A General Method of Solving Divide-and-Conquer
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A General Method of Solving Divide-and-Conquer 計計計計計計 Chapter05 計計計計 計計計計計
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A General Method of Solving Divide-and-Conquer. 計算方法設計 Chapter05 重要補充 唐傳義 老師. given. rewrite it into template. given. ,. Deduction-1. (1). (1). Example1-(1). Example 1 ： Solution ：. iteration and cancellation. 又. log n terms. ,. Deduction-2. 而. Deduction-3. - PowerPoint PPT Presentation
Transcript of A General Method of Solving Divide-and-Conquer
A General Method of Solving Divide-and-Conquer
計算方法設計Chapter05 重要補充
唐傳義老師
2
Deduction-1
)()2
()(
)1(
nfn
kTnT
T
)()2
(2)(
)1(
ngnn
TnT
T
pp
kp log pn
nfng
)()(
given
given
rewrite it into template
,
3
Example1-(1)
Example 1 ：
Solution ：3.0)
2(2)( nn
TnT
7.01
3.0
7.011
)(
1
)2
(2)(
nn
nng
p
nnn
TnT
(1)
3.0)2
(2)( nn
TnT
7.01
3.0
7.011
)(
1
)2
(2)(
nn
nng
p
nnn
TnT
(1)
3.0)2
(2)( nn
TnT
4
Deduction-2
而
)()2
(2)( ngnn
TnT pp
)()2
()]4
(2[2 ngnn
gnn
T pppp
)()]2
()2
()4
(2[2 ngnn
gnn
T pppp iteration and cancellation
)()2
()4
()8
(222 ngnn
gnn
gnn
T pppppp
)]2(...)4
()2
()([)1(2 log gn
gn
gngnT pnp 又 pnp nlog2
)]1()2(...)4
()2
()([ Tgn
gn
gngn p
log n terms
)](~)1([ ngTn p )2()(~log1
i
nigng
,
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Deduction-3
Ο ： upper bound ex: Ω ： lower bound ex: Θ ： exactly ex: ο ： exactly,even constant ex:
)(~ ng不同的 g(n) 對應不同的
)(ng )(~ ng
0),( qnq )1(
0,log jnj
0),( qnq ))(( ng
)(log1
log 1 nj
n jj
)(3 2nn )1(3 n)(3 nn )3(3 nn
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Example1-(2)
Example 1 ：
Solution ： From (1)
7.0)( nng
07.0 g
)1()(~ ng
nngTnnT p )](~)1([)(
1p
3.0)2
(2)( nn
TnT
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Example2
Example 2 ：
Solution ：
)()2
(7)( 2nn
TnT
72 p
7log p
7log2)( nng 07log2 ,
)]1()1([)( 7log TnnT
)( 7logn
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General Representation
In general
)()()(
)1(
nfc
nkTnT
T given
)()()( ngnc
nTcnT pp kp clog pn
nfng
)()( , ,
)]()1([)(log1
i
ni
p cgTnnTc
仍然是
只差 的constant
If g(n) is monotone
)(~ ng
clog
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Example3-- Pan’s Matrix
Example 3 ：
Solution ：
)()70
(*143640)( 2nn
TnT
)]1()1([)( TnnT p
)( pn
14364070 p
79.2143640log70 p
79.22)( nng
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More General Representation
More general
)())(()()(
)( 0
nfnbTnanT
nT given
,)()())(()()( ngnhnbTnanT ))(()()( nbhnanh
])(
)())((...)))((())(()([)()(
0
00
1
nh
nTnbgnbbgnbgngnhnT
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Example4
Example 4 ：
Solution ：
])2(
)2(4log...)log()log([log)( 4
121
h
TnnnnnT
nnnTnnT
T
log)()(
2)2(
21
21 時只考慮
in 22
2
1
)( nna 21
)( nnb ,
)()( 2
1
2
1
nhnnh
nnh )(
nnnTnnT log)()( 21
21
)1log2( nn
21
)( nnb 21 )( nnb
4)2(1 b
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Reference
Bentley, Haken & Saxe “A General Method of Solving
Divide-and-Conquer Recurrences”, SIGACT News Fall
1980,page36-44
