Homework 3 參考解答 1. First, let’s clarify what the function max( ( ), ( )) is. Let’s define the function ℎ( ) = max( ( ), ( )). Then ℎ( ) =
( ) if ( ) ≥ ( ), ( ) if ( ) < ( ).
Since ( ) and ( that ( ) ≥ 0 and Thus for ≥ , ( (
) are asymptotically nonnegative, there exists ( ) ≥ 0 for all ≥ . ) + ( ) ≥ ( ) ≥ 0 and ) + ( ) ≥ ( ) ≥ 0. Since for any particular , ℎ( ) is either ( ) or ( ), we have ( ) + ( ) ≥ ℎ( ) ≥ 0, which shows that ℎ( ) = max( ( ), ( )) ≤ ( ( ) + ( )) for all (with
such
≥
= 1 in the definition of Θ).
Similarly, since for any particular , ℎ( ) is the larger of ( ) and have for all ≥ , 0 ≤ ( ) ≤ ℎ( ) and 0 ≤ ( ) ≤ ℎ( ). Adding these two inequalities yields 0 ≤ ( ) + ( ) ≤ 2ℎ( ), or
( ), we
equivalently 0 ≤ ( ( ) + ( )) ≤ ℎ( ), which shows that ℎ( ) = max( ( ), ( )) ≥ (with
=
in the definition of Θ).
( ( ) + ( )) for all
≥
2. To show that ( + ) = Θ(
), we want to find constants
that 0 ≤ ≤( + ) ≤ for all Note that + ≤ +| | ≤2 when | | ≤ , and + ≥ −| | when | | ≤
≥ Thus, when
≥
,
,
> 0 such
.
.
≥ 2| |,
0≤
≤
Since
> 0, the inequality still holds when all parts are raised to the power
0≤(
) ≤ ( + ) ≤ (2 ) ,
0≤( ) Thus,
+
≤2 .
≤( + ) ≤2 =( ) ,
=2 ,
. = 2| | satisfy the definition.
:
3. (a) ( )=
−2 +1≤
exists one solution: Thus,
( )= (
, for all
>
,
>
,
= .
= 1, ).
(b) ( )=
−2 +1≥
= ,
exists one solution: Thus,
( ) = Ω(
, for all = 2.
).
(c) Let the running time be ( ). ( ) ≥ ( ) means that ( ) ≥ ( ) for some function ( ) in the set ( ). This statement holds for any running time ( ), since the function ( ) = 0 for all is in ( ), and running times are always nonnegative. Thus, the statement tells us nothing about the running time.
