When is an algorithm log n

Here's a single loop with O log n performance:. So, any algorithm where the number of required operations is on the order of the logarithm of the size of the input is O log n. Does it help to know from experience that algorithms that repeatedly partition their input typically have 'log n' as a component of their performance?

But don't look for the partitioning and jump to the conclusion that the algorithm's performance is O log n -- it might be something like O n log n , which is quite different. The idea is that an algorithm is O log n if instead of scrolling through a structure 1 by 1, you divide the structure in half over and over again and do a constant number of operations for each split.

Search algorithms where the answer space keeps getting split are O log n. An example of this is binary search , where you keep splitting an ordered array in half over and over again until you find the number. The typical examples are ones that deal with binary search. For example, a binary search algorithm is usually O log n. If you have a binary search tree , lookup, insert and delete are all O log n complexity. Any situation where you continually partition the space will often involve a log n component.

This is why many sorting algorithms have O nlog n complexity, because they often partition a set and sort as they go. More accurately traversing a tree from root to one not all!

However, these are all oversimplifications. If you see like a running time of 3,4,5,6 seconds or some multiple you can safely say it's O log N. If it's more like : 1,10,, seconds then it's probably O N. And if it's like 3,40,, seconds then it's O N log N. Sign up to join this community. The best answers are voted up and rise to the top.

That's why O log n algorithms are awesome. It is the number of times you can cut a log of length n repeatedly into b equal parts before reaching a section of size 1. Divide and conquer algorithms usually have a logn component to the running time. This comes from the repeated halving of the input. In the case of binary search, every iteration you throw away half of the input. It should be noted that in Big-O notation, log is log base 2. Edit: As noted, the log base doesn't matter, but when deriving the Big-O performance of an algorithm, the log factor will come from halving, hence why I think of it as base 2.

I would rephrase this as 'height of a complete binary tree is log n'. Figuring the height of a complete binary tree would be O log n , if you were traversing down step by step. Logarithm is essentially the inverse of exponentiation. So, if each 'step' of your function is eliminating a factor of elements from the original item set, that is a logarithmic time algorithm.

For the tree example, you can easily see that stepping down a level of nodes cuts down an exponential number of elements as you continue traversing. The popular example of looking through a name-sorted phone book is essentially equivalent to traversing down a binary search tree middle page is the root element, and you can deduce at each step whether to go left or right.

O log n is a bit misleading, more precisely it's O log 2 n , i. The height of a balanced binary tree is O log 2 n , since every node has two note the "two" as in log 2 n child nodes. So, a tree with n nodes has a height of log 2 n. Another example is binary search, which has a running time of O log 2 n because at every step you divide the search space by 2.

The logarithmic function is the inverse of the exponential function. Put another way, if your input grows exponentially rather than linearly, as you would normally consider it , your function grows linearly.

O log n running times are very common in any sort of divide-and-conquer application, because you are ideally cutting the work in half every time. If in each of the division or conquer steps, you are doing constant time work or work that is not constant-time, but with time growing more slowly than O log n , then your entire function is O log n. It's fairly common to have each step require linear time on the input instead; this will amount to a total time complexity of O n log n.

The running time complexity of binary search is an example of O log n. This is because in binary search, you are always ignoring half of your input in each later step by dividing the array in half and only focusing on one half with each step.

Each step is constant-time, because in binary search you only need to compare one element with your key in order to figure out what to do next irregardless of how big the array you are considering is at any point.

The running time complexity of merge sort is an example of O n log n. Thus, the total complexity is O n log n. Also by the change of base rule for logarithms, the only difference between logarithms of different bases is a constant factor. Simply put: At each step of your algorithm you can cut the work in half. Asymptotically equivalent to third, fourth, If you plot a logarithmic function on a graphical calculator or something similar, you'll see that it rises really slowly -- even more slowly than a linear function.

In lay terms, it means that the equation for time may have some other components: e. That's what bit O notation means: it means "what is the order of dominant term for any sufficiently large n". I can add something interesting, that I read in book by Kormen and etc. Now, imagine a problem, where we have to find a solution in a problem space.

This problem space should be finite. Now, if you can prove, that at every iteration of your algorithm you cut off a fraction of this space, that is no less than some limit, this means that your algorithm is running in O logN time. I should point out, that we are talking here about a relative fraction limit, not the absolute one.

The binary search is a classical example. But binary search is not the only such example. That means, your program is still running at O logN time, although significantly slower than the binary search. This is a very good hint in analyzing of recursive algorithms. It often can be proved that at each step the recursion will not use several variants, and this leads to the cutoff of some fraction in problem space. Every time we write an algorithm or code we try to analyze its asymptotic complexity.

It is different from its time complexity. Asymptotic complexity is the behavior of execution time of an algorithm while the time complexity is the actual execution time. But some people use these terms interchangeably.

Because time complexity depends on various parameters viz. Physical System 2. Programming Language 3. And much more Instead we take input size as the parameter because whatever the code is, the input is same.

So the execution time is a function of input size. Linear Search Given n input elements, to search an element in the array you need at most 'n' comparisons.

In other words, no matter what programming language you use, what coding style you prefer, on what system you execute it. In the worst case scenario it requires only n comparisons. The execution time is linearly proportional to the input size. And its not just search, whatever may be the work increment, compare or any operation its a function of input size.

So when you say any algorithm is O log n it means the execution time is log times the input size n. As the input size increases the work done here the execution time increases.

Hence proportionality. See as the input size increased the work done is increased and it is independent of any machine. And if you try to find out the value of units of work It's actually dependent onto those above specified parameters. It will change according to the systems and all. I can give an example for a for loop and maybe once grasped the concept maybe it will be simpler to understand in different contexts.

The complexity in O-notation of this program is O log n. Let's try to loop through it by hand n being somewhere between and excluding :.

Although n is somewhere between and , only 10 iterations take place. This is because the step in the loop grows exponentially and thus takes only 10 iterations to reach the termination. Now try to see it that way, if exponential grows very fast then logarithm grows inversely very slow.

Actually, if you have a list of n elements, and create a binary tree from that list like in the divide and conquer algorithm , you will keep dividing by 2 until you reach lists of size 1 the leaves. At the first step, you divide by 2. Searching for 4 yields 3 hits: 6, 3 then 4. In this article there is a quote: D. On the basis of the issues discussed here, I propose that members of SIGACT, and editors of computer science and mathematics journals, adopt notations as defined above, unless a better alternative can be found reasonably soon.

If you are looking for a intuition based answer I would like to put up two interpretations for you. Imagine a very high hill with a very broad base as well. To reach the top of the hill there are two ways: one is a dedicated pathway going spirally around the hill reaching at the top, the other: small terrace like carvings cut out to provide a staircase. Now if the first way is reaching in linear time O n , the second one is O log n.

Imagine an algorithm, which accepts an integer, n as input and completes in time proportional to n then it is O n or theta n but if it runs in time proportion to the number of digits or the number of bits in the binary representation on number then the algorithm runs in O log n or theta log n time. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Collectives on Stack Overflow.

Learn more. What does O log n mean exactly? Ask Question. Asked 11 years, 8 months ago. Active 1 year, 3 months ago. Viewed 1. Improve this question. Sisir 3, 2 2 gold badges 19 19 silver badges 27 27 bronze badges. Andreas Grech Andreas Grech k 98 98 gold badges silver badges bronze badges.

One thing I'm seeing in most answers is that they essentially describe "O something " means the running time of the algorithm grows in proportion to "something". Given that you asked for "exact meaning" of "O log n ", it's not true. That's the intuitive description of Big-Theta notation, not Big-O. O log n intuitively means the running time grows at most proportional to "log n": stackoverflow.

I always remember divide and conquer as the example for O log n — RichardOD. It's important to realize that its log base 2 not base This is because at each step in an algorithm, you remove half of your remaining choices. In computer science we almost always deal with log base 2 because we can ignore constants.

However there are some exceptions i. This works fine for small values of n , but is there a more efficient way? PDF - Download algorithm for free. Previous Next. So there must be some type of behavior that algorithm is showing to be given a complexity of log n. Let us see how it works. Since binary search has a best case efficiency of O 1 and worst case average case efficiency of O log n , we will look at an example of the worst case.

Consider a sorted array of 16 elements. You can see that after every comparison with the middle term, our searching range gets divided into half of the current range. So, for reaching one element from a set of 16 elements, we had to divide the array 4 times,.