MIT UNIVERSITY open course of introduction about algorithm divide and conquer (the summary of second and third lesson)

   this is the summary that I watch from tudou , It's deeply impressed me.I get a lot from it.

here is my summary share with you.the two lesson talked about two lessons with <divide and conquer algorithm>. this impressed me alot.

   the second lesson talk about  :

first   how to calculate the steps with divide and conquer algorithm   there are three methods 1 sabstitation 2recursion-tree-method 3 master method:

second  the meaning of each mathmatic signs.

   the third lesson talk about:

 several examples on divide and conquer algorithm. 1 powering a number .2 binary search 3.fibonacci  4 matrix multiplication (strassen algorithm)

 

second lesson:

  talk about the three sign   θ Ω Ο      

 θ          in this video , the professor talked about the sign  θ is this function. drop leading low order terms,lead constant  and ignore leading constant.and we usually use    θ() to describe a problem .like we usually describe the merge sort with θ(nlgn).  

for example .the steps is 3n³   + n²  +3. and the   θ  (n³ )   {drop leading low order terms,lead constant  and ignore leading constant}

Ω we use f(n)=Ω(g(n))  this is the above bound .  and  θ is talk about the below bound.

0<=c g(n)<=f(n)   c is constant.

0<f(n)<=c g(n)       ------------------------ review this things. to know better.

1 sabstitation    1 guess from exaclly

                        2 figure the constant 

                        3 solve the constant (verfiy the recurrence)

for example T(n)=4T(n/2)+n

guess:Tn=O(n³)

assume : T_k =ck³ 

T(n)=4c(n/2)³+n=1/2cn³ +n=cn³-(1/2cn³-n)                to prove it no bigger than cn³

{(1/2cn³-n)}is the residuals.

to prove it bigger than zero. only c>=2;

 

guess:Tn=O(n²)

assume : T_k =ck² is not right.

let us to guess

guess:Tn=O(n²)

assume : T_k =c1k² -c2k  is not right.

 T(n)=4T(n/2)+n

       =4(c1(n/2)²-c2*n/2)+n

       =c1n²-c2n-(-1+c2)n          c2>=1;        It is said that c1 is much larger than c2. from basic examples.

2 recursion tree :  this is nonrigorous  and we use is a sloppy way.

let's to solve a more complicated recursion.

                              draw a recursion tree

                                          n²                                                              sum up   n²

                             (n/4)²                       (n/2)²                                    sum  up  5/16n²

                 (n/16)²       (n/8)²     (n/8) ²      (n/4)²                           sum up   25/256n²

  sum up each lines time cost.

this is geometric numers.      sum them up.       you can use this as an example    1+1/2+1/4+1/8+----=2

                      this geometric ultimiate answer is 2.

so the sum is less than 2n².    and the   answer is θ (n²).

here is the key method .

3 master method :

this method is only apply to particular family of recursion.

T(n)=aT(n/b)+f(n)               f(n)is non-recursion work.

case 1:          f(n)=Θ(n ^lga/lgb-ε)             ε>0

          Tn= θ (n^lga/lgb)                                           small

case 2:         f(n)=Θ(n^lga/lgb*lg^kn)

                  T(n)=Θ(n^lga/lgb*lg^k+1n)                        equal

case 3:  f(n)=Ω ( n ^lga/lgb+ε)                               large

              af(n/b)<=(1-ε')f(n)

              Tn=Θ(f(n))

 T(n)=4T(n/2)+n                      n^lga/lgb=n²          suit to case one the answer is Θ(n²⁾

 T(n)=4T(n/2)+n²                     n²   =  n²       the answer is Θ(n²logn);

 T(n)=4T(n/2)+n³                                           the answer is Θ(n³);

 

 

the third lesson is divide and conquer

1 powering a numer is easy       the common method is Θ(n);

                                                 the improved method is  Θ(lg n);

2 binary search        is also easy          the answer is Θ(lg n);

3 fibonacci  can be used by recursion ,but it cost a lot of time and space to finish it . it 's Exponential time.

and we can use matrix multiplication     to calculate it.

we use [1,1;1,0];to calculate it.

it reprsent [fn-1,fn-2;fn-2,fn-3].

 [fn,fn-1;fn-1,fn-2]= [fn-1,fn-2;fn-2,fn-3]*[1,1;1,0];

this is the right answer.          Θ(lg n)

remember it can not change the position of two matrix 

              like a*b!=b*a

but to this problem it is right ,because a=b.this is an exception.                                

4 the matrix multiplication (strassen algorithm)

pseudocode               i 1~n

                                 j 1~n

                              cij=0

                               k 1~n

                        cij+=aik*bkj

this is  Θ(n³)

and we can use strassen algorithm.

is     Θ(n^lg7);

 you need to complete the matrix as the power of 2. is there are not a  matrix as the power of 2,add zero in it.

and use this steps.           

 

1969年斯特拉森利用分治策略并加上一些处理技巧设计出一种矩阵乘法。

 

　　为了得到2矩阵相乘的分治算法1、定义一个小问题，并指明那个问题如何进行乘法运算，2、确定一个大问题，进行划分，如何指明对这些小问题进行乘法运行。

 

　　设A和B是俩个n x n的矩阵，其中n可以写成2的幂。将A和B分别等分成4个小矩阵，此时如果把A和B都当成2x2矩阵来看,每个元素就是一个（n/2）x（n/2）矩阵，而A和B的成积就可以写成〔A11 A12〕[B11 B12] [C11 C12]

 

　　X =

 

　　[A21 A22] [B21 B22 ] [C21 C22]

 

　　其中 利用斯特拉森方法得到7个小矩阵，分别定义为：

 

　　m1=(A12=A22)(B21+B22)

 

　　m2=(A11+A22)(B11+B22)

 

　　m3=(A11-A21)(B11+B12)

 

　　M4=(A11+A12)B22

 

　　M5=A11(B12-B22)

 

　　M6=A22(B21-B11)

 

　　M7=(A21+A22)B11

 

　　矩阵m1~m7可以通过7次矩阵乘法、6次矩阵加法和4次矩阵减法计算得出，前述4个小矩阵可以由矩阵m1~m7，通过6次矩阵加法和2次矩阵减法得出，方法如下：

 

　　c11=m6+m1+m2-m4

 

　　c12=m5+m4

 

　　c21=m6+m7

 

　　c22=m5-m3+m2-m1

 

　　用上述方案解n=2;矩阵乘法；假定施特拉斯矩阵分割方案仅用于n>=8的矩阵乘法，而对于小于8的矩阵直接利用公式计算；n的值越大，斯拉特森方法更方便；设T(n)表示斯特拉森分治运算法所需时间，因为大的矩阵会被递归分成小矩阵直到每个矩阵的大小小于或等于k，所以T(n)的递归表达式为T(n)=d(n<=k);T(n)=7*t(n/2)+cn2(n平方）（n>k),其中cn2表示完成18次（n/2)(n/2)接矩阵的加减法，以及把大小为N的矩阵分割成小矩阵所需的时间；