1. and Structural Induction
Mathematics for Computer Science MIT 6.042J/18.062J
Recursive Definitions
and
Structural Induction
Copyright © Albert Meyer, 2002.
Prof. Albert Meyer & Dr. Radhika Nagpal

2. Build something from a simpler version of the same thing:
Recursive Definitions
Build something from a simpler version of the same thing:
Base case(s) that don’t depend on anything else
Induction (Construction) case(s) that depend on simpler cases

3. Example Definition: set E
Define set E  Z, recursively:
0  E Base Case
If n  E, then n + 2  E Induction case
If n  E, then -n  E
E : 0
, 0+2

4. Example Definition: set E
Define set E  Z, recursively:
0  E Base Case
If n  E, then n + 2  E Induction case If n  E, then -n  E
E : 0, 2, 2+2
, (2+2)+2

5. Example Definition: set E
Define set E  Z, recursively:
0  E
If n  E, then n + 2  E
If n  E, then -n  E
E : 0, 2, 4, 6, … All even numbers
Also, -2, -4, -6, …

6. Recursive Definition: Extremal Clause
So, E contains the even integers
Anything Else?
No!
0  E
If n  E, then n + 2  E
If n  E, then -n  E
That’s All!
Implicit part of recursive definition

7. Example Definition: set E
So E is exactly:
The Even Integers

8. l  S, (the empty string) axb bxa xy
Another Example
Define set of strings, S  {a, b}*
l  S, (the empty string)
If x, y  S, then the following are in S:
axb
bxa xy

9. Proof by Structural Induction on S
Lemma: Every x in S has an equal number of a’s and b’s.

10. Structural Induction on S
Proof:
Let
EQ ::= {strings with equal # of a’s and b’s}
P(x) ::= x  S  x  EQ
Base Case: x = l.
0 a’s and 0 b’s. OK

11. Structural Induction on S
Inductive Step
Assume: P(x) and P(y)
To Prove: P(axb), P(bxa), and P(xy)
Case 1: axb
has k + 1 a’s and b’s if x has k of each. OK

12. Structural Induction on S
Inductive Step
Assume: P(x) and P(y)
To Prove: P(axb), P(bxa), and P(xy)
Case 2: bxa SAME
Case 3: xy
has kx + ky of each. OK
So S  EQ

13. Structural Induction on S
Rest of Story:
Theorem: S = EQ
Show by strong induction on length of string in EQ. Proof in Notes 5.

14. In-Class Problem
Problem 1

15. Recursive Data Types
Rooted Binary Trees

16. Defined recursively as follows: A single node r is an RBT:
Rooted Binary Trees
Defined recursively as follows:
A single node r is an RBT:

17. If t, t’ are RBTs, then so are :
Rooted Binary Trees
If t, t’ are RBTs, then so are :
makeleft(t) t
makeright(t) t
makeboth(t,t’) t t’

18. A Rooted Binary Tree Example

19. Recursive Functions on RBT
Recursive def of set of nodes
nodes( )::= { },
nodes(makeleft(t))::= nodes(t) { }
nodes(makeright(t))::= nodes(t){ }
nodes(makeboth(t1, t2))::=
nodes(t1)  nodes(t2){ }
Remember, is not supposed to be in nodes(t)

20. Example
= {{ }, { }, { }, { }, { },
{ }, { }, { }, { } }
nodes( )

21. Recursive Functions on RBT
depth( )::= 0
depth(makeleft(t))::= 1 + depth(t)
depth(makeright(t))::= 1 + depth(t)
depth(makeboth(t1, t2))::=
1 + max(depth(t1),depth(t2))

22. Example
= 4
depth( )

23. Number of nodes vs Depth
Lemma: |nodes(t)| + 1  2depth(t)+1
Proof by Structural Induction
Base Case: t =
|nodes(t)| + 1 = 2 = 20+1 = 2depth(t)+1
OK!

24. Number of nodes vs Depth
Induction Case: makeleft(t)
Assume: |nodes(t)| + 1  2depth(t) +1
To Prove: |nodes(makeleft(t))| + 1  2depth(makeleft(t)) + 1
|nodes(makeleft(t))| + 1
= (|nodes(t)|+1) + 1 by def. of nodes
 (2depth(t)+1 + 1) + 1 by induction hyp.
= 2depth(t)  2depth(t)+1 + 2depth(t)+1
= 2depth(t)+1+1
= 2depth(makeleft(t))+1 by def. of depth

25. Number of nodes vs Depth
Induction Case: makeright(t)
Same

26. Number of nodes vs Depth
Induction Case: makeboth(t1, t2)
Assume: |nodes(t1)|+1  2depth(t1)+1
|nodes(t2)|+1  2depth(t2)+1
To Prove:
|nodes(makeboth(t1, t2))| +1
 2depth(makeboth(t1,t2))+1

27. |nodes(makeboth(t1,t2))| + 1
Proof of makeboth Case
|nodes(makeboth(t1,t2))| + 1
= (|nodes(t1)| + |nodes(t2)| + 1)+1 def. of nodes
= |nodes(t1)| +1 + (|nodes(t2)| +1)
 2depth(t1)+1 + 2depth(t2)+1 induction hyp.
 2max(depth(t1),depth(t2)) max(depth(t1),depth(t2))+1
= 2(max(depth(t1),depth(t2))+1)+1
= 2depth(makeboth(t1,t2)) def. of depth

28. In-Class Problem
Problems 2 & 3