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Section 6.1

Section 6.1. Recurrence Relations. Recursive definition of a sequence. Specify one or more initial terms Specify rule for obtaining subsequent terms from preceding terms We can use such definitions to solve counting problems that cannot easily be solved using techniques from chapter 4.

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Section 6.1

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  1. Section 6.1 Recurrence Relations

  2. Recursive definition of a sequence • Specify one or more initial terms • Specify rule for obtaining subsequent terms from preceding terms • We can use such definitions to solve counting problems that cannot easily be solved using techniques from chapter 4

  3. Recurrence Relations • When rule for finding subsequent terms from previous is used for solving counting problems, we call the rule a recurrence relation • Stated more formally: A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or more of the previous terms of the sequence a0, a1, … an-1 for all integers n with nn0, where n0 is non-negative

  4. Solutions • A sequence whose terms satisfy a recurrence relation is called a solution of the recurrence relation • Example 1: Let {an} be a sequence that satisfies the recurrence relation an = an-1- an-2 for n = 2, 3, 4 … • Suppose a0=3 and a1=5. What are a2 and a3? • a2 = a1 - a0 = 2, a3 = a2 - a1 = -3

  5. Example 2 • Find the first 5 terms of a sequence defined as follows: • recurrence relation: an = nan-1 + n2an-2 • initial condition: a0 = 1, a1 = 1 • Applying the rules: a2 = 2(1) + (2)21 = 6 a3 = 3(6) + (3)21 = 27 a4 = 4(27) + (4)26 = 204

  6. Example 2 • Determine whether {an} is a solution of the recurrence relation an = 2an-1-an-2 for n=2, 3, 4 … where an = 3n if an = 3n, then for n  2: 2an-1 - an-2 = 2[3(n-1)] - 3(n-2) = 2(3n - 3) - 3n + 6 = 6n - 6 - 3n + 6 = 3n So {an}, where an = 3n, is a solution

  7. Example 3 • Determine whether {an} is a solution of the recurrence relation an = 2an-1-an-2 for n=2, 3, 4 … where an = 2n: • By this rule, a0 = 20 = 1; a1 = 21 = 2; a2 = 22 = 4 • Applying original recurrence relation: an = 2a n-1 - a n-2 a2 = 2a1 - a0 substituting actual values: 4 = 2*2 - 1 4 = 3 not true, so {an} where an = 2n is not a solution

  8. Summary of Recurrence Relations • Initial conditions for a sequence specify terms that precede the first term where recurrence relation takes effect • Recurrence relation & initial conditions uniquely determine a sequence, providing a recursive definition of the sequence • Any term of a sequence can be found from initial conditions using recurrence relation a sufficient number of times (but there are better ways for computing certain classes of sequences)

  9. Example 4: modeling with recurrence relations • Suppose you invest $1,000 in a money market account that yields 7% per year (wow!) with interest compounded annually (oh). How much money will be in the account in 25 years? • Let Pn = amount in account after n years • Since Pn = amount after n-1 years + interest in nth year, {Pn} satisfies the recurrence relation: Pn = Pn-1 + .07(Pn-1) = 1.07(Pn-1)

  10. Example 4 continued • We are given P0 = 1000; so P1 = (1.07)(P0) P2 = (1.07)(P1) = (1.07)(1.07)(P0) = (1.07)2(P0) P3 = (1.07)(P2) = (1.07)(1.07)(1.07)(P0) = (1.07)3(P0) … thus Pn = 1.07(Pn-1) = (1.07)nP(0) so if n=25, P25 = (1.07)25(1000) = $5,427.43

  11. Example 5: is Fibonacci Italian for “bunny”? • Suppose 2 baby rabbits (aw!), a buck and a doe, are placed on an uninhabited island • They don’t breed until they’re 2 months old (we don’t know what they’re up to before that) • After they are 2 months old, each pair of rabbits produces another pair each month • How many rabbits are there after n months, assuming no rabbits die?

  12. Example 5 continued • Let fn = number of rabbit pairs after n months • At end of first month, f1 = 1 (too young to breed) • At end of second month, f2 = 1 (allow for gestation time!)

  13. Example 5 continued • To find the number of pairs after n months, add the number from the previous month (fn-1) to the number of newborn pairs (fn-2, since each newborn pair comes from a pair at least 2 months old) • So {fn} satisfies the recurrence relation: fn = f n-1 + f n-2 for n  3 with f1 = f2 = 1 • So the number of pairs of bunnies after n months is the nth Fibonacci number

  14. Example 6: how long before the world ends? • The Towers of Hanoi is an ancient puzzle, with the following characteritics: • You start with 3 pegs, one of which has a stack of disks on it (bird’s-eye view shown) • The disks are stacked by size, with the largest on the bottom and the smallest on top

  15. Example 6 continued • The goal is to move all disks to a second peg in order of size • Only one disk may be moved at a time • A larger disk can never be placed on top of a smaller disk • Let Hn = number of moves needed to solve a puzzle with n disks; set up a recurrence relation for sequence {Hn}

  16. Example 6 continued • Starting with n disks on peg 1, we can transfer n-1 disks to peg 3 using Hn-1 moves • It takes 1 move to transfer the largest disk to peg 2 • Then we can move the n-1 disks from peg 3 to peg 2 using another Hn-1 additional moves • So Hn = 2(Hn-1) + 1 with initial condition H1 = 1 (one disk can be transferred in one move)

  17. Example 6 continued • We can use an iterative approach to solve the relation: Hn = 2(Hn-1) + 1 = 2 (2Hn-2 + 1) = 22Hn-2 + 2 + 1 = 22(2Hn-3 + 1) + 2 + 1 = 23Hn-3 + 22 + 2 + 1 … = 2n-1H1 + 2n-2 + 2n-3 + … + 2 + 1 = 2n-1 + 2n-2 + 2n-3 + … + 2 + 1 (because H1 = 1) = 2n - 1 • So for 5 disks, it takes 25-1 or 31 moves to solve the puzzle

  18. Example 6 continued • The ancient legend says that the monks of Hanoi must solve a puzzle with 64 disks, taking 1 second to move each disk, and when they are finished the world will end. How long before they’re done? • 264 - 1 = 18,446,744,073,709,551,615 seconds • At 60 seconds per minute, 60 minutes per hour, 24 hours per day and 365.25 days per year, there are 31,557,600 seconds in a year • So the world should end something over 500 billion years since they started (but we don’t know when they started …)

  19. Example 7 • Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have consecutive 0s. How many such strings are there of length 5? • Let an denote the number of bit strings of length n that do not have 2 consecutive 0s; to obtain recurrence relation for {an}, note that: • by the sum rule, the # of bit strings of length n that do not have 2 consecutive 0s = # ending with 0 + # ending with 1 • assuming n  3, bit string has at least 3 bits

  20. Example 7 continued • Bit strings of length n ending with 1 that do not have 2 consecutive 0s are precisely the bit strings of length n-1 with no 2 consecutive 0s and with a 1 added to the end - there are an-1 such bit strings • Bit strings of length n ending with a 0 must have a 1 as their (n-1)th bit (otherwise there’d be 2 0s at the end) - bit strings of length n, ending with 0, without consecutive 0s, are precisely those bit strings of length n-2 with 10 appended to them - there are an-2 such strings

  21. Example 7 continued • So we conclude: an = an-1 + an-2 for n  3 Initial conditions: a1 = 2 since bit strings of length 1 have no consecutive bits and there are 2 possible values a2 = 3 since valid strings are 01, 11, 10 • a5 = a4 + a3; a4 = a3 + a2; a3 = a2 + a1 • a3 = 5; a4 = 8; so a5 = 13

  22. Catalan numbers • The number of ways to parenthesize the product of n+1 numbers to specify the order of multiplication is represented by Cn • For example, C3 = 5; since n=3, there are 4 terms: x0 * x1 * x2 * x3 and 5 ways to parenthesize: (x0 * x1) * (x2 * x3), ((x0 * x1) * x2) * x3, (x0 * (x1 * x2)) * x3, x0 * ((x1 * x2) * x3) and x0 * (x1 * (x2 * x3))

  23. Catalan numbers • However we insert parentheses, one multiplication operator remains outside all parentheses (and is therefore the last operation performed) • The final operator appears between some pair of the n+1 numbers - say xk and xk+1

  24. Catalan numbers • There are CkCn-k-1 ways to insert parentheses to determine order of multiplication of n-1 numbers when the final operator appears between xk and xk+1 because there are Ck ways to insert parentheses in x0*x1*…*xk and Cn-k-1 ways to insert parentheses in xk+1*xk+2*… *xn

  25. Catalan numbers • Since the final operator can appear between any 2 of the n+1 numbers, it follows that: Cn = C0Cn-1 + C1Cn-2 + … + Cn-2 C1 + Cn-1C0 = CkC n-k-1 with initial conditions C0=1, C1=1 • It can be shown that Cn = C(2n,n)/(n+1)

  26. Section 5.1 Recurrence Relations

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