问题
I have been noticing some very strange usage of O(1) in discussion of algorithms involving hashing and types of search, often in the context of using a dictionary type provided by the language system, or using dictionary or hash-array types used using array-index notation.
Basically, O(1) means bounded by a constant time and (typically) fixed space. Some pretty fundamental operations are O(1), although using intermediate languages and special VMs tends to distort ones thinking here (e.g., how does one amortize the garbage collector and other dynamic processes over what would otherwise be O(1) activities).
But ignoring amortization of latencies, garbage-collection, and so on, I still don't understand how the leap to assumption that certain techniques that involve some kind of searching can be O(1) except under very special conditions.
Although I have noticed this before, an example just showed up in the Pandincus question, "'Proper’ collection to use to obtain items in O(1) time in C# .NET?".
As I remarked there, the only collection I know of that provides O(1) access as a guaranteed bound is a fixed-bound array with an integer index value. The presumption is that the array is implemented by some mapping to random access memory that uses O(1) operations to locate the cell having that index.
For collections that involve some sort of searching to determine the location of a matching cell for a different kind of index (or for a sparse array with integer index), life is not so easy. In particular, if there are collisons and congestion is possible, access is not exactly O(1). And if the collection is flexible, one must recognize and amortize the cost of expanding the underlying structure (such as a tree or a hash table) for which congestion relief (e.g., high collision incidence or tree imbalance).
I would never have thought to speak of these flexible and dynamic structures as O(1). Yet I see them offered up as O(1) solutions without any identification of the conditions that must be maintained to actually have O(1) access be assured (as well as have that constant be negligibly small).
THE QUESTION: All of this preparation is really for a question. What is the casualness around O(1) and why is it accepted so blindly? Is it recognized that even O(1) can be undesirably large, even though near-constant? Or is O(1) simply the appropriation of a computational-complexity notion to informal use? I'm puzzled.
UPDATE: The Answers and comments point out where I was casual about defining O(1) myself, and I have repaired that. I am still looking for good answers, and some of the comment threads are rather more interesting than their answers, in a few cases.
回答1:
My understanding is that O(1) is not necessarily constant; rather, it is not dependent on the variables under consideration. Thus a hash lookup can be said to be O(1) with respect to the number of elements in the hash, but not with respect to the length of the data being hashed or ratio of elements to buckets in the hash.
The other element of confusion is that big O notation describes limiting behavior. Thus, a function f(N) for small values of N may indeed show great variation, but you would still be correct to say it is O(1) if the limit as N approaches infinity is constant with respect to N.
回答2:
The problem is that people are really sloppy with terminology. There are 3 important but distinct classes here:
O(1) worst-case
This is simple - all operations take no more than a constant amount of time in the worst case, and therefore in all cases. Accessing an element of an array is O(1) worst-case.
O(1) amortized worst-case
Amortized means that not every operation is O(1) in the worst case, but for any sequence of N operations, the total cost of the sequence is no O(N) in the worst case. This means that even though we can't bound the cost of any single operation by a constant, there will always be enough "quick" operations to make up for the "slow" operations such that the running time of the sequence of operations is linear in the number of operations.
For example, the standard Dynamic Array which doubles its capacity when it fills up requires O(1) amortized time to insert an element at the end, even though some insertions require O(N) time - there are always enough O(1) insertions that inserting N items always takes O(N) time total.
O(1) average-case
This one is the trickiest. There are two possible definitions of average-case: one for randomized algorithms with fixed inputs, and one for deterministic algorithms with randomized inputs.
For randomized algorithms with fixed inputs, we can calculate the average-case running time for any given input by analyzing the algorithm and determining the probability distribution of all possible running times and taking the average over that distribution (depending on the algorithm, this may or may not be possible due to the Halting Problem).
In the other case, we need a probability distribution over the inputs. For example, if we were to measure a sorting algorithm, one such probability distribution would be the distribution that has all N! possible permutations of the input equally likely. Then, the average-case running time is the average running time over all possible inputs, weighted by the probability of each input.
Since the subject of this question is hash tables, which are deterministic, I'm going to focus on the second definition of average-case. Now, we can't always determine the probability distribution of the inputs because, well, we could be hashing just about anything, and those items could be coming from a user typing them in or from a file system. Therefore, when talking about hash tables, most people just assume that the inputs are well-behaved and the hash function is well behaved such that the hash value of any input is essentially randomly distributed uniformly over the range of possible hash values.
Take a moment and let that last point sink in - the O(1) average-case performance for hash tables comes from assuming all hash values are uniformly distributed. If this assumption is violated (which it usually isn't, but it certainly can and does happen), the running time is no longer O(1) on average.
See also Denial of Service by Algorithmic Complexity. In this paper, the authors discuss how they exploited some weaknesses in the default hash functions used by two versions of Perl to generate large numbers of strings with hash collisions. Armed with this list of strings, they generated a denial-of-service attack on some webservers by feeding them these strings that resulted in the worst-case O(N) behavior in the hash tables used by the webservers.
回答3:
O(1) means constant time and (typically) fixed space
Just to clarify these are two separate statements. You can have O(1) in time but O(n) in space or whatever.
Is it recognized that even O(1) can be undesirably large, even though near-constant?
O(1) can be impractically HUGE and it's still O(1). It is often neglected that if you know you'll have a very small data set the constant is more important than the complexity, and for reasonably small data sets, it's a balance of the two. An O(n!) algorithm can out-perform a O(1) if the constants and sizes of the data sets are of the appropriate scale.
O() notation is a measure of the complexity - not the time an algorithm will take, or a pure measure of how "good" a given algorithm is for a given purpose.
回答4:
I can see what you're saying, but I think there are a couple of basic assumptions underlying the claim that look-ups in a Hash table have a complexity of O(1).
- The hash function is reasonably designed to avoid a large number of collisions.
- The set of keys is pretty much randomly distributed, or at least not purposely designed to make the hash function perform poorly.
The worst case complexity of a Hash table look-up is O(n), but that's extremely unlikely given the above 2 assumptions.
回答5:
Hashtables is a data structure that supports O(1) search and insertion.
A hashtable usually has a key and value pair, where the key is used to as the parameter to a function (a hash function) which will determine the location of the value in its internal data structure, usually an array.
As insertion and search only depends upon the result of the hash function and not on the size of the hashtable nor the number of elements stored, a hashtable has O(1) insertion and search.
There is one caveat, however. That is, as the hashtable becomes more and more full, there will be hash collisions where the hash function will return an element of an array which is already occupied. This will necesitate a collision resolution in order to find another empty element.
When a hash collision occurs, a search or insertion cannot be performed in O(1) time. However, good collision resolution algorithms can reduce the number of tries to find another suiteable empty spot or increasing the hashtable size can reduce the number of collisions in the first place.
So, in theory, only a hashtable backed by an array with an infinite number of elements and a perfect hash function would be able to achieve O(1) performance, as that is the only way to avoid hash collisions that drive up the number of required operations. Therefore, for any finite-sized array will at one time or another be less than O(1) due to hash collisions.
Let's take a look at an example. Let's use a hashtable to store the following (key, value) pairs:
(Name, Bob)(Occupation, Student)(Location, Earth)
We will implement the hashtable back-end with an array of 100 elements.
The key will be used to determine an element of the array to store the (key, value) pair. In order to determine the element, the hash_function will be used:
hash_function("Name")returns 18hash_function("Occupation")returns 32hash_function("Location")returns 74.
From the above result, we'll assign the (key, value) pairs into the elements of the array.
array[18] = ("Name", "Bob")
array[32] = ("Occupation", "Student")
array[74] = ("Location", "Earth")
The insertion only requires the use of a hash function, and does not depend on the size of the hashtable nor its elements, so it can be performed in O(1) time.
Similarly, searching for an element uses the hash function.
If we want to look up the key "Name", we'll perform a hash_function("Name") to find out which element in the array the desired value resides.
Also, searching does not depend on the size of the hashtable nor the number of elements stored, therefore an O(1) operation.
All is well. Let's try to add an additional entry of ("Pet", "Dog"). However, there is a problem, as hash_function("Pet") returns 18, which is the same hash for the "Name" key.
Therefore, we'll need to resolve this hash collision. Let's suppose that the hash collision resolving function we used found that the new empty element is 29:
array[29] = ("Pet", "Dog")
Since there was a hash collision in this insertion, our performance was not quite O(1).
This problem will also crop up when we try to search for the "Pet" key, as trying to find the element containing the "Pet" key by performing hash_function("Pet") will always return 18 initially.
Once we look up element 18, we'll find the key "Name" rather than "Pet". When we find this inconsistency, we'll need to resolve the collision in order to retrieve the correct element which contains the actual "Pet" key. Resovling a hash collision is an additional operation which makes the hashtable not perform at O(1) time.
回答6:
I can't speak to the other discussions you've seen, but there is at least one hashing algorithm that is guaranteed to be O(1).
Cuckoo hashing maintains an invariant so that there is no chaining in the hash table. Insertion is amortized O(1), retrieval is always O(1). I've never seen an implementation of it, it's something that was newly discovered when I was in college. For relatively static data sets, it should be a very good O(1), since it calculates two hash functions, performs two lookups, and immediately knows the answer.
Mind you, this is assuming the hash calcuation is O(1) as well. You could argue that for length-K strings, any hash is minimally O(K). In reality, you can bound K pretty easily, say K < 1000. O(K) ~= O(1) for K < 1000.
回答7:
There may be a conceptual error as to how you're understanding Big-Oh notation. What it means is that, given an algorithm and an input data set, the upper bound for the algorithm's run time depends on the value of the O-function when the size of the data set tends to infinity.
When one says that an algorithm takes O(n) time, it means that the runtime for an algorithm's worst case depends linearly on the size of the input set.
When an algorithm takes O(1) time, the only thing it means is that, given a function T(f) which calculates the runtime of a function f(n), there exists a natural positive number k such that T(f) < k for any input n. Essentially, it means that the upper bound for the run time of an algorithm is not dependent on its size, and has a fixed, finite limit.
Now, that does not mean in any way that the limit is small, just that it's independent of the size of the input set. So if I artificially define a bound k for the size of a data set, then its complexity will be O(k) == O(1).
For example, searching for an instance of a value on a linked list is an O(n) operation. But if I say that a list has at most 8 elements, then O(n) becomes O(8) becomes O(1).
In this case, it we used a trie data structure as a dictionary (a tree of characters, where the leaf node contains the value for the string used as key), if the key is bounded, then its lookup time can be considered O(1) (If I define a character field as having at most k characters in length, which can be a reasonable assumption for many cases).
For a hash table, as long as you assume that the hashing function is good (randomly distributed) and sufficiently sparse so as to minimize collisions, and rehashing is performed when the data structure is sufficiently dense, you can indeed consider it an O(1) access-time structure.
In conclusion, O(1) time may be overrated for a lot of things. For large data structures the complexity of an adequate hash function may not be trivial, and sufficient corner cases exist where the amount of collisions lead it to behave like an O(n) data structure, and rehashing may become prohibitively expensive. In which case, an O(log(n)) structure like an AVL or a B-tree may be a superior alternative.
回答8:
In general, I think people use them comparatively without regard to exactness. For example, hash-based data structures are O(1) (average) look up if designed well and you have a good hash. If everything hashes to a single bucket, then it's O(n). Generally, though one uses a good algorithm and the keys are reasonably distributed so it's convenient to talk about it as O(1) without all the qualifications. Likewise with lists, trees, etc. We have in mind certain implementations and it's simply more convenient to talk about them, when discussing generalities, without the qualifications. If, on the other hand, we're discussing specific implementations, then it probably pays to be more precise.
回答9:
HashTable looks-ups are O(1) with respect to the number of items in the table, because no matter how many items you add to the list the cost of hashing a single item is pretty much the same, and creating the hash will tell you the address of the item.
To answer why this is relevant: the OP asked about why O(1) seemed to be thrown around so casually when in his mind it obviously could not apply in many circumstances. This answer explains that O(1) time really is possible in those circumstances.
回答10:
Hash table implementations are in practice not "exactly" O(1) in use, if you test one you'll find they average around 1.5 lookups to find a given key across a large dataset
( due to to the fact that collisions DO occur, and upon colliding, a different location must be assigned )
Also, In practice, HashMaps are backed by arrays with an initial size, that is "grown" to double size when it reaches 70% fullness on average, which gives a relatively good addressing space. After 70% fullness collision rates grow faster.
Big O theory states that if you have a O(1) algorithm, or even an O(2) algorithm, the critical factor is the degree of the relation between input-set size and steps to insert/fetch one of them. O(2) is still constant time, so we just approximate it as O(1), because it means more or less the same thing.
In reality, there is only 1 way to have a "perfect hashtable" with O(1), and that requires:
- A Global Perfect Hash Key Generator
- An Unbounded addressing space.
( Exception case: if you can compute in advance all the permutations of permitted keys for the system, and your target backing store address space is defined to be the size where it can hold all keys that are permitted, then you can have a perfect hash, but its a "domain limited" perfection )
Given a fixed memory allocation, it is not plausible in the least to have this, because it would assume that you have some magical way to pack an infinite amount of data into a fixed amount of space with no loss of data, and that's logistically impossible.
So retrospectively, getting O(1.5) which is still constant time, in a finite amount of memory with even a relatively Naïve hash key generator, I consider pretty damn awesome.
Suffixory note Note I use O(1.5) and O(2) here. These actually don't exist in big-o. These are merely what people whom don't know big-o assume is the rationale.
If something takes 1.5 steps to find a key, or 2 steps to find that key, or 1 steps to find that key, but the number of steps never exceeds 2 and whether it takes 1 step or 2 is completely random, then it is still Big-O of O(1). This is because no matter how many items to you add to the dataset size, It still maintains the <2 steps. If for all tables > 500 keys it takes 2 steps, then you can assume those 2 steps are in fact one-step with 2 parts, ... which is still O(1).
If you can't make this assumption, then your not being Big-O thinking at all, because then you must use the number which represents the number of finite computational steps required to do everything and "one-step" is meaningless to you. Just get into your head that there is NO direct correlation between Big-O and number of execution cycles involved.
回答11:
O(1) means, exactly, that the algorithm's time complexity is bounded by a fixed value. This doesn't mean it's constant, only that it is bounded regardless of input values. Strictly speaking, many allegedly O(1) time algorithms are not actually O(1) and just go so slowly that they are bounded for all practical input values.
回答12:
Yes, garbage collection does affect the asymptotic complexity of algorithms running in the garbage collected arena. It is not without cost, but it is very hard to analyze without empirical methods, because the interaction costs are not compositional.
The time spent garbage collecting depends on the algorithm being used. Typically modern garbage collectors toggle modes as memory fills up to keep these costs under control. For instance, a common approach is to use a Cheney style copy collector when memory pressure is low because it pays cost proportional to the size of the live set in exchange for using more space, and to switch to a mark and sweep collector when memory pressure becomes greater, because even though it pays cost proportional to the live set for marking and to the whole heap or dead set for sweeping. By the time you add card-marking and other optimizations, etc. the worst case costs for a practical garbage collector may actually be a fair bit worse, picking up an extra logarithmic factor for some usage patterns.
So, if you allocate a big hash table, even if you access it using O(1) searches for all time during its lifetime, if you do so in a garbage collected environment, occasionally the garbage collector will traverse the entire array, because it is size O(n) and you will pay that cost periodically during collection.
The reason we usually leave it off of the complexity analysis of algorithms is that garbage collection interacts with your algorithm in non-trivial ways. How bad of a cost it is depends a lot on what else you are doing in the same process, so the analysis is not compositional.
Moreover, above and beyond the copy vs. compact vs. mark and sweep issue, the implementation details can drastically affect the resulting complexities:
- Incremental garbage collectors that track dirty bits, etc. can all but make those larger re-traversals disappear.
- It depends on whether your GC works periodically based on wall-clock time or runs proportional to the number of allocations.
- Whether a mark and sweep style algorithm is concurrent or stop-the-world
- Whether it marks fresh allocations black if it leaves them white until it drops them into a black container.
- Whether your language admits modifications of pointers can let some garbage collectors work in a single pass.
Finally, when discussing an algorithm, we are discussing a straw man. The asymptotics will never fully incorporate all of the variables of your environment. Rarely do you ever implement every detail of a data structure as designed. You borrow a feature here and there, you drop a hash table in because you need fast unordered key access, you use a union-find over disjoint sets with path compression and union by rank to merge memory-regions over there because you can't afford to pay a cost proportional to the size of the regions when you merge them or what have you. These structures are thought primitives and the asymptotics help you when planning overall performance characteristics for the structure 'in-the-large' but knowledge of what the constants are matters too.
You can implement that hash table with perfectly O(1) asymptotic characteristics, just don't use garbage collection; map it into memory from a file and manage it yourself. You probably won't like the constants involved though.
回答13:
I think when many people throw around the term "O(1)" they implicitly have in mind a "small" constant, whatever "small" means in their context.
You have to take all this big-O analysis with context and common sense. It can be an extremely useful tool or it can be ridiculous, depending on how you use it.
来源：https://stackoverflow.com/questions/332952/whats-up-with-o1