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Recurrence Relations (RRs)
A “Recurrence Relation” for a sequence {an} is an equation
that expresses an in terms of one or more of the previous
terms in the sequence (i.e., a0,a1,a2,…,an-1) for all n≥n0.
Examples:
a0=1, a1=3, a2=4;
for n ≥3, an= an-1+an-2-an-3
1,3,4,6,7,9,10,12,13,15,16,18,19,21,22,…
a0=2, a1=4, a2=3;
for n ≥3, an= an-1+an-2-an-3
2,4,3,5,4,6,5,7,6,8,7,9,8,10,9,11,…
a0=4, a1=3, a2=3;
for n ≥3, an= an-1+an-2-an-3
4,3,3,2,2,1,1,0,0,-1,-1,-2,-2,-3,-3,-4,-4,…
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9-AdvancedCounting -1
Solutions to Recurrence Relations
A sequence {an} is called a “solution” of the recurrence
relation if its terms satisfy the recurrence relation.
Examples:
For n ≥3, an= an-1+an-2-an-3; Initial conditions: a0=1, a1=3, a2=4.
Solution: 1,3,4,6,7,9,10,12,13,15,16,18,19,21,22,…
For n ≥3, an= an-1+an-2-an-3; Initial conditions: a0=2, a1=4, a2=3.
Solution: 2,4,3,5,4,6,5,7,6,8,7,9,8,10,9,11,…
For n ≥3, an= an-1+an-2-an-3; Initial conditions: a0=4, a1=3, a2=3.
Solution: 4,3,3,2,2,1,1,0,0,-1,-1,-2,-2,-3,-3,-4,-4,…
Every recurrence relationship has many solutions
each determined uniquely by its own initial conditions.
We say a function f:N→R is a “solution” to a recurrence
relation if the sequence {f(n)} is a solution to it.
Example: f(n)=5 ∙2n is a solution to the recurrence relation an=2∙an-1.
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9-AdvancedCounting -2
Solving Recurrence Relations
If an=4an-1 and a0=3, find a function f such that f(n)=an.
an=4an-1=42an-2=43an-3=...=4na0=34n.
If an=an-1+n and a0=4, find a function f such that f(n)=an.
an=an-1+n=an-2+n+(n-1) =an-2+n+(n-1)+(n-2)=...
=a0+n+(n-1)+(n-2)+...+1=4+n(n+1)/2=(n2+n+8)/2.
If an=2nan-1 and a0=5, find a function f such that f(n)=an.
an=2nan-1=22n(n-1)an-2=23n(n-1)an-3=...=2nn!a0=5 2nn!.
If an=2an-1+1 and a0=0, find a function f such that f(n)=an.
an=2an-1+1=22an-2+2+1=23an-3+4+2+1=24an-4+8+4+2+1=...
=2n-1+2n-2+2n-3+...+8+4+2+1=2n-1.
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9-AdvancedCounting -3
Modeling with Recurrence Relations
An initial deposit of P0 dollars deposited at 7% annual interest.
Pn=(1+0.07)Pn-1 is the value after n years. Pn=(1+0.07)nP0
Rabbits on an island. Each pair producing a new pair every month.
Fibonacci Numbers: f0=0, f1=1; for n ≥2, fn= fn-1+fn-2.
Tower of Hanoi
Disks of decreasing diameter on 1 of 3 pegs.
Move disks to another peg, always maintaining
decreasing disk diameters on each peg.
Hn = number of moves to transfer n disks from 1 peg to another.
H1=1; for n ≥2, Hn= Hn-1+1+Hn-1 = 2Hn-1+1.
1,3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,32767,65535,
Hn= 2n-1.
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9-AdvancedCounting -4
RRs for Counting Bit Strings
How many bit strings of length n do not contain 2
consecutive 0’s?
b1=2 ({0,1}), b2=3 ({01,10,11})
For n ≥2: 01counted or 1counted
bn = bn-2
+
bn-1 .
Recognize this?
How many bit strings of length n contain 2 consecutive 0’s?
b0=b1=0
For n ≥2: 00any or 01counted or 1counted
bn = 2n-2
+ bn-2
+ bn-1 .
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9-AdvancedCounting -5
Counting Code Words
How many strings of n digits have an even number of 0’s?
a1=9
For n ≥2:
Either the first digit isn’t 0 and the rest has an even
number of 0’s (there are 9an-1 of these)
or the first digit is 0 and the rest does not have an even
number of digits (there are 10n-1-an-1 of these)
an = 9an-1+(10n-1-an-1)=8an-1+10n-1
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9-AdvancedCounting -6
Catalan Numbers
Cn = number of ways to parenthesize the product of n+1 numbers.
C1=1: (1x2);
C2=2: (1x(2x3)), ((1x2)x3)
C3=5: (1x(2x(3x4))), (1x((2x3)x4)), ((1x2)x(3x4)), ((1x(2x3))x4), (((1x2)x3)x4)
C0=C1=1
Cn=C0Cn-1+C1Cn-2+C2Cn-3+…+Cn-3C2+Cn-2C1+Cn-1C0
Cn  k 0 Ck Cnk 1  C (2n, n) /( n  1) (Covered in Sec.7.4, Ex.41)
n 1
Cn = Number of binary
trees with n+1 leaves.
2
5
14
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9-AdvancedCounting -7
Solving Recurrence Relations
We’ve seen several examples of Recurrence Relations
an= a∙an-1
bn= n∙bn-1
fn =fn-2 +fn-1
bn=2n-2+bn-2+bn-1
bn=2n-3+bn-3+bn-2+bn-1 .
bn= bn-1+bn-2-bn-3
Hn=2Hn-1+1
In each case, many sequences satisfy the relationship and one also
needs to set/have initial conditions get a unique solution.
Goal: “closed form” expression for function, f, for the sequence, giving
the nth term in a formula that doesn’t depend on earlier ones.
Example: an=d*an instead of an= a∙an-1
Hn=2n-1 instead of Hn=2Hn-1+1
We’ll look at a class of useful recurrence relations where deriving a
closed form formula is easy.
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9-AdvancedCounting -8
Recurrence Relations
When c1, c2, …, ck are constants and ck≠0,
an= c1an-1+c2an-2+…+ckan-k +F(n)
is said to be a linear recurrence relation of degree k with
constant coefficients .
Linear = each aj appears only to the 1st power, no aj2 , aj3, . . .
degree k = k previous terms, ck≠0.
constant coefficients = cj are constants, not functions cj(n).
Additionally, if F(n)=0, the relation is called homogeneous.
Homogeneous linear recurrence relation of degree k with
constant coefficients:
an= c1an-1+c2an-2+…+ckan-k where ck≠0
We like these because we can solve them explicitly. 
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9-AdvancedCounting -9
Examples of Recurrence Relations
HLRRwCC =
Homogeneous linear recurrence relation with constant coefficients
an= a∙an-1
HLRRwCC of degree 1
an= (an-1)2
Not Linear
an= n∙an-1
Coefficients are not constant
fn =fn-2 +fn-1
HLRRwCC of degree 2
bn=2n-2+bn-2+bn-1 .
Not Homogeneous
bn=2n-3+bn-3+bn-2+bn-1 Not Homogeneous
an= an-1+an-2-an-4
HLRRwCC of degree 4
Hn=2Hn-1+1
Not Homogeneous
Remember:
A recurrence relation has many solutions.
Only when the initial conditions are specified is the solution unique.
For degree k, you need k contiguous initial conditions.
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9-AdvancedCounting -10
Solving Linear (degree 1) HLRRwCC
The relation is
an= c∙an-1
All solutions are of the form
an= d∙cn
The function is f (n)= d∙cn
With the initial condition a0 specified, the unique solution is
an= a0cn
We saw this in computing compound interest
An initial deposit of P0 dollars deposited at 7% annual interest.
Pn=(1+0.07)Pn-1 is the value after n years.
Pn=(1+0.07)nP0
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9-AdvancedCounting -11
Fibonacci Solved
Degree 1: f(n)=c*f(n-1); solution: f(n)=a*rn for some a and r
Degree 2: f(n)=c1*f(n-1)+c2*f(n-2); solution: ?
Consider Fibonacci numbers: F(n)=F(n-1)+F(n-2), and let’s try F(n)=rn.
It follows that r2=r+1.
There are 2 solutions to this quadratic equation: r1=(1+√5)/2, r2=(1-√5)/2
Therefore F1(n)=[(1+√5)/2]n and F2(n)=[(1-√5)/2]n are both solutions.
Observe: If F1,F2 are solutions then so is F(n) = d1*F1(n) + d2*F2(n)
Theorem (next slide): Every solution to this recurrence relation is of the form
F(n) = d1*F1(n) + d2*F2(n) for some d1,d2
For our initial conditions:
F(0)=0 and F(1)=1 we get d1+d2=0 and d1∙r1+d2∙r2=1
Therefore 1=d1∙(r1-r2 )=d1∙(√5). Therefore d1= 1/√5 and d2= -1/√5
Therefore the closed formula for Fibonacci numbers is:
((1  5 ) / 2) n  ((1  5 ) / 2) n
fn 
5
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9-AdvancedCounting -12
Solutions to HLRRwCC
We like HLRRwCC because we can solve them explicity.
That is, we can find functions f:N→R so that the
sequence {f(n)} solves the recurrence relation.
Here’s part of the reason why.
Lemma: If functions f and g are solutions to the HLRRwCC
an= c1an-1+c2an-2+c3an-3+…+ckan-k
then f+g is also a solution as is d∙f for any constant d.
Definitions:
The characteristic equation of this HLRRwCC is
rk= c1rk-1+c2rk-2+c3rk-3+…+ck-1r+ck
The solutions (roots) of this equation are called the
characteristic roots of the recurrence relation
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9-AdvancedCounting -13
(Fibonacci) Degree 2 HLRRwCC
1) Write out the characteristic equation. (For Fibonacci: r2=r+1)
2) Find the characteristic roots (r1 and r2). (r1=(1+√5)/2, r2=(1-√5)/2)
3) Any function of the form f(n)=d1∙r1 n +d2∙r2 n
(d1 and d2 constants) solves this recurrence relation.
4a) If the characteristic roots are distinct, we can pick d1 and d2 to
produce the required initial values.
4b) If there is only 1 characteristic root (r1=r2=r1),
then any function of the form f(n)=(d1+d2∙n)∙r n
(d1 and d2 constants) solves the recurrence relation.
Example:
a0=1, a1=4; for n ≥2, an= 4an-1-4an-2
1,4,12,32,80,192,448,
Characteristic Equation: r2=4r-4  r2-4r+4=0  r=2 (twice)
f(n)=(d1+d2∙n)∙2 n & 1=d1 & 4=2d1+2d2  d1=d2=1;
f(n)=(1+n)∙2n
Check: f(6)=(1+6)∙26=7∙64=448
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9-AdvancedCounting -14
All Solutions For Any HLRRwCC
Thrm: If rk= c1rk-1+c2rk-2+…+ck-1r+ck (ck≠0) has k distinct roots, r1,r2,…,rk,
then every solution of the recursion relation
an=c1an-1+c2an-2+…+ckan-k has the form
an= d1r1n+d2r2n+…+dkrkn for some d1, d2, …, dk
and every such sequence solves/satisfies the recursion relation.
If there are t distinct roots, each with multiplicity mi,
the sequences {an} solving the recursion relation are given by
an  (d1,0  d1,1n  ...  d1,m1 1n m11 )r1n
 (d 2,0  d 2,1n  ...  d 2, m2 1n m 21 )r2 n
 .....
 (dt ,0  dt ,1n  ...  dt ,mt 1n mt 1 )rt n
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9-AdvancedCounting -15
Degree 3 HLRRwCC Examples
a0=2, a1=5, a2=15; for n ≥3, an= 6∙an-1-11∙an-2- 6∙an-3
r3= 6∙r2-11∙r-6  r3-6∙r2+11∙r+6=0  (r-1)(r-2)(r-3)=0
General solution: an= x∙1n+y∙2n+z∙3n .
Initial values: 2=x+y+z; 5=x+2y+3z; 15=x+4y+9z  x=1; y=-1; z=2
Specific solution: an= 1-2n+2∙3n .
a0=1, a1=-2, a2=-1; for n ≥3, an= -3∙an-1-3∙an-2-an-3
r3= -3∙r2-3∙r-1  r3+3∙r2+3∙r+1=0  (r+1)3=0  r=-1 with multiplicity 3
General solution: an=(x+y∙n+z∙n2)∙(-1)n .
Initial values: 1=x; 2=x+y+z; -1=x+2y+4z  x=1; y=3; z=-2
Specific solution: an= (1+3n-2n2) (-1)n.
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9-AdvancedCounting -16