I am reviewing for a midterm test regarding Big-Oh runtime. One of the questions I have difficulty with is given the following recurrence and the question is asking for the Big-Oh notation.
T(n) = 2T(n/2) + (2n^2 + 3n + 5)
So by using the Master Theorem, where k > log_b (a), in this question, I am thinking k is 2 from (2n^2), a is 2 from 2T and b is 2 from (n/2). Therefore, the runtime of Master Theorem is when k > log_b (a), that is 2 > log_2 (2) = 1, then T(n) = O(n^2).
Is my thinking correct? I have never seen a quadratic runtime inside T(n) but I am fairly certain it is O(n^2) in this question.
Thank you for your input!
Yes, O(n^2) is correct. Actually there is a similar example in the Wikipedia article about the master theorem. This function could be anything and basically you just compare the depth and width of the recursion tree with the cost of this additional function and check what is dominating the complexity.
Related
I want to solve this recurrence with an accuracy of Θ:
T(n) = T(n/5) + T(7n/10) + Θ(n)
I can solving typical recurrence but I don't know what to do with this one as it doesn't match to any case of master theorem.
Any help or hint?
You can use the Akra-Bazzi generalization of the master theorem.
If the Theta(n) term were replaced by something smaller, like Theta(1) or Theta(sqrt(n)) in this case, you could simply find the value of alpha so that n^alpha = (n/5)^alpha + (7n/10)^alpha by factoring out n^alpha. However, if you do that, you get alpha = 0.84, and n^0.84 is asymptotically smaller than Theta(n), so the contribution from the Theta(n) terms dominate if you iterate the recursion until the arguments are small. The result is that T(n) is Theta(n), though with different constants than the Theta(n) in the recursion.
I have recently encountered a recurrence problem:
T(n) = 2*T(ceil((sqrt(n)))+1
T(1)=1;
I am unable to see this function terminate at all when I draw my recurrence tree. The general node form in the tree (n1/2i) becomes 1 only when 1/2i becomes 0. This means i should tend to infinity.
You're right that if sqrt is the ceiling of the square root, then you'll never reach 1 by repeatedly applying square roots. I'm going to assume that you meant to use the floor, which means that you will indeed eventually hit 1 as the recurrence unwinds.
In this case, your recurrence is more properly
T(1) = 1
T(n) = 2T(⌊√n⌋) + 1
A standard technique for solving recurrences involve square roots is to make a substitution. Let's define a new value k such that n = 2k. Notice that √n = (2k)1/2 = 2k/2. In other words, taking the square root of n is equivalent to halving the value of k. Because of this, we can convert the above recurrence, which involves square roots, into a new recurrence that will more closely match the form used by the Master Theorem and other recurrence-solving techniques. Specifically, let's define S(k) = T(2k). Then we get the recurrence
S(0) = 1
S(k) = 2S(⌊k / 2⌋) + 1
It's a lot easier to see how to solve this recurrence. Either by recognizing this recurrence from elsewhere or by using the Master Theorem, we get that S(k) = Θ(k). Now, we wanted to solve for T(n), so we can use the fact that S(k) = T(2k) = T(n). Since S(k) = Θ(k), we now see that T(n) = Θ(k). Since we chose k such that 2k = n, this means that k = lg n. Therefore, T(n) = Θ(log n), so the recurrence works out to Θ(log n).
Hope this helps!
I have to code to evaluate the value of following sequence :
( pow(1,k) + pow(2,k) + ... + pow(n,k) ) % MOD
for given value of n,k and MOD.
I have tried searching it on internet. I got an equation . It contains zeta functions and it seems difficult in implementation. I want any simple approach for implementing the same. Note that the value of n is large, so that we cannot simply use brute force to pass the time limit.
Newton's identities might be of help. Calculate the coefficients of the polynomial with 1..n as roots. That pretty trivial. Then use the identities.
It's just the first thing that comes to mind when I see sums of powers.
I think it is nicely compatible with modular arithmetics - there are only multiplications and additions.
I must admit, that Newton's identities are only the rearrangement of the terms, so not much speed gain here.
JUST USE PYTHON
k=input("Enter value for K: ")
n=input("Enter value for N: ")
mod=input("Enter value for MOD: ")
sum=0
for i in range(1,n+1):
sum+=pow(i,k)
result=sum % mod
print mod
May be this code is gonna help.
I agree that math.stackexchange.com is a better bet.
But here are random facts that, depending on parameters, may make the problem more manageable.
First, factor MOD, solve for each prime power factor, then use the Chinese Remainder Theorem to find the answer for MOD. Thus without loss of generality, you may assume that MOD is a prime power.
Next, note that 1^k + ... + MOD^k is always divisible by MOD. Therefore you can replace n by n mod MOD.
Next, if MOD = p^i and j is not divisible by p, then j^((p-1) * p^(i-1)) is 1 mod MOD, so we can reduce the size of k.
Of course if (k, n) < MOD and MOD is prime, this will not help you at all. (Which, depending on how this problem arises, may well be the case.)
(If k is small enough, there are explicit formulas that you can produce for the sum. But it seems that for you k can be large enough to make that approach intractable.)
I am trying to perform asymptotic analysis on the following recursive function for an efficient way to power a number. I am having trouble determining the recurrence equation due to having different equations for when the power is odd and when the power is even. I am unsure how to handle this situation. I understand that the running time is theta(logn) so any advice on how to proceed to this result would be appreciated.
Recursive-Power(x, n):
if n == 1
return x
if n is even
y = Recursive-Power(x, n/2)
return y*y
else
y = Recursive-Power(x, (n-1)/2)
return y*y*x
In any case, the following condition holds:
T(n) = T(floor(n/2)) + Θ(1)
where floor(n) is the biggest integer not greater than n.
Since floor doesn't have influence on results, the equation is informally written as:
T(n) = T(n/2) + Θ(1)
You have guessed the asymptotic bound correctly. The result could be proved using Substitution method or Master theorem. It is left as an exercise for you.
I've been working through a recent Computer Science homework involving recursion and big-O notation. I believe I understand this pretty well (certainly not perfectly, though!) But there is one question in particular that is giving me the most problems. The odd thing is that by looking it, it looks to be the most simple one on the homework.
Provide the best rate of growth using the big-Oh notation for the solution to the following recurrence?
T(1) = 2
T(n) = 2T(n - 1) + 1 for n>1
And the choices are:
O(n log n)
O(n^2)
O(2^n)
O(n^n)
I understand that big O works as an upper bound, to describe the most amount of calculations, or the highest running time, that program or process will take. I feel like this particular recursion should be O(n), since, at most, the recursion only occurs once for each value of n. Since n isn't available, it's either better than that, O(nlogn), or worse, being the other three options.
So, my question is: Why isn't this O(n)?
There's a couple of different ways to solve recurrences: substitution, recurrence tree and master theorem. Master theorem won't work in the case, because it doesn't fit the master theorem form.
You could use the other two methods, but the easiest way for this problem is to solve it iteratively.
T(n) = 2T(n-1) + 1
T(n) = 4T(n-2) + 2 + 1
T(n) = 8T(n-3) + 4 + 2 + 1
T(n) = ...
See the pattern?
T(n) = 2n-1⋅T(1) + 2n-2 + 2n-3 + ... + 1
T(n) = 2n-1⋅2 + 2n-2 + 2n-3 + ... + 1
T(n) = 2n + 2n-2 + 2n-3 + ... + 1
Therefore, the tightest bound is Θ(2n).
I think you have misunderstood the question a bit. It does not ask you how long it would take to solve the recurrence. It is asking what the big-O (the asymptotic bound) of the solution itself is.
What you have to do is to come up with a closed form solution, i. e. the non-recursive formula for T(n), and then determine what the big-O of that expression is.
The question is asking for the big-Oh notation for the solution to the recurrence, not the cost of calculation the recurrence.
Put another way: the recurrence produces:
1 -> 2
2 -> 5
3 -> 11
4 -> 23
5 -> 47
What big-Oh notation best describes the sequence 2, 5, 11, 23, 47, ...
The correct way to solve that is to solve the recurrence equations.
I think this will be exponential. Each increment to n makes the value to be twice as large.
T(2) = 2 * T(1) = 4
T(3) = 2 * T(2) = 2 * 4
...
T(x) would be the running time of the following program (for example):
def fn(x):
if (x == 1):
return # a constant time
# do the calculation for n - 1 twice
fn(x - 1)
fn(x - 1)
I think this will be exponential. Each increment to n brings twice as much calculation.
No, it doesn't. Quite on the contrary:
Consider that for n iterations, we get running time R. Then for n + 1 iterations we'll get exactly R + 1.
Thus, the growth rate is constant and the overall runtime is indeed O(n).
However, I think Dima's assumption about the question is right although his solution is overly complicated:
What you have to do is to come up with a closed form solution, i. e. the non-recursive formula for T(n), and then determine what the big-O of that expression is.
It's sufficient to examine the relative size of T(n) and T(n + 1) iterations and determine the relative growth rate. The amount obviously doubles which directly gives the asymptotic growth.
First off, all four answers are worse than O(n)... O(n*log n) is more complex than plain old O(n). What's bigger: 8 or 8 * 3, 16 or 16 * 4, etc...
On to the actual question. The general solution can obviously be solved in constant time if you're not doing recursion
( T(n) = 2^(n - 1) + 2^(n) - 1 ), so that's not what they're asking.
And as you can see, if we write the recursive code:
int T( int N )
{
if (N == 1) return 2;
return( 2*T(N-1) + 1);
}
It's obviously O(n).
So, it appears to be a badly worded question, and they are probably asking you the growth of the function itself, not the complexity of the code. That's 2^n. Now go do the rest of your homework... and study up on O(n * log n)
Computing a closed form solution to the recursion is easy.
By inspection, you guess that the solution is
T(n) = 3*2^(n-1) - 1
Then you prove by induction that this is indeed a solution. Base case:
T(1) = 3*2^0 - 1 = 3 - 1 = 2. OK.
Induction:
Suppose T(n) = 3*2^(n-1) - 1. Then
T(n+1) = 2*T(n) + 1 = 3*2^n - 2 + 1 = 3*2^((n+1)-1) - 1. OK.
where the first equality stems from the recurrence definition,
and the second from the inductive hypothesis. QED.
3*2^(n-1) - 1 is clearly Theta(2^n), hence the right answer is the third.
To the folks that answered O(n): I couldn't agree more with Dima. The problem does not ask the tightest upper bound to the computational complexity of an algorithm to compute T(n) (which would be now O(1), since its closed form has been provided). The problem asks for the tightest upper bound on T(n) itself, and that is the exponential one.