Discrete Mathematics and Its Applications with MathZone

ICS.6D: Fall 2007, Quiz 5 Solutions
1) [32 points] State whether each of the following is as a recursive definition of some function
from nonnegative integers to integers or not. Circle the right answer:
a) f(0) = 1, f(n) = f(n – 1) for n ≥ 1 Good one, f(n)=1 for all n≥0
b) f(0) = 1, f(n) = f(n – 2) for n ≥ 1 Failed, because f(1) is “defined” using f(-1)
c) f(0) = 1, f(n) = f(n – 2) for n ≥ 2 Failed, because f(1) is undefined.
d) f(0) = 1, f(1) = 1, f(n) = f(n – 2) for n ≥ 3 Failed, because f(2) is undefined
2) [36 points] State whether each of the following is a good recursive definition of the set of (all)
even positive integers , i.e. {2,4,6,8,10,…}, or not. Circle the right answer:
a) 2 ∈ S, and if n ∈ S then n+2 ∈ S. Good One
b) 2 ∈ S, and if n ∈ S then 2*n ∈ S. Failed, because e.g. 6 is not in this set S.
(Btw, this is a good recursive definition but of a different set, namely S = {2 k | k ≥ 1}.)
c) 2 ∈ S, and if n
1
∈ S and n
2
∈ S then (n
1
+n
2
) ∈ S. Good One. Note that we can have n
1
=n
2
.
3) [32 points]
Here are definitions of (1) a string, (2) a string inverse, and (3) a string bit-wise xor:
1) λ is a string, and s•b is a string for any string s and bit b .
2) inv(λ) = λ , and inv(s•b) = b•inv(s) for any string s and bit b .
3) xor(λ) = 0 , and xor(s•b) = (b+xor(s) mod 2)
Fact 4) xor(b•s) = (b+xor(s) mod 2) for any string s and bit b ,
Using these definitions show by structural induction that xor
(
inv(s)
)
= xor(s) for any string s :
a) [6 points] Base case . First, write the string which you need to consider: ___ Empty string, i.e. λ
Now prove that the claim is true for this string:
xor(inv( λ )) = xor( λ ) because inv( λ ) = λ by the base case of definition (2).
b) [26 points] Inductive case . First, write a string which you need to consider: ___ s•b _______
Now prove that the claim is true for this string assuming it holds for its component(s):
xor(inv( s•b )) = xor
(
b• inv( s )
)
% by recursive case of definition (2)