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Section 8.3
231
Since 4
=
n
#
1 and
n
2, we set
f
=
L~J
=
1 and m
=
L~J
=
2. Since 30
=
x
>
am,
we
replace
n
by
n
 m
=
2 and change the list to 30,31.
Since
n
=
2,
x
a1
and
x
=
a2,
we output "true" and stop.
This search required four comparisons of
x
with an element in the list; a linear search would have
used ten comparisons.
(c) Since
n
=
11
1 and
n
2, the algorithm sets
f
=
L
1i
J
=
3 and m
=
L
2;
J
=
7. Since
7
=
a3
<
x
=
16
~
a7
=
17, the algorithm replaces
n
by m 
f
=
4 and changes the list to
10, 12, 15, 17.
Since 4
=
n
1 and
n
2, the algorithm sets
f
=
L~J
=
1 and m
=
L~
=
2. Since
x
=
16
>
a2
=
12, the algorithm replaces
n
with
n
 m
=
2 and changes the list to 15, 17.
Since
n
=
2 and
x
and
x
the algorithm outputs "false" and stops.
This search required five comparisons of
x
with other elements; a linear search would have re
quired eleven.
6. (a) [BB] To find the complement of the set
A
=
{a1, a2,
... ,
ad
withrespectto
U
= {I,
... , 100},
Step 1:
For i
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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