bài tập cấu trúc rời rạc homework6 solutions

COM S 330 — Homework 06 — Solutions
Type your answers to the following questions and submit a PDF file to Blackboard. One page per problem.
Problem 1. [10pts] Consider the following sequence definitions. Write out the first 10 terms (n ∈ {0, . . . , 9})
of each sequence.
a. [3pts] a0 = 2 and an = 2an−1 − 1 for n ≥ 1.
n 0

1

2

3

4

5

6

7

8

9

an

2

3

5

9

17

35

69

137

273

545

b. [3pts] b0 = 0 and bn = bn−1 +

1
2n

for n ≥ 1.

n 0

1

2

3

4

5

6

7

8

9

0

1
2

3
4

7
8

15
16

31
32

63
64

127
128

255
256

511
512

bn

c. [4pts] c0 = 1, c1 = 2, and cn = 2(cn−1 − cn−2 ) for n ≥ 2.
n

0

1

2

3

cn

1

2

2

0 −4

4

1

5

6

7

8

9

−8

−8

0

16

32

COM S 330 — Homework 06 — Solutions
Problem 2. [5pts] Let {dn }∞
n=0 be a sequence with the recurrence relation dn = 2dn−1 −dn−2 for n ≥ 2. (We
do not specify the terms d0 , d1 .) Use backward reasoning to motivate the closed form dn = nd1 − (n − 1)d0 .
We will apply the recurrence relation several times to deduce a pattern.

dn = 2dn−1 − dn−2
= 2(2dn−2 − dn−3 ) − dn−2
= (4 − 1)dn−2 − 2dn−3
= 3dn−2 − 2dn−3
= 3(2dn−3 − dn−4 ) − 2dn−3
= (6 − 2)dn−3 − 3dn−4
= 4dn−3 − 3dn−4
= 4(2dn−4 − dn−5 ) − 3dn−4
= 5dn−4 − 4dn−5
..
.
= kdn−(k−1) − (k − 1)dn−k
..
.

For k ≥ 1.

= ndn−(n−1) − (n − 1)dn−n = nd1 − (n − 1)d0 .

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COM S 330 — Homework 06 — Solutions
Problem 3. [10pts] Mathematical induction can be used to prove things that can be proven directly. In

the two problems below, you will essentially prove the induction step of an induction proof.
a. [5pts] For n ≥ 0, let P (n) denote the sentence “The 2 × n chessboard can be tiled using dominoes.” Prove
“P (n) → P (n + 1).”
Proof. Let C be the 2 × (n + 1) chessboard. Place a domino vertically on the far-right edge, leaving a 2 × n
chessboard of open squares. Since the 2 × n chessboard can be tiled using dominoes, that tiling can complete
the tiling of C. Therefore, the 2 × (n + 1) chessboard can be tiled using dominoes.
b. [5pts] For n ≥ 0, let Q(n) denote the sentence “The 3 × 2n chessboard can be tiled using dominoes.”
Prove “Q(n) → Q(n + 1).”
Proof. Let C be the 3 × 2(n + 1) chessboard. Place two dominoes vertically in the top-right corner, one tile
covering the positions (1, 2n + 2), (2, 2n + 2) and the other covering positions (1, 2n + 1), (2, 2n + 1). Then
place a domino horizontally in the bottom-right corner, covering the positions (3, 2n + 2), (3, 2n + 1). The
remaining squares form a 3 × 2n chessboard. Since the 3 × 2n chessboard can be tiled using dominoes, this
tiling completes the tiling of C. Therefore, the 3 × 2(n + 1) chessboard can be tiled using dominoes.

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COM S 330 — Homework 06 — Solutions
Problem 4. [10pts] Define a sequence {fn }∞
n=0 as f0 = 1 and for n ≥ 1, fn =
fn =

Fn+1
Fn+2 ,

where

{Fn }∞
n=0

1
1+fn−1 .

Prove that for n ≥ 0,

is the Fibonacci sequence.

Proof. We use induction.
Case n = 0: f0 = 1 =

1
1

=

F1
F2 .

(Induction Hypothesis) Assume that fn =

Fn+1
Fn+2

for some n ≥ 0.

Case n + 1: Consider fn+1 .
fn+1 =
=
=
=

=

1
1 + fn
1
1+

Fn+1
Fn+2

By recurrence relation of fn+1
By Induction Hypothesis

1
Fn+1 +Fn+2
Fn+2

1
Fn+3
Fn+2

By recurrence relation Fn+3 = Fn+2 + Fn+1 , since n + 3 ≥ 2.

Fn+2
.
Fn+3

Therefore, by induction we have that fn =

Fn+1

Fn+2

for all n ≥ 0.

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COM S 330 — Homework 06 — Solutions
Problem 5. [10pts] Let {An }1
n=1 be an arbitrary sequence of non-empty sets. Recall the n-fold Cartesian
n
product i=1 Ai is the set of n-tuples (a1 , . . . , an ) where each ai is an element in Ai . The iterated product is
the set sequence {Bn }1
n=1 where B1 = A1 , and for n ≥ 2, Bn = Bn 1 × An . (Thus, B1 = A1 , B2 = A1 × A2 ,
B3 = (A1 × A2 ) × A3 , etc.) Use induction to prove that for all n ≥ 2, there is a bijection fn from n-fold
n
Cartesian product i=1 Ai to the iterated product Bn = Bn 1 × An . [Note: the n-fold Cartesian product
n
i=1 Ai consists of n-tuples (a1 , . . . , an ), while the iterated product Bn consists of ordered pairs (b, an )
where b ∈ Bn 1 and an ∈ An .]
Proof. We use induction on n ≥ 1.
Case n = 1: Note that A1 = B1 , so the identity map f1 : A1 → B1 defined as f1 (x) = x is a bijection
between A1 and B1 .
(Induction Hypothesis) Assume that for some n ≥ 1 we have a bijection fn :

n
i=1

Ai → B n .
n

Case n + 1: Note that by the Induction Hypothesis there exists a bijection fn : i=1 Ai → Bn . We den+1
fine a function fn+1 : i=1 Ai → Bn+1 as follows: For an arbitrary (n + 1)-tuple (x1 , . . . , xn , xn+1 ) in
n+1
i=1 Ai , let fn+1 (x1 , . . . , xn , xn+1 ) = (fn (x1 , . . . , xn ), xn+1 ). We claim that fn+1 is a bijection. First, since
n
fn (x1 , . . . , xn ) ∈ Bn for all (x1 , . . . , xn ) ∈ i=1 Ai (by Induction Hypothesis), we have that (fn (x1 , . . . , xn ), xn+1 )
n+1
is an element of Bn × An+1 for every (x1 , . . . , xn+1 ) ∈ i=1 Ai . Therefore, fn+1 is a well-defined function
with codomain Bn+1 = Bn × An+1 .
n+1

We claim fn+1 is an injection: Let (x1 , . . . , xn , xn+1 ) and (y1 , . . . , yn , yn+1 ) be elements of i=1 Ai where
fn+1 (x1 , . . . , xn+1 ) = fn+1 (y1 , . . . , yn+1 ). Since the first coordinate in these function outputs are equal, we
have that fn (x1 , . . . , xn ) = fn (y1 , . . . , yn ). Since fn is an injection, we have that (x1 , . . . , xn ) = (y1 , . . . , yn ).
Since the second coordinate in the function outputs are equal, xn+1 = yn+1 . Therefore, fn+1 is an injection.
We claim fn+1 is a surjection: Let (y, z) ∈ Bn × An+1 = Bn+1 be any element of the codomain. Since fn
n
is surjective, there exists an n-tuple (x1 , . . . , xn ) ∈ i=1 Ai where fn (x1 , . . . , xn ) = y. Since z ∈ An+1 , note
that fn+1 (x1 , . . . , xn , z) = (y, z) and therefore (y, z) is in the range of fn+1 and hence fn+1 is surjective.
By induction, there exists a bijection fn :

n
i=1

Ai → Bn for every n ≥ 1.

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