If there are 2 algorthims that calculate the same result with different complexities, will O(log n) always be faster? If so please explain. BTW this is not an assignment question.
No. If one algorithm runs in N/100 and the other one in (log N)*100, then the second one will be slower for smaller input sizes. Asymptotic complexities are about the behavior of the running time as the input sizes go to infinity.
No, it will not always be faster. BUT, as the problem size grows larger and larger, eventually you will always reach a point where the O(log n) algorithm is faster than the O(n) one.
In real-world situations, usually the point where the O(log n) algorithm would overtake the O(n) algorithm would come very quickly. There is a big difference between O(log n) and O(n), just like there is a big difference between O(n) and O(n^2).
If you ever have the chance to read Programming Pearls by Jon Bentley, there is an awesome chapter in there where he pits a O(n) algorithm against a O(n^2) one, doing everything possible to give O(n^2) the advantage. (He codes the O(n^2) algorithm in C on an Alpha, and the O(n) algorithm in interpreted BASIC on an old Z80 or something, running at about 1MHz.) It is surprising how fast the O(n) algorithm overtakes the O(n^2) one.
Occasionally, though, you may find a very complex algorithm which has complexity just slightly better than a simpler one. In such a case, don't blindly choose the algorithm with a better big-O -- you may find that it is only faster on extremely large problems.
For the input of size n, an algorithm of O(n) will perform steps proportional to n, while another algorithm of O(log(n)) will perform steps roughly log(n).
Clearly log(n) is smaller than n hence algorithm of complexity O(log(n)) is better. Since it will be much faster.
More Answers from stackoverflow
Related
When trying to properly understand Big-O, I am wondering whether it's true that O(n log n) algorithms are always better than all O(n^2) algorithms.
Are there any particular situations where O(n^2) would be better?
I've read multiple times that in sorting for example, a O(n^2) algorithm like bubble sort can be particularly quick when the data is almost sorted, so would it be quicker than a O(n log n) algorithm, such as merge sort, in this case?
No, O(n log n) algorithms are not always better than O(n^2) ones.
The Big-O notation describes an upper bound of the asymptotic behavior of an algorithm, i.e. for n that tends towards infinity.
In this definition you have to consider some aspects:
The Big-O notation is an upper bound of the algorithm complexity, meaning that for some inputs (like the one you mentioned about sorting algorithms) an algorithm with worst Big-O complexity may actually perform better (bubble sort runs in O(n) for an already sorted array, while mergesort and quicksort takes always at least O(n log n));
The Big-O notation only describes the class of complexity, hiding all the constant factors that in real case scenarios may be relevant. For example, an algorithm that has complexity 1000000 x that is in class O(n) perform worst than an algorithm with complexity 0.5 x^2 (class O(n^2)) for inputs smaller than 2000000. Basically the Big-O notation tells you that for big enough input n, the O(n) algorithms will perform better than O(n^2), but if you work with small inputs you may still prefer the latter solution.
O(n log n) is better than O(n2) asymptotically.
Big-O, Big-Theta, Big-Omega, all those measure the asymptotic behavior of functions, i.e., how functions behave when their argument goes toward a certain limit.
O(n log n) functions grow slower than O(n2) functions, that's what Big-O notation essentially says. However, this does not mean that O(n log n) is always faster. It merely means that at some point, the O(n log n) function will always be cheaper for an ever-rising value of n.
In that image, f(n) = O(g(n)). Note that there is a range where f(n) is actually more costly than g(n), even though it is bounded asymptotically by g(n). However, when talking limits, or asymptotics for that matter, f(n) outperforms g(n) "in the long run," so to say.
In addition to #cadaniluk's answer:
If you restrict the inputs to the algorithms to a very special type, this also can effect the running time. E.g. if you run sorting algorithms only on already sorted lists, BubbleSort will run in linear time, but MergeSort will still need O(n log n).
There are also algorithms that have a bad worst-case complexity, but a good average case complexity. This means that there are bad input instances such that the algorithm is slow, but in total it is very unlikely that you have such a case.
Also never forget, that Big-O notation hides constants and additive functions of lower orders. So an Algorithm with worst-case complexity O(n log n) could actually have a complexity of 2^10000 * n * log n and your O(n^2) algorithm could actually run in 1/2^1000 n^2. So for n < 2^10000 you really want to use the "slower" algorithm.
Here is a practical example.
The GCC implementations of sorting functions have O(n log n) complexity. Still, they employ O(n^2) algorithms as soon as the size of the part being sorted is less than some small constant.
That's because for small sizes, they tend to be faster in practice.
See here for some of the internal implementation.
Example of when I would choose O(n) algorithm over O(1) algorithm if for all of n, O(1) is faster than O(n)
Often, real data lends itself to algorithms with worse time complexities. For example, bubble sort, which runs in O(n^2) time is often faster on almost sorted data. Oftentimes, the constant factors might make an algorithm too slow to be practical. Remember, big-O deals with things that are more efficient in the limit, rather than in the immediate case. An algorithm that is O(1) with a constant factor of 10000000 will be significantly slower than an O(n) algorithm with a constant factor of 1 for n < 10000000.
One example is the O(1) algorithm consumes lots of memory while the O(n) one does not. And memory is more important for you compare to performance.
This question has appeared in my algorithms class. Here's my thought:
I think the answer is no, an algorithm with worst-case time complexity of O(n) is not always faster than an algorithm with worst-case time complexity of O(n^2).
For example, suppose we have total-time functions S(n) = 99999999n and T(n) = n^2. Then clearly S(n) = O(n) and T(n) = O(n^2), but T(n) is faster than S(n) for all n < 99999999.
Is this reasoning valid? I'm slightly skeptical that, while this is a counterexample, it might be a counterexample to the wrong idea.
Thanks so much!
Big-O notation says nothing about the speed of an algorithm for any given input; it describes how the time increases with the number of elements. If your algorithm executes in constant time, but that time is 100 billion years, then it's certainly slower than many linear, quadratic and even exponential algorithms for large ranges of inputs.
But that's probably not really what the question is asking. The question is asking whether an algorithm A1 with worst-case complexity O(N) is always faster than an algorithm A2 with worst-case complexity O(N^2); and by faster it probably refers to the complexity itself. In which case you only need a counter-example, e.g.:
A1 has normal complexity O(log n) but worst-case complexity O(n^2).
A2 has normal complexity O(n) and worst-case complexity O(n).
In this example, A1 is normally faster (i.e. scales better) than A2 even though it has a greater worst-case complexity.
Since the question says Always it means it is enough to find only one counter example to prove that the answer is No.
Example for O(n^2) and O(n logn) but the same is true for O(n^2) and O(n)
One simple example can be a bubble sort where you keep comparing pairs until the array is sorted. Bubble sort is O(n^2).
If you use bubble sort on a sorted array, it will be faster than using other algorithms of time complexity O(nlogn).
You're talking about worst-case complexity here, and for some algorithms the worst case never happen in a practical application.
Saying that an algorithm runs faster than another means it run faster for all input data for all sizes of input. So the answer to your question is obviously no because the worst-case time complexity is not an accurate measure of the running time, it measures the order of growth of the number of operations in a worst case.
In practice, the running time depends of the implementation, and is not only about this number of operations. For example, one has to care about memory allocated, cache-efficiency, space/temporal locality. And obviously, one of the most important thing is the input data.
If you want examples of when the an algorithm runs faster than another while having a higher worst-case complexity, look at all the sorting algorithms and their running time depending of the input.
You are correct in every sense, that you provide a counter example to the statement. If it is for exam, then period, it should grant you full mark.
Yet for a better understanding about big-O notation and complexity stuff, I will share my own reasoning below. I also suggest you to always think the following graph when you are confused, especially the O(n) and O(n^2) line:
Big-O notation
My own reasoning when I first learnt computational complexity is that,
Big-O notation is saying for sufficient large size input, "sufficient" depends on the exact formula (Using the graph, n = 20 when compared O(n) & O(n^2) line), a higher order one will always be slower than lower order one
That means, for small input, there is no guarantee a higher order complexity algorithm will run slower than lower order one.
But Big-O notation tells you an information: When the input size keeping increasing, keep increasing....until a "sufficient" size, after that point, a higher order complexity algorithm will be always slower. And such a "sufficient" size is guaranteed to exist*.
Worst-time complexity
While Big-O notation provides a upper bound of the running time of an algorithm, depends on the structure of the input and the implementation of the algorithm, it may generally have a best complexity, average complexity and worst complexity.
The famous example is sorting algorithm: QuickSort vs MergeSort!
QuickSort, with a worst case of O(n^2)
MergeSort, with a worst case of O(n lg n)
However, Quick Sort is basically always faster than Merge Sort!
So, if your question is about Worst Case Complexity, quick sort & merge sort maybe the best counter example I can think of (Because both of them are common and famous)
Therefore, combine two parts, no matter from the point of view of input size, input structure, algorithm implementation, the answer to your question is NO.
In the Princeton tutorial on Coursera the lecturer explains the common order-of-growth functions that are encountered. He says that linear and linearithmic running times are "what we strive" for and his reasoning was that as the input size increases so too does the running time. I think this is where he made a mistake because I have previously heard him refer to a linear order-of-growth as unsatisfactory for an efficient algorithm.
While he was speaking he also showed a chart that plotted the different running times - constant and logarithmic running times looked to be more efficient. So was this a mistake or is this true?
It is a mistake when taken in the context that O(n) and O(n log n) functions have better complexity than O(1) and O(log n) functions. When looking typical cases of complexity in big O notation:
O(1) < O(log n) < O(n) < O(n log n) < O(n^2)
Notice that this doesn't necessarily mean that they will always be better performance-wise - we could have an O(1) function that takes a long time to execute even though its complexity is unaffected by element count. Such a function would look better in big O notation than an O(log n) function, but could actually perform worse in practice.
Generally speaking: a function with lower complexity (in big O notation) will outperform a function with greater complexity (in big O notation) when n is sufficiently high.
You're missing the broader context in which those statements must have been made. Different kinds of problems have different demands, and often even have theoretical lower bounds on how much work is absolutely necessary to solve them, no matter the means.
For operations like sorting or scanning every element of a simple collection, you can make a hard lower bound of the number of elements in the collection for those operations, because the output depends on every element of the input. [1] Thus, O(n) or O(n*log(n)) are the best one can do.
For other kinds of operations, like accessing a single element of a hash table or linked list, or searching in a sorted set, the algorithm needn't examine all of the input. In those settings, an O(n) operation would be dreadfully slow.
[1] Others will note that sorting by comparisons also has an n*log(n) lower bound, from information-theoretic arguments. There are non-comparison based sorting algorithms that can beat this, for some types of input.
Generally speaking, what we strive for is the best we can manage to do. But depending on what we're doing, that might be O(1), O(log log N), O(log N), O(N), O(N log N), O(N2), O(N3), or (or certain algorithms) perhaps O(N!) or even O(2N).
Just for example, when you're dealing with searching in a sorted collection, binary search borders on trivial and gives O(log N) complexity. If the distribution of items in the collection is reasonably predictable, we can typically do even better--around O(log log N). Knowing that, an algorithm that was O(N) or O(N2) (for a couple of obvious examples) would probably be pretty disappointing.
On the other hand, sorting is generally quite a bit higher complexity--the "good" algorithms manage O(N log N), and the poorer ones are typically around O(N2). Therefore, for sorting an O(N) algorithm is actually very good (in fact, only possible for rather constrained types of inputs), and we can pretty much count on the fact that something like O(log log N) simply isn't possible.
Going even further, we'd be happy to manage a matrix multiplication in only O(N2) instead of the usual O(N3). We'd be ecstatic to get optimum, reproducible answers to the traveling salesman problem or subset sum problem in only O(N3), given that optimal solutions to these normally require O(N!).
Algorithms with a sublinear behavior like O(1) or O(Log(N)) are special in that they do not require to look at all elements. In a way this is a fallacy because if there are really N elements, it will take O(N) just to read or compute them.
Sublinear algorithms are often possible after some preprocessing has been performed. Think of binary search in a sorted table, taking O(Log(N)). If the data is initially unsorted, it will cost O(N Log(N)) to sort it first. The cost of sorting can be balanced if you perform many searches, say K, on the same data set. Indeed, without the sort, the cost of the searches will be O(K N), and with pre-sorting O(N Log(N)+ K Log(N)). You win if K >> Log(N).
This said, when no preprocessing is allowed, O(N) behavior is ideal, and O(N Log(N)) is quite comfortable as well (for a million elements, Lg(N) is only 20). You start screaming with O(N²) and worse.
He said those algorithms are what we strive for, which is generally true. Many algorithms cannot possibly be improved better than logarithmic or linear time, and while constant time would be better in a perfect world, it's often unattainable.
constant time is always better because the time (or space) complexity doesn't depend on the problem size... isn't it a great feature? :-)
then we have O(N) and then Nlog(N)
did you know? problems with constant time complexity exist!
e.g.
let A[N] be an array of N integer values, with N > 3. Find and algorithm to tell if the sum of the first three elements is positive or negative.
What we strive for is efficiency, in the sense of designing algorithms with a time (or space) complexity that does not exceed their theoretical lower bound.
For instance, using comparison-based algorithms, you can't find a value in a sorted array faster than Omega(Log(N)), and you cannot sort an array faster than Omega(N Log(N)) - in the worst case.
Thus, binary search O(Log(N)) and Heapsort O(N Log(N)) are efficient algorithms, while linear search O(N) and Bubblesort O(N²) are not.
The lower bound depends on the problem to be solved, not on the algorithm.
Yes constant time i.e. O(1) is better than linear time O(n) because the former is not depending on the input-size of the problem. The order is O(1) > O (logn) > O (n) > O (nlogn).
Linear or linearthimic time we strive for because going for O(1) might not be realistic as in every sorting algorithm we atleast need a few comparisons which the professor tries to prove with his decison Tree- comparison analysis where he tries to sort three elements a b c and proves a lower bound of nlogn. Check his "Complexity of Sorting" in the Mergesort lecture.
I have this question on a practice test and I'm not sure of when code would run quicker on O(n*n) over O(log n).
Big oh notation gives upper bounds. Not more.
If algorithm A is O(n ^ 2), it could require exactly n ^ 2 steps.
If algorithm B is O(log n), it could require exactly 10000 * log n steps.
Algorithm A is a lot faster than algorithm B for small n.
Remember that Big-O is the upper bound. It's quite possible that because of constants that under smaller input sizes the O(n^2) algorithm can run faster than O(log n). It could be entirely possible that in most cases the n^2 can also run faster and that algorithm happens to run in n^2 only because of certain input sets that cause it to have to do a lot of work.
I am retracting my previous answer of never because technically it is possible for a O(n*n) algorithm to be faster than a O(log n) algorithm, though highly improbably. See my discussion with Jesus under his answer for more details. The graph below shows that an algorithm that has a time complexity of exactly log n is always faster than an algorithm that has a time complexity of exactly n*n.