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Recurrence Relations (RRs)

Recurrence Relations (RRs). A “ Recurrence Relation ” for a sequence {a n } is an equation that expresses a n in terms of one or more of the previous terms in the sequence (i.e., a 0 ,a 1 ,a 2 ,…,a n-1 ) for all n ≥n 0 . Examples: a 0 =1, a 1 =3, a 2 =4; for n ≥3, a n = a n-1 +a n-2 -a n-3

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Recurrence Relations (RRs)

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  1. 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,… UCI ICS/Math 6A, Summer 2007

  2. 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. UCI ICS/Math 6A, Summer 2007

  3. 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. UCI ICS/Math 6A, Summer 2007

  4. 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. UCI ICS/Math 6A, Summer 2007

  5. 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 . UCI ICS/Math 6A, Summer 2007

  6. 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 UCI ICS/Math 6A, Summer 2007

  7. 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(Covered in Sec.7.4, Ex.41) Cn = Number of binary trees with n+1 leaves. 2 5 14 UCI ICS/Math 6A, Summer 2007

  8. 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. UCI ICS/Math 6A, Summer 2007

  9. 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. • Homogeneouslinear 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.  UCI ICS/Math 6A, Summer 2007

  10. Examples of Recurrence Relations • HLRRwCC = Homogeneous linear recurrence relation with constant coefficients • an= a∙an-1 HLRRwCC of degree 1 • an= (an-1)2Not 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. UCI ICS/Math 6A, Summer 2007

  11. Solving Linear (degree 1) HLRRwCC • The relation is an= c∙an-1 • All solutions are of the form an= d∙cnThe 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 UCI ICS/Math 6A, Summer 2007

  12. 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: UCI ICS/Math 6A, Summer 2007

  13. Solutions to HLRRwCC • We like HLRRwCC because we can solve them explicity.That is, we can find functions f:N→R so that thesequence {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+ckThe solutions (roots) of this equation are called thecharacteristic roots of the recurrence relation UCI ICS/Math 6A, Summer 2007

  14. (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)∙rn (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)∙2n & 1=d1 & 4=2d1+2d2d1=d2=1; f(n)=(1+n)∙2n Check: f(6)=(1+6)∙26=7∙64=448 UCI ICS/Math 6A, Summer 2007

  15. 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 UCI ICS/Math 6A, Summer 2007

  16. 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. UCI ICS/Math 6A, Summer 2007

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