Section 3.3

Report
Section 3.4
Recursive Definitions
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Recursion
• Recursion is the process of defining an
object in terms of itself
• Technique can be used to define sequences,
functions and sets
• To recursively define a sequence:
– give first term
– give rule for defining subsequent terms from
previous terms
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Recursively Defined Functions
• Specify f(0)
• Give a rule for finding f from f’s value at
smaller integers
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Example 1
• Define f recursively by:
f(0) = 3
f(n+1) = 2(f(n)) + 3
• Find f(1), f(2), f(3) and f(4)
f(1) = 2(f(0)) + 3 = 9
f(2) = 2(f(1)) + 3 = 21
f(3) = 2(f(2)) + 3 = 45
f(4) = 2(f(3)) + 3 = 93
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Example 2: give a recursive
definition of the factorial function
• Define f(0): f(0) = 1
• Define rule for f(n+1) from f(n):
– Since (n+1)! = n!(n+1)
– then f(n+1) = (n+1)(f(n))
• For example, to find f(5) = 5! Do:
f(5) = 5(f(4)) = 5(4(f(3))) = 5(4(3(f(2)))) =
5(4(3(2(f(1))))) = 5(4(3(2(1(f(0))))))) =
5*4*3*2*1*1 = 120
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Example 3
• Give recursive definition for the sum of the
first n positive integers
• Define f(0) = 0
• Then f(n+1) = n+1+f(n)
• Thus f(4), for example, is:
4+f(3) = 4+3+f(2) = 4+3+2+f(1) = 4+3+2+1+f(0)
= 4+3+2+1+0 = 10
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Example 4
• Give recursive definition of an where a is a
non-zero real number and n is a nonnegative integer
• Define a0 = 1
• Then define an+1 = a(an)
• So x5 = x(x4) = x(x(x3)) = x(x(x(x2))) =
x(x(x(x(x1)))) = x(x(x(x(x(x0) =
x*x*x*x*x*1
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Example 5: summation
• Give a recursive definition of:
n
 ak
k=0
• Basis:
n
 ak = a0
k=0
• Inductive:
n+1
n
k=0
k=0
 ak = an+1 +  ak
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Example 6
• Give recursive definition of Pm(n), product
of integer m and non-negative integer n
• Basis: Pm(0) = 0
• Inductive: Pm(n+1) = Pm(n) + m
• So, for example, 4*3 = P4(3) = P4(2)+4 =
P4(1)+4+4 = P4(0)+4+4+4 = 0+4+4+4 = 12
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Fibonacci numbers
• Fibonacci numbers are defined by:
f0 = 0, f1 = 1, fn = fn-1 + fn-2 where n=2,3,4 …
So f2 = f1 + f0 = 1
f3 = f2 + f1 = 2
f4 = f3 + f2 = 3
f5 = f4 + f3 = 5
f6 = f5 + f4 = 8
etc.
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Example 7: Fibonacci numbers
• The recursive definition of Fibonacci
numbers can be used to prove many
properties of these numbers. For example:
fn > n-2 where  = (1 + 5) / 2 whenever n  3
Let P(n) = fn > n-2 ; we wish to show that P(n) is
true for n  3
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Example 7
• Note that:
 < 2, 2 = f3 (f3 > 3-2)
2 = ((1+5)/2)((1+5)/2) = (1+25+5)/4 =
(2(5+3))/(2*2) = (3 + 5) / 2
(3 + 5) / 2 < 3, 3 = f4 (f4 > 4-2)
• So P(n) is true for n=3 and n=4
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Example 7
• Assume P(k) is true, so fk > k-2 for all
integers k with 3 <= k <= n, with n  4
• Show P(n+1) is true: fn+1 > n-1
• By the quadratic formula: x=(-bb2-4ac)/2a
 is a solution for x2-x-1=0
so 2 =  + 1
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Example 7
• Thus, n-1 = 2(n-3) = (+1) n-3 =
(n-3) +1(n-3) = (n-2 + n-3)
• By the inductive hypothesis, if n5, it
follows that fn-1 > n-3, fn > n-2
• So we have:
fn+1 = (fn + fn-1) > (n-2 + n-3) = n-1, completing
the proof
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Example 8
• Prove that f1+f3+ … +f2n-1 = f2n whenever n
is a positive integer
• Basis: f1 = f2*1 true, since f1 = f2 = 1
• We want to prove:
f1+f3+ … f2n-1+f2n+1=f2(n+1)
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Example 8
• Assume P(n) is true; then:
f1+f3+ … f2n-1+f2n+1 = f2n + f2(n+1)
f2n + f2(n+1) = f2n+2
• Recall the definition of Fibonacci numbers:
fn = fn-1 + fn-2
• So the theorem is true
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Recursively defined sets
• Sets may be recursively defined as follows:
– an initial collection of elements is given
– rules used to construct elements of the set from
other elements are given
• Such sets are well-defined, and theorems
about them can be proved using their
recursive definitions
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Example 9
• S is recursively defined by:
3S
(x+y)  S if x  S and y  S
• Show that S is the set of all positive integers
divisible by 3
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Example 9: Proof
• Let A = {x | x is divisible by 3} (or, x=3n)
• To prove A=S, show that AS and SA
• Proving AS by mathematical induction:
– P(n): 3n belongs to S
– Basis: 3*1  S - true by definition of S
– Inductive: Since we already know 3  S (by
definition), and (3n+3)  S (also by definition)
and 3n+3 = 3(n+1), AS is proven
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Example 9: Proof
• To prove SA, use recursive definition of
S:
– Basis: 3  S (by definition)
– Since 3 = 3*1, all elements specified by the
definition’s basis step are divisible by 3
• To complete proof, must show all integers
generated by using second part of definition
of S are in A
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Example 9: Proof
• To do this, show that x+y is in A when xS
and yS and xA and yA
• If x  A and y  A, 3|x and 3|y - so 3|(x+y)
• This completes the proof
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Well-formed Formulae
• Common application of the recursive definition
of sets
• Example: well-formed formulae of variables,
numerals and operators from {+,-,*,/,} are
defined by:
– x is a well-formed formula is x is a numeral or
variable;
– (f+g), (f-g), (f*g), (f/g) and (fg) are well-formed
formulae if f and g are well-formed formulae
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Application of well-formed
formulae
• Using this definition, we can define any
infix expression
• For example, x and 5 are well-formed
formulae
• So (x+5), (x*5), etc. are well-formed
formulae
• And (x+5) / (x*5) is a well-formed formula
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Example 10
• Well-formed formulae for compound
propositions involving T, F, variables and
operators {,,,,} are defined by:
– T, F and p (where p is a propositional variable) are
well-formed formulae;
– (p), (p  q), (p  q), (p q), (p  q) are wellformed formulae is p and q are well-formed
– Thus, we could build any compound propositional
expression in the same way as with infix arithmetic
expressions
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Recursive definition of strings
• The set * of strings over the alphabet 
can be recursively defined by:
–   * where  is the empty string and
– wx  * whenever w  * and x  
• Translation:
– empty strings belong to the set of strings
– can produce new strings by concatenating
strings from the set of strings with symbols
from the alphabet
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Recursive definition of strings
• Can also recursively define the length of
strings; if l(w) is the length of string w,
then:
– l() = 0
– l(wx) = l(w)+1 if w  * and x  
• Can use these definitions in proofs
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Example 11
• Prove that l(xy) = l(x) + l(y) where x  * and
y  *
• Basis: P()
– show l(x) = l(x)+l() for x*
– since l() = 0, l(x)+l() = l(x), so l(x) is true
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Example 11
• Inductive step:
– Assume P(y) is true; show that this implies
P(ya) is true where a  
– Need to show l(xya) = l(x) + l(ya)
– By recursive definition of l(w), we have
l(xya)=l(xy+1) and l(ya) = l(y)+1
– By inductive hypothesis, l(xy)=l(x)+l(y)
– So we conclude l(xya)=l(x)+l(y)+1 = l(x)+l(ya)
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Section 3.3
Recursive Definitions
- ends -
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