其二，杠杆的长度。假设阿基米德体重100公斤，有100公斤的力气。地球的质量是6.0×1021公斤。支点选在月亮，月亮离地球3.8×105公里。根据杠杆原理，那杠杆长度就得有2 ×1024公里。光速是3×105公里每秒。阿基米德爬过去最少需要 6.7×1018秒，也就是2×1011年。目前地球的年龄大约46亿年（4.6×109）。月球寿命大约4.3×109年。等阿基米德的幽魂爬到杠杆的另一端，月亮和地球的幽魂又在哪里呢？
其三，阿基米德的游魂站在哪里撬呢？这杠杆太长了，有2×1011光年长。银河系的直径大约10万光年。真是零头都不够啊。据说宇宙也就260亿（2.6×1010）光年吧。可怜的阿基米德，伟大的物理学家，的幽魂，却只能置身宇宙之外，努力地撬遥远的一个宇宙里的，也许最不出名的一颗小小的蓝色星球，的幽魂。
http://www.math.ucsb.edu/~cooper/puz1.html谁谁有兴趣就做做。我只做过第1道和第9道。
蚂蚁这题很有意思。应该是一个微积分的题。上面的解答算一个模拟解吧。假设先拉橡皮筋，蚂蚁后走：/**
* MATH PUZZLE 01
*
* An ant starts at one end of a rubber band and walks along it at a speed
* of 1 inch per second. The rubber band is 10 inches long, and is being
* stretched uniformly at a rate of 20 inches per second. Does the ant ever
* reach the other end of the rubber band?. Give reasons.
*/
public void testMathPuzzle01() {
long t = 0;
final long MINUTE = 60, HOUR = MINUTE * 60, DAY = HOUR * 24, YEAR = DAY * 365;
double rubber = 10;
double walked = 0;
while (walked < rubber) {
// After 1 second
t++;
rubber = rubber + 20;
walked = walked * rubber / (rubber - 20) + 1;
if (t % YEAR == 0) {
System.out.println("" + t / YEAR + " years has been flying by");
}
}
System.out.println("Time used:" + t / YEAR + " years, " + t % YEAR
/ DAY + " days, " + t % DAY / HOUR + " hours," + t % HOUR
/ MINUTE + " minutes," + t % MINUTE + " seconds");
}
walked = walked * rubber / (rubber - 20) + 1;算出蚂蚁需要大约16年才能爬到橡皮筋另一头。或者蚂蚁先走，再拉橡皮筋：
walked = (walked + 1) * rubber / (rubber - 20) ;可以算出蚂蚁需要2年多一点可以爬到另一头。早一步就节省12年。这道题也印证了抢占先机的重要性。
实质上是大家常讲的三个门的题。门后面可能藏别的东西。/**
* MATH PUZZLE 09
*
* In a TV show, a prize is hidden behind one of 3 closed doors. The
* contestant tries to guess where the prize is. After the contestant
* chooses a door, the host of the show (who knows where the prize is) opens
* one of the 2 remaining doors, to reveal that the prize is not behind that
* door. The host then gives the contestant the opportunity to change her
* guess. Should she?
*/
public void testMathPuzzle09() {
final int SIZE = 100000;
boolean[] doors = new boolean[3];
Random random = new Random();
int insist = 0;
int reselect = 0;
for (int i = 0; i < SIZE; i++) {
// Init doors
doors[0] = doors[1] = doors[2] = false;
int index = random.nextInt(3);
doors[index] = true;
int guess = random.nextInt(3);
// does not change her guess
if (doors[guess] == true) {
insist++;
}
// change her guess
if (doors[guess] == false) {
reselect++;
}
}
System.out.println("Not change will get:" + insist);
System.out.println("Change will get:" + reselect);
}
Structure通过空格来展示. Sequence里的项用"-"来代表, Map里的键值对用":"分隔.
这几乎就是所有的语法了.
name: John SmithJohn今年37岁, 有一个幸福的四口之家. 两个孩子Jimmy 和Jenny活泼可爱. 妻子Jane年轻美貌. 如果深入研究的话还可能发现一些社会问题.
age: 37
spouse:
name: Jane Smith
age: 25
children:
- name: Jimmy Smith
age: 15
- name: Jenny Smith
age 12
public class Person {
private String name;
private int age;
private Person sponse;
private Person[] children;
// setXXX, getXXX方法略.
}
Person john = new Person();
john.setAge(37);
john.setName("John Smith");
Person sponse = new Person();
sponse.setName("Jane Smith");
sponse.setAge(25);
john.setSponse(sponse);
Person[] children = {new Person(), new Person()};
children[0].setName("Jimmy Smith");
children[0].setAge(15);
children[1].setName("Jenny Smith");
children[1].setAge(12);
john.setChildren(children);
File dumpfile = new File("John.yaml");下面我们看看John.yaml是什么样子:
Yaml.dump(john, dumpfile);
--- !yaml.test.internal.Person其中!yaml.test.internal.Person是一些类型的信息. load的时候需要用.
age: 37
children: !yaml.test.internal.Person[]
- !yaml.test.internal.Person
age: 15
name: Jimmy Smith
- !yaml.test.internal.Person
age: 12
name: Jenny Smith
name: John Smith
sponse: !yaml.test.internal.Person
age: 25
name: Jane Smith
Person john2 = (Person) Yaml.load(dumpfile);我们最初给的John.yaml并没有类型的信息. 可以用下面的方法load进来:
assertEquals(john, john2);
Person john2 = (Person) Yaml.loadType(dumpfile, Person.class);我们还可以用下面的方法dump出最初的John.jaml:
Yaml.dump(john,dumpfile, true);
YamlEncoder enc = new YamlEncoder(new FileOutputStream(dumpfile));下面再把这十个对象一个一个读出来:
for(int i=0; i<10; i++){
john.setAge(37+i);
enc.writeObject(john);
enc.flush();
}
enc.close();
YamlDecoder dec = new YamlDecoder(new FileInputStream(dumpfile));
int age = 37;
while(true){
try{
john = (Person) dec.readObject();
assertEquals(age, john.getAge());
age++;
}catch(EOFException eofe){
break;
}
}
class Person把John装配起来:
attr_accessor :name, :age, :sponse, :children
def initialize(name, age, sponse=nil, children=nil)
@name = name
@age = age
@sponse = sponse
@children = children
end
end
jane = Person.new("Jane Smith", 25)Dump出John到John.yaml:
children = [Person.new("Jimmy Smith", 15), Person.new("Jenny Smith", 12)]
john = Person.new("John Smith", 37, jane, children)
File.open('John.yaml', 'w') do |os|我们看看Dump的结果:
YAML::dump(john, os)
end
--- !ruby/object:Person仔细观察会发现和JYaml Dump出来的还是有些区别的.
age: 37
children:
- !ruby/object:Person
age: 15
children:
name: Jimmy Smith
sponse:
- !ruby/object:Person
age: 12
children:
name: Jenny Smith
sponse:
name: John Smith
sponse: !ruby/object:Person
age: 25
children:
name: Jane Smith
sponse:
john2 = File.open('John.yaml') { |is| YAML::load(is) }
public class YamlDataObject {YamlDataObject里用到的类型, 都是JYaml能够支持的. 其中, uRI和name用来描述DataObject自身. attributes来描述DataObject里的值. 这些值可能为DataObject, 也可能是JYaml所支持的其他类型. 我们希望用YamlDataObject来担当DataObject的角色, 因此YamlDataObject和DataObject要能互相转化. 增加两个API:
private String uRI;
private String name;
private Map attributes;
}
private static YamlDataObject newInstance(DataObject dataobject){...上面两个之所以是private的, 是因为我们打算遵照JYaml的习惯, 增加两个接口, dump和load, 这样newInstance和toDataObject只在内部使用了. 下面是dump和load的定义:
private DataObject toDataObject() {...
public static void dump(DataObject dataobject, File dumpfile) throws FileNotFoundException{...下面是一小段测试代码:
public static Object load(File file) throws FileNotFoundException{...
// ... create dataobject ...具体的实现细节就不谈了. 读者可以作为练习题, 熟悉一下这两个数据整合方案.
File bodump=new File("bodump.yaml");
YamlDataObject.dump(dataobject,bodump);
assertTrue(bodump.exists());
DataObject dataobject2 = (DataObject) YamlDataObject.load(bodump);
// check result
assertNotNull(dataobject2);
for(int i=0; i<5; i++){
assertEquals(dataobject.get(i),dataobject2.get(i));
}
