I am new to Big-O notation. While reading I came across an example :
Qus : Find upper bound for f(n) = n^2 + 1
Sol : n^2 + 1 <= 2n^2 for all n >= 1
so f(n) = O(n^2) with c = 2 and n0 = 1
after this I came across second example :
Qus : Find upper bound for f(n) = n
Sol : n <= n^2 for all n >=1
so f(n) = O(n^2) with c = 1 and n0 = 1
I am confused that according to Big-O notation if f(n) = n then f(n) = O(n) so I want to know if the following solution to second example is right ?
Sol : n <= 2n for all n >=1
so f(n) = O(n) with c = 2 and n0 = 1
A complement of those comments:
Start from the definition of Big O
we can say that f(x) is O(g(x)) if and only if there exist a positive
number B and a non-negative real number b such that:
|f(x)|is less or equal to B|g(x)| for all numbers x that large than b
Therefore, we could say that for f(n)=n f(n)=O(n),
since there exists a number B (for instance "2") make that n<=Bn.
(n<=2n, for all the n=>0)
Related
Here are the steps that are used to prove the above
|f(x)| = |4x^2 – 5x + 3|
<= |4x^2|+ |- 5x| + |3|
<= 4x^2 + 5x + 3, for all x > 0
<= 4x^2 + 5x^2 + 3x^2, for all x > 1
<= 12x^2, for all x > 1
Hence we conclude that f(x) is O(x^2)
I referred this But it does not help
Can someone explain the above proof step by step?
Why the absolute value of f(x) is taken ?
Why and how were all the term replaced by x^2 term?
Preparations
We start by loosely stating the definition of a function or algorithm f being in O(g(n)):
If a function f is in O(g(n)), then c · g(n) is an upper
bound on f(n), for some non-negative constant c such that f(n) ≤ c · g(n)
holds, for sufficiently large n (i.e. , n ≥ n0 for some constant
n0).
Hence, to show that f ∈ O(g(n)), we need to find a set of (non-negative) constants (c, n0) that fulfils
f(n) ≤ c · g(n), for all n ≥ n0, (+)
We note, however, that this set is not unique; the problem of finding the constants (c, n0) such that (+) holds is degenerate. In fact, if any such pair of constants exists, there will exist an infinite amount of different such pairs.
Analysis
For common convention, we'll analyse your example using variable name n rather than x.
f(n) = 4n^2 - 5n + 3 (++)
Now, for your example, we may assume, without loss of generality (since we're studying asymptotic complexity: function/algorithm behavior for "large" n) that n > n0 where n0 > 0. This would correspond to the analysis you've shown in your question analyzing absolute values of x. Anyway, given this assumption, the following holds:
f(n) = 4n^2 - 5n + 3 < 4n^2 + 3, for all n > n0
Now let, again without loss of generality, n0 equal 2 (we could choose any value, but lets choose 2 here). For n0 = 2, naturally n^2 > 3 holds for n > n0, which means the following holds:
f(n) = 4n^2 - 5n + 3 < 4n^2 + 3 < 4n^2 + n^2, for all n > n0 = 2
f(n) < 5n^2, for all n > n0 = 2
Now choose c = 5 and let g(n) = n^2:
f(n) < c · g(n), for all n > n0,
with c = 5, n0 = 2, g(n) = n^2
Hence, from (+), we've shown that f as defined in (++) is in O(g(n)) = O(n^2).
I have this:
a) f(n) = n
b) f(n) = 1045n
c) f(n) = n2 + 70
d) f(n) = 7n + 3
e) f(n) = Cn + D (where C and D are both constants)
f) f(n) = 8
g) f(n) = n3 + n + 1
h) f(n) = 4n + 2log n + 5
I want to check if the Big O notation of them is O(n).
How can I determinate it?
And how to find the Big-O notation for the functions below:
a) f(n) = 3n3 + n
b) f(n) = 3 log n + 5n
c) f(n) = 3n2 + 5n + 4
d) f(n) = 3n3 + n2 + 5n + 99
f(x) is O(g(x)) if there exists a constant c such that f(x) < c*g(x)
You should look at the "biggest" asymptoticall factor in your function (highest exponent or something like that) and that would be your big O
Example: f(n) = 2^n + n^5 + n^2 + n*log(n) + n
This function has 5 different factors that influence big O, sorted from biggest to smallest, so this one would be O(2^n). Drop the 2^n and now f(n) is O(n^5).
Constants are O(1).
Hope I explained it well
Generally the Big O notation of a function is measured by the largest power of n that appears. In your case, this would be n², since the only other factor is the 70 which is constant.
Edit: Original post only contained the function f(n) = n² + 70.
See this answer.
In short, there's no set way to determine Big O results. Strictly speaking any function which eventually (for some n) will always be bigger than your function, is Big O of it. In practice you're looking for as tight a bound as you can get. If the only components of the function are polynomial in n, then the Big O will just be the largest power of n that appears (or rather, n to that power). Another useful thing to know is that log n is of a lower order than n (but higher than constant).
O( sqrt(n) ) = O(n) ?
We should find c and n0 verifying :
0 ( sqrt(n) ) < c*n ; c>0 and n>n0
How to find c and n0 ? or should i find another idea ?
Thanks
For n > 1, we have √n > 1, hence we have the following inequality:
√n < √n * √n = n, for any n > 1.
So we can take c = 1 and n0 = 2 to prove that √n = O(n).
Remark
Strictly speaking, you should avoid writing down something like O(√n) = O(n). Big-O notation is for describing the asymptotic upper bound of a function, but O(√n) is not a function.
O(√n) = O(n) is an abuse of notation, it really means the following:
If f is a function such that f(n) = O(√n), then f(n) = O(n).
In our case, if for any function f we have f(n) = O(√n), since √n < n for any n > 1, clearly we have f(n) < c * √n < c * n for any n > 1, and hence f(n) = O(n).
How they are calculating Big O notation?
Big O notation examples:
Question:
Find upper bound for f(n) = n⁴ + 100n² + 50
Solution:
n⁴ + 100n² + 50 <= 2n⁴, for all n >= 11
n⁴ + 100n² + 50 = O(n⁴) with c = 2 and n0 = 11
I am unable to understand - how they are calculating Big O notation?
How come c = 2 and n0 = 11
Question:
Find upper bound for f(n) = 2n³ - 2n²
Solution:
2n³ - 2n² <= 2n³ , for all n >= 1
2n³ - 2n² = O(2n³) with c = 2 and n0 =1
How come n >= 1?
Question:
f(n) = n
Solution :
n <= n , for all n >= 1
N = O(n) with c = 1 and n0 = 1
Question 1:
Definition of big O notation is (note that the wikipedia definition uses x0 and M, there is no real difference, just different signs for the same things).
f(n) is said to be in O(g(n)) if there are constants n0 and c such that for all n>n0 - f(n) <= c*g(n).
In your example, since n^4+100n^2+50 <= 2n^4 for all n>11, it is basically same as saying the above with c=2,n0=11 - thus, by definition of big O notation, you could say n^4+100n^2+50 is in O(n^4)
Question 2:
In here, n>=1 because the claim is true for all positive values of n - note that 2n^3-2n^2 will always be smaller than 2n^3 - for all positive values of n.
I am going over the Big-Oh notation, and I have a problem understanding the solution to this question:
Is 2n + 10 ≡ O(n)?
Can we find c and n0?
2n + 10 <= cn
(c-2)n >= 10
n >= 10/(c-2)
Pick c = 3 and n0 = 10
There is also a graph used in this example:
I am confused as to how c = 3 and how n0 = 10? Can somebody please enlighten me?
I would solve 2n + 10 <= cn differently.
2n + 10 <= cn
2 + 10/n <= c //divide by n
c >= 2 + 10/n
Clearly c has to be bigger than 2. Now take your favorite number bigger than 2. Uhm. c=2.718. This gives
2n + 10 <= 2.718*n
10 <= 0.718*n //subtract 2n
13.93 <= n
Thus, 2n + 10 <= c*n. If we take c=2.718 and n bigger than 15. (15 because it's bigger than the limit, 13.93, and I like 15.)
Any c bigger than 2 works, c=100000000000000000000000 is fine too (but, it costs much ink and paper to write down.)
It might have been easier to take c=3. That would give
2n + 10 <= 3*n
10 <= n //subtract 2n
This makes 3 the most logical / natural solution.
To say a function f(n) is O(n) means you can find c and n0 such that for all n >= n0, f(n) <= cn.
To verify this in your case: if n >= 10, then:
f(n) = 2n + 10
<= 3n // because 10 <= n
= cn
So f(n) <= cn for all n >= 10, so f(n) is an O(n) function.
Note that other values for c and n0 work; you just need to find one pair.
In Big-O notation you drop numbers added and multiplied.
And also use the largest power.
10*N^3+ 23*N^2 + 43*N + 154 = O(N^3)
let f(n) = 2n+10
as per the Big-Oh notation,
f(n)<= c*g(n) ----->eq1(let)
2n+10 <= 2n+n
2n+10 <= 3n for n>=10 //here the constant 10 is replaced by n ---->eq2
comparing eq1 with eq2
c=3 and g(n)=n for value of n=10
f(n)= O(g(n)) = O(n)