function. An ideal hash function is one that causes a nice spread on its elements. In particular, hash functions are usually chosen so that the probability that any two elements collide is low. Under this assumption, it's possible to formally prove that the expected number of elements to check is O(1). If you search online for "universal family of hash functions," you'll probably find some good proofs of this result.
As for using placeholders - there are several different ways to implement a hash table. The approach you're using is called "closed addressing" or "hashing with chaining," and in that approach there's little reason to use placeholders. However, other hashing strategies exist as well. One common family of approaches is called "open addressing" (the most famous of which is linear probing hashing), and in those setups placeholder elements are necessary to avoid false negative lookups. Searching online for more details on this will likely give you a good explanation about why.
As for how this differs from HashMap, the HashMap is just one possible implementation of a map abstraction backed by a hash table. Java's HashMap does support nulls, while other approaches don't.
The lookup time wouldn't be O(n) because not all items need to be searched, it also depends on the number of buckets. More buckets would decrease the probability of a collision and reduce the chain length.
The number of buckets can be kept as a constant factor of the number of entries by resizing the hash table as needed. Along with a hash function that evenly distributes the values, this keeps the expected chain length bounded, giving constant time lookups.
The hash tables used by hashmaps and hashsets are the same except they store different values. A hashset will contain references to a single value, and a hashmap will contain references to a key and a value. Hashsets can be implemented by delegating to a hashmap where the keys and values are the same.
A lot has been written here about open hash tables, but some fundamental points are missed.
Practical implementations generally have O(1) lookup and delete because they guarantee buckets won't contain more than a fixed number of items (the load factor). But this means they can only achieve amortized O(1) time for insert because the table needs to be reorganized periodically as it grows.
(Some may opt to reorganize on delete, also, to shrink the table when the load factor reaches some bottom threshold, gut this only affect space, not asymptotic run time.)
Reorganization means increasing (or decreasing) the number of buckets and re-assigning all elements into their new bucket locations. There are schemes, e.g. extensible hashing, to make this a bit cheaper. But in general it means touching each element in the table.
Reorganization, then, is O(n). How can insert be O(1) when any given one may incur this cost? The secret is amortization and the power of powers. When the table is grown, it must be grown by a factor greater than one, two being most common. If the table starts with 1 bucket and doubles each time the load factor reaches F, then the cost of N reorganizations is
F + 2F + 4F + 8F ... (2^(N-1))F = (2^N - 1)F
At this point the table contains (2^(N-1))F elements, the number in the table during the last reorganization. I.e. we have done (2^(N-1))F inserts, and the total cost of reorganization is as shown on the right. The interesting part is the average cost per element in the table (or insert, take your pick):
(2^N - 1)F 2^N
---------- ~= ------- = 2
(2^(N-1))F 2^(N-1)
That's where the amortized O(1) comes from.
One additional point is that for modern processors, linked lists aren't a great idea for the bucket lists. With 8-byte pointers, the overhead is meaningful. More importantly, heap-allocated nodes in a single list will almost never be contiguous in memory. Traversing such a list kills cache performance, which can slow things down by orders of magnitude.
Arrays (with an integer count for number of data-containing elements) are likely to work out better. If the load factor is small enough, just allocate an array equal in size to the load factor at the time the first element is inserted in the bucket. Otherwise, grow these element arrays by factors the same way as the bucket array! Everything will still amortize to O(1).
To delete an item from such a bucket, don't mark it deleted. Just copy the last array element to the location of the deleted one and decrement the element count. Of course this won't work if you allow external pointers into the hash buckets, but that's a bad idea anyway.
When working with dictionaries in Python, this page says that the time complexity of iterating through the element of the dictionary is O(n), where n is the largest size the dictionary has been.
However, I don't think that there is an obvious way to iterate through the elements of a hash table. Can I assume good performance of dict.iteritems() when iterating through element of a hash table, without too much overhead?
Since dictionaries are used a lot in Python, I assume that this is implemented in some smart way. Still, I need to make sure.
If you look at the notes on Python's dictionary source code, I think the relevant points are the following:
Those methods (iteration and key listing) loop over every potential entry
How many potential entries will there be, as a function of largest size (largest number of keys ever stored in that dictionary)? Look at the following two sections in the same document:
Maximum dictionary load in PyDict_SetItem. Currently set to 2/3
Growth rate upon hitting maximum load. Currently set to *2.
This would suggest that the sparsity of a dictionary is going to be somewhere around 1/3~2/3 (unless growth rate is set to *4, then it's 1/6~2/3). So basically you're going to be checking upto 3 (or 6 if *4) potential entries for every key.
Of course, whether it's 3 entries or 1000, it's still O(n) but 3 seems like a pretty acceptable constant factor.
Incidentally here are the rest of the source & documentation, including that of the DictObject:
http://svn.python.org/projects/python/trunk/Objects/
I'm making an abstract algebra library in Python, and one of the things it does it takes the Cayley table (think of it as an abstract "multiplication" table, which doesn't have to obey the standard rules for multiplication or addition), and uses it to prove whether or not certain identities or properties hold for the binary operator defined by the Cayley table.
The computations for any of these procedures basically boils down to:
Use the values in the Cayley table to get the result of some abstract binary operation
Compare the resulting values to see if they are equal
Do all of the above multiple times in a loop.
This can get very cpu intensive with the sheer number of operations that need to be done, especially if you want to do this for all permutations of cayley tables with size n*n (O(n^2!) for that, I think). The good thing is, this is highly paralelizable, as I have algorithms that can calculate the mth permutation of an n*n cayley table in about O(n) time, and thus the set of all Cayley table permutations can be split near evenly, and multiple processes can be assigned to work on a different subset of the problem in parallel.
Is process like this (replacing values from lookup tables, comparing results) suited to open CL, or any other GPU library for that matter?
If you could change the
Compare the resulting values to see if they are equal
part to something like
subtract one from another, save the result and check for a zero from host side.
if the result will decide for just an addition(+) operation, even something looks like
subtract one from another, multiply (1-signum(result)) with the adder value, then add.
can be faster than cpu. Because gpu memory is generally marginally faster than host memory when read/written in non-random.