1. Opracowanie językowe dr inż. J. Jarnicki
Internet Engineering
Czesław Smutnicki
Discrete Mathematics – Recursive Equations

2. CONTENTS
Recursive equations
Generating functions

3. HANOI TOWERS A B C

4. HANOI TOWERS cont. lr(n) – the number of moves (for n discs).
n=0, lr(0)=0
n=1
ALG(1)
K1: move disc on C. lr(1)=1
n=2
ALG(2)
K1: small on B;
K2: large on C
K3: small on C; lr(2)=3
n=3
ALG(3)
K1: small on C;
K2: medium on B
K3: small on B
K4: large on C
K5: small on A
K6: medium on C
K7: small on C lr(3)=7
n=4
ALG(4)
KROK 1: 3 upper discs from A on B (alg. ALG(3)) lr(3)
KROK 2: large na C 1 move
KROK 3: 3 upper discs from B on C (alg. ALG(3)) lr(3)
lr(4)=2lr(3)+1

5. HANOI TOWERS cont. Generally
lr(n)=2lr(n-1)+1=2n-1 with initial condition lr(0)=0. n 1 2 3 4 5 6 7 8
lr(n) 15 31 64
127
255

6. FIBONACCI NUMBERS Fibonacci numbers 1 2 3 5 8 13 21 34 55 89 ...
Recursive formula F1 = 1; F2 = 1;
Fn = Fn-1 + Fn-2 , dla n > 2
b) Analytically

7. RABBITS M R RA
pary
okresy

8. CIRCLES Similarly for n=3 we have: 1 + 3 + 3 + 1 = 8 = 23
1 = = 2 = = 4 = 22 2 1 4 3
Similarly
for n=3 we have: = 8 = 23
for n=4 we have: = 16 = 24

9. SOLUTION METHOD. CASE STUDY I
Given recursive sequence:
(1) an = 2 an-1 + 1, n > 1 with initial value a1=1
The sequence starts from: 1, 3, 7, 15, 31,
Find algebraic formula for the n-th element of the sequence.
For n-1 we have:
a n-1 =2 an , which we substitute into equation (1).
Such substitution we continue until we obtain a1, so
an = 2·an = 2·(2·an-2 + 1) + 1 = 22·an =
= 22·(2·an-3 + 1) = 23·an =
= 2kan-k + 2k-1 + 2k =
= 2n-1 + 2n = 2n – 1

10. SOLUTION METHOD. CASE STUDY I. cont
This is the sum of a geometric sequence
with the ratio
q=ai+1/ai =2 and a1=1; so
S= (1-2n)/(1-2) = - (1-2n) = 2n –1.
Therefore an=2n –1. We can check this below. n
an=2an-1 + 1
an = 2n - 1 1 2 3 4 5
2*1 + 1 = 3
2*3 + 3 = 7
2*7 + 1 = 15
2* = 31
22 – 1=3
23 –1 =7
24 – 1=15
25 – 1=31

11. SOLUTION METHOD. CASE STUDY II.
Consider the equation:
sn = asn bsn-2
where values s0 and s1 are known, and a,b - are certain constants
Case a)
either a = 0 or b = 0
If b=0 then sn = asn-1 for n  1.
Therefore s1 = as0 , s2 = as1 = a2 s0, ... and finally sn = ans0.
If a = 0 then sn = bsn-2 for n  2.
Therefore s2=bs0 , s4=bs2 =b2s0 and finally … s2n =bns0 n  N
Similarly s3 = bs1 , s5 = bs3 = b2s1 and finally s2n+1 = bns1 n  N

12. SOLUTION METHOD. CASE STUDY II. cont
Case b)
a ≠ 0 or b ≠ 0
has characteristics equation
x2 - ax - b = 0
This equation follows from the supposition for the given constant c.
Hence dividing by C we obtain
then r is the root of the equation x2 - ax - b = 0

13. Thank you for your attention
DISCRETE MATHEMATICS
Czesław Smutnicki