Discrrete mathematics for computer science structural induction warmup

Structural Induction
Warm-Up


“Balanced String of Parentheses”
• Base case:
– The empty string εis balanced

• Constructor rules:
– C1: If x is balanced then so is (x), that is,
the result of writing a “(“, then x, then
“)”
– C2: If x and y are balanced then so is xy


A String of Parentheses is Balanced iff it
Satisfies the Counting Rule (SCR)
• Say that a string s∈{ ), ( }* SCR iff
starting from 0 at the left end of the
string, adding 1 for each (,
subtracting 1 for each ), gives 0 at
the end without ever going negative.
• Theorem: A string of parentheses is
balanced iff it SCR.


If a string of parentheses x is balanced, then x SCR.

• Structural Induction Base case: x=ε. Then
count begins and ends 0 so x SCR.
• Induction case 1: x = (y) where y is balanced.

Then y SCR (why?). Then the count for x is +1
after first (, +1 again after the last character of
y, stays positive in between, and ends at 0
after the final ).
• Induction case 2: x = yz where y and z are
balanced. Then y and z SCR and the count goes
from 0, to 0 after y, to 0 after z without ever
going negative.


If x SCR then x is balanced.
• Proof by strong induction on |x|. Suppose x SCR.
• Base case: |x|=0. Then x=εand x is balanced.
• Induction step. Fix n, suppose |x| = n+1>0 and for
all m≤n and any y of length m, if y SCR then y is
balanced.
• Case 1. The count never reaches 0 except at the end.
Then x=(y) where y SCR (why?). But |y|=|x|-2 and by
IH y is balanced. By construction rule 1 so is x.
• Case 2. x=yz where the y and z are nonempty and
the count goes to 0 after y. Then y and z are shorter
than x and each SCR, so each is balanced by IH.
Then x is balanced by construction rule 2.