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Mastering Recurrence Relations: Sequences, Solutions, and Applications

Dive deep into the world of recurrence relations, understanding how to express sequence elements based on previous terms. Learn to solve and identify solutions through examples and models like Tower of Hanoi and financial growth. Explore bit-strings, Catalan numbers, and binary trees, mastering advanced counting techniques.

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Mastering Recurrence Relations: Sequences, Solutions, and Applications

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  1. Chapter 6Advanced Counting 6.1 Recurrence Relations

  2. Recurrence Relations Definition: An equation that expresses the element an of the sequence {an} in terms of one or more previous terms of the sequence a0,...,an-1. A sequence is a solution of the recurrence relation if its terms satisfy the recurrence relation. Example: Consider the recurrence relation an = 2a(n-1) –a(n-2), n in N Consider the sequence: an=3n. Is it a solution?  2x3x(n-1)-3x(n-2) =3n YES! (a0 = 0, a1 = 3, a2 = 6,...) Consider the sequence an=5. Is it a solution?  2x5 – 5 = 5 YES! (a0 = 5, a1 = 5, ...) we see that to sequences can be a solution of the same recurrence relation since the initial conditions are also very important. The initial conditions plus the recurrence relation provide a unique recursive definition of a sequence.

  3. Modelling Let’s say you put $1 dollar in the bank now. How much will you receive in 50 years if the interest is 4%? B(0) = 1. B(n) = B(n-1) + 0.04xB(n-1) (add the interest). = 1.04 x B(n-1). = 1.04^2 B(n-2) = 1.04^50 B(0) = $ 7.1

  4. Tower of Hanoi How many moves does it take to solve the Hanoi puzzle? Let Hn be the number of moves to solve a problem with n disks. -We first move the top n-1 disks to peg 2 in H(n-1) moves. -We then move the largest disk to peg 3 in 1 move. -We then move the n-1 disks on top of the largest disk in another H(n-1) moves:  Total: Hn = 2xH(n-1)+1 -Basis Step: 1 disk = 1 move: H(1)=1. -Hn = 2(2H(n-2)+1)+1 = 2^2H(n-2)+2^1 + 1 .... =2^(n-1)H1 + 2^(n-2) + 2^(n-3)+...+2+1 =2^n-1 (by sums of geometric series 3.1).

  5. Examples How many bit-strings of length n with no consecutive 0’s?  an is number of bit-strings with no 2 consecutive 0’s in it. How can we generate a bit-string of length n from bit-strings of length n-1? Valid bit-strings fall into 2 classes, ones that end with a “0” and ones that end with a “1”. Ones that end in a “1” could be generated by adding a one to a valid bit-string of any kind: a(n-1) ways. Ones that end with a “0” came about by appending a smaller bit-string of length n-1 that ended with a “1” (otherwise 2 zeros appear), which must have come about in turn by appending a “1” to any valid bit-string of length n-2: a(n-2) ways. Total: an = a(n-1) + a(n-2) n>2. a1 = 2, a2 = 3. Recognize it? 2,3,5,8,13.....?

  6. Catalan Numbers In how many ways Cn can we put parentheses around n+1 symbols to indicate in which order they should be processed?  Each string x0 x1 x2 ... xn is divided into 2 sub-strings: e.g. ((x0 x1) x2) (x3 x4) Assume break occurs at after position k, then there are Ck x Cn-k-1 ways to put parenthesis around the first & second sub-string. Thus the total is: Cn = C0 Cn-1 + C1Cn-2+ ... +Cn-1C0 with C0=1 and C1 = 1. (derivation later) number of extended binary trees with n internal nodes. 2 5 14

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