Recursive Definitions
and Structural Induction
CS/APMA 202
Rosen section 3.4
Aaron Bloomfield
1
Recursion
Recursion means defining something,
such as a function, in terms of itself


For example, let f(x) = x!
We can define f(x) as f(x) = x * f(x-1)
2
Recursion example
Rosen, section 3.4, question 1

a)
Find f(1), f(2), f(3), and f(4), where f(0) = 1
Let f(n+1) = f(n) + 2
f(1) = f(0) + 2 = 1 + 2 = 3
f(2) = f(1) + 2 = 3 + 2 = 5
f(3) = f(2) + 2 = 5 + 2 = 7
f(4) = f(3) + 2 = 7 + 2 = 9
b)
Let f(n+1) = 3f(n)
f(1) = 3 * f(0) = 3*1 = 3
f(2) = 3 * f(1) = 3*3 = 9
f(3) = 3 * f(2) = 3*9 = 27
f(4) = 3 * f(3) = 3*27 = 81
3
Recursion example
Rosen, section 3.4, question 1

c)
Find f(1), f(2), f(3), and f(4), where f(0) = 1
Let f(n+1) = 2f(n)
f(1) = 2f(0) = 21 = 2
f(2) = 2f(1) = 22 = 4
f(3) = 2f(2) = 24 = 16
f(4) = 2f(3) = 216 = 65536
d)
Let f(n+1) = f(n)2 + f(n) + 1
f(1) = f(0)2 + f(0) + 1 = 12 + 1 + 1 = 3
f(2) = f(1)2 + f(0) + 1 = 32 + 3 + 1 = 13
f(3) = f(2)2 + f(0) + 1 = 132 + 13 + 1 = 183
f(4) = f(3)2 + f(0) + 1 = 1832 + 183 + 1 = 33673
4
Fractals
A fractal is a pattern that uses recursion

The pattern itself repeats indefinitely
5
Fractals
6
Fibonacci sequence
Definition of the Fibonacci sequence
Non-recursive:

1  5   1  5 
F ( n) 
Recursive:
or:
F(n) = F(n-1) + F(n-2)
F(n+1) = F(n) + F(n-1)
n


n
5  2n
Must always specify base case(s)!


F(1) = 1, F(2) = 1
Note that some will use F(0) = 1, F(1) = 1
7
Fibonacci sequence in Java
long Fibonacci
if ( (n ==
return
else
return
}
(int n) {
1) || (n == 2) )
1;
Fibonacci (n-1) + Fibonacci (n-2);
long Fibonacci2 (int n) {
return (long) ((Math.pow((1.0+Math.sqrt(5.0)),n)Math.pow((1.0-Math.sqrt(5.0)),n)) /
(Math.sqrt(5) * Math.pow(2,n)));
}
8
Recursion definition
From “The Hacker’s Dictionary”:
recursion n. See recursion.
tail recursion.
See also
9
Bad recursive definitions
Consider:



f(0) = 1
f(n) = 1 + f(n-2)
What is f(1)?
Consider:



f(0) = 1
f(n) = 1+f(-n)
What is f(1)?
10
Defining sets via recursion
Same two parts:


Base case (or basis step)
Recursive step
Example: the set of positive integers


Basis step: 1  S
Recursive step: if x  S, then x+1  S
11
Defining sets via recursion
Rosen, section 3.4, question 24: give recursive
definitions for:
a)
The set of odd positive integers


b)
The set of positive integer powers of 3


c)
1S
If x  S, then x+2  S
3S
If x  S, then 3*x  S
The set of polynomials with integer coefficients


0S
If p(x)  S, then p(x) + cxn  S

c  Z, n  Z and n ≥ 0
12
Defining strings via recursion
Terminology


 is the empty string: “”
 is the set of all letters: { a, b, c, …, z }
The set of letters can change depending on the problem
We can define a set of strings * as follows




Base step:   *
If w  * and x  , then wx  *
Thus, * s the set of all the possible strings that can
be generated with the alphabet
Is this countably infinite or uncountably infinite?
13
Defining strings via recursion
Let  = { 0, 1 }
Thus, * is the set of all binary numbers


Or all binary strings
Or all possible computer files
14
String length via recursion
How to define string length recursively?


Basis step: l() = 0
Recursive step: l(wx) = l(w) + 1 if w  * and x

Example: l(“aaa”)





l(“aaa”) = l(“aa”) + 1
l(“aa”) = l(“a”) + 1
l(“a”) = l(“”) + 1
l(“”) = 0
Result: 3
15
Today’s demotivators
16
Strings via recursion example
Rosen, section 3.4, question 38: Give a
recursive definition for the set of string that
are palindromes

We will define set P, which is the set of all
palindromes
Basis step:   P

Second basis step: x  P when x  
Recursive step: xpx  P if x   and p  P
17
Strings and induction example
This requires structural induction, which
will be covered later in this slide set
18
Recursion pros
Easy to program
Easy to understand
19
Recursion cons
Consider the recursive Fibonacci generator
How many recursive calls does it make?











F(1): 1
F(2): 1
F(3): 3
F(4): 5
F(5): 9
F(10): 109
F(20): 13,529
F(30): 1,664,079
F(40): 204,668,309
F(50): 25,172,538,049
F(100): 708,449,696,358,523,830,149  7 * 1020
At 1 billion recursive calls per second (generous), this would take
over 22,000 years
But that would also take well over 1012 Gb of memory!
20
Trees
Rooted trees:

A graph containing nodes and edges
Cannot contain a cycle!
Cycle not allowed in a tree
21
Rooted trees
Recursive definition:


Basis step: A single vertex r is a rooted tree
Recursive step:
Let T1, T2, …, Tn be rooted trees
Form a new tree with a new root r that contains an edge
to the root of each of the trees T1, T2, …, Tn
22
(Extended) Binary trees
Recursive definition


Basis step: The empty set is an extended binary tree
Recursive step:
Let T1, and T2 be extended binary trees
Form a new tree with a new root r
Form a new tree such that T1 is the left subtree,
and T2 is the right subtree
23
Full binary trees
Recursive definition


Basis step: A full binary tree consisting only of the
vertex r
Recursive step:
Let T1, and T2 be extended binary trees
Form a new tree with a new root r
Form a new tree T such that T1 is the left subtree,
and T2 is the right subtree
 This is denoted by T = T1∙T2
Note the only difference between a regular
binary tree and a full one is the basis step
24
Binary tree height
h(T) denotes the height of tree T
Recursive definition:


Basis step: The height of a tree with
only one node r is 0
Recursive step:
Let T1 and T2 be binary trees
The binary tree T = T1∙T2 has height
h(T) = 1 + max ( h(T1), h(T2) )
This definition can be generalized to
non-binary trees
25
Binary tree size
n(T) denotes the number of vertices in tree T
Recursive definition:



Basis step: The number of vertices of an
empty tree is 0
Basis step: The number of vertices of a tree
with only one node r is 1
Recursive step:
Let T1 and T2 be binary trees
The number of vertices in binary tree T = T1∙T2 is:
n(T) = 1 + n(T1) + n(T2)
This definition can be generalized to nonbinary trees
26
A bit of humor: Computer
terminology
27
Recursion vs. induction
Consider
factorial:

the
recursive
definition
for
f(0) = 1
Base case

f(n) = n * f(n-1)
The “step”
Sort of like induction
28
Recursion vs. induction
Rosen, section 3.4, example 7 (page 262)
Consider the set of all integers that are multiples
of 3


{ 3, 6, 9, 12, 15, … }
{ x | x = 3k and k  Z+ }
Recursive definition:


Basis step: 3  S
Recursive step: If x  S and y  S, then x+y  S
29
Recursion vs. induction
Proof via induction: prove that S contains all the integers
that are divisible by 3


Let A be the set of all ints divisible by 3
Show that S = A
Two parts:

Show that S  A
Let P(n) = 3n  S
Base case: P(1) = 3*1  S

By the basis step of the recursive definition
Inductive hypothesis: assume P(k) = 3*k  S is true
Inductive step: show that P(k+1) = 3*(k+1) is true





3*(k+1) = 3k+3
3k  S by the inductive hypothesis
3  S by the base case
Thus, 3k+3  S by the recursive definition
Show that A  S
Done in the text, page 267 (not reproduced here)
30
What did we just do?
Notice what we did:



Showed the base case
Assumed the inductive hypothesis
For the inductive step, we:
Showed that each of the “parts” were in S

The parts being 3k and 3
Showed that since both parts were in S, by the
recursive definition, the combination of those parts
is in S

i.e., 3k+3  S
This is called structural induction
31
Structural induction
A more convenient form of induction for
recursively defined “things“
Used in conjunction with the recursive definition
Three parts:


Basis step: Show the result holds for the elements in
the basis step of the recursive definition
Inductive hypothesis: Assume that the statement is
true for some existing elements
Usually, this just means assuming the statement is true

Recursive step: Show that the recursive definition
allows the creation of a new element using the
existing elements
32
Tree structural induction example
Rosen, section 3.4, question 43
Show that n(T) ≥ 2h(T) + 1
Basis step: Let T be the full binary
tree of just one node r





h(T) = 0
n(T) = 1
n(T) ≥ 2h(T) + 1
1 ≥ 2*0 + 1
1≥1
34
Tree structural induction example
Show that n(T) ≥ 2h(T) + 1
Inductive hypothesis:

Let T1 and T2 be full binary trees
Assume that n(T1) ≥ 2h(T1) + 1 for some tree T1
Assume that n(T2) ≥ 2h(T2) + 1 for some tree T2
Recursive step:

Let T = T1 ∙ T2
Here the ∙ operator means creating a new tree with a root note r and
subtrees T1 and T2
New element is T

By the definition of height and size, we know:
n(T) = 1 + n(T1) + n(T2)
h(T) = 1 + max ( h(T1), h(T2) )

Therefore:
n(T) = 1 + n(T1) + n(T2)
≥ 1 + 2h(T1) + 1 + 2h(T2) + 1
≥ 1 + 2*max ( h(T1), h(T2) )
the sum of two non-neg #’s is at least
as large as the larger of the two
= 1 + 2*h(T)

Thus, n(T) ≥ 2h(T) + 1
35
String structural induction example
Rosen, section 3.4, question 32
Part (a): Give the definition for ones(s), which counts the
number of ones in a bit string s
Let  = { 0, 1 }
Basis step: ones() = 0
Recursive step: ones(wx) = ones(w) + x


Where x   and w  *
Note that x is a bit: either 0 or 1
36
String structural induction example
Part (b): Use structural induction to prove that ones(st) = ones(s) +
ones(t)
Basis step: t = 

ones (s∙) = ones(s) = ones(s)+0 = ones(s) + ones()
Inductive hypothesis: Assume ones(s∙t) = ones(s) + ones(t)
Recursive step: Want to show that ones(s∙t∙x) = ones(s) + ones(t∙x)








Where s, t  * and x  
New element is ones(s∙t∙x)
ones (s∙t∙x) = ones ((s∙t)∙x))
= x+ones(s∙t)
= x + ones(s) + ones(t)
= ones(s) + (x + ones(t))
= ones(s) + ones(t∙x)
Proven!
by associativity of concatenation
by recursive definition
by inductive hypothesis
by commutativity and assoc. of +
by recursive definition
37
Quick survey

a)
b)
c)
d)
I feel I understand structural induction…
Very well
With some review, I’ll be good
Not really
Not at all
38
Human stupidity
39
Induction methods compared
Used for
Assumption
Weak
mathematical
Strong
Mathematical
Structural
Usually
formulae
Usually formulae not
provable via
mathematical induction
Only things defined
via recursion
Assume P(1), P(2), …,
P(k)
Assume statement is
true for some "old"
elements
Assume P(k)
Statement is true for
some "new" elements
created with "old"
elements
What to
prove
True for
P(k+1)
True for P(k+1)
Step 1
called
Base case
Base case
Basis step
Step 3
called
Inductive step
Inductive step
Recursive step
40
Induction types compared
Show that F(n) < 2n


Where F(n) is the nth Fibonacci number
Actually F(n) < 20.7*n, but we won’t prove that here
Fibonacci definition:


Basis step: F(1) = 1 and F(2) = 1
Recursive step: F(n) = F(n-1) + F(n-2)
Base case (or basis step): Show true for F(1)
and F(2)


F(1) = 1 < 21 = 2
F(2) = 1 < 22 = 4
41
Via weak mathematical induction
Inductive hypothesis: Assume F(k) < 2k
Inductive step: Prove F(k+1) < 2k+1



F(k+1) = F(k) + F(k-1)
We know F(k) < 2k by the inductive hypothesis
Each term is less than the next, therefore
F(k) > F(k-1)
Thus, F(k-1) < F(k) < 2k


Therefore, F(k+1) = F(k) + F(k-1) < 2k + 2k = 2k+1
Proven!
42
Via strong mathematical induction
Inductive hypothesis: Assume F(1) < 21,
F(2) < 22, …, F(k-1) < 2k-1, F(k) < 2k
Inductive step: Prove F(k+1) < 2k+1





F(k+1) = F(k) + F(k-1)
We know F(k) < 2k by the inductive hypothesis
We know F(k-1) < 2k-1 by the inductive
hypothesis
Therefore, F(k) + F(k-1) < 2k + 2k-1 < 2k+1
Proven!
43
Via structural induction
Inductive hypothesis: Assume F(n) < 2n
Recursive step:



Show true for “new element”: F(n+1)
We know F(n) < 2n by the inductive
hypothesis
Each term is less than the next, therefore
F(n) > F(n-1)
Thus, F(n-1) < F(n) < 2n


Therefore, F(n) + F(n-1) < 2n + 2n = 2n+1
Proven!
44
Another way via structural induction
Inductive hypothesis: Assume F(n) < 2n and
F(n-1) < 2n-1

The difference here is we are using two “old”
elements versus one, as in the last slide
Recursive step:






Show true for “new element”: F(n+1)
F(n+1) = F(n) + F(n-1)
We know F(n) < 2n by the inductive hypothesis
We know F(n-1) < 2n-1 by the inductive hypothesis
Therefore, F(n) + F(n-1) < 2k + 2k-1 < 2k+1
Proven!
45
But wait!
In this example, the structural induction proof
was essentially the same as the weak or strong
mathematical induction proof

It’s hard to find an example that works well for all of
the induction types
Structural induction will work on some recursive
problems which weak or strong mathematical
induction will not

Trees, strings, etc.
46
A bit of humor…
47
Section 3.4, question 8
Give the recursive definition of the following sequences

Note that many answers are possible!
a) an = 4n – 2
 Terms: 2, 6, 10, 14, 16, etc.
 a1 = 2
 an = an-1 + 4
b) an = 1 + (-1)n
 Terms: 0, 2, 0, 2, 0, 2, etc.
 a1 = 0, a2 = 2
 an = an-2
c) an = n(n+1)
 Terms: 2, 6, 12, 20, 30, 42, etc.
 a1 = 2
 an = an-1 + 2*n
d) an = n2
 Terms: 1, 4, 9, 16, 25, 36, 49, etc.
 a1 = 1
 an = an-1 + 2n - 1
48
Section 3.4, question 12
Show that f12 + f22 + f32 + … + fn2 = fnfn+1
Base case: n = 1


f12 = f1f2
12 = 1*1
Inductive hypothesis: Assume

f12 + f22 + f32 + … + fk2 = fkfk+1
Inductive step: Prove

f12 + f22 + f32 + … + fk2 + fk+12 = fk+1fk+2
49
Section 3.4, question 12
Inductive hypothesis: Assume

f12 + f22 + f32 + … + fk2 = fkfk+1
Inductive step: Prove

f12 + f22 + f32 + … + fk2 + fk+12 = fk+1fk+2

fkfk+1 + fk+12 = fk+1fk+2


fkfk+1 + fk+12 = fk+1 (fk + fk+1)
fkfk+1 + fk+12 = fkfk+1 + fk+12
50
Section 3.4, question 13
Show that f1 + f2 + f3 + … + f2n-1 = f2n
Base case: n = 1


f1 = f2*1
1=1
Inductive hypothesis: Assume

f1 + f2 + f3 + … + f2k-1 = f2k
Inductive step: Prove


f1 + f2 + f3 + … + f2k-1 + f2(k+1)-1 = f2(k+1)
f1 + f2 + f3 + … + f2k-1 + f2k+1 = f2k+2
51
Section 3.4, question 13
Inductive hypothesis: Assume

f1 + f2 + f3 + … + f2k-1 = f2k
Inductive step: Prove

f1 + f2 + f3 + … + f2k-1 + f2k+1 = f2k+2

f2k + f2k+1 = f2k+2

True by definition of f2k+2
52
Section 3.4, question 22
Show that the set S defined by


Basis step: 1  S
Recursive step: s + t  S when s  S and t  S
is the set of positive integers:

Z+ = { 1, 2, 3, … }
Note the (somewhat recursive) definition of the positive integers:


1 is a positive integer
For any arbitrary n that is a positive integer, n+1 is also a positive
integer
Proof by structural induction
Basis step: 1  S and 1  Z+
Inductive hypothesis: Assume k  S
Recursive step: Show k+1  S



k  S by the inductive hypothesis
1  S by the base case
k+1  S by the recursive step of the recursive definition above
53
Section 3.4, question 35
Give a recursive definition of the reversal of a
string
Basis step: R = 

Note that the superscripted R means reversal of a
string
Recursive step: Consider a string w  *

Rewrite w as vy where v  * and y  
v is the first n-1 characters in w
y is the last character in w

wR = y(vR)
Parentheses are for our benefit
54
Quick survey

a)
b)
c)
d)
I felt I understood the material in this slide set…
Very well
With some review, I’ll be good
Not really
Not at all
55
Quick survey

a)
b)
c)
d)
The pace of the lecture for this slide set was…
Fast
About right
A little slow
Too slow
56
Quick survey

a)
b)
c)
d)
How interesting was the material in this slide
set? Be honest!
Wow! That was SOOOOOO cool!
Somewhat interesting
Rather borting
Zzzzzzzzzzz
57
Today’s demotivators
58