MIT 算法导论公开课笔记——分而治之——寻峰器（peak finder）
Efficient programmer
Scalability可伸缩性
Classic data structures
Modules 模块
Trees 树sorting排序一万亿个数字
复杂性
RSA加密 SLL
Graphs 图最基本的数据结构：魔方任务
使用特定算法编码
最短路径问题
Advanced topics
一个具体问题:
Peak finder:一维
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1 2 3 4 5 6 7 8 9
Peak:高峰——position 2 is a peak if and only if b>=a and b>=c;position 9 is a peak iff i>=h;
Problem:find a peak if it is exists;
Straightforward algorithm 简单的算法
Start from left：
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1 2 3 4 5 n/2 n-1 n
Look at n/2 elements
Worst case complexity theta n
一维寻峰器，如何降低渐进复杂度？
二分法：分而治之 把一个一维数组分成几个小数组
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1 2 ……… n/2-1 n/2 n/2+1 …… n
分而治之—Divide & Conquer
Look at n/2 position
If a[n/2] <a[n/2-1],then only look at left half;1………n/2-1 to look for a peak;
Else if a[n/2]<a[n/2+1],then only look at right half;n/2+1………n to look for a peak;
Else n/2 position is a peak;
(Argue that algorithm is correct)
T(n)=T(n/2)+theta(1)
Base case :T(1)=theta(1)
T(n)=theta(1)+………+theta(1)=theta(log2n) 与上面的递归算法相对应的复杂度的递归定义
Log2n times
2D version
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n行rows m列colomns
a is a 2D-peak iff a >=b,a>=d,a>=c,a>=e;
贪婪的上升算法：Greedy Ascent algorithm:本质上是pick a direction
Theat(nm)complexity
Theta(n^2)if m=n;
Binary search:
Pick middle coloumn j=m/2;
Find a 1D-peak at (i,j);
Use (I,j) as a start to find a 1D-peak on row i ;
Efficient but incorrect
Problem :2D-peak may not exist on row i ;
AHempt#2
Pick middle column j=m/2;
Find global max on column j at (i,j)
Compare (i,j-1),(i,j),(i,j+1)
Pick left cols if (i,j-1)>(i,j) similarly for the right ;
If(i,j)>=(i,j-1),(i,j+1)=>(i,j) is a 2D-peak
Solve the new problem with half the number of cols
When you have a single col,find the global max <—done
T(n,m)=T(n,m/2) +theta(n)
分解 找到global max
T(n,1)=theta(n);
T(n,m)= theta(n)+………+theta(n)
Log2m times//theta(n)相加log2m次；