Question

Somewhat similar to fibonacci sequence

Running time of an algorithm is given by

T (n) =T (n-1)+T(n-2)+T(n-3) if n > 3

= n otherwise the order of this algorithm is?

if calculated by induction method then

T(n) = T(n-1) + T(n-2) + T(n-3) 

Let us assume T(n) to be some function aⁿ
then aⁿ = a^n-1 + a^n-2 + a^n-3
=> a³ = a² + a + 1

which give complex solutions also roots of above equation according to my calculations are

a = 1.839286755
a = 0.419643 - i ( 0.606291) 
a = 0.419643 + i ( 0.606291) 

Now, how can I proceed further or is there any other method for this?

Answer 1

If I remember correctly, when you have determined the roots of the characteristic equation, then the T(n) can be the linear combination of the powers of those Roots

T(n)=A1*root1^n+A2*root2^n+A3*root3^n

So I guess the maximum complexity here will be
(maxroot)^n where maxroot is the maximum absolute value of your roots. So for your case it is ~ 1.83^n