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 bitwise 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)
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 Fall '07
 Jarecki
 Recursion, Recurrence relation, 2k, 5digit

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