4.Divide-and-Conquer

4.Divide-and-Conquer

Hsu, Lih-Hsing

Chapter 4 P.2

Computer Theory Lab.

Instruction Divide Conquer Combine

Chapter 4 P.3


Recurrences -- if n=1 ,

if n ＞ 1. Substitution method Recursion-tree method Master method

T(n)=aT(n /b)+f (n), where a≧1, b ＞ 1, and f (n) is a given function.

T n aT n b f n( ) ( / ) ( )

)()2/(2

)1()(

nnTnT

Chapter 4 P.4


Technicalities in recurrences Boundary conditions represent anoth

er class of details that we typically ignore.

When we state and solve recurrences, we often omit floors, ceilings, and boundary conditions.

Chapter 4 P.5


4.1 The maximum-subarray problem

Chapter 4 P.6


Chapter 4 P.7


A brute-force solution Take time.Can we do better?

A transformation Maximum subarray

)( 2n

Chapter 4 P.8


Chapter 4 P.9


FIND-MAX-CROSSING-SUBARRAY(A,low,mid,high)

1 left-sun = –∞2 sum = 03 for I = mid downto low4 sum = sum ＋ A[i]5 if sum ＞ left-sum6 left-sum = sum7 max-left = I8 right-sun = –∞

Chapter 4 P.10


9 sum = 010 for j = mid ＋ 1 to high11 sum = sum ＋ A[j]12 if sum ＞ right-sum13 right-sum = sum14 max-right = j15 return(max-left,max-right,left-sum

＋ right-sum)

Chapter 4 P.11


FIND-MAXIMUM-SUBARRAY(A,low,high)1 if high == low //base case:only one elem

ent2 return(low,high,A[low])3 else mid =4 (left-low,left-high,left-sum) = FIND-MAXI

MUM-SUBBARY(A,low,mid)5 (right-low,right-high,right-sum)= FIND-

MAXIMUM-SUBBARY (A,mid+1,high)

2/highlow

Chapter 4 P.12


6 (cross-low,cross-high,cross-sum) = FIND-MAX-CROSSING-SUBARRAY (A,low,mid,high)

7 if left-sum≧right-sum and left-sum ≧ cross-sum

8 return(left-low,left-high,left-sum)9 elseif right-sum≧left-sum and right-sum

≧ cross-sum10 return(right-low,right-high,right-sum)11 else return(cross-low,cross-high,cross-su

m)

Chapter 4 P.13


Analyzing the divide-and conquer algorithm

if n=1 if n ＞ 1

)()2/(2

)1()(

nnTnT

Chapter 4 P.14


4.2 Strassen`s algorithm for matrix multiplication If A = (aij) and B = (bij) are square n×n

matrices. C = A‧B Cij =

n

kkjik ba

1

Chapter 4 P.15


SQUARE-MATRIX-MULTIPLY(A, B)1 n = A.rows2 let C be a new n×n matrix3 for i = 1 to n4 for j = 1 to n5 cij=06 for k = 1 to n7 cij = cij + aik‧bkj

8 return C

Chapter 4 P.16


A simple divide-and-conquer algorithm , ,

＝ ‧ C11 = A11‧B11 ＋ A12‧B21 ，

C12 = A11‧B12 ＋ A12‧B22 ，C21 = A21‧B11 ＋ A22‧B21 ，C22 = A21‧B12 ＋ A22‧B22 。

2221

1211

AA

AA

2221

1211

BB

BBB

2221

1211

AA

AAA

2221

1211

CC

CCC

2221

1211

CC

CC

2221

1211

BB

BB

Chapter 4 P.17


SQUARE-MATRIX-MULTIPLY-RECURSIVE(A, B)1 n=A.rows2 let C be a new n×n matrix3 if n == 14 c11=a11‧b115 else partition A, B, and C as in equations6 c11=SQUARE-MATRIX-MULTIPLY-RECURSIVE

(A11, b11) + SQUARE-MATRIX-MULTIPLY- RECURSIVE (A12, B21)

7 c12=SQUARE-MATRIX-MULTIPLY-RECURSIVE (A11, b12) + SQUARE-MATRIX-MULTIP

LY- RECURSIVE (A12, B22)

Chapter 4 P.18


8 c21=SQUARE-MATRIX-MULTIPLY-RECURSIVE (A21, b11) + SQUARE-MATRIX-M

ULTIPLY- RECURSIVE (A22, B21)9 c22=SQUARE-MATRIX-MULTIPLY-RECURSIVE

(A21, b12) + SQUARE-MATRIX-MULTIPLY- RECURSIVE (A22, B22)

10 return C

if n = 1, if n ＞ 1.

)()2/(8

)1()( 2nnTnT

Chapter 4 P.19


Strassen`s method(1/2)1 Divide the input matrices A and B and out

put matrix C into n/2×n/2 submatrices. This step takes time by index calculation.

2 Creat 10 matrices S1, S2,…,S10, each of which is n/2×n/2 and is the sum or difference of two matrices created step 1. We can creat all 10 matrices in time.

)1(

)( 2n

Chapter 4 P.20


Strassen`s method(2/2)3 Using the submatrices created in step 1 a

nd the 10 matrices created in step 2, recursively compute seven matrix products P1,P2,…,P7. Each matrix Pi is n/2 × n/2.

4 Compute the desired submatrices C11,C12,C21,C22 of the result matrix C by adding and subtracting various combinations of the Pi matrices. We can compute all four submatrices in time.

)( 2n

Chapter 4 P.21


Strassen`s algorithm(1/7) if n = 1, if n ＞ 1.

)()2/(7

)1()( 2nnTnT

Chapter 4 P.22


Strassen`s algorithm(2/7) S1 = B12 － B22 S2 = A11 ＋ A12 S3 = A21 ＋ A22 S4 = B21 － B11 S5 = A11 ＋ A22 S6 = B11 ＋ B22 S7 = A12 － A22 S8 = B21 ＋ B22 S9 = A11 － A21 S10 = B11 ＋ B12

Chapter 4 P.23


Strassen`s algorithm(3/7) P1 = A11 ． S1 = A11 ． B12 － A11 ． B22 ，

P2 = S2 ． B22 = A11 ． B22 － A12 ． B22 ，P3 = S3 ． B11 = A21 ． B11 － A22 ． B11 ，P4 = A22 ． S4 = A22 ． B21 － A22 ． B11 ，P5=S5 ． S6=A11 ． B11 ＋ A11 ． B22 ＋ A22 ． B11 ＋

A22 ． B22

P6=S7 ． S8=A12 ． B21 ＋ A12 ． B22 － A22 ． B21 －A22 ． B22

P7=S9 ． S10=A11 ． B11 ＋ A11 ． B12 － A21 ． B11 －A21 ． B12

Chapter 4 P.24


Strassen`s algorithm(4/7) We start with

C11 = P5 ＋ P4 － P2 ＋ P6

A11 ． B11 ＋ A11 ． B22 ＋ A22 ． B11 ＋ A22 ． B22

－ A22 ． B11 ＋ A22 ． B21

－ A11 ． B22 － A12 ． B22

－ A22 ． B22 － A22 ． B21 ＋ A12 ． B22 ＋ A12 ． B21

___________________________________________________________________________________________

A11 ． B11 ＋ A12 ． B21

Chapter 4 P.25


Strassen`s algorithm(5/7) Similarly, we set

C12 = P1 ＋ P2

A11 ． B12 － A11 ． B22 ＋ A11 ． B22 ＋ A12 ． B22

A11 ． B12 ＋ A12 ． B22

Chapter 4 P.26


Strassen`s algorithm(6/7) Setting

C21 = P3 ＋ P4

A21 ． B11 ＋ A22 ． B11

－ A22 ． B11 ＋ A22 ． B21

A21 ． B11 ＋ A22 ． B21

Chapter 4 P.27


Strassen`s algorithm(7/7) Finally, we set

C22 = P5 ＋ P1 － P3 － P73

A11 ． B11 ＋ A11 ． B22 ＋ A22 ． B11 ＋ A22 ． B22

－ A11 ． B22 ＋ A11 ． B12

－ A22 ． B11 － A21 ． B11

－ A11 ． B11 － A11 ． B12 ＋ A21 ． B11 ＋ A21 ．B12

A22 ． B22 ＋ A21 ． B12

Chapter 4 P.28


4.3 The substitution method : Mathematical induction The substitution method for

solving recurrence entails two steps:

1. Guess the form of the solution. 2. Use mathematical induction to find

the constants and show that the solution works.

Chapter 4 P.29


Example

(We may omit the initial condition later.)

Guess Assume

T n T n n

T

( ) ( / )

( )

2 2

1 1

)log()( nnOnT

T n c n n( / ) / log /2 2 2

Chapter 4 P.30


Initial condition However,

T n c n n n cnn

n

cn n cn n cn n c .

( ) / log / ) log

log log log )

2( 2 22

2 1 (if

)(01log)1(1 cnT

4 2 2 4 T cn c( ) log ( )if

Chapter 4 P.31


Making a good guess

We guess

Making guess provides loose upper bound and lower bound. Then improve the gap.

T n T n n( ) ( / ) 2 2 17

T n O n n( ) ( log )

Chapter 4 P.32


Subtleties

Guess Assume

However, assume

T n T n T n( ) ( / ) ( / ) 2 2 1

T n O n( ) ( )

T n cn( )

cncnncncnT 112/2/)(

dcnnT )(

)1 Choose(12

1)2/()2/()(

ddcndcn

dncdncnT

Chapter 4 P.33


Exercise 4.3-6

Show that the solution to T(n) = 2T(n2 + 17) + n is O(n lg n)

Solution:

assume a > 0, b > 0, c > 0 and T(n) ≦ an lg n – blg n - c

T(n) ≦ 2[(n2 + 17)lg(

n2 +17) - blg(

n2 + 17)-c]+n

≦ (an + 34a)lg(n2 +17) - 2blg(

n2 +17) - 2c + n

≦ anlg(n2 +17) + anlg21/a + (34a-2b)lg(

n2 +17) - 2c

≦ anlg(n) 21/a+ (34a-2b)lg(n) - 2c

Chapter 4 P.34


anlg(n) 21/a+ (34a-2b)lg(n) - 2c

n≧ n2 +17，n≧ 34

n≧ (n2 +17) 21/a ∵， 21/2≦ 1.5∴ n ≧ 12

34a-2b≦ -b，b ≧ 34a

c > 0，-c > -2c

T(n) ≦ anlgn - blgn - c，T(n) ≦ anlgn

T(n) = O(nlgn)

Chapter 4 P.35


Avoiding pitfalls

Assume Hence

(WRONG!) You cannot find such a c.

T n T n n

T

( ) ( / )

( )

2 2

1 1

T n O n( ) ( )

T n cn( )

constant) a is Since(

)()2/(2)(

c

nOncnnncnT

Chapter 4 P.36


Changing variables nnTnT lg)(2)(

Let m nlg .

T T mm m( ) ( )/2 2 2 2

Then S m S m m( ) ( / ) 2 2 .

S m O m m

T n T S m O m m

O n n

m

( ) ( lg )

( ) ( ) ( ) ( lg )

(lg lg lg )

2

Chapter 4 P.37


4.4 the Recursion-tree method

)()4/(3)( 2nnTnT

Chapter 4 P.38


Chapter 4 P.39


).(1)16/3(

1)16/3(

16

3

)(16

3...

16

3

16

3)(

3log2log

1log

0

3log2

3log21log

22

22

4

4

4

4

4

4

ncn

ncn

ncncncncnnT

n

n

i

i

n

The cost of the entire tree

Chapter 4 P.40


)(

)(13

16

)16/3(1

1

16

3

16

3)(

2

3log2

3log2

3log2

0

1log

0

3log2

4

4

4

4

4

nO

ncn

ncn

ncn

ncnnT

i

i

n

i

i

Chapter 4 P.41


substitution methodWe want to Show that T(n) ≤ dn2 for some

constant d > 0. using the same constant c > 0 as before, we have

Where the last step holds as long as d (16/13)c.

,

16

3

)4/(3

4/3

)4/(3)(

2

22

22

22

2

dn　　　

cndn　　　

cnnd　　　

cnnd　　　

cnnT　nT

Chapter 4 P.42


cnnTnTnT )3/2()3/()(

Chapter 4 P.43


substitution method

,lg

)3/23(lglg

)2lg)3/2(3lg)3/2(3lg)3/((lg

))2/3lg()3/2(3lg)3/((lg

))2/3lg()3/2(lg)3/2(()3lg)3/(lg)3/((

)3/2lg()3/2()3/lg()3/(

)3/2()3/()(

ndn　　　

cndnndn　　　

cnnnndndn　　　

cnnndndn　　　

cnndnndndnnd　　　

cnnndnnd　　　

cnnTnTnT

as long as d c/(lg3 – (2/3)).

Chapter 4 P.44


4.5 The master method Theorem 4.1 (((MMMaaasssttteeerrr ttthhheeeooorrreeemmm)))

Let a 1 and b 1 be constants, let f n( ) be a function, and T n( )

be defined on the nonnegative integers by the recurrence T n aT n b f n( ) ( / ) ( )

where we interpret n b/ mean either n b/ or n b/ .

1. If f n O n b a( ) ( )log for some constant 0 , then T n n b a( ) ( )log .

2. If )()( log abnnf , then T n n nb a( ) ( log )log .

3. If )()( log abnnf for some constant 0 , and if )()/( ncfbnaf for

some constant c 1 and all sufficiently large　n, then T n f n( ) ( ( )) .

Proof. (In section 4.6 by recursive tree)

Chapter 4 P.45


T n T n n( ) ( / ) 9 3

a b f n n

n n f n O n

9 3

3 39 2 9 1

, , ( )

, ( ) ( )log log

Case T n n1 2 ( ) ( )

T n T n( ) ( / ) 2 3 1

a b f n

n n f n

1, 3 2 1

13 21 0

/ , ( )

( ),log /

Case T n n2 ( ) (log )

Chapter 4 P.46


T n T n n n( ) ( / ) log 3 4

)()(),(

log)(,4,33log793.03log 44

nnfnOn

nnnfba

Case

af(n/b)= (n) (

n)

nn=cf(n)

c= n

T n n n

3

34 4

34

34

Check

for and sufficiently large

log log

,

( ) ( log )

Chapter 4 P.47


Note that the three cases do not cover all the possibilities for f(n). There is a gap between cases 1 and 2 when f(n) is smaller than but not polynomially smaller. Similarly, there is a gap between cases 2 and 3 when f(n) is larger than but not polynomially larger.

abnlog

abnlog

Chapter 4 P.48


The master method does not apply to the recurrence

even though it has the proper form: a = 2, b=2, f(n)= n lgn, and It might seem that case 3 should apply, since f(n)= n lgn is asymptotically larger than

The problem is that it is not polynomially larger.

,lg)2/(2)( nnnTnT

.log nn ab

.log nn ab

Chapter 4 P.49


Exercise 4.3-9Solve the recurrence T(n)=3T( ) ＋ logn by making a change of variables. Your solution should be asymptotically tight. Do not worry about whether values are integral.

n

Chapter 4 P.50


Exercise 4.5-1Use the master method to give tight asympttotic bounds for the following recurrences.a. T(n) = 2T(n/4)+1.b. T(n) = 2T(n/4)+ .c. T(n) = 2T(n/4)+n.d. T(n) = 2T(n/4)+n2.

n

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