Recurrence relations

A recurrence relation for the sequence a0 , a1 , is an equation that relates an to certain of its predecessors a0,a1, , an-1 .

e.g. Fibonacci sequence :

fn = fn-1 + fn-2 for n ³ 3

with initial condition : f1 = 1 , f2 = 2

Example 1 : Tower of Hanoi :

The Tower of Hanoi is a puzzle consisting of three pegs mounted on a board and n disks of various sizes with holes in their centers. It is assumed that if a disk is on a peg, only a disk of smaller diameter can be placed on top of the first disk. Given all the disks stacked on one peg , the problem is to transfer the disks to another peg by moving one disk at a time.

 

Let Cn be the number of disk movements needed to solve the n-disk Tower of Hanoi problem, then

Cn = 2 C n-1 + 1       for n > 1 with initial condition : C1 = 1

 

 

Example 2: Ackermann's Function :

A(m,0) = A(m-1,1) m = 1,2,.

A(m,n) = A(m-1 , A(m,n-1) ) m = 1,2,. and n = 1,2,

With the initial conditions :

A(0,n ) = n + 1 , n = 0, 1, .

 

Ackermann's function is a fast growing function.

 

 

Solving Recurrence relations

  1. Iteration method :
  2. To solve a recurrence relation involving a0 , a1 . by iteration , one uses the recurrence relation to replace the nth an in terms of certain of its

    predecessors a n-1, a0. Then successively use the recurrence relation to replace each of a n-1, by certain of their predecessors. Continue in this fashion until an explicit formula is obtained.

     

    Example 5.2.1

    Solve an = a n-1 + 3 subject to the initial condition a1 = 2.

    an = a n-1 + 3 = a n-2 + 3 + 3 = an = a n-3 + 3 + 3 + 3 = a n-3 + 3*3

    =

    = a n-(n-1) + 3*(n-1) = 2 + 3*(n-1) = 3*n -1

     

    Example 5.2.4 :

    Solve cn = 2 c n-1 + 1 with initial condition c1 = 1 .

    cn = 2 c n-1 + 1 = 2 *(2 c n-2 + 1) + 1 = 2 2 c n-2 + 2 + 1

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

    = . = 2 k c n-k + 2 k-1 + .. + 2 + 1

    = 2 n-1c n-(n-1) + 2 n-2 + . + 2 + 1

    = 2 n-1 + 2 n-2 + . + 2 + 1 = 2 n - 1

     

     

    Definition :

    A linear homogeneous recurrence relation of order k with constant coefficients is a recurrence relation of the form :

    an = c1a n-1 + c2 a n-2 + . + ck a n-k with ck ¹ 0     --- eq.1

    A linear homogeneous recurrence relation of order k with constant coefficients, together with the k initial conditions

    a0 = C0 , a1 = C1, .., a k-1 = C k-1

    uniquely defines a sequence a0, a1, .

    To solve :

    1. substitute each ak by tk into the recurrence relation :

    an = c1a n-1 + c2 a n-2 + . + ck a n-k

  3. find the roots for the resulting characteristic equation :
  4. tn = c1t n-1 + c2 t n-2 + . + ck t n-k

    (tk - c1t k-1 - c2 t k-2 -. - ck )t n-k = 0       eq. 2

    r1 , r2 , .. , rk are the k distinct non-trivial roots

  5. then the general solution to the recurrence relation eq.1 is of the form

an = b1(r1)n + b2(r 2)n + + bk(rk)n

where bj can be determined by solving the k linear equations derived from the k initial conditions:

a0 = C0 è C0 = b1(r1)0 + b2(r2)0 + + bk(rk)0 = b1 + b2+ + b k

a1 = C1 è C1 = b1(r1)1 + b2(r2) 1 + + bk(rk)1 = b1r1+ b2 r2 + + bkrk

.

a k-1 = C k-1 è Ck-1 = b1(r1)k-1 + b2(r2 )k-1 + + bk(rk)k-1

 

Special cases : when k = 2 , eq.2 reduces to a quadratic equation :

t2 - c1t - c2 = 0

and always can be solved using the quadratic formula :

r = ( c1 ± ( (c1)2 - 4 c2) 1/2 )/2

Then

an = A(r1)n + B(r2)n     if r1 ¹ r2

and  an = A(r)n + Bn(r)n     if r1 = r2 = r