9-AdvancedCounting -1

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 nn0.
• Examples
• a01, a13, a24for n 3, an an-1an-2-an-3
• 1,3,4,6,7,9,10,12,13,15,16,18,19,21,22,
• a02, a14, a23for n 3, an an-1an-2-an-3
• 2,4,3,5,4,6,5,7,6,8,7,9,8,10,9,11,
• a04, a13, a23for n 3, an an-1an-2-an-3
• 4,3,3,2,2,1,1,0,0,-1,-1,-2,-2,-3,-3,-4,-4,

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-1an-2-an-3 Initial conditions
  a01, a13, a24.Solution 1,3,4,6,7,9,10,12,13,1
  5,16,18,19,21,22,
• For n 3, an an-1an-2-an-3 Initial conditions
  a02, a14, a23. Solution 2,4,3,5,4,6,5,7,6,8,
  7,9,8,10,9,11,
• For n 3, an an-1an-2-an-3 Initial conditions
  a04, a13, a23. 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 fN?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 an2an-1.

3
Solving Recurrence Relations

• If an4an-1 and a03, find a function f such that
  f(n)an.
• an4an-142an-243an-3...4na03?4n.
• If anan-1n and a04, find a function f such
  that f(n)an.
• anan-1nan-2n(n-1) an-2n(n-1)(n-2)...a0
  n(n-1)(n-2)...14n(n1)/2(n2n8)/2.
• If an2nan-1 and a05, find a function f such
  that f(n)an.
• an2nan-122n(n-1)an-223n(n-1)an-3...2nn!a05?
  2nn!.
• If an2an-11 and a00, find a function f such
  that f(n)an.
• an2an-1122an-22123an-342124an-48421
  ...2n-12n-22n-3...84212n-1.

4
Modeling with Recurrence Relations

• An initial deposit of P0 dollars deposited at 7
  annual interest.
• Pn(10.07)Pn-1 is the value after n years.
  Pn(10.07)nP0
• Rabbits on an island. Each pair producing a new
  pair every month.
• Fibonacci Numbers f00, f11 for n 2, fn
  fn-1fn-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.
• H11 for n 2, Hn Hn-11Hn-1 2Hn-11.
  1,3,7,15,31,63,127,255,511,1023,2047,4095,8191,163
  83,32767,65535,
• Hn 2n-1.

5
RRs for Counting Bit Strings

• How many bit strings of length n do not contain 2
  consecutive 0s? b12 (0,1), b23
  (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 0s? b0b10
• For n 2 00any or 01counted or
  1counted bn 2n-2 bn-2
  bn-1 .

6
Counting Code Words

• How many strings of n digits have an even number
  of 0s?
• a19
• For n 2
• Either the first digit isnt 0 and the rest has
  an even number of 0s (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-110n-1

7
Catalan Numbers

• Cn number of ways to parenthesize the product
  of n1 numbers. C11 (1x2)
  C22 (1x(2x3)), ((1x2)x3)
• C35 (1x(2x(3x4))), (1x((2x3)x4)),
  ((1x2)x(3x4)), ((1x(2x3))x4), (((1x2)x3)x4)
• C0C11
• CnC0Cn-1C1Cn-2C2Cn-3Cn-3C2Cn-2C1Cn-1C
  0
  (Covered in Sec.7.4, Ex.41)

Cn Number of binary trees with n1 leaves.
2
5
14
8
Solving Recurrence Relations

• Weve seen several examples of Recurrence
  Relations
• an aan-1 bn nbn-1
  fn fn-2 fn-1
• bn2n-2bn-2bn-1 bn2n-3bn-3bn-2bn-1
  .
• bn bn-1bn-2-bn-3 Hn2Hn-11
• 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 doesnt depend on earlier ones.
• Example andan instead of an aan-1
• Hn2n-1 instead of Hn2Hn-11
• Well look at a class of useful recurrence
  relations where deriving a closed form formula is
  easy.

9
Recurrence Relations

• When c1, c2, , ck are constants and ck?0, an
  c1an-1c2an-2ckan-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 coefficientsan
  c1an-1c2an-2ckan-k where ck?0
• We like these because we can solve them
  explicitly. ?

10
Examples of Recurrence Relations

• HLRRwCC Homogeneous linear recurrence relation
  with constant coefficients
• an aan-1 HLRRwCC of
  degree 1
  
• an (an-1)2 Not Linear
  
• an nan-1 Coefficients
  are not constant
• fn fn-2 fn-1 HLRRwCC
  of degree 2
• bn2n-2bn-2bn-1 . Not Homogeneous
• bn2n-3bn-3bn-2bn-1 Not Homogeneous
• an an-1an-2-an-4 HLRRwCC of degree 4
• Hn2Hn-11 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.

11
Solving Linear (degree 1) HLRRwCC

• The relation is an can-1
• All solutions are of the form an dcn
  The function is f (n) dcn
• 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(10.07)Pn-1 is the value
  after n years. Pn(10.07)nP0

12
Fibonacci Solved

• Degree 1 f(n)cf(n-1) solution f(n)arn
  for some a and r
• Degree 2 f(n)c1f(n-1)c2f(n-2) solution
  ?
• Consider Fibonacci numbers F(n)F(n-1)F(n-2),
  and lets try F(n)rn.
• It follows that r2r1.
• There are 2 solutions to this quadratic equation
  r1(1v5)/2, r2(1-v5)/2
• Therefore F1(n)(1v5)/2n and
  F2(n)(1-v5)/2n are both solutions.
• Observe If F1,F2 are solutions then so is F(n)
  d1F1(n) d2F2(n)
• Theorem (next slide) Every solution to this
  recurrence relation is of the form F(n)
  d1F1(n) d2F2(n) for some d1,d2
• For our initial conditions
• F(0)0 and F(1)1 we get d1d20 and
  d1r1d2r21
• Therefore 1d1(r1-r2 )d1(v5). Therefore d1
  1/v5 and d2 -1/v5
• Therefore the closed formula for Fibonacci
  numbers is

13
Solutions to HLRRwCC

• We like HLRRwCC because we can solve them
  explicity.That is, we can find functions fN?R
  so that thesequence f(n) solves the recurrence
  relation.Heres part of the reason why.
• Lemma If functions f and g are solutions to the
  HLRRwCC an c1an-1c2an-2c3an-3ckan-k
  then fg is also a solution as is df for any
  constant d.
• Definitions The characteristic equation of this
  HLRRwCC is rk c1rk-1c2rk-2c3rk-3ck-1r
  ck The solutions (roots) of this equation are
  called the characteristic roots of the
  recurrence relation

14
(Fibonacci) Degree 2 HLRRwCC

• 1) Write out the characteristic equation. (For
  Fibonacci r2r1)
• 2) Find the characteristic roots (r1 and r2).
  (r1(1v5)/2, r2(1-v5)/2)
• 3) Any function of the form f(n)d1r1 n d2r2
  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
  (r1r2r1), then any function of the form
  f(n)(d1d2n)r n (d1 and d2 constants) solves
  the recurrence relation.
• Example
• a01, a14 for n 2, an 4an-1-4an-2
  1,4,12,32,80,192,448,
• Characteristic Equation r24r-4 ? r2-4r40
  ? r2 (twice)
• f(n)(d1d2n)2 n 1d1 42d12d2 ?
  d1d21
• f(n)(1n)2n
• Check f(6)(16)26764448

15
All Solutions For Any HLRRwCC

• Thrm If rk c1rk-1c2rk-2ck-1rck (ck?0) has
  k distinct roots, r1,r2,,rk,
• then every solution of the recursion
  relation
• anc1an-1c2an-2ckan-k has the
  form
• an d1r1nd2r2ndkrkn 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

16
Degree 3 HLRRwCC Examples

• a02, a15, a215 for n 3, an 6an-1-11an-2-
  6an-3
• r3 6r2-11r-6 ? r3-6r211r60 ?
  (r-1)(r-2)(r-3)0
• General solution an x1ny2nz3n .
• Initial values 2xyz 5x2y3z 15x4y9z
  ? x1 y-1 z2
• Specific solution an 1-2n23n .
• a01, a1-2, a2-1 for n 3, an
  -3an-1-3an-2-an-3
• r3 -3r2-3r-1 ? r33r23r10 ? (r1)30
  ? r-1 with multiplicity 3
• General solution an(xynzn2)(-1)n .
• Initial values 1x 2xyz -1x2y4z ? x1
  y3 z-2
• Specific solution an (13n-2n2) (-1)n.