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Chapter 8
Recursion
8.1
Recursively Defined Sequences
Recursion
• A Sequence can be defined as:
– informally be providing a few terms to
demonstrate the pattern, i.e. 3, 5, 7, ….
– give an explicit
formula for it nth term, i.e.
n
an 

1
n 1
– or, by recursion which requires a recurrence
relation.
Defining Recursion
• Recursion requires
• recurrence relation relates later terms in the sequence to
earlier terms and
• initial conditions, values of the first few terms of the
sequence.
– Example b0, b1, b2, … For all integers k>= 2,
1. bk = bk-1 + bk-2 (recurrence relation)
2. b0 = 1, b1 = 3 (initial conditions)
b2 = b2-1 + b2-2 = b1 + b0 = 3 + 1 = 4
b3 = b3-1 + b3-2 = b2 + b1 = 4 + 3 = 7
b5 = b5-1 + b5-2 = b4 + b3 = 11 + 7 = 18
Recursion
• Definition:
– A recurrence relation for a sequence a0, a1, a2,… is
a formula that relates each term ak to ceratin of its
predecessors ak-1, ak-2, … , ak-I, where I is an integer
and k is any integer greater than or equal to i.
– The initial conditions for such a recurrence
relation specify the values of a0, a1, a2,…,ai-1, if I is
a fixed integer, or a0, a1, a2,…,am, where m is an
integer with m>=0, if i depends on k.
Example
• Computing terms
– Recursive sequence ck, for all integers k >=2, find
c2, c3, c4
1. ck = ck-1 + k ck-2 + 1 (recurrence relation)
2. c0 = 1 and c1 = 2 (initial conditions)
c2 = c2-1 + 2*c2-2 + 1 = c1 + 2*c0 + 1 = 2 + 2*1 + 1 = 5
c3 = c3-1 + 3*c3-2 + 1 = c2 + 3*c1 + 1 = 5 + 3*2 + 1 = 12
c4 = c4-1 + 4*c4-2 + 1 = c3 + 4*c2 + 1 = 10 + 4*12 + 1 = 59
Equivalent Recursion
• There is more than one way to setup a recursive
sequence!
– for all k>=1, sk = 3sk-1 – 1
– for all k>=0, sk+1 = 3sk – 1
• Are the two sequences equivalent?
– show the results of the sequence for terms:
• starting at k = 1: s1 = 3s0 – 1, s2 = 3s1 – 1, s3 = 3s2 – 1
• starting at k = 0: s0+1 = 3s0 – 1, s1+1 = 3s1 – 1, …
– change first term to second by adjusting first k
• for all k>=1, sk = 3sk-1 – 1 => k start at 0 (k≥0) sk+1 =3sk – 1
• same as the second form.
Example
• Show that sequence given satisfies a recurrence
relation:
– 1, -1!, 2!, -3!, 4!, …, (-1)nn!, …, for n≥0 is equivalent to sk =
(-k)sk-1 for k≥1
– General term of the seq sn starting with s0 = 1, then sn = (1)nn! for n≥0
• substitute for k and k-1: sk = (-1)kk! , sk-1 = (-1)k-1(k-1)!
(-k)sk-1 = (-k)[(-1)k-1(k-1)!] (sk defined above)
= (k)(-1)(-1)k-1(k-1)!
= (k)(-1)k(k-1)!
= (-1)k(k)(k-1)!
= (-1)k k!
= sk
Solving Recursive Problems
• To solve a problem recursively means to find a
way to break it down into smaller
subproblems each having the same form as
the original problem.
Towers of Hanoi
• Problem statement: eight
disks with holes in the
center that are stacked
from largest diameter to
smallest on the first of
three poles. Move the
stacked disk from one pole
to another.
• Rules: a larger disk cannot
be placed on top of a
smaller disk at any time.
Towers of Hanoi
• Recursive Solution
1.
2.
3.
Transfer the top k-1 disks
from pole A to pole B.
(Note: k>2 requires a
number of moves.)
Move the bottom disk
from pole A to pole C.
Transfer the top k-1 disks
from pole B to pole C.
(Again, if k>2, execution
of this step will require
more than one move.)
Towers of Hanoi
min
moves to
transfer a
tower of k
disks from
A to C
=
min num
of moves
to go from
position a
to
position b
+
min num
of moves
to go from
position b
to
position c
+
min num
of moves
to go from
position c
to
position d
For each integer n≥1, mn = min num of moves to transfer a tower of n disks from one pole
to another.
position a to position b = mk-1, position b to c = 1 move, position c to position d = mk-1 moves.
mk = mk-1 + 1 + mk-1 = 2mk-1 + 1, integers k≥2
initial condition: m1 = [move one disk to another pole] = 1
Towers of Hanoi
1. mk = 2mk-1 + 1 (recurrence relation)
2. m1 = 1 (initial condition)
m2 = 2m1 + 1 = 3
m3 = 2m2 + 1 = 7
m4 = 2m3 + 1 = 15
…
m6 = 2m5 + 1 = 63
Fibonacci
• Leonardo of Pisa was the greatest mathematician
of the 13th century.
• Proposed the following problem:
– A single pair of rabbits (male/female) is born at the
beginning of a year. Assuming the following
conditions:
1. Rabbit pairs are not fertile during their first month of life
but thereafter give birth to one new male/female pair at
the end of every month.
2. No rabbits die
– How many rabbits will there be at the end of the
year?
Fibonacci
• To solve the problem you can hand compute
for each of the 12 months.
– m1 = 1, m2 = 1, m3 = 2, m4 = 3, m5 = 5, … m11 =
144, m12 = 233
num of
pairs alive
at end of
month k
=
num of
pairs alive
at end of
month k-1
=
num of
pairs alive
at month
k-1
+
num of
pairs born
at end of
month k
+
num of
pairs alive
at month
k-2
Fibonacci
–
–
–
–
for n ≥1, Fn= num of pairs alive at month n
F0 = the initial number of pairs
F0 = 1
Fk = Fk-1 + Fk-2
• Fibonacci Sequence
1. Fk = Fk-1 + Fk-2 (recurrence relation)
2. F0 = 1, F1 = 1 (initial condition)
F2 = F1 + F0 = 1+1 = 2
F3 = F2 + F1 = 2+1 = 3
F4 = F3 + F2 = 3+2 = 5
…
F12 = F11 + F10 = 144 + 89 = 233
Compound Interest
• Compute compound interest on principle
value using recursive sequence.
• $100,000 principle, earning 4% per year, how
much would you have in 21 years.
amt in
account end
of a year
=
amt in account
at end of
previous year
=
amt in account
at end of
previous year
+
interest earned
on account
during year
+
(0.04)* amt in
account previous
year
An = amt in account at end of year n
A0 = initial amount (principle)
Ak = Ak-1 + (0.04)*Ak-1
Ak = (1.04)*Ak-1
Compound Interests
• Compounding multiple times a year
– interest is broken up of over the year based on the
number of compounding periods
– example 3% compounded quarterly
• 3%/4 = .03/4 = 0.0075 (interest rate per period)
• interest earned during kth period = Pk-1(i/m)
• Pk = Pk-1 + Pk-1 (i/m) = Pk-1(1 + i/m)
Example
• Given $10,000, How much will the account be
work at the end of one year with a interest
rate of 3% compounded quarterly?
– Pk = Pk-1 (1+0.0075) = Pk-1(1.0075), integers k≥1
– P0 = 10,000
– P1 = 1.0075*P0=1.0075*10,000 =10,075.00
– P2 = 10,150.56
– P3 = 10,226.69
– P4 = 10,303.39

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