I have the following worked out:
T(n) = T(n - 1) + n = O(n^2)
Now when I work this out I find that the bound is very loose. Have I done something wrong or is it just that way?
You need also a base case for your recurrence relation.
T(1) = c
T(n) = T(n-1) + n
To solve this, you can first guess a solution and then prove it works using induction.
T(n) = (n + 1) * n / 2 + c - 1
First the base case. When n = 1 this gives c as required.
For other n:
T(n)
= (n + 1) * n / 2 + c - 1
= ((n - 1) + 2) * n / 2 + c - 1
= ((n - 1) * n / 2) + (2 * n / 2) + c - 1
= (n * (n - 1) / 2) + c - 1) + (2 * n / 2)
= T(n - 1) + n
So the solution works.
To get the guess in the first place, notice that your recurrence relationship generates the triangular numbers when c = 1:
T(1) = 1:
*
T(2) = 3:
*
**
T(3) = 6:
*
**
***
T(4) = 10:
*
**
***
****
etc..
Intuitively a triangle is roughly half of a square, and in Big-O notation the constants can be ignored so O(n^2) is the expected result.
Think of it this way:
In each "iteration" of the recursion you do O(n) work.
Each iteration has n-1 work to do, until n = base case. (I'm assuming base case is O(n) work)
Therefore, assuming the base case is a constant independant of n, there are O(n) iterations of the recursion.
If you have n iterations of O(n) work each, O(n)*O(n) = O(n^2).
Your analysis is correct. If you'd like more info on this way of solving recursions, look into Recursion Trees. They are very intuitive compared to the other methods.
The solution is pretty easy for this one. You have to unroll the recursion:
T(n) = T(n-1) + n = T(n-2) + (n - 1) + n =
= T(n-3) + (n-2) + (n-1) + n = ... =
= T(0) + 1 + 2 + ... + (n-1) + n
You have arithmetic progression here and the sum is 1/2*n*(n-1). Technically you are missing the boundary condition here, but with any constant boundary condition you see that the recursion is O(n^2).
Looks about right, but will depend on the base case T(1). Assuming you will do n steps to get T(n) to T(0) and each time the n term is anywhere between 0 and n for an average of n/2 so n * n/2 = (n^2)/2 = O(n^2).
Related
I need help in solving T(n) = T(n/4) + T(n/3) + 2n using iteration method (recursion tree. I am thinking it would be either Θ(2n) or Θ(n)?
It's straightforward. We have two following inequalities:
T(n) > 2T(n/4) + 2n
and
T(n) < 2T(n/3) + 2n
Now, try to find upper bound and lower bound by expansion. Based on both cases, you will find that T(n) = Theta(n).
For example, for T'(n) = 2T(n/3) + 2n we have the following expansion:
T'(n) = 2T(n/3) + 2n = 2^2 T(n/3^2) + (1 + 2/3) * 2n
By induction we can show that:
T'(n) = 2^log_3(n) T(0) + (1 + 2/3 + 2^2/3^2 + ...) * 2n
< n + 6 * n = 7n
Because 2^log_3(n) < 2^log_2(n) = n and (1 + 2/3 + 2^2/3^2 + ...) is a geometric sum with factor 2/3. Hence, the sum will be 1/(1-2/3) = 3 when n goes to infinity.
You can do the same analysis for the lower bound of T(n).
Therefore, as c1 * n <= T(n) <= c_2 * n, we can conclude that T(n) is in Theta(n).
could someone help me solve this recurrence?
T(n) = 8T(n/2) + n^2 is T(n) = O(n^3) using the substitution method. Considering T(1)=1
One way to approach this recurrence relation, is rsolve from sympy, Python's symbolic math library:
from sympy import Function, rsolve, log
from sympy.abc import k, n
f = Function('f')
g = Function('g')
# T(n) = 8T(n/2) + n^2 T(1) = 1
T = f(n) - 8 * f(n / 2) - n ** 2 - 3
Tk = T.subs({n: 2 ** k, f(n): g(k), f(n / 2): g(k - 1)})
s = rsolve(Tk, g(k), {g(0): 1})
print("solution for k:", s.cancel())
print("solution for n:", s.subs({k: log(n, 2)}).simplify().cancel())
for k in range(0, 11):
print(f"k={k}, n={2 ** k}, T(n)={(17 * n ** 3 - 3) // 7 - n ** 2}")
This outputs the solution (17*n^3 - 3)/7 - n^2.
Note that the function is only defined for n a power of 2.
The first values for n=1,2,4,8,... are 1, 15, 139, 1179, 9691, 78555, 632539, 5076699, 40679131, 325695195, 2606610139
Just try to expand the relation and keep continue by the induction:
T(n) = 8 T(n/2) + n^2 =
8(8T(n/4) + (n/2)^2) + n^2 = 8^2 T(n/2^2) + 8 (n/2)^2 + n^2
As each time n is divided by 2 each time, if n = 2^k, the length of the series k = log(n):
T(n) = n^2 + 8 (n/2)^2 + 8^2 (n/2^2)^2 + ... + 8^log(n) T(1)
= sum_{i=0}^{log(n)} 8^i (n/2^i)^2
Rewrite the relation by 2^2i = 4^i:
T(n) = sum_{i=0}^{log(n)} 8^i/4^i n^2 = n^2 * sum_{i=0}^{log(n)} 2^i
= n^2 (2^(log(n)+1) - 1) = \Theta(n^3)
Notice that 2^log(n) = n.
Please can anyone help me with this:
Solve using iteration Method T (n) = T (n - 1) + (n - 1)
And prove that T (n) ∈Θ (n²)
Please, if you can explain step by step I would be grateful.
I solved an easy way :
T (n) = T (n - 1) + (n - 1)-----------(1)
//now submit T(n-1)=t(n)
T(n-1)=T((n-1)-1)+((n-1)-1)
T(n-1)=T(n-2)+n-2---------------(2)
now submit (2) in (1) you will get
i.e T(n)=[T(n-2)+n-2]+(n-1)
T(n)=T(n-2)+2n-3 //simplified--------------(3)
now, T(n-2)=t(n)
T(n-2)=T((n-2)-2)+[2(n-2)-3]
T(n-2)=T(n-4)+2n-7---------------(4)
now submit (4) in (2) you will get
i.e T(n)=[T(n-4)+2n-7]+(2n-3)
T(n)=T(n-4)+4n-10 //simplified
............
T(n)=T(n-k)+kn-10
now, assume k=n-1
T(n)=T(n-(n-1))+(n-1)n-10
T(n)=T(1)+n^2-n-10
According to the complexity 10 is constant
So , Finally O(n^2)
T(n) = T(n - 1) + (n - 1)
= (T(n - 2) + (n - 2)) + (n - 1)
= (T(n - 3) + (n - 3)) + (n - 2) + (n - 1)
= ...
= T(0) + 1 + 2 + ... + (n - 3) + (n - 2) + (n - 1)
= C + n * (n - 1) / 2
= O(n2)
Hence for sufficient large n, we have:
n * (n - 1) / 3 ≤ T(n) ≤ n2
Therefore we have T(n) = Ω(n²) and T(n) = O(n²), thus T(n) = Θ (n²).
T(n)-T(n-1) = n-1
T(n-1)-T(n-2) = n-2
By substraction
T(n)-2T(n-1)+T(n-2) = 1
T(n-1)-2T(n-2)+T(n-3) = 1
Again, by substitution
T(n)-3T(n-1)+3T(n-2)-T(n-3) = 0
Characteristic equation of the recursion is
x^3-3x^2+3x-1 = 0
or
(x-1)^3 = 0.
It has roots x_1,2,3 = 1,
so general solution of the recursion is
T(n) = C_1 1^n + C_2 n 1^n + C_3 n^2 1^n
or
T(n) = C_1 + C_2 n + C_3 n^2.
So,
T(n) = Θ(n^2).
I need to solve the recurrence relation for
T(n) = T(n - 1) * T(n - 2) + c where T(0) = 1 and T(1) = 2.
The algorithm computes 2f(n) where f(n) is the nth Fibonacci number. I assume that T(n - 2) is approximately equal to T(n - 1) and then I rewrite the relation as
T(n) = T(n - 1)^2 + c
and finality I reach to complexity of O(n). This the algorithm:
Power(n):
if n == 0 then x = 1;
else if n == 1 then x = 2;
else x = Power(n - 1) * Power(n - 2);
return x;
Is it true that the recurrence relation T(n) = T(n - 1) * T(n - 2) + c has O(n) complexity?
You are confusing multiplication in the algorithm with multiplication in the recurrence relation. You calculate Power(n - 1) and Power(n - 2) and multiply them, but this does not take T(n - 1) * T(n - 2) time. These values are computed separately, and then multiplied, taking T(n - 1) + T(n - 2) time, assuming constant-time multiplication.
The recurrence then becomes T(n) = T(n - 1) + T(n - 2) + c. We'll solve this by substitution. I'll skip choosing an n0; our induction hypothesis will be T(n) = d * 2^n - 1. (We need the -1 to allow of to later cancel the c; this is called strengthening the induction hypothesis, see e.g. here).
T(n) = T(n - 1) + T(n - 2) + c
= d * (2^(n - 1) - 1) + d * (2^(n - 2) - 1) + c (substitute by IH)
= 3d * 2^(n - 2) - 2d + c
<= d * 2^n - 2d + c (3 < 2^2)
<= d * 2^n (when d > c / 2)
Conclusion: T(n) ∈ O(2n).
Bonus fact: the tight bound is actually Θ(φn), where φ is the golden ratio. See this answer for an explanation.
First, the recurrence that models the running time is T(n) = T(n-1) + T(n-2) + C for T(1) = T(0) = 1, for some C>0. This is because you make two recursive calls, one on n-1 and the other on n-2.
The recurrence you have is precisely the Fibonacci recurrence if we forget the additional constant (which contributed a factor C in the end). The exact solution is this (image source: Wikipedia):
where
and
You could say that T(n) = Theta(1.618... ^ n).
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I am having some issues on how to solve recurrence relations.
T(n) = T(n/2) + log2(n), T(1) = 1, where n is a power of 2
This is a homework problem, so don't just give me the answer. I was just wondering how to start the problem.
In class we went over the Master theorem. But I don't think that would be the best way to solve this particular relation.
I don't really know how to start the problem... should I just be going
T(n) = T(n/2) + log_base2(n)
T(n/2) = [T(n/4)+log_base2(n/2)]
T(n) = [T(n/4)+log_base2(n/2)] + log_base2(n)
And just keep working my way down to get something I can see makes a basic equation?
This recurrence solves to Θ((log n)2). Here are two ways to see this.
Some Substitutions
If you know that n is a perfect power of two (that is, n = 2k), you can rewrite the recurrence as
T(2k) = T(2k-1) + k
Let's define a new recurrence S(k) = T(2k). Then we get that
S(k) = S(k - 1) + k
If we expand out this recurrence, we get that
S(k) = S(k - 1) + k
= S(k - 2) + (k - 1) + k
= S(k - 3) + (k - 2) + (k - 1) + k
= S(k - 4) + (k - 3) + (k - 2) + (k - 1) + k
...
= S(0) + 1 + 2 + 3 + ... + k
= S(0) + Θ(k2)
Assuming S(0) = 1, then this recurrence solves to Θ(k2).
Since S(k) = T(2k) = T(n), we get that T(n) = Θ(k2) = Θ(log2 n).
Iterating the Recurrence
Another option here is to expand out a few terms of the recurrence and to see if any nice patterns emerge. Here’s what we get:
T(n) = T(n / 2) + lg n
= T(n / 4) + lg (n / 2) + lg n
= T(n / 8) + lg (n / 4) + lg (n / 2) + lg n
...
Eventually, after lg n layers, this recurrence bottoms out and we’re left with this expression:
lg n + lg (n / 2) + lg (n / 4) + ... + lg (n / 2lg n)
Using properties of logarithms, we can rewrite this as
lg n + (lg n - 1) + (lg n - 2) + (lg n - 3) + ... + (lg n - lg n)
Or, written in reverse, this is the sum
0 + 1 + 2 + 3 + ... + lg n
That sum is Gauss’s sum up to lg n, which evaluates to (lg n)(lg n + 1) / 2 = Θ((log n)2).
Hope this helps!
If n is a power of 2 then you can just expand out the recurrence and solve exactly, using that lg(a/b) = lg(a) - lg(b).
T(n) = lg(n) + lg(n/2) + lg(n/4) + ... + lg(1) + 1
= (lg(n) - 0) + (lg(n) - 1) .... + (lg(n) - lg(n)) + 1
= lg(n)*lg(n) - lg(n)*(lg(n)+1)/2 + 1
= lg(n)*lg(n)/2 - lg(n)/2 + 1
This can be done with the Akra-Bazzi theorem. See the third example in http://people.mpi-inf.mpg.de/~mehlhorn/DatAlg2008/NewMasterTheorem.pdf.
This can be solved with Master theorem. Your a=1 and b=2 and f(n) = log(n). Then c = log2(1) = 0. Because of your c and f(n) you fall into the second case (where k=1).
So the solution is Θ(log2 n)