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236
Solutions to Exercises
19. As mentioned in the text and illustrated in Problem 26, the maximum number of comparisons occurs
when just one element remains in one list at the time the other is emptied. Thus an example exists for
any s and
t.
Let /:.;2: bl :::;
b2
:::;
... :::;
bt
be any ordered list of length
t.
Then choose /:.;1:
al
:::;
a2
:::;
... :::;
as
with
asl
<
bt
<
as.
This ensures that
as
is the only remaining element at the time that
/:.;2
is emptied.
20. The most efficient sorting algorithm is
0
(n
log
n).
Thus sorting and then applying a binary search is
o
(
n
log
n
+
log
n)
=
0
(
n
log
n).
Since
n
::::5
n
log
n,
this is less efficient than a linear search.
21. [BB] The answer is min { s,
t}.
It
is impossible to have fewer than this number of comparisons since
until min { s,
t}
of comparisons have been made, elements remain in each list. To see that min { s,
t}
can be achieved, consider ordered lists
aI, a2,
... , as
and bl,
b2,
... , b
t
where s
<
t
and
as
<
bl.
After s comparisons, the first list is empty.
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 Summer '10
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 Graph Theory

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