下面的打印语句将打印“hello world”。有人能解释一下吗?
System.out.println(randomString(-229985452) + " " + randomString(-147909649));
randomString()如下所示:
public static String randomString(int i)
{
Random ran = new Random(i);
StringBuilder sb = new StringBuilder();
while (true)
{
int k = ran.nextInt(27);
if (k == 0)
break;
sb.append((char)('`' + k));
}
return sb.toString();
}
我写了一个快速程序来寻找这些种子:
import java.lang.*;
import java.util.*;
import java.io.*;
public class RandomWords {
public static void main (String[] args) {
Set<String> wordSet = new HashSet<String>();
String fileName = (args.length > 0 ? args[0] : "/usr/share/dict/words");
readWordMap(wordSet, fileName);
System.err.println(wordSet.size() + " words read.");
findRandomWords(wordSet);
}
private static void readWordMap (Set<String> wordSet, String fileName) {
try {
BufferedReader reader = new BufferedReader(new FileReader(fileName));
String line;
while ((line = reader.readLine()) != null) {
line = line.trim().toLowerCase();
if (isLowerAlpha(line)) wordSet.add(line);
}
}
catch (IOException e) {
System.err.println("Error reading from " + fileName + ": " + e);
}
}
private static boolean isLowerAlpha (String word) {
char[] c = word.toCharArray();
for (int i = 0; i < c.length; i++) {
if (c[i] < 'a' || c[i] > 'z') return false;
}
return true;
}
private static void findRandomWords (Set<String> wordSet) {
char[] c = new char[256];
Random r = new Random();
for (long seed0 = 0; seed0 >= 0; seed0++) {
for (int sign = -1; sign <= 1; sign += 2) {
long seed = seed0 * sign;
r.setSeed(seed);
int i;
for (i = 0; i < c.length; i++) {
int n = r.nextInt(27);
if (n == 0) break;
c[i] = (char)((int)'a' + n - 1);
}
String s = new String(c, 0, i);
if (wordSet.contains(s)) {
System.out.println(s + ": " + seed);
wordSet.remove(s);
}
}
}
}
}
我现在已经在后台运行了它,但它已经找到了足够的词来描述经典的穿山甲:
import java.lang.*;
import java.util.*;
public class RandomWordsTest {
public static void main (String[] args) {
long[] a = {-73, -157512326, -112386651, 71425, -104434815,
-128911, -88019, -7691161, 1115727};
for (int i = 0; i < a.length; i++) {
Random r = new Random(a[i]);
StringBuilder sb = new StringBuilder();
int n;
while ((n = r.nextInt(27)) > 0) sb.append((char)('`' + n));
System.out.println(sb);
}
}
}
(视频演示。)
第727295876页、第128911页、第1611659页、第235516779页。
这里的每个人都做了很好的工作,解释了代码是如何工作的,并展示了如何构建自己的示例,但这里有一个信息理论的答案,说明了为什么我们可以合理地期望存在暴力搜索最终会找到的解决方案。
26个不同的小写字母组成了我们的字母表∑。为了允许生成不同长度的单词,我们进一步添加了终止符,以生成扩展的字母表∑':=∑⑪{⑪}。
设α是符号,X是∑′上均匀分布的随机变量。获得该符号P(X=α)及其信息内容I(α)的概率由下式给出:
P(X=α)=1/|∑'|=1/27I(α)=-log₂[P(X=α)]=-log₂(1/27)=对数₂(27)
对于一个词ω∈∑*和它的∈终止的对应词ω':=ω·∈(∑')*,我们有
I(ω):=I(ω')=|ω'|*log₂(27)=(|ω|+1)*log₂(27)
由于伪随机数生成器(PRNG)是用32位种子初始化的,因此我们可以预期长度最多为
λ=地板[32/log₂(27)] - 1 = 5
由至少一个种子产生。即使我们要搜索一个6个字符的单词,我们仍有41.06%的成功率。不太破旧。
对于7个字母,我们看到接近1.52%,但我在尝试之前没有意识到这一点:
#include <iostream>
#include <random>
int main()
{
std::mt19937 rng(631647094);
std::uniform_int_distribution<char> dist('a', 'z' + 1);
char alpha;
while ((alpha = dist(rng)) != 'z' + 1)
{
std::cout << alpha;
}
}
参见输出:http://ideone.com/JRGb3l