Solutions to Homework One
CSE 101
1.
Exercise 0.1.
(a) Θ; (b)
O
; (c) Θ; (d) Θ; (l)
O
; (m)
O
; (n) Θ; (o) Ω; (p)
O
.
2.
Exercise 0.3.
(a) Base cases
n
= 6 and
n
= 7 are easy to check individually. Now assume the statement holds upto
some value of
n
greater than 6; we need to extend it to
n
+ 1.
F
n
+1
=
F
n
+
F
n
−
1
≥
2
0
.
5
n
+ 2
0
.
5(
n
−
1)
(induction hypothesis)
≥
2
·
2
0
.
5(
n
−
1)
=
2
0
.
5(
n
+1)
.
Note: for this problem (and some of the others below), we need
two
base cases. Do you see why?
(b) We’ll show, again by induction, that
F
n
≤
2
cn
for
c
= log
2
((1 +
√
5)
/
2)
≈
0
.
694.
The base cases
n
= 0
,
1 are easy to check. As before, assume the statement is true upto some
n >
0; we’ll show it for
n
+ 1:
F
n
+1
=
F
n
+
F
n
−
1
≤
2
cn
+ 2
c
(
n
−
1)
=
2
c
(
n
+1)
·
(2
−
c
+ 2
−
2
c
)
,
and the particular value of
c
chosen satisFes 2
−
c
+2
−
2
c
= 1. (We found
c
by solving this quadratic
equation.)
(c) Combined with (b).
3.
Exercise 3.5.
function reverse
(
G
)
Input:
directed graph
G
= (
V,E
)
in adjacency list format
Output:
the reverse of
G
, also in adjacency list format
create
G
R
= (
V,E
R
)
, with edgeset
E
R
initially empty
for each
u
∈
V
:
for each
(
u,w
)
∈
E
:
add edge
(
w,u
)
to
E
R
return
G
R
±or each edge (
u,w
)
∈
E
, adding (
w,u
) to
E
R
takes
O
(1) time: the edge is simply added to the front
of the adjacency list for node
w
. Thus the total running time is
O
(

V
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 Paturi
 Algorithms, Graph Theory, Vodd, Veven, valid coloring

Click to edit the document details