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This questions came up on a homework assignment. I cannot fathom why? It seems like you would always want to choose the algorithm that produces the best run time.
Big O and Big Theta notation only imply that for arbitrarily large input sizes, the performance tends towards some limit. For example, the function 99999999n is O(n) but the function (1/9999999999)n^2 is O(n^2). However, for any input of reasonable size (not infinitely large), the O(n^2) function is obviously likely to be faster.
In other words, if you can make assumptions about your input data, there are some cases where a generally worse algorithm may perform better.
A real world example of the above concept is sorting - there are some algorithms which perform in O(n) time if the array is already sorted (bubble sort). If you know a lot of your arrays already are sorted you may choose to use bubble sort over merge sort for this reason.
Another corner case where you might want to not use a more time-efficient algorithm is space efficiency. Maybe you are programming on an embedded device with very little RAM. You would rather use less memory and waste slightly more time than be as perfectly time-efficient as possible.
Are there are any cases where you would prefer O(log n) time complexity to O(1) time complexity? Or O(n) to O(log n)?
Do you have any examples?
There can be many reasons to prefer an algorithm with higher big O time complexity over the lower one:
most of the time, lower big-O complexity is harder to achieve and requires skilled implementation, a lot of knowledge and a lot of testing.
big-O hides the details about a constant: algorithm that performs in 10^5 is better from big-O point of view than 1/10^5 * log(n) (O(1) vs O(log(n)), but for most reasonable n the first one will perform better. For example the best complexity for matrix multiplication is O(n^2.373) but the constant is so high that no (to my knowledge) computational libraries use it.
big-O makes sense when you calculate over something big. If you need to sort array of three numbers, it matters really little whether you use O(n*log(n)) or O(n^2) algorithm.
sometimes the advantage of the lowercase time complexity can be really negligible. For example there is a data structure tango tree which gives a O(log log N) time complexity to find an item, but there is also a binary tree which finds the same in O(log n). Even for huge numbers of n = 10^20 the difference is negligible.
time complexity is not everything. Imagine an algorithm that runs in O(n^2) and requires O(n^2) memory. It might be preferable over O(n^3) time and O(1) space when the n is not really big. The problem is that you can wait for a long time, but highly doubt you can find a RAM big enough to use it with your algorithm
parallelization is a good feature in our distributed world. There are algorithms that are easily parallelizable, and there are some that do not parallelize at all. Sometimes it makes sense to run an algorithm on 1000 commodity machines with a higher complexity than using one machine with a slightly better complexity.
in some places (security) a complexity can be a requirement. No one wants to have a hash algorithm that can hash blazingly fast (because then other people can bruteforce you way faster)
although this is not related to switch of complexity, but some of the security functions should be written in a manner to prevent timing attack. They mostly stay in the same complexity class, but are modified in a way that it always takes worse case to do something. One example is comparing that strings are equal. In most applications it makes sense to break fast if the first bytes are different, but in security you will still wait for the very end to tell the bad news.
somebody patented the lower-complexity algorithm and it is more economical for a company to use higher complexity than to pay money.
some algorithms adapt well to particular situations. Insertion sort, for example, has an average time-complexity of O(n^2), worse than quicksort or mergesort, but as an online algorithm it can efficiently sort a list of values as they are received (as user input) where most other algorithms can only efficiently operate on a complete list of values.
There is always the hidden constant, which can be lower on the O(log n) algorithm. So it can work faster in practice for real-life data.
There are also space concerns (e.g. running on a toaster).
There's also developer time concern - O(log n) may be 1000× easier to implement and verify.
I'm surprised nobody has mentioned memory-bound applications yet.
There may be an algorithm that has less floating point operations either due to its complexity (i.e. O(1) < O(log n)) or because the constant in front of the complexity is smaller (i.e. 2n2 < 6n2). Regardless, you might still prefer the algorithm with more FLOP if the lower FLOP algorithm is more memory-bound.
What I mean by "memory-bound" is that you are often accessing data that is constantly out-of-cache. In order to fetch this data, you have to pull the memory from your actually memory space into your cache before you can perform your operation on it. This fetching step is often quite slow - much slower than your operation itself.
Therefore, if your algorithm requires more operations (yet these operations are performed on data that is already in cache [and therefore no fetching required]), it will still out-perform your algorithm with fewer operations (which must be performed on out-of-cache data [and therefore require a fetch]) in terms of actual wall-time.
In contexts where data security is a concern, a more complex algorithm may be preferable to a less complex algorithm if the more complex algorithm has better resistance to timing attacks.
Alistra nailed it but failed to provide any examples so I will.
You have a list of 10,000 UPC codes for what your store sells. 10 digit UPC, integer for price (price in pennies) and 30 characters of description for the receipt.
O(log N) approach: You have a sorted list. 44 bytes if ASCII, 84 if Unicode. Alternately, treat the UPC as an int64 and you get 42 & 72 bytes. 10,000 records--in the highest case you're looking at a bit under a megabyte of storage.
O(1) approach: Don't store the UPC, instead you use it as an entry into the array. In the lowest case you're looking at almost a third of a terabyte of storage.
Which approach you use depends on your hardware. On most any reasonable modern configuration you're going to use the log N approach. I can picture the second approach being the right answer if for some reason you're running in an environment where RAM is critically short but you have plenty of mass storage. A third of a terabyte on a disk is no big deal, getting your data in one probe of the disk is worth something. The simple binary approach takes 13 on average. (Note, however, that by clustering your keys you can get this down to a guaranteed 3 reads and in practice you would cache the first one.)
Consider a red-black tree. It has access, search, insert, and delete of O(log n). Compare to an array, which has access of O(1) and the rest of the operations are O(n).
So given an application where we insert, delete, or search more often than we access and a choice between only these two structures, we would prefer the red-black tree. In this case, you might say we prefer the red-black tree's more cumbersome O(log n) access time.
Why? Because the access is not our overriding concern. We are making a trade off: the performance of our application is more heavily influenced by factors other than this one. We allow this particular algorithm to suffer performance because we make large gains by optimizing other algorithms.
So the answer to your question is simply this: when the algorithm's growth rate isn't what we want to optimize, when we want to optimize something else. All of the other answers are special cases of this. Sometimes we optimize the run time of other operations. Sometimes we optimize for memory. Sometimes we optimize for security. Sometimes we optimize maintainability. Sometimes we optimize for development time. Even the overriding constant being low enough to matter is optimizing for run time when you know the growth rate of the algorithm isn't the greatest impact on run time. (If your data set was outside this range, you would optimize for the growth rate of the algorithm because it would eventually dominate the constant.) Everything has a cost, and in many cases, we trade the cost of a higher growth rate for the algorithm to optimize something else.
Yes.
In a real case, we ran some tests on doing table lookups with both short and long string keys.
We used a std::map, a std::unordered_map with a hash that samples at most 10 times over the length of the string (our keys tend to be guid-like, so this is decent), and a hash that samples every character (in theory reduced collisions), an unsorted vector where we do a == compare, and (if I remember correctly) an unsorted vector where we also store a hash, first compare the hash, then compare the characters.
These algorithms range from O(1) (unordered_map) to O(n) (linear search).
For modest sized N, quite often the O(n) beat the O(1). We suspect this is because the node-based containers required our computer to jump around in memory more, while the linear-based containers did not.
O(lg n) exists between the two. I don't remember how it did.
The performance difference wasn't that large, and on larger data sets the hash-based one performed much better. So we stuck with the hash-based unordered map.
In practice, for reasonable sized n, O(lg n) is O(1). If your computer only has room for 4 billion entries in your table, then O(lg n) is bounded above by 32. (lg(2^32)=32) (in computer science, lg is short hand for log based 2).
In practice, lg(n) algorithms are slower than O(1) algorithms not because of the logarithmic growth factor, but because the lg(n) portion usually means there is a certain level of complexity to the algorithm, and that complexity adds a larger constant factor than any of the "growth" from the lg(n) term.
However, complex O(1) algorithms (like hash mapping) can easily have a similar or larger constant factor.
The possibility to execute an algorithm in parallel.
I don't know if there is an example for the classes O(log n) and O(1), but for some problems, you choose an algorithm with a higher complexity class when the algorithm is easier to execute in parallel.
Some algorithms cannot be parallelized but have so low complexity class. Consider another algorithm which achieves the same result and can be parallelized easily, but has a higher complexity class. When executed on one machine, the second algorithm is slower, but when executed on multiple machines, the real execution time gets lower and lower while the first algorithm cannot speed up.
Let's say you're implementing a blacklist on an embedded system, where numbers between 0 and 1,000,000 might be blacklisted. That leaves you two possible options:
Use a bitset of 1,000,000 bits
Use a sorted array of the blacklisted integers and use a binary search to access them
Access to the bitset will have guaranteed constant access. In terms of time complexity, it is optimal. Both from a theoretical and from a practical point view (it is O(1) with an extremely low constant overhead).
Still, you might want to prefer the second solution. Especially if you expect the number of blacklisted integers to be very small, as it will be more memory efficient.
And even if you do not develop for an embedded system where memory is scarce, I just can increase the arbitrary limit of 1,000,000 to 1,000,000,000,000 and make the same argument. Then the bitset would require about 125G of memory. Having a guaranteed worst-case complexitity of O(1) might not convince your boss to provide you such a powerful server.
Here, I would strongly prefer a binary search (O(log n)) or binary tree (O(log n)) over the O(1) bitset. And probably, a hash table with its worst-case complexity of O(n) will beat all of them in practice.
My answer here Fast random weighted selection across all rows of a stochastic matrix is an example where an algorithm with complexity O(m) is faster than one with complexity O(log(m)), when m is not too big.
A more general question is if there are situations where one would prefer an O(f(n)) algorithm to an O(g(n)) algorithm even though g(n) << f(n) as n tends to infinity. As others have already mentioned, the answer is clearly "yes" in the case where f(n) = log(n) and g(n) = 1. It is sometimes yes even in the case that f(n) is polynomial but g(n) is exponential. A famous and important example is that of the Simplex Algorithm for solving linear programming problems. In the 1970s it was shown to be O(2^n). Thus, its worse-case behavior is infeasible. But -- its average case behavior is extremely good, even for practical problems with tens of thousands of variables and constraints. In the 1980s, polynomial time algorithms (such a Karmarkar's interior-point algorithm) for linear programming were discovered, but 30 years later the simplex algorithm still seems to be the algorithm of choice (except for certain very large problems). This is for the obvious reason that average-case behavior is often more important than worse-case behavior, but also for a more subtle reason that the simplex algorithm is in some sense more informative (e.g. sensitivity information is easier to extract).
People have already answered your exact question, so I'll tackle a slightly different question that people may actually be thinking of when coming here.
A lot of the "O(1) time" algorithms and data structures actually only take expected O(1) time, meaning that their average running time is O(1), possibly only under certain assumptions.
Common examples: hashtables, expansion of "array lists" (a.k.a. dynamically sized arrays/vectors).
In such scenarios, you may prefer to use data structures or algorithms whose time is guaranteed to be absolutely bounded logarithmically, even though they may perform worse on average.
An example might therefore be a balanced binary search tree, whose running time is worse on average but better in the worst case.
To put my 2 cents in:
Sometimes a worse complexity algorithm is selected in place of a better one, when the algorithm runs on a certain hardware environment. Suppose our O(1) algorithm non-sequentially accesses every element of a very big, fixed-size array to solve our problem. Then put that array on a mechanical hard drive, or a magnetic tape.
In that case, the O(logn) algorithm (suppose it accesses disk sequentially), becomes more favourable.
There is a good use case for using a O(log(n)) algorithm instead of an O(1) algorithm that the numerous other answers have ignored: immutability. Hash maps have O(1) puts and gets, assuming good distribution of hash values, but they require mutable state. Immutable tree maps have O(log(n)) puts and gets, which is asymptotically slower. However, immutability can be valuable enough to make up for worse performance and in the case where multiple versions of the map need to be retained, immutability allows you to avoid having to copy the map, which is O(n), and therefore can improve performance.
Simply: Because the coefficient - the costs associated with setup, storage, and the execution time of that step - can be much, much larger with a smaller big-O problem than with a larger one. Big-O is only a measure of the algorithms scalability.
Consider the following example from the Hacker's Dictionary, proposing a sorting algorithm relying on the Multiple Worlds Interpretation of Quantum Mechanics:
Permute the array randomly using a quantum process,
If the array is not sorted, destroy the universe.
All remaining universes are now sorted [including the one you are in].
(Source: http://catb.org/~esr/jargon/html/B/bogo-sort.html)
Notice that the big-O of this algorithm is O(n), which beats any known sorting algorithm to date on generic items. The coefficient of the linear step is also very low (since it's only a comparison, not a swap, that is done linearly). A similar algorithm could, in fact, be used to solve any problem in both NP and co-NP in polynomial time, since each possible solution (or possible proof that there is no solution) can be generated using the quantum process, then verified in polynomial time.
However, in most cases, we probably don't want to take the risk that Multiple Worlds might not be correct, not to mention that the act of implementing step 2 is still "left as an exercise for the reader".
At any point when n is bounded and the constant multiplier of O(1) algorithm is higher than the bound on log(n). For example, storing values in a hashset is O(1), but may require an expensive computation of a hash function. If the data items can be trivially compared (with respect to some order) and the bound on n is such that log n is significantly less than the hash computation on any one item, then storing in a balanced binary tree may be faster than storing in a hashset.
In a realtime situation where you need a firm upper bound you would select e.g. a heapsort as opposed to a Quicksort, because heapsort's average behaviour is also its worst-case behaviour.
Adding to the already good answers.A practical example would be Hash indexes vs B-tree indexes in postgres database.
Hash indexes form a hash table index to access the data on the disk while btree as the name suggests uses a Btree data structure.
In Big-O time these are O(1) vs O(logN).
Hash indexes are presently discouraged in postgres since in a real life situation particularly in database systems, achieving hashing without collision is very hard(can lead to a O(N) worst case complexity) and because of this, it is even more harder to make them crash safe (called write ahead logging - WAL in postgres).
This tradeoff is made in this situation since O(logN) is good enough for indexes and implementing O(1) is pretty hard and the time difference would not really matter.
When n is small, and O(1) is constantly slow.
When the "1" work unit in O(1) is very high relative to the work unit in O(log n) and the expected set size is small-ish. For example, it's probably slower to compute Dictionary hash codes than iterate an array if there are only two or three items.
or
When the memory or other non-time resource requirements in the O(1) algorithm are exceptionally large relative to the O(log n) algorithm.
when redesigning a program, a procedure is found to be optimized with O(1) instead of O(lgN), but if it's not the bottleneck of this program, and it's hard to understand the O(1) alg. Then you would not have to use O(1) algorithm
when O(1) needs much memory that you cannot supply, while the time of O(lgN) can be accepted.
This is often the case for security applications that we want to design problems whose algorithms are slow on purpose in order to stop someone from obtaining an answer to a problem too quickly.
Here are a couple of examples off the top of my head.
Password hashing is sometimes made arbitrarily slow in order to make it harder to guess passwords by brute-force. This Information Security post has a bullet point about it (and much more).
Bit Coin uses a controllably slow problem for a network of computers to solve in order to "mine" coins. This allows the currency to be mined at a controlled rate by the collective system.
Asymmetric ciphers (like RSA) are designed to make decryption without the keys intentionally slow in order to prevent someone else without the private key to crack the encryption. The algorithms are designed to be cracked in hopefully O(2^n) time where n is the bit-length of the key (this is brute force).
Elsewhere in CS, Quick Sort is O(n^2) in the worst case but in the general case is O(n*log(n)). For this reason, "Big O" analysis sometimes isn't the only thing you care about when analyzing algorithm efficiency.
There are plenty of good answers, a few of which mention the constant factor, the input size and memory constraints, among many other reasons complexity is only a theoretical guideline rather than the end-all determination of real-world fitness for a given purpose or speed.
Here's a simple, concrete example to illustrate these ideas. Let's say we want to figure out whether an array has a duplicate element. The naive quadratic approach is to write a nested loop:
const hasDuplicate = arr => {
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
if (arr[i] === arr[j]) {
return true;
}
}
}
return false;
};
console.log(hasDuplicate([1, 2, 3, 4]));
console.log(hasDuplicate([1, 2, 4, 4]));
But this can be done in linear time by creating a set data structure (i.e. removing duplicates), then comparing its size to the length of the array:
const hasDuplicate = arr => new Set(arr).size !== arr.length;
console.log(hasDuplicate([1, 2, 3, 4]));
console.log(hasDuplicate([1, 2, 4, 4]));
Big O tells us is that the new Set approach will scale a great deal better from a time complexity standpoint.
However, it turns out that the "naive" quadratic approach has a lot going for it that Big O can't account for:
No additional memory usage
No heap memory allocation (no new)
No garbage collection for the temporary Set
Early bailout; in a case when the duplicate is known to be likely in the front of the array, there's no need to check more than a few elements.
If our use case is on bounded small arrays, we have a resource-constrained environment and/or other known common-case properties allow us to establish through benchmarks that the nested loop is faster on our particular workload, it might be a good idea.
On the other hand, maybe the set can be created one time up-front and used repeatedly, amortizing its overhead cost across all of the lookups.
This leads inevitably to maintainability/readability/elegance and other "soft" costs. In this case, the new Set() approach is probably more readable, but it's just as often (if not more often) that achieving the better complexity comes at great engineering cost.
Creating and maintaining a persistent, stateful Set structure can introduce bugs, memory/cache pressure, code complexity, and all other manner of design tradeoffs. Negotiating these tradeoffs optimally is a big part of software engineering, and time complexity is just one factor to help guide that process.
A few other examples that I don't see mentioned yet:
In real-time environments, for example resource-constrained embedded systems, sometimes complexity sacrifices are made (typically related to caches and memory or scheduling) to avoid incurring occasional worst-case penalties that can't be tolerated because they might cause jitter.
Also in embedded programming, the size of the code itself can cause cache pressure, impacting memory performance. If an algorithm has worse complexity but will result in massive code size savings, that might be a reason to choose it over an algorithm that's theoretically better.
In most implementations of recursive linearithmic algorithms like quicksort, when the array is small enough, a quadratic sorting algorithm like insertion sort is often called because the overhead of recursive function calls on increasingly tiny arrays tends to outweigh the cost of nested loops. Insertion sort is also fast on mostly-sorted arrays as the inner loop won't run much. This answer discusses this in an older version of Chrome's V8 engine before they moved to Timsort.
What is the use of Big-O notation in computer science if it doesn't give all the information needed?
For example, if one algorithm runs at 1000n and one at n, it is true that they are both O(n). But I still may make a foolish choice based on this information, since one algorithm takes 1000 times as long as the other for any given input.
I still need to know all the parts of the equation, including the constant, to make an informed choice, so what is the importance of this "intermediate" comparison? I end up loosing important information when it gets reduced to this form, and what do I gain?
What does that constant factor represent? You can't say with certainty, for example, that an algorithm that is O(1000n) will be slower than an algorithm that's O(5n). It might be that the 1000n algorithm loads all data into memory and makes 1,000 passes over that data, and the 5n algorithm makes five passes over a file that's stored on a slow I/O device. The 1000n algorithm will run faster even though its "constant" is much larger.
In addition, some computers perform some operations more quickly than other computers do. It's quite common, given two O(n) algorithms (call them A and B), for A to execute faster on one computer and B to execute faster on the other computer. Or two different implementations of the same algorithm can have widely varying runtimes on the same computer.
Asymptotic analysis, as others have said, gives you an indication of how an algorithm's running time varies with the size of the input. It's useful for giving you a good starting place in algorithm selection. Quick reference will tell you that a particular algorithm is O(n) or O(n log n) or whatever, but it's very easy to find more detailed information on most common algorithms. Still, that more detailed analysis will only give you a constant number without saying how that number relates to real running time.
In the end, the only way you can determine which algorithm is right for you is to study it yourself and then test it against your expected data.
In short, I think you're expecting too much from asymptotic analysis. It's a useful "first line" filter. But when you get beyond that you have to look for more information.
As you correctly noted, it does not give you information on the exact running time of an algorithm. It is mainly used to indicate the complexity of an algorithm, to indicate if it is linear in the input size, quadratic, exponential, etc. This is important when choosing between algorithms if you know that your input size is large, since even a 1000n algorithm well beat a 1.23 exp(n) algorithm for large enough n.
In real world algorithms, the hidden 'scaling factor' is of course important. It is therefore not uncommon to use an algorithm with a 'worse' complexity if it has a lower scaling factor. Many practical implementations of sorting algorithms are for example 'hybrid' and will resort to some 'bad' algorithm like insertion sort (which is O(n^2) but very simple to implement) for n < 10, while changing to quicksort (which is O(n log(n)) but more complex) for n >= 10.
Big-O tells you how the runtime or memory consumption of a process changes as the size of its input changes. O(n) and O(1000n) are both still O(n) -- if you double the size of the input, then for all practical purposes the runtime doubles too.
Now, we can have an O(n) algorithm and an O(n2) algorithm where the coefficient of n is 1000000 and the coefficient of n2 is 1, in which case the O(n2) algorithm would outperform the O(n) for smaller n values. This doesn't change the fact, however, that the second algorithm's runtime grows more rapidly than the first's, and this is the information that big-O tells us. There will be some input size at which the O(n) algorithm begins to outperform the O(n2) algorithm.
In addition to the hidden impact of the constant term, complexity notation also only considers the worst case instance of a problem.
Case in point, the simplex method (linear programming) has exponential complexity for all known implementations. However, the simplex method works much faster in practice than the provably polynomial-time interior point methods.
Complexity notation has much value for theoretical problem classification. If you want some more information on practical consequences check out "Smoothed Analysis" by Spielman: http://www.cs.yale.edu/homes/spielman
This is what you are looking for.
It's main purpose is for rough comparisons of logic. The difference of O(n) and O(1000n) is large for n ~ 1000 (n roughly equal to 1000) and n < 1000, but when you compare it to values where n >> 1000 (n much larger than 1000) the difference is miniscule.
You are right in saying they both scale linearly and knowing the coefficient helps in a detailed analysis but generally in computing the difference between linear (O(cn)) and exponential (O(cn^x)) performance is more important to note than the difference between two linear times. There is a larger value in the comparisons of runtime of higher orders such as and Where the performance difference scales exponentially.
The overall purpose of Big O notation is to give a sense of relative performance time in order to compare and further optimize algorithms.
You're right that it doesn't give you all information, but there's no single metric in any field that does that.
Big-O notation tells you how quickly the performance gets worse, as your dataset gets larger. In other words, it describes the type of performance curve, but not the absolute performance.
Generally, Big-O notation is useful to express an algorithm's scaling performance as it falls into one of three basic categories:
Linear
Logarithmic (or "linearithmic")
Exponential
It is possible to do deep analysis of an algorithm for very accurate performance measurements, but it is time consuming and not really necessary to get a broad indication of performance.
For instance, suppose I have an algorithm that's O(n) and an algorithm that's an amortized O(n). Is it fair to say that in strictly big oh terms, the non-amortized algorithm will always be as fast or faster than the amortized algorithm? Or is there ever any reason to prefer the amortized version (ignoring things like code simplicity or ease of implementation)?
Big O notation only tells you how your code scales, not how fast it will be for finite N.
The difference between amortized and non-amortized is important if you care about worst-case performance or latency (such as real-time or interactive systems). If you only care about average throughput, though, they're for all practical purposes the same. Amortized analysis is good enough even in some very performance-critical situations like scientific computing and large-scale data mining.
Or is there ever any reason to prefer the amortized version
Smaller constants.
The main difference between a O(n) algorithm and an amortized O(n) algorithm is that you don't know anything about the worst-case behavior of the amortized O(n) algorithm. In practice, that doesn't matter very often: If your algorithm is being run many times, then you can count on the law of averages to balance out a few bad cases, and if your algorithm isn't being run very many times, then it's unlikely that you'll ever hit the worst case at all.
The only cases where the word "amortized" should matter are cases where you can't accept occasional bad performance for some reason. For example, in GUI applications, you would happily give up some average-case performance in return for a guarantee that you'll never bog down and stop responding while the user is sitting there and getting bored. In such applications, you want to make sure that even the worst-case behavior is good for any code that could cause the GUI to stop responding.
Still, most of the time, you don't need to worry about amortized O(n) versus O(n), and you can worry instead about what the constant factors are (as others have already said).
Big O notation tells you about how your algorithm changes with growing input.
It is not a shortcut for profiling your code.
It is possible that a better algorithm is O(n^2) for all n in your program, because there is a constant in an O(n) that is larger.
So your choice of algorithms really depends on which algorithm is faster for your input size. I guess the answer to your question is to profile both algorithms in your program and then decide which is faster.
A classic example of the need for amortized algorithm would be std::vector for which insert is amortized O(1).
Some reasons to use amortized algorithms:
Much more efficient average case.
Easier Implementation.
No worst case guarantee algorithm exists.
Strictly speaking big-O isn't precise enough of a measure to so be able to say that an algorithm with O(n) is faster than one with a amortized O(n). Think about it. If you drop down a level of fidelity in your complexity analysis, the constant factors may be significantly different and make the amortized version faster.
Also, the impact of amortization tends to depend on use. For example, if you're using a hash table, the impact of the amortization will be largely dependent on your ratio of get to add operations. So if you add 1000 entries, then do a billion lookups, the fact that you had to rehash a couple times doesn't make much of a difference. But if you're constantly adding entries the cost of rehashing might add up.
That's why amortization is different than worst case. Big-O reflects the worst case, while amortization lets you say "one in every x will take a hit, and x is sufficiently large that it doesn't matter." Also, if you consider examples like insertion into a hash table, x grows according to some constant. So if you have a hash table that starts with 100 buckets and doubles every rehash, then the rehashes are asymptotically going to get farther and farther apart. In addition, the absolute worst case complexity for an algorithm with an amortized complexity is dependent on previous calls, while in measures like average case the calls are considered independent.
Whenever I consider algorithms/data structures I tend to replace the log(N) parts by constants. Oh, I know log(N) diverges - but does it matter in real world applications?
log(infinity) < 100 for all practical purposes.
I am really curious for real world examples where this doesn't hold.
To clarify:
I understand O(f(N))
I am curious about real world examples where the asymptotic behaviour matters more than the constants of the actual performance.
If log(N) can be replaced by a constant it still can be replaced by a constant in O( N log N).
This question is for the sake of (a) entertainment and (b) to gather arguments to use if I run (again) into a controversy about the performance of a design.
Big O notation tells you about how your algorithm changes with growing input. O(1) tells you it doesn't matter how much your input grows, the algorithm will always be just as fast. O(logn) says that the algorithm will be fast, but as your input grows it will take a little longer.
O(1) and O(logn) makes a big diference when you start to combine algorithms.
Take doing joins with indexes for example. If you could do a join in O(1) instead of O(logn) you would have huge performance gains. For example with O(1) you can join any amount of times and you still have O(1). But with O(logn) you need to multiply the operation count by logn each time.
For large inputs, if you had an algorithm that was O(n^2) already, you would much rather do an operation that was O(1) inside, and not O(logn) inside.
Also remember that Big-O of anything can have a constant overhead. Let's say that constant overhead is 1 million. With O(1) that constant overhead does not amplify the number of operations as much as O(logn) does.
Another point is that everyone thinks of O(logn) representing n elements of a tree data structure for example. But it could be anything including bytes in a file.
I think this is a pragmatic approach; O(logN) will never be more than 64. In practice, whenever terms get as 'small' as O(logN), you have to measure to see if the constant factors win out. See also
Uses of Ackermann function?
To quote myself from comments on another answer:
[Big-Oh] 'Analysis' only matters for factors
that are at least O(N). For any
smaller factor, big-oh analysis is
useless and you must measure.
and
"With O(logN) your input size does
matter." This is the whole point of
the question. Of course it matters...
in theory. The question the OP asks
is, does it matter in practice? I
contend that the answer is no, there
is not, and never will be, a data set
for which logN will grow so fast as to
always be beaten a constant-time
algorithm. Even for the largest
practical dataset imaginable in the
lifetimes of our grandchildren, a logN
algorithm has a fair chance of beating
a constant time algorithm - you must
always measure.
EDIT
A good talk:
http://www.infoq.com/presentations/Value-Identity-State-Rich-Hickey
about halfway through, Rich discusses Clojure's hash tries, which are clearly O(logN), but the base of the logarithm is large and so the depth of the trie is at most 6 even if it contains 4 billion values. Here "6" is still an O(logN) value, but it is an incredibly small value, and so choosing to discard this awesome data structure because "I really need O(1)" is a foolish thing to do. This emphasizes how most of the other answers to this question are simply wrong from the perspective of the pragmatist who wants their algorithm to "run fast" and "scale well", regardless of what the "theory" says.
EDIT
See also
http://queue.acm.org/detail.cfm?id=1814327
which says
What good is an O(log2(n)) algorithm
if those operations cause page faults
and slow disk operations? For most
relevant datasets an O(n) or even an
O(n^2) algorithm, which avoids page
faults, will run circles around it.
(but go read the article for context).
This is a common mistake - remember Big O notation is NOT telling you about the absolute performance of an algorithm at a given value, it's simply telling you the behavior of an algorithm as you increase the size of the input.
When you take it in that context it becomes clear why an algorithm A ~ O(logN) and an algorithm B ~ O(1) algorithm are different:
if I run A on an input of size a, then on an input of size 1000000*a, I can expect the second input to take log(1,000,000) times as long as the first input
if I run B on an input of size a, then on an input of size 1000000*a, I can expect the second input to take about the same amount of time as the first input
EDIT: Thinking over your question some more, I do think there's some wisdom to be had in it. While I would never say it's correct to say O(lgN) == O(1), It IS possible that an O(lgN) algorithm might be used over an O(1) algorithm. This draws back to the point about absolute performance above: Just knowing one algorithm is O(1) and another algorithm is O(lgN) is NOT enough to declare you should use the O(1) over the O(lgN), it's certainly possible given your range of possible inputs an O(lgN) might serve you best.
You asked for a real-world example. I'll give you one. Computational biology. One strand of DNA encoded in ASCII is somewhere on the level of gigabytes in space. A typical database will obviously have many thousands of such strands.
Now, in the case of an indexing/searching algorithm, that log(n) multiple makes a large difference when coupled with constants. The reason why? This is one of the applications where the size of your input is astronomical. Additionally, the input size will always continue to grow.
Admittedly, these type of problems are rare. There are only so many applications this large. In those circumstances, though... it makes a world of difference.
Equality, the way you're describing it, is a common abuse of notation.
To clarify: we usually write f(x) = O(logN) to imply "f(x) is O(logN)".
At any rate, O(1) means a constant number of steps/time (as an upper bound) to perform an action regardless of how large the input set is. But for O(logN), number of steps/time still grows as a function of the input size (the logarithm of it), it just grows very slowly. For most real world applications you may be safe in assuming that this number of steps will not exceed 100, however I'd bet there are multiple examples of datasets large enough to mark your statement both dangerous and void (packet traces, environmental measurements, and many more).
For small enough N, O(N^N) can in practice be replaced with 1. Not O(1) (by definition), but for N=2 you can see it as one operation with 4 parts, or a constant-time operation.
What if all operations take 1hour? The difference between O(log N) and O(1) is then large, even with small N.
Or if you need to run the algorithm ten million times? Ok, that took 30minutes, so when I run it on a dataset a hundred times as large it should still take 30minutes because O(logN) is "the same" as O(1).... eh...what?
Your statement that "I understand O(f(N))" is clearly false.
Real world applications, oh... I don't know.... EVERY USE OF O()-notation EVER?
Binary search in sorted list of 10 million items for example. It's the very REASON we use hash tables when the data gets big enough. If you think O(logN) is the same as O(1), then why would you EVER use a hash instead of a binary tree?
As many have already said, for the real world, you need to look at the constant factors first, before even worrying about factors of O(log N).
Then, consider what you will expect N to be. If you have good reason to think that N<10, you can use a linear search instead of a binary one. That's O(N) instead of O(log N), which according to your lights would be significant -- but a linear search that moves found elements to the front may well outperform a more complicated balanced tree, depending on the application.
On the other hand, note that, even if log N is not likely to exceed 50, a performance factor of 10 is really huge -- if you're compute-bound, a factor like that can easily make or break your application. If that's not enough for you, you'll frequently see factors of (log N)^2 or (logN)^3 in algorithms, so even if you think you can ignore one factor of (log N), that doesn't mean you can ignore more of them.
Finally, note that the simplex algorithm for linear programming has a worst case performance of O(2^n). However, for practical problems, the worst case never comes up; in practice, the simplex algorithm is fast, relatively simple, and consequently very popular.
About 30 years ago, someone developed a polynomial-time algorithm for linear programming, but it was not initially practical because the result was too slow.
Nowadays, there are practical alternative algorithms for linear programming (with polynomial-time wost-case, for what that's worth), which can outperform the simplex method in practice. But, depending on the problem, the simplex method is still competitive.
The observation that O(log n) is oftentimes indistinguishable from O(1) is a good one.
As a familiar example, suppose we wanted to find a single element in a sorted array of one 1,000,000,000,000 elements:
with linear search, the search takes on average 500,000,000,000 steps
with binary search, the search takes on average 40 steps
Suppose we added a single element to the array we are searching, and now we must search for another element:
with linear search, the search takes on average 500,000,000,001 steps (indistinguishable change)
with binary search, the search takes on average 40 steps (indistinguishable change)
Suppose we doubled the number of elements in the array we are searching, and now we must search for another element:
with linear search, the search takes on average 1,000,000,000,000 steps (extraordinarily noticeable change)
with binary search, the search takes on average 41 steps (indistinguishable change)
As we can see from this example, for all intents and purposes, an O(log n) algorithm like binary search is oftentimes indistinguishable from an O(1) algorithm like omniscience.
The takeaway point is this: *we use O(log n) algorithms because they are often indistinguishable from constant time, and because they often perform phenomenally better than linear time algorithms.
Obviously, these examples assume reasonable constants. Obviously, these are generic observations and do not apply to all cases. Obviously, these points apply at the asymptotic end of the curve, not the n=3 end.
But this observation explains why, for example, we use such techniques as tuning a query to do an index seek rather than a table scan - because an index seek operates in nearly constant time no matter the size of the dataset, while a table scan is crushingly slow on sufficiently large datasets. Index seek is O(log n).
You might be interested in Soft-O, which ignores logarithmic cost. Check this paragraph in Wikipedia.
What do you mean by whether or not it "matters"?
If you're faced with the choice of an O(1) algorithm and a O(lg n) one, then you should not assume they're equal. You should choose the constant-time one. Why wouldn't you?
And if no constant-time algorithm exists, then the logarithmic-time one is usually the best you can get. Again, does it then matter? You just have to take the fastest you can find.
Can you give me a situation where you'd gain anything by defining the two as equal? At best, it'd make no difference, and at worst, you'd hide some real scalability characteristics. Because usually, a constant-time algorithm will be faster than a logarithmic one.
Even if, as you say, lg(n) < 100 for all practical purposes, that's still a factor 100 on top of your other overhead. If I call your function, N times, then it starts to matter whether your function runs logarithmic time or constant, because the total complexity is then O(n lg n) or O(n).
So rather than asking if "it matters" that you assume logarithmic complexity to be constant in "the real world", I'd ask if there's any point in doing that.
Often you can assume that logarithmic algorithms are fast enough, but what do you gain by considering them constant?
O(logN)*O(logN)*O(logN) is very different. O(1) * O(1) * O(1) is still constant.
Also a simple quicksort-style O(nlogn) is different than O(n O(1))=O(n). Try sorting 1000 and 1000000 elements. The latter isn't 1000 times slower, it's 2000 times, because log(n^2)=2log(n)
The title of the question is misleading (well chosen to drum up debate, mind you).
O(log N) == O(1) is obviously wrong (and the poster is aware of this). Big O notation, by definition, regards asymptotic analysis. When you see O(N), N is taken to approach infinity. If N is assigned a constant, it's not Big O.
Note, this isn't just a nitpicky detail that only theoretical computer scientists need to care about. All of the arithmetic used to determine the O function for an algorithm relies on it. When you publish the O function for your algorithm, you might be omitting a lot of information about it's performance.
Big O analysis is cool, because it lets you compare algorithms without getting bogged down in platform specific issues (word sizes, instructions per operation, memory speed versus disk speed). When N goes to infinity, those issues disappear. But when N is 10000, 1000, 100, those issues, along with all of the other constants that we left out of the O function, start to matter.
To answer the question of the poster: O(log N) != O(1), and you're right, algorithms with O(1) are sometimes not much better than algorithms with O(log N), depending on the size of the input, and all of those internal constants that got omitted during Big O analysis.
If you know you're going to be cranking up N, then use Big O analysis. If you're not, then you'll need some empirical tests.
In theory
Yes, in practical situations log(n) is bounded by a constant, we'll say 100. However, replacing log(n) by 100 in situations where it's correct is still throwing away information, making the upper bound on operations that you have calculated looser and less useful. Replacing an O(log(n)) by an O(1) in your analysis could result in your large n case performing 100 times worse than you expected based on your small n case. Your theoretical analysis could have been more accurate and could have predicted an issue before you'd built the system.
I would argue that the practical purpose of big-O analysis is to try and predict the execution time of your algorithm as early as possible. You can make your analysis easier by crossing out the log(n) terms, but then you've reduced the predictive power of the estimate.
In practice
If you read the original papers by Larry Page and Sergey Brin on the Google architecture, they talk about using hash tables for everything to ensure that e.g. the lookup of a cached web page only takes one hard-disk seek. If you used B-tree indices to lookup you might need four or five hard-disk seeks to do an uncached lookup [*]. Quadrupling your disk requirements on your cached web page storage is worth caring about from a business perspective, and predictable if you don't cast out all the O(log(n)) terms.
P.S. Sorry for using Google as an example, they're like Hitler in the computer science version of Godwin's law.
[*] Assuming 4KB reads from disk, 100bn web pages in the index, ~ 16 bytes per key in a B-tree node.
As others have pointed out, Big-O tells you about how the performance of your problem scales. Trust me - it matters. I have encountered several times algorithms that were just terrible and failed to meet the customers demands because they were too slow. Understanding the difference and finding an O(1) solution is a lot of times a huge improvement.
However, of course, that is not the whole story - for instance, you may notice that quicksort algorithms will always switch to insertion sort for small elements (Wikipedia says 8 - 20) because of the behaviour of both algorithms on small datasets.
So it's a matter of understanding what tradeoffs you will be doing which involves a thorough understanding of the problem, the architecture, & experience to understand which to use, and how to adjust the constants involved.
No one is saying that O(1) is always better than O(log N). However, I can guarantee you that an O(1) algorithm will also scale way better, so even if you make incorrect assumptions about how many users will be on the system, or the size of the data to process, it won't matter to the algorithm.
Yes, log(N) < 100 for most practical purposes, and No, you can not always replace it by constant.
For example, this may lead to serious errors in estimating performance of your program. If O(N) program processed array of 1000 elements in 1 ms, then you are sure it will process 106 elements in 1 second (or so). If, though, the program is O(N*logN), then it will take it ~2 secs to process 106 elements. This difference may be crucial - for example, you may think you've got enough server power because you get 3000 requests per hour and you think your server can handle up to 3600.
Another example. Imagine you have function f() working in O(logN), and on each iteration calling function g(), which works in O(logN) as well. Then, if you replace both logs by constants, you think that your program works in constant time. Reality will be cruel though - two logs may give you up to 100*100 multiplicator.
The rules of determining the Big-O notation are simpler when you don't decide that O(log n) = O(1).
As krzysio said, you may accumulate O(log n)s and then they would make a very noticeable difference. Imagine you do a binary search: O(log n) comparisons, and then imagine that each comparison's complexity O(log n). If you neglect both you get O(1) instead of O(log2n). Similarly you may somehow arrive at O(log10n) and then you'll notice a big difference for not too large "n"s.
Assume that in your entire application, one algorithm accounts for 90% of the time the user waits for the most common operation.
Suppose in real time the O(1) operation takes a second on your architecture, and the O(logN) operation is basically .5 seconds * log(N). Well, at this point I'd really like to draw you a graph with an arrow at the intersection of the curve and the line, saying, "It matters right here." You want to use the log(N) op for small datasets and the O(1) op for large datasets, in such a scenario.
Big-O notation and performance optimization is an academic exercise rather than delivering real value to the user for operations that are already cheap, but if it's an expensive operation on a critical path, then you bet it matters!
For any algorithm that can take inputs of different sizes N, the number of operations it takes is upper-bounded by some function f(N).
All big-O tells you is the shape of that function.
O(1) means there is some number A such that f(N) < A for large N.
O(N) means there is some A such that f(N) < AN for large N.
O(N^2) means there is some A such that f(N) < AN^2 for large N.
O(log(N)) means there is some A such that f(N) < AlogN for large N.
Big-O says nothing about how big A is (i.e. how fast the algorithm is), or where these functions cross each other. It only says that when you are comparing two algorithms, if their big-Os differ, then there is a value of N (which may be small or it may be very large) where one algorithm will start to outperform the other.
you are right, in many cases it does not matter for pracitcal purposes. but the key question is "how fast GROWS N". most algorithms we know of take the size of the input, so it grows linearily.
but some algorithms have the value of N derived in a complex way. if N is "the number of possible lottery combinations for a lottery with X distinct numbers" it suddenly matters if your algorithm is O(1) or O(logN)
Big-OH tells you that one algorithm is faster than another given some constant factor. If your input implies a sufficiently small constant factor, you could see great performance gains by going with a linear search rather than a log(n) search of some base.
O(log N) can be misleading. Take for example the operations on Red-Black trees.
The operations are O(logN) but rather complex, which means many low level operations.
Whenever N is the amount of objects that is stored in some kind of memory, you're correct. After all, a binary search through EVERY byte representable by a 64-bit pointer can be achieved in just 64 steps. Actually, it's possible to do a binary search of all Planck volumes in the observable universe in just 618 steps.
So in almost all cases, it's safe to approximate O(log N) with O(N) as long as N is (or could be) a physical quantity, and we know for certain that as long as N is (or could be) a physical quantity, then log N < 618
But that is assuming N is that. It may represent something else. Note that it's not always clear what it is. Just as an example, take matrix multiplication, and assume square matrices for simplicity. The time complexity for matrix multiplication is O(N^3) for a trivial algorithm. But what is N here? It is the side length. It is a reasonable way of measuring the input size, but it would also be quite reasonable to use the number of elements in the matrix, which is N^2. Let M=N^2, and now we can say that the time complexity for trivial matrix multiplication is O(M^(3/2)) where M is the number of elements in a matrix.
Unfortunately, I don't have any real world problem per se, which was what you asked. But at least I can make up something that makes some sort of sense:
Let f(S) be a function that returns the sum of the hashes of all the elements in the power set of S. Here is some pesudo:
f(S):
ret = 0
for s = powerset(S))
ret += hash(s)
Here, hash is simply the hash function, and powerset is a generator function. Each time it's called, it will generate the next (according to some order) subset of S. A generator is necessary, because we would not be able to store the lists for huge data otherwise. Btw, here is a python example of such a power set generator:
def powerset(seq):
"""
Returns all the subsets of this set. This is a generator.
"""
if len(seq) <= 1:
yield seq
yield []
else:
for item in powerset(seq[1:]):
yield [seq[0]]+item
yield item
https://www.technomancy.org/python/powerset-generator-python/
So what is the time complexity for f? As with the matrix multiplication, we can choose N to represent many things, but at least two makes a lot of sense. One is number of elements in S, in which case the time complexity is O(2^N), but another sensible way of measuring it is that N is the number of element in the power set of S. In this case the time complexity is O(N)
So what will log N be for sensible sizes of S? Well, list with a million elements are not unusual. If n is the size of S and N is the size of P(S), then N=2^n. So O(log N) = O(log 2^n) = O(n * log 2) = O(n)
In this case it would matter, because it's rare that O(n) == O(log n) in the real world.
I do not believe algorithms where you can freely choose between O(1) with a large constant and O(logN) really exists. If there is N elements to work with at the beginning, it is just plain impossible to make it O(1), the only thing that is possible is move your N to some other part of your code.
What I try to say is that in all real cases I know off you have some space/time tradeoff, or some pre-treatment such as compiling data to a more efficient form.
That is, you do not really go O(1), you just move the N part elsewhere. Either you exchange performance of some part of your code with some memory amount either you exchange performance of one part of your algorithm with another one. To stay sane you should always look at the larger picture.
My point is that if you have N items they can't disappear. In other words you can choose between inefficient O(n^2) algorithms or worse and O(n.logN) : it's a real choice. But you never really go O(1).
What I try to point out is that for every problem and initial data state there is a 'best' algorithm. You can do worse but never better. With some experience you can have a good guessing of what is this intrisic complexity. Then if your overall treatment match that complexity you know you have something. You won't be able to reduce that complexity, but only to move it around.
If problem is O(n) it won't become O(logN) or O(1), you'll merely add some pre-treatment such that the overall complexity is unchanged or worse, and potentially a later step will be improved. Say you want the smaller element of an array, you can search in O(N) or sort the array using any common O(NLogN) sort treatment then have the first using O(1).
Is it a good idea to do that casually ? Only if your problem asked also for second, third, etc. elements. Then your initial problem was truly O(NLogN), not O(N).
And it's not the same if you wait ten times or twenty times longer for your result because you simplified saying O(1) = O(LogN).
I'm waiting for a counter-example ;-) that is any real case where you have choice between O(1) and O(LogN) and where every O(LogN) step won't compare to the O(1). All you can do is take a worse algorithm instead of the natural one or move some heavy treatment to some other part of the larger pictures (pre-computing results, using storage space, etc.)
Let's say you use an image-processing algorithm that runs in O(log N), where N is the number of images. Now... stating that it runs in constant time would make one believe that no matter how many images there are, it would still complete its task it about the same amount of time. If running the algorithm on a single image would hypothetically take a whole day, and assuming that O(logN) will never be more than 100... imagine the surprise of that person that would try to run the algorithm on a very large image database - he would expect it to be done in a day or so... yet it'll take months for it to finish.