Chapter 6 Advanced Counting

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Chapter 6Advanced Counting

• 6.1 Recurrence Relations

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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
an3n. Is it a solution? ? 2x3x(n-1)-3x(n-2) 3n
YES! (a0 0, a1 3, a2 6,...) Consider the
sequence an5. 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.
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Modelling
Lets 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).
1.042 B(n-2) 1.0450 B(0)
7.1
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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 22H(n-2)21
1 .... 2(n-1)H1 2(n-2) 2(n-3)...21 2n
-1 (by sums of geometric series 3.1).
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Examples
How many bit-strings of length n with no
consecutive 0s? ? an is number of bit-strings
with no 2 consecutive 0s 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) ngt2. a1 2, a2
3. Recognize it? 2,3,5,8,13.....?
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Catalan Numbers
In how many ways Cn can we put parentheses around
n1 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 C01 and C1 1.
(derivation later)
number of extended binary trees with n internal
nodes.
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