I have tried determining the running time given by a recurrence relation, but my result is not correct.
Recurrence
T(n) = c + T(n-1) if n >= 1
= d if n = 0
My attempt
I constructed this recursion tree:
n
|
n-1
|
n-2
|
n-3
|
n-4
|
n-5
|
|
|
|
|
|
Till we get 1
Now at level i, the size of the sub problem should be, n-i
But at last we want a problem of size 1. Thus, at the last level, n-i=1 which gives, i=n-1.
So the depth of the tree becomes n-1 and the height becomes n-1+1= n.
Now the time required to solve this recursion = height of the tree*time required at each level which is :
n+(n-1)+(n-2)+(n-3)+(n-4)+(n-5)+ ...
==> (n+n+n+n+n+ ... )-(1+2+3+4+5+ ... )
==> n - (n(n+1)/2)
Now the time taken = n* ((n-n2)/2) which should give the order to be n2, but that is not the correct answer.
Now at level i, the size of the sub problem should be, n-i
Yes, that is correct. But you're assuming, that the runtime equals the sum of all the subproblem sizes. Just think about it, already summing the first two levels gives n + (n - 1) = 2n - 1, why would the problem size increase? Disclaimer: A bit handwavy and not an entirely accurate statement.
What the formula actually says
T(n) = c + T(n-1)
The formula says, solving it for some n takes the same time it takes to solve it for a problem size that is one less, plus an additional constant c: c + T(n - 1)
Another way to put the above statement is this: Given the problem takes some time t for a certain problem size, it will take t + c for a problem size, that is bigger by one.
We know, that at a problem size of n = 0, this takes time d. According to the second statement, for a size of one more, n = 1, it will take d + c. Applying our rule again, it thus takes d + c + c for n = 2. We conclude, that it must take d + n*c time for any n.
This is not a proof. To actually prove this, you must use induction as shown by amit.
A correct recursion tree
Your recursion tree only lists the problem size. That's pretty much useless, I'm afraid. Instead, you need to list the runtime for said problem size.
Every node in the tree corresponds to a certain problem size. What you write into that node is the additional time it takes for the problem size. I.e. you sum over all the descendants of a node plus the node itself to get the runtime for a certain problem size.
A graphical representation of such a tree would look like this
Tree Corresponding problem size
c n
|
c n - 1
|
c n - 2
|
c n - 3
.
.
.
|
c 2
|
c 1
|
d 0
Formalizing: As already mentioned, the label of a node is the additional runtime it takes to solve for that problem size, plus all its descendants. The uppermost node represents a problem size of n, bearing the label c because that's in addition to T(n-1), to which it is connected using a |.
In a formula, you would only write this relation: T(n) = c + T(n-1). Given that tree, you can see how this applies to every n>=1. You could write this down like this:
T(n) = c + T(n - 1) # This means, `c` plus the previous level
T(n - 1) = c + T(n - 2) # i.e. add the runtime of this one to the one above^
T(n - 2) = c + T(n - 3)
...
T(n - (n - 2)) = c + T(1)
T(n - (n - 1)) = c + T(0)
T(0) = d
You can now expand the terms from bottom to top:
T(n - (n - 1)) = c + T(0)
T(0) = d
T(n - (n - 2)) = c + T(1)
T(n - (n - 1)) = c + d
T(0) = d
T(n - (n - 3)) = c + T(2)
T(n - (n - 2)) = c + (c + d)
T(n - (n - 1)) = c + d
T(0) = d
T(n - (n - 4)) = c + T(3)
T(n - (n - 3)) = c + (2*c + d)
T(n - (n - 2)) = c + (c + d)
...
T(n) = c + T(n - 1)
T(n - 1) = c + ((n-2)c + d)
T(n) = c + (n-1)c + d = n*c + d
T(n - 1) = (n-1)c + d
Summing 1 to n
n+(n-1)+(n-2)+(n-3)+(n-4)+(n-5)+ ...
==> (n+n+n+n+n+ ... )-(1+2+3+4+5+ ... )
==> n - (n(n+1)/2)
From the first line to the second line, you have reduced your problem from summing 1 to n to, well, summing 1 to n-1. That's not very helpful, because you're stuck with the same problem.
I'm not sure what you did on the third line, but your transition from the first to the second is basically correct.
This would have been the correct formula:
T(n) = c + T(n-1)
= c + (c + T(n-2))
= ...
= c*i + T(n-i)
= ...
= c*n + T(0)
= c*n + d
If we assume c,d are constants - it gets you O(n)
To prove it mathematically - one can use mathematical induction
For each k < n assume T(n) = c*n + d
Base is T(0) = 0*n + d = d, which is correct for n < 1
T(n) = c + T(n-1) (*)
= c + (n-1)*c + d
= c*n + d
(*) is the induction hypothesis, and is valid since n-1 < n
The complexity would be O(n).
As you described the functions converts the problem for input n into problem for (n-1) by using a constant operation 'c'.
So moving down the recursion tree we will have in total n levels, and at each step we require some constant operation 'c'.
So there will be total c*n operations resulting into the complexity O(n).
Related
I am struggling with writing the formula that describes the recursive nature of the foo method.
The problem is that as far as I can tell, since every time n is divided with 2,
the binary tree formula should apply here.
This says that when in each call we divide the data we get a formula like this:
And then if we analyze for 2 so :
We get:
Which means that C(N) = log(N + 1), namely O(logN)
That all makes sense and seems to be the right choice for the foo method but it cant be because for
n = 8 I would get 3 + 1 iterations that are not n + 1 = 8 + 1 = 9 iterations
So here is your code:
void foo(int n) {
if (n == 1) System.out.println("Last line I print");
if (n > 1) {
System.out.println("I am printing one more line");
foo(n/2);
}
}
We can write a recurrence relation down for its runtime T as a function of the value of the parameter passed into it, n:
T(1) = a, a constant
T(n) = b + T(n/2), b constant, n > 1
We can write out some values of T(n) for various values of n to see if a pattern emerges:
n T(n)
---------
1 a
2 a + b
4 a + 2b
8 a + 3b
...
2^k a + kb
So for n = 2^k, T(n) = a + kb. We can solve for k in terms of n as follows:
n = 2^k <=> k = log(n)
Then we recover the expression T(n) = a + blog(n). We can verify this expression works easily:
a + blog(1) = a, as required
a + blog(n) = b + (a + blog(n/2))
= b + (a + b(log(n) - 1)
= b + a + blog(n) - b
= a + blog(n), as required
You can also use mathematical induction to do the same thing.
I am trying to solve the reccurence of the quicksort algorithm by the substitution method:
I can not find any way to proof that this will lead to . What further steps do I have to make to make this work?
The worst case of quicksort is when you choose a pivot element which is the minimum or maximum element from the array, so that all remaining elements go to one side of the partition, and the other side of the partition is empty. In this case, the non-empty partition has a size of (n - 1), and it takes linear time (kn for some constant k > 0) to do the partitioning itself, so the recurrence relation is
T(n) = T(n - 1) + T(0) + kn
If we guess that T(n) = an² + bn + c for some constants a, b, c, then we can substitute:
an² + bn + c = [ a(n - 1)² + b(n - 1) + c ] + [ c ] + kn
where the two square-bracketed terms are T(n - 1) and T(0) respectively. By expanding the brackets and equating coefficients, we get
an² = an²
bn = -2an + bn + kn
c = a - b + 2c
It follows that there is a family of solutions, parameterised by c = T(0), where a = k/2 and b = k/2 + c. This family of solutions can be written exactly as
T(n) = (k/2) n² + (k/2 + c) n + c
which is not just O(n²), but Ө(n²), meaning the running time is a quadratic function, not merely bounded above by a quadratic function. Note that the actual value of c doesn't change the asymptotic behaviour of the function, so long as k > 0 (i.e. the partitioning step does take a positive amount of time).
I have found an answer to my question, the continuation of the previous equation is,
This is true if
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).
I am refreshing on Master Theorem a bit and I am trying to figure out the running time of an algorithm that solves a problem of size n by recursively solving 2 subproblems of size n-1 and combine solutions in constant time.
So the formula is:
T(N) = 2T(N - 1) + O(1)
But I am not sure how can I formulate the condition of master theorem.
I mean we don't have T(N/b) so is b of the Master Theorem formula in this case b=N/(N-1)?
If yes since obviously a > b^k since k=0 and is O(N^z) where z=log2 with base of (N/N-1) how can I make sense out of this? Assuming I am right so far?
ah, enough with the hints. the solution is actually quite simple. z-transform both sides, group the terms, and then inverse z transform to get the solution.
first, look at the problem as
x[n] = a x[n-1] + c
apply z transform to both sides (there are some technicalities with respect to the ROC, but let's ignore that for now)
X(z) = (a X(z) / z) + (c z / (z-1))
solve for X(z) to get
X(z) = c z^2 / [(z - 1) * (z-a)]
now observe that this formula can be re-written as:
X(z) = r z / (z-1) + s z / (z-a)
where r = c/(1-a) and s = - a c / (1-a)
Furthermore, observe that
X(z) = P(z) + Q(z)
where P(z) = r z / (z-1) = r / (1 - (1/z)), and Q(z) = s z / (z-a) = s / (1 - a (1/z))
apply inverse z-transform to get that:
p[n] = r u[n]
and
q[n] = s exp(log(a)n) u[n]
where log denotes the natural log and u[n] is the unit (Heaviside) step function (i.e. u[n]=1 for n>=0 and u[n]=0 for n<0).
Finally, by linearity of z-transform:
x[n] = (r + s exp(log(a) n))u[n]
where r and s are as defined above.
so relabeling back to your original problem,
T(n) = a T(n-1) + c
then
T(n) = (c/(a-1))(-1+a exp(log(a) n))u[n]
where exp(x) = e^x, log(x) is the natural log of x, and u[n] is the unit step function.
What does this tell you?
Unless I made a mistake, T grows exponentially with n. This is effectively an exponentially increasing function under the reasonable assumption that a > 1. The exponent is govern by a (more specifically, the natural log of a).
One more simplification, note that exp(log(a) n) = exp(log(a))^n = a^n:
T(n) = (c/(a-1))(-1+a^(n+1))u[n]
so O(a^n) in big O notation.
And now here is the easy way:
put T(0) = 1
T(n) = a T(n-1) + c
T(1) = a * T(0) + c = a + c
T(2) = a * T(1) + c = a*a + a * c + c
T(3) = a * T(2) + c = a*a*a + a * a * c + a * c + c
....
note that this creates a pattern. specifically:
T(n) = sum(a^j c^(n-j), j=0,...,n)
put c = 1 gives
T(n) = sum(a^j, j=0,...,n)
this is geometric series, which evaluates to:
T(n) = (1-a^(n+1))/(1-a)
= (1/(1-a)) - (1/(1-a)) a^n
= (1/(a-1))(-1 + a^(n+1))
for n>=0.
Note that this formula is the same as given above for c=1 using the z-transform method. Again, O(a^n).
Don't even think about Master's Theorem. You can only use Masther's Theorem when you're given master's theorem when b > 1 from the general form T(n) = aT(n/b) + f(n).
Instead, think of it this way. You have a recursive call that decrements the size of input, n, by 1 at each recursive call. And at each recursive call, the cost is constant O(1). The input size will decrement until it reaches 1. Then you add up all the costs that you used to make the recursive calls.
How many are they? n. So this would take O(2^n).
Looks like you can't formulate this problem in terms of the Master Theorem.
A good start is to draw the recursion tree to understand the pattern, then prove it with the substitution method. You can also expand the formula a couple of times and see where it leads.
See also this question which solves 2 subproblems instead of a:
Time bound for recursive algorithm with constant combination time
May be you could think of it this way
when
n = 1, T(1) = 1
n = 2, T(2) = 2
n = 3, T(3) = 4
n = 4, T(4) = 8
n = 5, T(5) = 16
It is easy to see that this is a geometric series 1 + 2+ 4+ 8 + 16..., the sum of which is
first term (ratio^n - 1)/(ratio - 1). For this series it is
1 * (2^n - 1)/(2 - 1) = 2^n - 1.
The dominating term here is 2^n, therefore the function belongs to Theta(2^n). You could verify it by doing a lim(n->inf) [2^n / (2^n - 1)] = +ve constant.
Therefore the function belongs to Big Theta (2^n)
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).