I can clearly see than N^2 is bounded by c2^N, but how do i prove it by using formal definition of big-O. I can simply prove it by M.I.
Here is my attempt..
By definition, there for any n>n0, there exist a constant C which
f(n) <= Cg(n)
where
f(n) = n^2
and
g(n) = 2^n
Should I take log to both side and solve for C?
and one more question about fibonacci sequence, i wanna solve the recurrence relation.
int fib(int n){
if(n<=1) return n;
else return fib(n-1) + fib(n-2);
The equation is..
T(n) = T(n-1)+T(n-2)+C // where c is for the adding operation
so
T(n) = T(n-2) + 2T(n-3) + T(n-4) + 3c
and one more
T(n) = T(n-3) + 3T(n-4) + 3T(n-5) + T(n-6) + 6c
then i started to get lost to form the general equation i
The pattern is somehow like pascal triangle?
t(n) = t(n-i) + aT(n-i-1) + bT(n-i-2) + ... + kT(n-i-i) + C
As you point out, to see if f(x) ϵ O(g(x)) you need to find...
...some c > 0 and
...some x0
such that f(x) < c·g(x) for all x > x0.
In this case, you can pick c = 1 and x0 = 2. What you need to prove is that
x2 < 2x for all x > 2
At this point you can log both sides (since if log(x) > log(y), then x > y.) Assuming you're using base-2 log you get the following
log(x2) < log(2x)
and by standard laws of logarithms, you get
2·log(x) < x·log(2)
Since log(2) = 1 this can be written as
2·log(x) < x
If we set x = 2, we get
2·log(2) = 2
and since x grows faster than log(x) we know that 2·log(x) < x holds for all x > 2.
For the most part, the accepted answer (from aioobe) is correct, but there is an important correction that needs to be made.
Yes, for x=2, 2×log(x) = x or 2×log(2) = 2 is correct, but then he incorrectly implies that 2×log(x) < x is true for ALL x>2, which is not true.
Let's take x=3, so the equation becomes: 2×log(3) < 3 (an invalid equation).
If you calculate this, you get: 2×log(3) ≈ 3,16993 which is greater than 3.
You can clearly see this if you plot f(x) = x2 and g(x) = 2x or if you plot f(x)= 2×log(x) and g(x) = x (if c=1).
Between x=2 and x=4, you can see that g(x) will dip below f(x). It is only when x ≥ 4, that f(x) will remain ≤ c×g(x).
So to get the correct answer, you follow the steps described in aioobe's answer, but you plot the functions to get the last intersection where f(x) = c×g(x). The x at that intersection is your x0 (together with the choosen c) for which the following is true: f(x) ≤ c×g(x) for all x ≥ x0.
So for c=1 it should be: for all x≥4, or x0=4
To improve upon the accepted answer:
You have to prove that x^2 < 2^x for all x > 2
Taking log on both sides, we have to prove that:
2·log(x) < x for all x > 2
Thus we have to show the function h(x)=x-2·log(x)>0 for all x>2
h(2)=0
Differentiating h(x) with respect to x, we get h'(x)= 1 - 1/(x·ln(2))
For all x>2, h'(x) is always greater than 0, thus h(x) keeps increasing and since h(2)=0,
it is hence proved that h(x) > 0 for all x > 2,
or x^2 < 2^x for all x > 2
Related
My Algorithms textbook has the following excerpt:
I am struggling to understand their proof that there exists a tight bound IF the limit as n goes to infinity of the ratio of two functions is a constant.
Specifically, where do they get 0.5c and 2c from?
My thoughts: A tight bound means that a function T(n) is bounded above by f(n) and below by g(n). Now lets say T(n) = n^2, f(n) = an^2, and g(n) = bn^2. Then we know the tightbound of T(n) is Theta(n^2) since the ratio of f(n) and g(n) is a constant, a/b.
The formal definition of the statement "lim w(x) = c as x -> infinity" is the following:
For all epsilon > 0, there exists some N such that for all x > N, |w(x) - c| < epsilon.
Now we are given that lim f(x) / g(x) = c as x -> infinity, and that c > 0. Then c / 2 > 0.
Consider epsilon = c / 2. Then epsilon > 0, so there exists some N such that for all x > N, we have |f(x) / g(x) - c| < epsilon = c / 2. This is equivalent to saying -c/2 < f(x) / g(x) - c < c / 2, which is in turn equivalent to saying c/2 < f(x) / g(x) < 3c / 2.
Now since for all x > N, we have c/2 < f(x) / g(x), then (since we always assume that f and g are positive valued) we can conclude that for all x > N, f(x) > g(x) c/2. Thus, we have shown that f(x) = Omega(g(x)).
And similarly, since for all x > N, we have f(x) / g(x) < 3/2 c, we see that for all x > N, f(x) < g(x) (3/2 c). Then we have shown that f(x) = O(g(x)).
Thus, we see that f(x) = Theta(g(x)), as required.
I came across this question. To prove whether the following statement was true or falseLet f(n) = n + log n, then f(n) = O(log^2 n).I'm unsure as to how I can go about proving or disproving whether log^2n is the upper bound for n or not. Could someone help me construct a proof for the same.
Consider the function
g(x) = x(ln x)^2 ; x > 0
This function is positive and increasing for 0 < x < e^(-2).
To see why this is true, let's calculate its derivative:
g'(x) = 1*(ln x)^2 + x*2(ln x)/x
basically because the derivative of ln x is 1/x. Then
g'(x) = (ln x)((ln x) + 2)
which is positive for 0 < x < e^(-2), since both factors are negative in that interval.
This proves that g(x) is positive and increasing in the interval (0, e^(-2)). Therefore, there exists a positive constant c such that
g(x) > c ; if x is small enough
which implies that
g(1/n) > c ; if n is large enough
or
(1/n)(ln n)^2 > c
or
n < (1/c)(ln n)^2 = O((ln n)^2)
and since ln n is also O((ln n)^2) we get
n + (ln n) = O((ln n)^2)
as we wanted to see.
I am studying big O notation from this book.
The deffinition of big O notation is:
We say that f (x) is O(g(x)) if there are constants C and k such that |f (x)| ≤ C|g(x)| whenever x > k.
Now here is the first example:
EXAMPLE 1 Show that f (x) = x^2 + 2x + 1 is O(x^2).
Solution: We observe that we can readily estimate the size of f (x) when x > 1 because x 1. It follows that
0 ≤ x^2 + 2x + 1 ≤ x^2 + 2x^2 + x^2 = 4x^2
whenever x > 1. Consequently, we can take C = 4 and k = 1 as witnesses to show that f (x) is O(x^2). That is, f (x) = x^2 + 2x + 1
1. (Note that it is not necessary to use absolute values here because all functions in these equalities are positive when x is positive.)
I honestly don't know how they got c = 4, looks like they jump straight to the equation manipulation and my algebra skills are pretty weak. However, I found another way through [The accepted answer to this question])What is an easy way for finding C and N when proving the Big-Oh of an Algorithm?) that says to add all coefficients to find c if k = 1. So x^2+2x+1 = 1+2+1 = 4.
Now for k = 2, I'm completely lost:
Alternatively, we can estimate the size of f (x) when x > 2. When x > 2, we have 2x ≤ x^2 and 1 ≤ x^2. Consequently, if x > 2, we have
0 ≤ x^2 + 2x + 1 ≤ x^2 + x^2 + x^2 = 3x^2.
It follows that C = 3 and k = 2 are also witnesses to the relation f (x) is O(x^2).
Can anyone explain what is happening? What method are they using?
First alternative:
C=4?
The C=4 come from the inequalities
0 ≤ x^2 + 2x + 1 ≤ x^2 + 2x^2 + x^2 = 4x^2 = C*x^2, with C=4 (+)
The second inequality in (+) is true for all x greater than 1, since, term by term
2x < 2x^2, given x>1
1 < x^2, given x>1
k = 1?
From above, we've shown that (+) holds as long as x is larger than 1, i.e.
(+) is true given x > k, with k=1
Second alternative:
k=2?
By the statement, we want to study f(x) for x larger than 2, i.e.
Study f(x) for x > k, k=2
Given x > 2, it's apparent that
0 ≤ x^2 + 2x + 1 ≤ x^2 + x^2 + x^2 = 3x^2 = C*x^2, with C=3 (++)
since, for x>2, we have
2x = x^2 given x=2 ==> 2x < x^2 given x>2
for x=2, 1 < x^2 = 4, so 1 < x^2 for all x>2
Both examples show that f(x) is O(x^2). By using your constants C and k, recall that then Big-O notation for f(x) can be summarized as something along the lines
... we can say that f(x) is O(g(x)) if we can find a constant C such
that |f(x)| is less than C|g(x)| or all x larger than k, i.e., for all
x>k. (*)
This, by no means, implies that we need to find a unique set of (C, k) to prove that some f(x) is some O(g(x)), just some set (C, k) such that (*) above holds.
See e.g. the following link for some reference on how to specify the asymptotic behaviour of a function:
https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-o-notation
I want to find out the time complexity of the program using recurrence equations.
That is ..
int f(int x)
{
if(x<1) return 1;
else return f(x-1)+g(x);
}
int g(int x)
{
if(x<2) return 1;
else return f(x-1)+g(x/2);
}
I write its recurrence equation and tried to solve it but it keep on getting complex
T(n) =T(n-1)+g(n)+c
=T(n-2)+g(n-1)+g(n)+c+c
=T(n-3)+g(n-2)+g(n-1)+g(n)+c+c+c
=T(n-4)+g(n-3)+g(n-2)+g(n-1)+g(n)+c+c+c+c
……………………….
……………………..
Kth time …..
=kc+g(n)+g(n-1)+g(n-3)+g(n-4).. .. . … +T(n-k)
Let at kth time input become 1
Then n-k=1
K=n-1
Now i end up with this..
T(n)= (n-1)c+g(n)+g(n-1)+g(n-2)+g(n-3)+….. .. g(1)
I ‘m not able to solve it further.
Any way if we count the number of function calls in this program , it can be easily seen that time complexity is exponential but I want proof it using recurrence . how can it be done ?
Explanation in Anwer 1, looks correct , similar work I did.
The most difficult task in this code is to write its recursion equation. I have drawn another diagram , I identified some patterns , I think we can get some help form this diagram what could be the possible recurrence equation.
And I came up with this equation , not sure if it is right ??? Please help.
T(n) = 2*T(n-1) + c * logn
Ok, I think I have been able to prove that f(x) = Theta(2^x) (note that the time complexity is the same). This also proves that g(x) = Theta(2^x) as f(x) > g(x) > f(x-1).
First as everyone noted, it is easy to prove that f(x) = Omega(2^x).
Now we have the relation that f(x) <= 2 f(x-1) + f(x/2) (since f(x) > g(x))
We will show that, for sufficiently large x, there is some constant K > 0 such that
f(x) <= K*H(x), where H(x) = (2 + 1/x)^x
This implies that f(x) = Theta(2^x), as H(x) = Theta(2^x), which itself follows from the fact that H(x)/2^x -> sqrt(e) as x-> infinity (wolfram alpha link of the limit).
Now (warning: heavier math, perhap cs.stackexchange or math.stackexchange is better suited)
according to wolfram alpha (click the link and see series expansion near x = infinity),
H(x) = exp(x ln(2) + 1/2 + O(1/x))
And again, according to wolfram alpha (click the link (different from above) and see the series expansion for x = infinity), we have that
H(x) - 2H(x-1) = [1/2x + O(1/x^2)]exp(x ln(2) + 1/2 + O(1/x))
and so
[H(x) - 2H(x-1)]/H(x/2) -> infinity as x -> infinity
Thus, for sufficiently large x (say x > L) we have the inequality
H(x) >= 2H(x-1) + H(x/2)
Now there is some K (dependent only on L (for instance K = f(2L))) such that
f(x) <= K*H(x) for all x <= 2L
Now we proceed by (strong) induction (you can revert to natural numbers if you want to)
f(x+1) <= 2f(x) + f((x+1)/2)
By induction, the right side is
<= 2*K*H(x) + K*H((x+1)/2)
And we proved earlier that
2*H(x) + H((x+1)/2) <= H(x+1)
Thus f(x+1) <= K * H(x+1)
Using memoisation, both functions can easily be computed in O(n) time. But the program takes at least O(2^n) time, and thus is a very inefficient way of computing f(n) and g(n)
To prove that the program takes at most O(2+epsilon)^n time for any epsilon > 0:
Let F(n) and G(n) be the number of function calls that are made in evaluating f(n) and g(n), respectively. Clearly (counting the addition as 1 function call):
F(0) = 1; F(n) = F(n-1) + G(n) + 1
G(1) = 1; G(n) = F(n-1) + G(n/2) + 1
Then one can prove:
F and G are monotonic
F > G
Define H(1) = 2; H(n) = 2 * H(n-1) + H(n/2) + 1
clearly, H > F
for all n, H(n) > 2 * H(n-1)
hence H(n/2) / H(n-1) -> 0 for sufficiently large n
hence H(n) < (2 + epsilon) * H(n-1) for all epsilon > 0 and sufficiently large n
hence H in O((2 + epsilon)^n) for any epsilon > 0
(Edit: originally I concluded here that the upper bound is O(2^n). That is incorrect,as nhahtdh pointed out, but see below)
so this is the best I can prove.... Because G < F < H they are also in O((2 + epsilon)^n) for any epsilon > 0
Postscript (after seeing Mr Knoothes solution): Because i.m.h.o a good mathematical proof gives insight, rather than lots of formulas, and SO exists for all those future generations (hi gals!):
For many algorithms, calculating f(n+1) involves twice (thrice,..) the amount of work for f(n), plus something more. If this something more becomes relatively less with increasing n (which is often the case) using a fixed epsilon like above is not optimal.
Replacing the epsilon above by some decreasing function ε(n) of n will in many cases (if ε decreases fast enough, say ε(n)=1/n) yield an upper bound O((2 + ε(n))^n ) = O(2^n)
Let f(0)=0 and g(0)=0
From the function we have,
f(x) = f(x - 1) + g(x)
g(x) = f(x - 1) + g(x/2)
Substituting g(x) in f(x) we get,
f(x) = f(x-1) + f(x -1) + g(x/2)
∴f(x) = 2f(x-1) + g(x/2)
Expanding this we get,
f(x) = 2f(x-1)+f(x/2-1)+f(x/4-1)+ ... + f(1)
Let s(x) be a function defined as follows,
s(x) = 2s(x-1)
Now clearly f(x)=Ω(s(x)).
The complexity of s(x) is O(2x).
Therefore function f(x)=Ω(2x).
I think is clear to see that f(n) > 2n, because f(n) > h(n) = 2h(n-1) = 2n.
Now I claim that for every n, there is an ε such that:
f(n) < (2+ε)n, to see this, let do it by induction, but to make it more sensible at first I'll use ε = 1, to show f(n) <= 3n, then I'll extend it.
We will use strong induction, suppose for every m < n, f(m) < 3m then we have:
f(n) = 2[f(n-1) + f(n/2 -1) + f(n/4 -1)+ ... +f(1-1)]
but for this part:
A = f(n/2 -1) + f(n/4 -1)+ ... +f(1-1)
we have:
f(n/2) = 2[f(n/2 -1) + f(n/4 -1)+ ... +f(1-1]) ==>
A <= f(n/2) [1]
So we can rewrite f(n):
f(n) = 2f(n-1) + A < 2f(n-1) +f(n/2),
Now let back to our claim:
f(n) < 2*3^(n-1) + 2*3^(n/2)==>
f(n) < 2*3^(n-1) + 3^(n-1) ==>
f(n) < 3^n. [2]
By [2], proof of f(n)∈O(3n) is completed.
But If you want to extend this to the format of (2+ε)n, just use 1 to replace the inequality, then we will have
for ε > 1/(2+ε)n/2-1 → f(n) < (2+ε)n.[3]
Also by [3] you can say that for every n there is an ε such that f(n) < (2+ε)n actually there is constant ε such that for n > n0, f(n)∈O((2+ε)n). [4]
Now we can use wolfarmalpha like #Knoothe, by setting ε=1/n, then we will have:
f(n) < (2+1/n)n which results on f(n) < e*2n, and by our simple lower bound at start we have: f(n)∈ Θ(2^n).[5]
P.S: I didn't calculate epsilon exactly, but you can do it with pen and paper simply, I think this epsilon is not correct, but is easy to find it, and if is hard tell me is hard, and I'll write it.
(log n)^k = O(n)? For k greater or equal to 1.
My professor presented us with this statement in class, however I am not sure what it means for a function to a have a time complexity of O(n). Even stuff like n^2 = O(n^2), how can a function f(x) have a run time complexity?
As for the statement how does it equal O(n) rather than O((logn)^k)?
(log n)^k = O(n)?
Yes. The definition of big-Oh is that a function f is in O(g(n)) if there exist positive constants N and c, such that for all n > N: f(n) <= c*g(n). In this case f(n) is (log n)^k and g(n) is n, so if we insert that into the definition we get: "there exist constants N and c, such that for all n > N: (log n)^k <= c*n". This is true so (log n)^k is in O(n).
how can a function f(x) have a run time complexity
It doesn't. Nothing about big-Oh notation is specific to run-time complexity. Big-Oh is a notation to classify the growth of functions. Often the functions we're talking about measure the run-time of certain algorithms, but we can use big-Oh to talk about arbitrary functions.
f(x) = O(g(x)) means f(x) grows slower or comparably to g(x).
Technically this is interpreted as "We can find an x value, x_0, and a scale factor, M, such that this size of f(x) past x_0 is less than the scaled size of g(x)." Or in math:
|f(x)| < M |g(x)| for all x > x_0.
So for your question:
log(x)^k = O(x)? is asking : is there an x_0 and M such that
log(x)^k < M x for all x>x_0.
The existence of such M and x_0 can be done using various limit results and is relatively simple using L'Hopitals rule .. however it can be done without calculus.
The simplest proof I can come up with that doesn't rely on L'Hopitals rule uses the Taylor series
e^z = 1 + z + z^2/2 + ... = sum z^m / m!
Using z = (N! x)^(1/N) we can see that
e^(x^(1/N)) = 1 + (N! x)^(1/N) + (N! x)^(2/N)/2 + ... (N! x)^(N/N)/N! + ...
For x>0 all terms are positive so, keeping only the Nth term we get that
e^((N! x)^(1/N)) = N! x / N! + (...)
= x + (...)
> x for x > 0
Taking logarithms of both sides (since log is monotonic increasing), then raising to Nth power (also monotonic increasing since N>0)
(N! x)^(1/N) > log x for x > 0
N! x > (log x)^n for x > 0
Which is exactly the result we need, (log x)^N < M x for some M and all x > x_0, with M = N! and x_0=0