Discrrete mathematics for computer science structural induction warmup

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 , s

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