Hi I am having a tough time showing the run time of these three algorithms for T(n). Assumptions include T(0)=0.
1) This one i know is close to Fibonacci so i know it's close to O(n) time but having trouble showing that:
T(n) = T(n-1) + T(n-2) +1
2) This on i am stumped on but think it's roughly about O(log log n):
T(n) = T([sqrt(n)]) + n. n greater-than-or-equal to 1. sqrt(n) is lower bound.
3) i believe this one is in roughly O(n*log log n):
T(n) = 2T(n/2) + (n/(log n)) + n.
Thanks for the help in advance.
T(n) = T(n-1) + T(n-2) + 1
Assuming T(0) = 0 and T(1) = a, for some constant a, we notice that T(n) - T(n-1) = T(n-2) + 1. That is, the growth rate of the function is given by the function itself, which suggests this function has exponential growth.
Let T'(n) = T(n) + 1. Then T'(n) = T'(n-1) + T'(n-2), by the above recurrence relation, and we have eliminated the troublesome constant term. T(n) and U(n) differ by a constant factor of 1, so assuming they are both non-decreasing (they are) then they will have the same asymptotic complexity, albeit for different constants n0.
To show T'(n) has asymptotic growth of O(b^n), we would need some base cases, then the hypothesis that the condition holds for all n up to, say, k - 1, and then we'd need to show it for k, that is, cb^(n-2) + cb^(n-1) < cb^n. We can divide through by cb^(n-2) to simplify this to 1 + b <= b^2. Rearranging, we get b^2 - b - 1 > 0; roots are (1 +- sqrt(5))/2, and we must discard the negative one since we cannot use a negative number as the base for our exponent. So for b >= (1+sqrt(5))/2, T'(n) may be O(b^n). A similar thought experiment will show that for b <= (1+sqrt(5))/2, T'(n) may be Omega(b^n). Thus, for b = (1+sqrt(5))/2 only, T'(n) may be Theta(b^n).
Completing the proof by induction that T(n) = O(b^n) is left as an exercise.
T(n) = T([sqrt(n)]) + n
Obviously, T(n) is at least linear, assuming the boundary conditions require T(n) be nonnegative. We might guess that T(n) is Theta(n) and try to prove it. Base case: let T(0) = a and T(1) = b. Then T(2) = b + 2 and T(4) = b + 6. In both cases, a choice of c >= 1.5 will work to make T(n) < cn. Suppose that whatever our fixed value of c is works for all n up to and including k. We must show that T([sqrt(k+1)]) + (k+1) <= c*(k+1). We know that T([sqrt(k+1)]) <= csqrt(k+1) from the induction hypothesis. So T([sqrt(k+1)]) + (k+1) <= csqrt(k+1) + (k+1), and csqrt(k+1) + (k+1) <= c(k+1) can be rewritten as cx + x^2 <= cx^2 (with x = sqrt(k+1)); dividing through by x (OK since k > 1) we get c + x <= cx, and solving this for c we get c >= x/(x-1) = sqrt(k+1)/(sqrt(k+1)-1). This eventually approaches 1, so for large enough n, any constant c > 1 will work.
Making this proof totally rigorous by fixing the following points is left as an exercise:
making sure enough base cases are proven so that all assumptions hold
distinguishing the cases where (a) k + 1 is a perfect square (hence [sqrt(k+1)] = sqrt(k+1)) and (b) k + 1 is not a perfect square (hence sqrt(k+1) - 1 < [sqrt(k+1)] < sqrt(k+1)).
T(n) = 2T(n/2) + (n/(log n)) + n
This T(n) > 2T(n/2) + n, which we know is the recursion relation for the runtime of Mergesort, which by the Master theorem is O(n log n), s we know our complexity is no less than that.
Indeed, by the master theorem: T(n) = 2T(n/2) + (n/(log n)) + n = 2T(n/2) + n(1 + 1/(log n)), so
a = 2
b = 2
f(n) = n(1 + 1/(log n)) is O(n) (for n>2, it's always less than 2n)
f(n) = O(n) = O(n^log_2(2) * log^0 n)
We're in case 2 of the Master Theorem still, so the asymptotic bound is the same as for Mergesort, Theta(n log n).
Related
I think the complexity of this recursion is O(n^2/3)` by change variable and induction. but I'm not sure. Is this solution correct?
This is a fascinating recurrence and it does not solve to Θ(n). Rather, it appears to solve to Θ(n2/3).
To give an intuition for why this isn't likely to be Θ(n), let's imagine that we're dealing with a really, really large value of n. Then since
T(n) = (nT(√n) + n)1/2
under the assumption that T(√n) ≈ √n, we'd get that
T(n) = (n√n + n)1/2
= (n3/2 + n)1/2
≈ n3/4.
In other words, assuming that T(n) = Θ(n) would give us a different value of T(n) as n gets large.
On the other hand, let's assume that T(n) = Θ(n2/3). Then the same calculation gives us that
T(n) = (nT(n) + n)1/2
= (n · n2/3 + n)1/2
≈ (n4/3)1/2
= n2/3,
which is consistent with itself.
To validate this, I wrote a short program that printed out different values of T(n) given different inputs and plotted the results. Here's the version of T(n) that I wrote up:
double T(double n) {
if (n <= 2) return n;
return sqrt(n * T(sqrt(n)) + n);
}
I decided to use 2 as a base case, since repeatedly taking square roots will never let n drop to one. I also decided to use real-valued arguments rather than discrete integer values just to make the math easier.
If you plot the values of T(n), you get this curve:
.
This doesn't look like what I'd expect from a linear plot. To figure out what this was, I plotted it on a log/log plot, which has the nice property that all polynomial functions get converted to straight lines whose slope is equal to the exponent. Here's the result:
I consulted my Handy Neighborhood Regression Software and asked it to determine the slope of this line. Here's what it gave back:
Slope: 0.653170918815869
R2: 0.999942627574643
That's a very good fit, and the slope of 0.653 is pretty close to 2/3. So that's more empirical evidence supporting that the recurrence solves to Θ(n2/3).
All that's left to do now is to work out the math. We'll solve this recurrence using a series of substitutions.
First, I'm generally not that comfortable working with exponents in the way that this recurrence uses them, so let's take the log of both sides. (Throughout this exposition, I'll use lg n to mean log2 n).
lg T(n) = lg (nT(√n) + n)1/2
= (1/2) lg (nT(√n) + n)
= (1/2) lg(T(√n) + 1) + (1/2)lg n
≈ (1/2) lg T(√n) + (1/2) lg n
Now, let's define S(n) = lg T(n). Then we have
S(n) = lg T(n)
≈ (1/2) lg T(√ n) + (1/2) lg n
= (1/2) S(√ n) + (1/2) lg n
That's a lot easier to work with, though we still have the problem of the recurrence shrinking by powers each time. To address this, let's do one more substitution, which is a fairly common one when working with these sorts of expressions. Let's define R(n) = S(2n). Then we have that
R(n) = S(2n)
≈ (1/2)S(√2n) + (1/2) lg 2n
= (1/2)S(2n/2) + (1/2) n
= (1/2) R(n / 2) + (1/2) n
Great! All that's left to do now is to solve R(n).
Now, there is a slight catch here. We could immediately use the Master Theorem to conclude that R(n) = Θ(n). The problem with this is that just knowing that R(n) = Θ(n) won't allow us to determine what T(n) is. Specifically, let's suppose that we just know R(n) = Θ(n). Then we could say that
S(n) = S(2lg n) = R(lg n) = Θ(log n)
to get that S(n) = Θ(log n). However, we get stuck when trying to solve for T(n) in terms of S(n). Specifically, we know that
T(n) = 2S(n) = 2Θ(log n),
but we cannot go from this to saying that T(n) = Θ(n). The reason is that the hidden coefficient in the Θ(log n) is significant here. Specifically, if S(n) = k lg n, then we have that
2k lg n = 2lg nk = nk,
so the leading coefficient of the logarithm will end up determining the exponent on the polynomial. As a result, when solving R, we need to determine the exact coefficient of the linear term, which translates into the exact coefficient of the logarithmic term for S.
So let's jump back to R(n), which we know is
R(n) ≈ (1/2) R(n/2) + (1/2)n.
If we iterate this a few times, we see this pattern:
R(n) ≈ (1/2) R(n/2) + (1/2)n
≈ (1/2)((1/2) R(n/4) + (1/4)n) + (1/2)n
≈ (1/4)R(n/4) + (1/8)n + (1/2)n
≈ (1/4)((1/2)R(n/8) + n/8) + (1/8)n + (1/2)n
≈ (1/8)R(n/8) + (1/32)n + (1/8)n + (1/2)n.
The pattern appears to be that, after k iterations, we get that
R(n) ≈ (1/2k)R(n/2k) + n(1/2 + 1/8 + 1/32 + 1/128 + ... + 1/22k+1).
This means we should look at the sum
(1/2) + (1/8) + (1/32) + (1/128) + ...
This is
(1/2)(1 + 1/4 + 1/16 + 1/64 + ... )
which, as the sum of a geometric series, solves to
(1/2)(4/3)
= 2/3.
Hey, look! It's the 2/3 we were talking about earlier. This means that R(n) works out to approximately (2/3)n + c for some constant c that depends on the base case of the recurrence. Therefore, we see that
T(n) = 2S(n)
= 2S(2lg n)
= 2R(lg n)
≈ 2(2/3)lg n + c
= 2lg n2/3 + c
= 2c 2lg n2/3
= 2c n2/3
= Θ(n2/3)
Which matches the theoretically predicted and empirically observed values from earlier.
This was a very fun problem to work through and I'll admit I'm surprised by the answer! I am a bit nervous, though, that I may have missed something when going from
lg T(n) = (1/2) lg (T(√n) + 1) + (1/2) lg n
to
lg T(n) ≈ (1/2) lg T(√ n) + (1/2) lg n.
It's possible that this +1 term actually introduces some other term into the recurrence that I didn't recognize. For example, is there an O(log log n) term that arises as a result? That wouldn't surprise me, given that we have a recurrence that shrinks by a square root. However, I've done some simple data explorations and I'm not seeing any terms in there that look like there's a double log involved.
Hope this helps!
We know that:
T(n) = sqrt(n) * sqrt(T(sqrt(n)) + 1)
Hence:
T(n) < sqrt(n) * sqrt(T(sqrt(n)) + T(sqrt(n)))
1 is replaced by T(sqrt(n)). So,
T(n) < sqrt(2) * sqrt(n) * sqrt(T(sqrt(n))
Now, to find an upper bound we need to solve the following recurrent relation:
G(n) = sqrt(2n) * sqrt(G(sqrt(n))
To solve this, we need to expand it (suppose n = 2^{2^k} and T(1) = 1):
G(n) = (2n)^{1/2} * (2n)^{1/8} * (2n)^{1/32} * ... * (2n)^(1/2^k) =>
G(n) = (2n)^{1/2 + 1/8 + 1/32 + ... + 1/2^k} =
If we take a factor 1/2 from 1/2 + 1/8 + 1/32 + ... + 1/2^k we will have 1/2 * (1 + 1/4 + 1/8 + ... + 1/2^{k-1}).
As we know that 1 + 1/4 + 1/8 + ... + 1/2^{k-1} is a geometric series with a ratio 1/4, it is equal to 4/3 at infinity. Therefore G(n) = Theta(n^{2/3}) and T(n) = O(n^{2/3}).
Notice that as sqrt(n) * sqrt(T(sqrt(n)) < T(n), we can show similar to the previous case that T(n) = Omega(n^{2/3}). It means T(n) = Theta(n^{2/3}).
i want to know what the Time Complexity of my recursion method :
T(n) = 2T(n/2) + O(1)
i saw a result that says it is O(n) but i don't know why , i solved it like this :
T(n) = 2T(n/2) + 1
T(n-1) = 4T(n-1/4) + 3
T(n-2) = 8T(n-2/8) + 7
...... ………….. ..
T(n) = 2^n+1 T (n/2^n+1) + (2^n+1 - 1)
I think you have got the wrong idea about recursive relations. You can think as follows:
If T(n) represents the value of function T() at input = n then the relation says that output is one more double the value at half of the current input. So for input = n-1 output i.e. T(n-1) will be one more than double the value at half of this input, that is T(n-1) = 2*T((n-1)/2) + 1
The above kind of recursive relation should be solved as answered by Yves Daoust. For more examples on recursive relations, you can refer this
Consider that n=2^m, which allows you to write
T(2^m)=2T(2^(m-1))+O(1)
or by denoting S(m):= T(2^m),
S(m)=2 S(m-1) + O(1),
2^m S(m)=2 2^(m-1)S(m-1) + 2^(m-1) O(1)
and finally,
R(m) = R(m-1) + 2^(m-1) O(1).
Now by induction,
R(m) = R(0) + (2^m-1) O(1),
T(n) = S(m) = 2^(1-m) T(2^m) + (2 - 2^(m-1)) O(1) = 2/n T(n) + (2 - n/2) O(1).
There are a couple of rules that you might need to remember. If you can remember these easy rules then Master Theorem is very easy to solve recurrence equations. The following are the basic rules which needs to be remembered
case 1) If n^(log b base a) << f(n) then T(n) = f(n)
case 2) If n^(log b base a) = f(n) then T(n) = f(n) * log n
case 3) 1) If n^(log b base a) >> f(n) then T(n) = n^(log b base a)
Now, lets solve the recurrence using the above equations.
a = 2, b = 2, f(n) = O(1)
n^(log b base a) = n = O(n)
This is case 3) in the above equations. Hence T(n) = n^(log b base a) = O(n).
How to get big-O for this?
T(N) = 2T(N − 1) + N, T(1) = 2
I got two variants of answer O(2^N) or O(N^2), but I am not sure how to solve it correctly
Divide T(N) by 2^N and name the result:
S(N) = T(N)/2^N
From the definition of T(N) we get
S(N) = S(N-1) + N/2^N (eq.1)
meaning that S(N) increases, but quickly converges to a constant (since N/2^N -> 0). So,
T(N)/2^N -> constant
or
T(N) = O(2^N)
Detailed proof
In the comment below Paul Hankin suggests how to complete the proof. Take eq.1 and sum from N=2 to N=M
sum_{N=2}^M S(N) = sum_{N=2}^M S(N-1) + sum_{N=2}^M N/2^N
= sum_{N=1}{M-1} S(N) + sum_{N=1}^{M-1} (N-1)/2^{N-1}
thus, after canceling terms with indexes N = 2, 3, ..., M-1, we get
S(M) = S(1) + sum_{N=1}^M N/2^N - M/2^M
and since the series on the right converges (because its terms are bounded by 1/N^2 for N>>1 which is known to converge), S(M) converges to a finite constant.
It's a math problem and Leandro Caniglia is right.
let b(n) = T(n) / 2^n
thus b(n) = b(n-1) + n / 2^n = b(n-2) + n / 2^n + (n-1) / 2^(n-1) ....
i / 2^i is less than 1 for every integer i
So the sum of them has limit and must smaller than some constant.
thus b(n) < C.
thus T(n) < 2^n * C.
It is obvious that T(n) >= 2^n.
So T(n) is O(2^n)
Check by plugging the answer in the equation.
2^N = 2.2^(N-1) + N = 2^N + N
or
N^2 = 2 (N-1)^2 + N
Keeping only the dominant terms, you have
2^N ~ 2^N
or
N^2 ~ 2 N^2.
Conclude.
From what I have studied: I have been asked to determine the complexity of a function with respect to another function. i.e. Given f(n) and g(n), determine O(f(n(). In such cases, I substitute values, compare both of them and arrive at a complexity - using O(), Theta and Omega notations.
However, in the substitution method for solving recurrences, every standard document has the following lines:
• [Assume that T(1) = Θ(1).]
• Guess O(n3) . (Prove O and Ω separately.)
• Assume that T(k) ≤ ck3 for k < n .
• Prove T(n) ≤ cn3 by induction.
How am I supposed to find O and Ω when nothing else (apart from f(n)) is given? I might be wrong (I, definitely am), and any information on the above is welcome.
Some of the assumptions above are with reference to this problem: T(n) = 4T(n/2) + n
, while the basic outline of the steps is for all such problems.
That particular recurrence is solvable via the Master Theorem, but you can get some feedback from the substitution method. Let's try your initial guess of cn^3.
T(n) = 4T(n/2) + n
<= 4c(n/2)^3 + n
= cn^3/2 + n
Assuming that we choose c so that n <= cn^3/2 for all relevant n,
T(n) <= cn^3/2 + n
<= cn^3/2 + cn^3/2
= cn^3,
so T is O(n^3). The interesting part of this derivation is where we used a cubic term to wipe out a linear one. Overkill like that is often a sign that we could guess lower. Let's try cn.
T(n) = 4T(n/2) + n
<= 4cn/2 + n
= 2cn + n
This won't work. The gap between the right-hand side and the bound we want is is cn + n, which is big Theta of the bound we want. That usually means we need to guess higher. Let's try cn^2.
T(n) = 4T(n/2) + n
<= 4c(n/2)^2 + n
= cn^2 + n
At first that looks like a failure as well. Unlike our guess of n, though, the deficit is little o of the bound itself. We might be able to close it by considering a bound of the form cn^2 - h(n), where h is o(n^2). Why subtraction? If we used h as the candidate bound, we'd run a deficit; by subtracting h, we run a surplus. Common choices for h are lower-order polynomials or log n. Let's try cn^2 - n.
T(n) = 4T(n/2) + n
<= 4(c(n/2)^2 - n/2) + n
= cn^2 - 2n + n
= cn^2 - n
That happens to be the exact solution to the recurrence, which was rather lucky on my part. If we had guessed cn^2 - 2n instead, we would have had a little credit left over.
T(n) = 4T(n/2) + n
<= 4(c(n/2)^2 - 2n/2) + n
= cn^2 - 4n + n
= cn^2 - 3n,
which is slightly smaller than cn^2 - 2n.
I have this recurrence:
T(n)= 2T(n/2) + (n-1)
My try is as follow:
the tree is like this:
T(n) = 2T(n/2) + (n-1)
T(n/2) = 2T(n/4) + ((n/2)-1)
T(n/4) = 2T(n/8) + ((n/4)-1)
...
the hight of the tree : (n/(2h))-1 = 1 ⇒ h = lg n - 1 = lg n - lg 2
the cost of the last level : 2h = 2lg n - lg 2 = (1/2) n
the cost of all levels until level h-1 : Σi=0,...,lg(2n) n - (2i-1), which is a geometric series and equals (1/2)((1/2)n-1)
So, T(n) = Θ(n lg n)
my question is: Is that right?
No, it isn't. You have the cost of the last level wrong, so what you derived from that is also wrong.
(I'm assuming you want to find the complexity yourself, so no more hints unless you ask.)
Edit: Some hints, as requested
To find the complexity, one usually helpful method is to recursively apply the equation and insert the result into the first,
T(n) = 2*T(n/2) + (n-1)
= 2*(2*T(n/4) + (n/2-1)) + (n-1)
= 4*T(n/4) + (n-2) + (n-1)
= 4*T(n/4) + 2*n - 3
= 4*(2*T(n/8) + (n/4-1)) + 2*n - 3
= ...
That often leads to a closed formula you can prove via induction (you don't need to carry out the proof if you have enough experience, then you see the correctness without writing down the proof).
Spoiler: You can look up the complexity in almost any resource dealing with the Master Theorem.
This can be easily solved with Masters theorem.
You have a=2, b=2, f(n) = n - 1 = O(n) and therefore c = log2(2) = 1. This falls into the first case of Master's theorem, which means that the complexity is O(n^c) = O(n)