COMP232 - Mathematics for Computer Science
Tutorial 10
Ali Moallemi
moa [email protected]
Iraj Hedayati
h [email protected]
Concordia University, Fall 2015
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
1 / 16
Table of Contents
1
5.3 Recursive Definitions and Structural Induction
Exercise 3
Exercise 6
Exercise 8
Exercise 13
Exercise 20
Exercise 21
Exercise 28
Exercise 42
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
2 / 16
Exercise 3
Find f (2), f (3), f (4), and f (5) if f is defined recursively by f (0) = −1,
f (1) = 2, and for n = 1, 2, . . .
a) f (n + 1) = f (n) + 3f (n − 1).
Answer:
f (2) = f (1) + 3f (0) = 2 + 3 · (−1) = −1
f (3) = f (2) + 3f (1) = −1 + 6 = 5
f (4) = f (3) + 3f (2) = 5 + 3 · (−1) = 2
f (5) = f (4) + 3f (3) = 2 + 3 · 5 = 17
b) f (n + 1) = f (n)2 f (n − 1).
Answer:
f (2) = f (1)2 f (0) = 22 · (−1) = −4
f (3) = f (2)2 f (1) = (−4)2 · 2 = 32
f (4) = f (3)2 f (2) = 322 · (−4) = −4096
f (5) = f (4)2 f (3) = (−4096)2 · 32 = 536, 870, 912
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
3 / 16
Exercise 3 - Cont...
Find f (2), f (3), f (4), and f (5) if f is defined recursively by f (0) = −1,
f (1) = 2, and for n = 1, 2, . . .
c) f (n + 1) = 3f (n)2 − 4f (n − 1)2 .
Answer:
f (2) = 3 · f (1)2 − 4f (0)2 = 3 · 22 − 4(−1)2 = 8
f (3) = 3 · f (2)2 − 4f (1)2 = 3 · 82 − 4(2)2 = 176
f (4) = 3 · f (3)2 − 4f (2)2 = 3 · 1762 − 4(8)2 = 92, 672
f (5) = 3 · f (4)2 − 4f (3)2 = 3 · 92.6272 − 4(176)2 2 = 25, 764, 174, 848
d) f (n + 1) = f (n − 1)/f (n).
Answer:
f (2) = f (0)/f (1) = −1/2
f (3) = f (1)/f (2) = −21 = −4
2
f (4) = f (2)/f (3) =
f (5) = f (3)/f (4) =
Ali Moallemi, Iraj Hedayati
− 12
−4
−4
1
8
= 18
= −32
COMP232 - Mathematics for Computer Science
4 / 16
Exercise 6
Determine whether each of these proposed definitions is a valid recursive
definition of a function f from the set of nonnegative integers to the set of
integers. If f is well defined, find a formula for f (n) when n is a
nonnegative integer and prove that your formula is valid.
a) f (0) = 1, f (n) = −f (n − 1) for n ≥ 1
Answer: Is Valid
f (n) = (−1)n
proof.
Basis step: f (0) = (−1)0 = 1
Inductive: if f (k) = (−1)k then,f (k + 1) = (−1)k+1 = (−1) · (−1)k
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
5 / 16
Exercise 6 - Cont...
Determine whether each of these proposed definitions is a valid recursive
definition of a function f from the set of nonnegative integers to the set of
integers. If f is well defined, find a formula for f (n) when n is a
nonnegative integer and prove that your formula is valid.
b) f (0) = 1, f (1) = 0, f (2) = 2, f (n) = 2f (n − 3) for n ≥ 3
Answer: Is Valid
0
if
n mod 3 = 1
f (n) =
dn/3e
2
otherwise
proof.
Basis step: f (0) = 2d0/3e = 20 = 1, f (1) = 0, f (2) = 2d2/3e = 21 = 2
Inductive: if f (k) = 2dk/3e then, f (k − 2) = 2d(k−2)/3e and
f (k + 1) = 2d(k+1)/3e = 2 · 2d(k−2)/3e
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
6 / 16
Exercise 6 - Cont...
Determine whether each of these proposed definitions is a valid recursive
definition of a function f from the set of nonnegative integers to the set of
integers. If f is well defined, find a formula for f (n) when n is a
nonnegative integer and prove that your formula is valid.
c) f (0) = 0, f (1) = 1, f (n) = 2f (n + 1) for n ≥ 2
Answer: Is not Valid
Because value of f (n) depends on n + 1
d) f (0) = 0, f (1) = 1, f (n) = 2f (n − 1) for n ≥ 1
Answer: Is not Valid
Because f (1) = 1 in basis step and is f (1) = 2f (0) = 0 in inductive
step.
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
7 / 16
Exercise 6 - Cont...
Determine whether each of these proposed definitions is a valid recursive
definition of a function f from the set of nonnegative integers to the set of
integers. If f is well defined, find a formula for f (n) when n is a
nonnegative integer and prove that your formula is valid.
e) f (0) = 2, f (n) = f (n − 1) if n is odd and n ≥ 1 and f (n) =
2f (n − 2) if n ≥ 2
Answer: Is Valid f (n) = 2d(n+1)/2e
proof.
Basis step: f (0) = 21 = 2
Inductive: if f (k) = 2d(k+1)/2e then,f (k + 1) = 2d(k+2)/2e
I
I
If k + 1 is odd, f (k + 1) = 2d(k+2)/2e = 2d(k+1)/2e = f (k)
If k + 1 ≥ 2, f (k + 1) = 2d(k+2)/2e = 2 · 2dk/2e = 2f (k − 1)
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
8 / 16
Exercise 8
Give a recursive definition of the sequence {an }, n = 1, 2, 3, · · · if
a) an = 4n − 2.
Answer: f (1) = 2, f (n) = f (n − 1) + 4
b) an = 1 + (−1)n .
Answer: f (1) = 0, f (n) = 2 − f (n − 1)
c) an = n(n + 1).
Answer: f (1) = 2, f (n) = 2n + f (n − 1)
d) an = n2 .
Answer: f (1) = 1, f (n) = 2n − 1 + f (n − 1)
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
9 / 16
Exercise 13
Prove that f1 + f3 + . . . + f2n−1 = f2n when n is a positive integer.
Answer: (fn is nth Fibonancci number)
Let P(n) be f1 + f3 + . . . + f2n−1 = f2n
Basis step: P(1) is true because f1 = 1 = f2 .
Inductive step: Assume that P(k) is true. Then
f1 + f3 + . . . + f2k−1 + f2k+1 = f2k + f2k+1 = f2k+2 = f2(k+1)
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
10 / 16
Exercise 20
Give a recursive definition of the functions max and min so that
max(a1 , a2 , ..., an ) and min(a1 , a2 , ..., an ) are the maximum and minimum
of the n numbers a1 , a2 , ..., an , respectively.
answer:
max(a) = a
max(a1 , a2 , ..., an ) = max(max(a1 , a2 , ..., an−1 ), an )
min(a) = a
min(a1 , a2 , ..., an ) = min(min(a1 , a2 , ..., an−1 ), an )
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
11 / 16
Exercise 21
Let a1 , a2 , ..., an , and b1 , b2 , ..., bn be real numbers. Use the recursive
definitions that you gave in Exercise 20 to prove these.
a) max(−a1 , −a2 , ..., −an ) = min(a1 , a2 , ..., an )
answer: Proof by induction. Basis step: For n = 1,
max(−a1 ) = −a1 = −min(a1 ). For n = 2, there are two cases. If
a2 ≥ a1 , then −a1 ≥ −a2 , so max(−a1 , −a2 ) = −a1 = −min(a1 , a2 ).
If a2 < a1 , then −a1 < −a2 , so
max(−a1 , −a2 ) = −a2 = −min(a1 , a2 ).
Inductive step: Assume it is true for k with k ≥ 2. Then
max(−a1 , −a2 , ..., −ak , −ak+1 ) = max(max(−a1 , ..., −ak ), −ak+1 ) =
max(−min(a1 , ..., ak ), −ak+1 ) = −min(min(a1 , ..., ak ), ak+1 ) =
−min(a1 , ..., ak+1 ).
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
12 / 16
Exercise 21 Cont...
b) max(a1 + b1 , a2 + b2 , ..., an + bn ) ≤
max(a1 , a2 , ..., an ) + max(b1 , b2 , ..., bn )
answer: Proof by mathematical induction. Basis step: For n = 1, the
result is the identity a1 + b1 = a1 + b1 . For n = 2, first consider the
case in which a1 + b1 ≥ a2 + b2 . Then
max(a1 + b1 , a2 + b2 ) = a1 + b1 . Also note that a1 ≤ max(a1 , a2 ) and
b1 ≤ max(b1 , b2 ), so a1 + b1 ≤ max(a1 , a2 ) + max(b1 , b2 ). Therefore,
max(a1 + b1 , a2 + b2 ) = a1 + b1 ≤ max(a1 , a2 ) + max(b1 , b2 ).The
case with a1 + b1 < a2 + b2 is similar.
Inductive step: Assume that the result is true for k. Then
max(a1 + b1 , a2 + b2 , ..., ak + bk , ak+1 + bk+1 ) =
max(max(a1 + b1 , a2 + b2 , ..., ak + bk ), ak+1 + bk+1 ) ≤
max(max(a1 , a2 , ..., ak ) + max(b1 , b2 , ..., bk ), ak+1 + bk+1 ) ≤
max(max(a1 , a2 , ..., ak ), ak+1 ) + max(max(b1 , b2 , ..., bk ), bk+1 ) =
max(a1 , a2 , ..., ak , ak+1 ) + max(b1 , b2 , ..., bk , bk+1 ).
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
13 / 16
Exercise 21 Cont...
c) min(a1 + b1 , a2 + b2 , ..., an + bn ) ≥
min(a1 , a2 , ..., an ) + min(b1 , b2 , ..., bn )
answer: Proof by mathematical induction. Basis step: For n = 1, the
result is the identity a1 + b1 = a1 + b1 . For n = 2, first consider the
case in which a1 + b1 ≥ a2 + b2 . Then
min(a1 + b1 , a2 + b2 ) = a1 + b1 . Also note that a1 ≥ min(a1 , a2 ) and
b1 ≥ min(b1 , b2 ), so a1 + b1 ≥ min(a1 , a2 ) + min(b1 , b2 ). Therefore,
min(a1 + b1 , a2 + b2 ) = a1 + b1 ≥ min(a1 , a2 ) + min(b1 , b2 ).The case
with a1 + b1 > a2 + b2 is similar.
Inductive step: Assume that the result is true for k. Then
min(a1 + b1 , a2 + b2 , ..., ak + bk , ak+1 + bk+1 ) =
min(min(a1 + b1 , a2 + b2 , ..., ak + bk ), ak+1 + bk+1 ) ≥
min(min(a1 , a2 , ..., ak ) + min(b1 , b2 , ..., bk ), ak+1 + bk+1 ) ≥
min(min(a1 , a2 , ..., ak ), ak+1 ) + min(min(b1 , b2 , ..., bk ), bk+1 ) =
min(a1 , a2 , ..., ak , ak+1 ) + min(b1 , b2 , ..., bk , bk+1 ).
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
14 / 16
Exercise 28
Give a recursive definition of each of these sets of ordered pairs of positive
integers. [Hint: Plot the points in the set in the plane and look for lines
containing points in the set.]
a) S = {(a, b)|a ∈ Z+ , b ∈ Z+ , and a + b is odd }
answer: (1, 2) ∈ S, (2, 1) ∈ S
If (a, b) ∈ S then (a + 2, b), (a, b + 2) and (a + 1, b + 1) ∈ S.
b) S = {(a, b)|a ∈ Z+ , b ∈ Z+ , and a|b}
answer: (a, a) ∈ S for all a ∈ Z+
If (a, b) ∈ S then (a, b + a) ∈ S.
c) S = {(a, b)|a ∈ Z+ , b ∈ Z+ , and 3|a + b}
answer: (1, 2) ∈ S, (2, 1) ∈ S
If (a, b) ∈ S then (a + 3, b), (a, b + 3), (a + 2, b + 1) and
(a + 1, b + 2) ∈ S.
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
15 / 16
Exercise 42
Show that (w R )i = (w i )R whenever w is a string and i is a nonnegative
integer; that is, show that the ith power of the reversal of a string is the
reversal of the ith power of the string.
answer: Proof by mathematical induction. Basis step: For i = 1, the
result is the identity (w R )1 = w R = (w 1 )R .For Inductive step: Assume
that the result is true for k.
Then
(w R )k+1 = (w R )k w R = (w k )R w R = (w · w k )R = (w k+1 )R .
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
16 / 16