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What is a plain English explanation of "Big O" notation?
(43 answers)
Closed 9 years ago.
Can someone please explain to me what this means:
Definition: Given functions f(n) and g(n), then we say that
f(n) is O( g(n) )
if and only if there exist positive constants c and n0 such that
f(n) <= c g(n) for all n => n0
It basically means, that for large enough n and ignoring constant factors, f(n) does not grow faster than g(n).
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Having trouble with a homework problem on time complexity, how do you properly proof the equation. Everything I've done so far leads me to dead ends.
Question as listed:
Let f(n) and g(n) be non-negative functions such that f(n) is O(g(n)) and g(n) is
O(f(n)). Use the definition of “big Oh” to prove that f(n) − g(n) is O(f(n)).
Without outright giving you the answer to your homework, I'll rather push you to the right place.
1. Prove that f(n) = Θ(g(n)) iff g(n) = Θ(f(n))
2. http://web.cse.ohio-state.edu/~lai.1/780-class-notes/2.math.pdf
Here are some notes to read over and after that working out the proof shouldn't be hard.
Also, I'd ask this question on the math stack exchange and not stack overflow.
This question already has an answer here:
Show that g(n) is O(g(n)) for each of the following [closed]
(1 answer)
Closed 4 years ago.
I don't get how to show it---I take the log of both sides, and?
This question is to prove that f(n) is O(g(n)), which I know how to do for things that have the same base. not as much for this.
2^(sqrt(log(n)) is O(n(^4/3))
For sufficiently large n, sqrt(log(n)) is positive and bounded from above by log(n). Since 2^x is monotonically increasing, 2^sqrt(log(n)) is bounded from above by 2^log(n) = n. Moreover, for large n, n is clearly bounded from above by n^(4/3). Therefore the original function itself is bounded from above by n^(4/3) as well.
This question already has answers here:
What is a plain English explanation of "Big O" notation?
(43 answers)
Closed 5 years ago.
I have just started "Cracking the Coding Interview" by Gayle Macdowell. In this BigO topic, It says we should drop the non-dominant term.
O(n^2 + n) becomes O(n^2) and O(n + log n) becomes O(n).
Well, I understand that. If we suppose the value of n to be some large number then we can ignore the smaller result since, it will be comparatively much more smaller than the larger one.
But, in this case how can O(5*2^n + 1000n^100) become O(2^n)) ?
Isn't n^100 dominant than 2 ^n ?
n^100, or n raised to any constant, does not dominate 2^n.
This question already has answers here:
What is the difference between Θ(n) and O(n)?
(9 answers)
Closed 5 years ago.
I am taking an online class on algorithms and I had the following quiz. I got it wrong and am trying to understand the reason for the answer.
Which of the following is O(n^3)?
a) 11n + 151 gn + 100
b) 1/3 n^2
c) 25000 n^3
d) All of the above.
The correct answer is (d) all of the above. The reason is that Big-O notation provides only the upper bound on the growth rate of function as n gets large.
I am not sure why the answer is not (c). For example, the upper bound on (b) is less than n^3.
The reason is that formally, big-O notation is an asymptotic upper bound.
So 1/3*n^2 is O(n^2), but it is also O(n^3) and also O(2^n).
While in every-day conversion about complexity O(...) is used as a tight (both upper and lower bound), the theta-notation, or Θ(...) is the technically correct term for this.
For more info see What is the difference between Θ(n) and O(n)?
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Closed 8 years ago.
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I'm confused as to how f(n) can be O(g(n)), theta(g(n)) and omega(g(n)). Could someone help explain?
In fact each function that is of theta g(n) will be of O(g(n)) and omega(g(n)). The simplified definition is that f(n) is in Theta(g(n)) if it grows precisely as fast as g(n), while f(n) is in O(g(n)) if it grows no faster than g(n) and is in Omega(g(n)) if it grows no slower than g(n)(all there definitions hold for sufficiently large n). Thus when the speed at which f(n) and g(n) are the same both the conditions for omega and O hold.
As for why f(n) is in Theta(g(n)) - try dividing the two functions and analyzing the fraction when n grows to infinity.
The clearest and straightforward way to solve such a question, you should use the Limit Method Process like the following: