In binary search since we keep dividing by half the algorithm will be log2(n) but it states that is is log(n) in many sources. Does log(n) in programming equivalent to log2(n)? What is the difference?
Yes, in programming in general, log refers to log2 as it is the most common logarithmic complexity.
It's also important to note that constants are ignored, so for example 3n and 8n both have complexity O(n), although O(3n) is a bit faster, they will both get slower at same rate.
For your example particularly, log2(n) and log9(n) can be both represented as ln(n)/ln(2) and ln(n)/ln(9), therefore, removing constant factors you will get both as ln(n).
When we have O(N + logN) we drop logN as a non-dominant term.
When we have O(N + M) we have to keep both terms because these terms are unrelated.
What happens if we have something like O(N + logM)? On one hand, terms are still unrelated. On the other hand even if M is very big and N is very small N will sooner or later surpass logM so maybe N should be considered the dominant term. Should logM be dropped in this case or not?
Because we use big-O notation, we're interested in the upper bound. For the 1st case O(N + logN) we can omit logN because it grows more slowly than N. See it on the graph.
The last two cases should keep both terms because we can't surely say in the common case what's grows slowly or faster.
Talking about Big O notations, if one algorithm time complexity is O(N) and other's is O(2N), which one is faster?
The definition of big O is:
O(f(n)) = { g | there exist N and c > 0 such that g(n) < c * f(n) for all n > N }
In English, O(f(n)) is the set of all functions that have an eventual growth rate less than or equal to that of f.
So O(n) = O(2n). Neither is "faster" than the other in terms of asymptotic complexity. They represent the same growth rates - namely, the "linear" growth rate.
Proof:
O(n) is a subset of O(2n): Let g be a function in O(n). Then there are N and c > 0 such that g(n) < c * n for all n > N. So g(n) < (c / 2) * 2n for all n > N. Thus g is in O(2n).
O(2n) is a subset of O(n): Let g be a function in O(2n). Then there are N and c > 0 such that g(n) < c * 2n for all n > N. So g(n) < 2c * n for all n > N. Thus g is in O(n).
Typically, when people refer to an asymptotic complexity ("big O"), they refer to the canonical forms. For example:
logarithmic: O(log n)
linear: O(n)
linearithmic: O(n log n)
quadratic: O(n2)
exponential: O(cn) for some fixed c > 1
(Here's a fuller list: Table of common time complexities)
So usually you would write O(n), not O(2n); O(n log n), not O(3 n log n + 15 n + 5 log n).
Timothy Shield's answer is absolutely correct, that O(n) and O(2n) refer to the same set of functions, and so one is not "faster" than the other. It's important to note, though, that faster isn't a great term to apply here.
Wikipedia's article on "Big O notation" uses the term "slower-growing" where you might have used "faster", which is better practice. These algorithms are defined by how they grow as n increases.
One could easily imagine a O(n^2) function that is faster than O(n) in practice, particularly when n is small or if the O(n) function requires a complex transformation. The notation indicates that for twice as much input, one can expect the O(n^2) function to take roughly 4 times as long as it had before, where the O(n) function would take roughly twice as long as it had before.
It depends on the constants hidden by the asymptotic notation. For example, an algorithm that takes 3n + 5 steps is in the class O(n). So is an algorithm that takes 2 + n/1000 steps. But 2n is less than 3n + 5 and more than 2 + n/1000...
It's a bit like asking if 5 is less than some unspecified number between 1 and 10. It depends on the unspecified number. Just knowing that an algorithm runs in O(n) steps is not enough information to decide if an algorithm that takes 2n steps will complete faster or not.
Actually, it's even worse than that: you're asking if some unspecified number between 1 and 10 is larger than some other unspecified number between 1 and 10. The sets you pick from being the same doesn't mean the numbers you happen to pick will be equal! O(n) and O(2n) are sets of algorithms, and because the definition of Big-O cancels out multiplicative factors they are the same set. Individual members of the sets may be faster or slower than other members, but the sets are the same.
Theoretically O(N) and O(2N) are the same.
But practically, O(N) will definitely have a shorter running time, but not significant. When N is large enough, the running time of both will be identical.
O(N) and O(2N) will show significant difference in growth for small numbers of N, But as N value increases O(N) will dominate the growth and coefficient 2 becomes insignificant. So we can say algorithm complexity as O(N).
Example:
Let's take this function
T(n) = 3n^2 + 8n + 2089
For n= 1 or 2, the constant 2089 seems to be the dominant part of function but for larger values of n, we can ignore the constants and 8n and can just concentrate on 3n^2 as it will contribute more to the growth, If the n value still increases the coefficient 3 also seems insignificant and we can say complexity is O(n^2).
For detailed explanation refer here
O(n) is faster however you need to understand that when we talk about Big O, we are measuring the complexity of a function/algorithm, not its speed. And we measure this complexity asymptotically. In lay man terms, when we talk about asymptotic analysis, we take immensely huge values for n. So if you plot the graph for O(n) and O(2n), the values will stay in some particular range from each other for any value of n. They are much closer compared to the other canonical forms like O(nlogn) or O(1), so by convention we approximate the complexity to the canonical form O(n).
Okay so I have this project I have to do, but I just don't understand it. The thing is, I have 2 algorithms. O(n^2) and O(n*log2n).
Anyway, I find out in the project info that if n<100, then O(n^2) is more efficient, but if n>=100, then O(n*log2n) is more efficient. I'm suppose to demonstrate with an example using numbers and words or drawing a photo. But the thing is, I don't understand this and I don't know how to demonstrate this.
Anyone here that can help me understand how this works?
Good question. Actually, I always show these 3 pictures:
n = [0; 10]
n = [0; 100]
n = [0; 1000]
So, O(N*log(N)) is far better than O(N^2). It is much closer to O(N) than to O(N^2).
But your O(N^2) algorithm is faster for N < 100 in real life. There are a lot of reasons why it can be faster. Maybe due to better memory allocation or other "non-algorithmic" effects. Maybe O(N*log(N)) algorithm requires some data preparation phase or O(N^2) iterations are shorter. Anyway, Big-O notation is only appropriate in case of large enough Ns.
If you want to demonstrate why one algorithm is faster for small Ns, you can measure execution time of 1 iteration and constant overhead for both algorithms, then use them to correct theoretical plot:
Example
Or just measure execution time of both algorithms for different Ns and plot empirical data.
Just ask wolframalpha if you have doubts.
In this case, it says
n log(n)
lim --------- = 0
n^2
Or you can also calculate the limit yourself:
n log(n) log(n) (Hôpital) 1/n 1
lim --------- = lim -------- = lim ------- = lim --- = 0
n^2 n 1 n
That means n^2 grows faster, so n log(n) is smaller (better), when n is high enough.
Big-O notation is a notation of asymptotic complexity. This means it calculates the complexity when N is arbitrarily large.
For small Ns, a lot of other factors come in. It's possible that an algorithm has O(n^2) loop iterations, but each iteration is very short, while another algorithm has O(n) iterations with very long iterations. With large Ns, the linear algorithm will be faster. With small Ns, the quadratic algorithm will be faster.
So, for small Ns, just measure the two and see which one is faster. No need to go into asymptotic complexity.
Incidentally, don't write the basis of the log. Big-O notation ignores constants - O(17 * N) is the same as O(N). Since log2N is just ln N / ln 2, the basis of the logarithm is just another constant and is ignored.
Let's compare them,
On one hand we have:
n^2 = n * n
On the other hand we have:
nlogn = n * log(n)
Putting them side to side:
n * n versus n * log(n)
Let's divide by n which is a common term, to get:
n versus log(n)
Let's compare values:
n = 10 log(n) ~ 2.3
n = 100 log(n) ~ 4.6
n = 1,000 log(n) ~ 6.9
n = 10,000 log(n) ~ 9.21
n = 100,000 log(n) ~ 11.5
n = 1,000,000 log(n) ~ 13.8
So we have:
n >> log(n) for n > 1
n^2 >> n log(n) for n > 1
Anyway, I find out in the project info that if n<100, then O(n^2) is
more efficient, but if n>=100, then O(n*log2n) is more efficient.
Let us start by clarifying what is Big O notation in the current context. From (source) one can read:
Big O notation is a mathematical notation that describes the limiting
behavior of a function when the argument tends towards a particular
value or infinity. (..) In computer science, big O notation is used to classify algorithms
according to how their run time or space requirements grow as the
input size grows.
Big O notation does not represent a function but rather a set of functions with a certain asymptotic upper-bound; as one can read from source:
Big O notation characterizes functions according to their growth
rates: different functions with the same growth rate may be
represented using the same O notation.
Informally, in computer-science time-complexity and space-complexity theories, one can think of the Big O notation as a categorization of algorithms with a certain worst-case scenario concerning time and space, respectively. For instance, O(n):
An algorithm is said to take linear time/space, or O(n) time/space, if its time/space complexity is O(n). Informally, this means that the running time/space increases at most linearly with the size of the input (source).
and O(n log n) as:
An algorithm is said to run in quasilinear time/space if T(n) = O(n log^k n) for some positive constant k; linearithmic time/space is the case k = 1 (source).
Mathematically speaking the statement
Which is better: O(n log n) or O(n^2)
is not accurate, since as mentioned before Big O notation represents a set of functions. Hence, more accurate would have been "does O(n log n) contains O(n^2)". Nonetheless, typically such relaxed phrasing is normally used to quantify (for the worst-case scenario) how a set of algorithms behaves compared with another set of algorithms regarding the increase of their input sizes. To compare two classes of algorithms (e.g., O(n log n) and O(n^2)) instead of
Anyway, I find out in the project info that if n<100, then O(n^2) is
more efficient, but if n>=100, then O(n*log2n) is more efficient.
you should analyze how both classes of algorithms behaves with the increase of their input size (i.e., n) for the worse-case scenario; analyzing n when it tends to the infinity
As #cem rightly point it out, in the image "big-O denote one of the asymptotically least upper-bounds of the plotted functions, and does not refer to the sets O(f(n))"
As you can see in the image after a certain input, O(n log n) (green line) grows slower than O(n^2) (orange line). That is why (for the worst-case) O(n log n) is more desirable than O(n^2) because one can increase the input size, and the growth rate will increase slower with the former than with the latter.
First, it is not quite correct to compare asymptotic complexity mixed with N constraint. I.E., I can state:
O(n^2) is slower than O(n * log(n)), because the definition of Big O notation will include n is growing infinitely.
For particular N it is possible to say which algorithm is faster by simply comparing N^2 * ALGORITHM_CONSTANT and N * log(N) * ALGORITHM_CONSTANT, where ALGORITHM_CONSTANT depends on the algorithm. For example, if we traverse array twice to do our job, asymptotic complexity will be O(N) and ALGORITHM_CONSTANT will be 2.
Also I'd like to mention that O(N * log2N) which I assume logariphm on basis 2 (log2N) is actually the same as O(N * log(N)) because of logariphm properties.
We have two way to compare two Algo
->first way is very simple compare and apply limit
T1(n)-Algo1
T2(n)=Alog2
lim (n->infinite) T1(n)/T2(n)=m
(i)if m=0 Algo1 is faster than Algo2
(ii)m=k Both are same
(iii)m=infinite Algo2 is faster
*Second way pretty simple as compare to 1st there you just take a log of both but do not neglet multi constant
Algo 1=log n
Algo 2=sqr(n)
keep log n =x
Any poly>its log
O(sqr(n))>o(logn)
I am a mathematician so i will try to explain why n^2 is faster than nlogn for small values of n , with a simple limit , while n-->0 :
lim n^2 / nlogn = lim n / logn = 0 / -inf = 0
so , for small values of n ( in this case "small value" is n existing in [1,99] ) , the nlogn is faster than n^2 , 'cause as we see limit = 0 .
But why n-->0? Because n in an algorithm can take "big" values , so when n<100 , it is considered like a very small value so we can take the limit n-->0.
what would the time complexity for f(n)=n^3/logn be? And for a double series (2 sums)? I know it is either polynomial or polylog.
f(n) = n^3 / log(n) is obviously Θ(n^3/log(n)), which is also o(n^3), but ω(n^2). Therefore the function does not grow faster than polynomial, but faster than polylogarithmic (Because all polylogarithmic functions grow slower than quadratic and actually slower than any power).
In fact for all ε > 0: f(n) in ω(n^(3-ε)).
Proofs for all of these statements should be pretty simple.
The second question depends on the sum limits and inner term.