2.1 Insertion sort
Pseudocode is used to specify a given algorithm in English. Issues of data abstraction,
modularity, and error handling are often ignored in order to convey the essence of the
algorithm more concisely.
伪代码与真实代码的区别只是伪代码更加清晰、简洁。最清晰的表达方式就是直接用英语。
为了简洁地表达算法的核心内容，数据抽象、模块化和异常处理等问题常常被忽略。
We use loop invariants to help us understand why an algorithm is correct.
Initialization: It is true prior to the first iteration of the loop.
Maintenance: If it is true before an iteration of the loop, it remains true before the next iteration.
Termination: When the loop terminates, the invariant shows that the algorithm is correct.
The first two properties are similar to mathematical induction which you prove a base case and an
inductive step. The difference is we apply the inductive step infinitely but here we stop when the
loop terminates.
我们用循环不变量来证明算法的正确性。前两项与数学归纳法相同，不一样的也是最重要的是第三项。
数学归纳法中，归纳步骤是无穷的。但在这，当循环停止时“归纳”就停止了。
Pseudocode conventions
Indentation indicates block structure instead of conventional indicators such as begin and end statement.
The constructs while, for and repeat-until and if-else have interpretations similar to those in C/Java/Python.
The symbol "//" indicates that the remainder of the line is a comment.
We pass parameters to a procedure by value.
The boolean operators "and" and "or" are short circuiting.
用缩进来表示代码结构，而不是begin/end语句，从而保证伪代码的简洁。
while, for, repeat-until和if-else语句与C, C++, Java, Python, Pascal中的语句类似。
// 后是注释。
函数参数按值传递。
and和or运算是短路的。
2.2 Analyzing algorithms
We shall assume a generic one-processor, random-access machine(RAM) model of computation as
our implementation technology. The RAM model contains instructions(arithmetic, control) and data type
(integer,float) commonly found in real computers. Each such instruction takes a constant amount of time.
What if a RAM had an instruction that sorts? What if we could store huge amounts of data in one word and
operate on it all in constant time? These are unrealistic scenario.
我们需要假定一种随机访问模型RAM来帮助我们分析算法的性能。RAM与现实计算机拥有相似的指令和数据
类型。每条指令都耗费等量的时间。像排序指令和任意大字长都是不现实的，破坏了RAM模型。
There are some instructions in real computers but not listed above. Is exponentiation a constant-time instruction?
In general case, no: it takes several instructions to compute but many computers have a shift left instruction, which
in constant time shifts the bits of an integer by k positions to the left. We will avoid such gray areas in the RAM model.
And we do not attempt to model the memory hierarchy. Models that include the memory hierarchy are more complex
than the RAM model. Moreover, RAM-model analyses are usually excellent predictors of performance.
像求k次方问题，在有的计算机中需要执行几条指令，在有的计算机中则有左移运算符。我们要避开这种灰色区域。
我们不去尝试将内存结构模型化。包含内存结构的模型比RAM复杂得多，通常基于RAM模型的分析就能出色的预测性能。
The mathematical tools required may include combinatorics, probability theory, algebraic dexterity, and ability to
identify the most significant terms in a formula.
所需数学工具：组合数学、概率论、代数、方程式。
Analysis of insertion sort
We need to define the terms "running time" and "size of input" more carefully.
The best notion for input size depends on the problem being studied.
For many problem such as sorting, the most natural measure is the number of items in the input (the array size).
For many other problems, such as multiplying two integers, the best measure is the total number of bits.
If the input to an algorithm is a graph, it is more appropriate to describe with two integers - vertices and edges.
The running time of an algorithm is the number of primitive operations or "steps" executed.
输入规模可以是数组长度，整数的二进制位数，图形的左边等。运行时间是执行了多少步或多少条基本操作。
设n = A.length，循环头要比循环体多执行一次检测，用tj表示内层while循环的循环次数。
最好情况是数组已经排好序了，那么内层while循环不会执行，只是进行n-1次检测，即tj=1。
所以运行时间可以表示为an+b，即n的线性函数(linear function)。
最坏情况是数组是倒序的，则tj=j。运行时间表示为an²+bn+c，是n的二次函数(quadratic function)。
We shall usually concentrate on finding only the worst-case running time. We give three reasons:
1.The worst-case running time of an algorithm gives us an upper bound on the running time for any input.
2.For some algorithms, the worst case occurs fairly often. For example, in search a database for a
particular piece of information, the worst case will often occur when the information is not in the database.
3.The "average case" is often roughly as bad as the worst case. How long does it take to determine where
in subarray A[1..j-1] to insert element A[j]? On average, half the elements in A[1..j-1] are less than A[j], so
tj is about j/2. The average-case running time turns out to be a quadratic function too.
我们通常仅仅关注最坏情况下的运行时间，原因有三：
1.最坏情况运行时间是任何输入执行时间的上界；
2.对某些算法，最坏情况经常发生。如查找数据库中一段信息经常会是找不到的信息。
3.平均情况通常跟最坏情况一样糟糕。插入排序的平均情况执行时间仍然是二次函数。
Order of growth
One more simplifying abstraction: consider only the leading term of a formula (e.g., an²).
Due to constant factors and lower-order terms, an algorithm whose running time has a higher order
of growth might take less time for small inputs than an algorithm whose running time has a lower
order of growth. But for large enough inputs, lower order of growth will run more quickly.
除了RAM模型对算法分析的简化外，另一个简化是我们只考虑方程的最高项。
由于常数和低阶项的影响，对于小规模输入，高阶增长的算法可能会比低阶的执行时间更短。
但当输入规模足够大时，低阶的会运行得快得多。
2.3 Designing algorithms
For insertion sort, we used an incremental approach: having sorted the subarray A[1..j-1],
we inserted the single element A[j] into its proper place, yielding the sorted subarray A[1..j].
Now we will use divide-and-conquer to design a sorting algorithm whose worst-case running
time is much less than insertion sort. One advantage of divide-and-conquer algorithms is that
their running time are often easily determined.
对于插入排序，我们使用了增量法。现在我们要用分治法来设计一种比插入排序快得多的排序
算法。用分治法来设计算法的好处是算法的运行时间很容易判断。
分治法的三个步骤，以及与归并排序的对应：
Divide the problem into a number of subproblems that are smaller instances of the same problem
(Divide the n-element sequence to be sorted into two subsequences of n/2 elements each).
Conquer the subproblems by solving them recursively. If the subproblem sizes are small enough,
however, just solve the subproblems in a straightforward manner.
(Sort the two subsequences recursively using merge sort).
Combine the solutions to the subproblems into the solution for original problem.
(Merge the two sorted subsequences to produce the sorted answer).
The recursion "bottoms out" when sequence to be sorted has length 1, in which case there is no
work to be done. So the key operation is the merging of two sorted sequences in the "combine" step.
递归的最底层是当序列长度为1的时候。所以归并排序的关键是第三步合并两个已经排好序的序列。
We place on the bottom of each pile a sentinel card, which contains a special value that we use to
simplify our code. Here, we use infinity as the sentinel value.
用哨兵牌来简化代码。
设T(n)是当输入规模为n时的归并排序运行时间。
当n足够小时，T(n)可在常数时间内完成。
否则，假设原问题可以划分为a个子问题，每个规模是原来的1/b。对于归并排序a和b都是2。
D(n)是划分问题成子问题的时间，C(n)是合并子问题结果的时间。
Divide: just computes the middle of the subarray, which takes constant time. Thus, D(n) = Θ(1).
Conquer: We recursively solve two subproblems, each of size n/2, which contribute 2T(n/2) to the running time.
Combine: already noted that merge on an n-element subarray takes time Θ(n), and so C(n) = Θ(n).
设c代表解决规模为1的问题所需的时间，D(n) = Θ(1) = c, C(n) = Θ(n) = cn.
树高是lgn + 1，每层子问题解决时间的和都是cn，所以总的时间是cnlgn + cn，即Θ(nlgn)
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