Question

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?

Answer 1

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).