Ash*_*ral 6 java apache random probability birthday-paradox
我需要编写一个随机字符串生成类,它从31个字符的字符集和一些字母表中生成7char字符串(10 + 26-5,省略5个元音).简单的数学给出了一组31 ^ 7种可能的组合~275亿.我对bday悖论有疑问,我进行了一些测试,重复数量呈指数级增长.我可以做些什么来避免这种情况吗?
At 1 million, duplicates encountered till now = 19
At 2 million, duplicates encountered till now = 69
At 3 million, duplicates encountered till now = 157
At 4 million, duplicates encountered till now = 280
At 5 million, duplicates encountered till now = 470
At 6 million, duplicates encountered till now = 662
At 7 million, duplicates encountered till now = 896
At 8 million, duplicates encountered till now = 1185
At 9 million, duplicates encountered till now = 1500
At 10 million, duplicates encountered till now = 1823
At 11 million, duplicates encountered till now = 2204
At 12 million, duplicates encountered till now = 2584
At 13 million, duplicates encountered till now = 3020
At 14 million, duplicates encountered till now = 3527
At 15 million, duplicates encountered till now = 4110
At 16 million, duplicates encountered till now = 4683
At 17 million, duplicates encountered till now = 5284
At 18 million, duplicates encountered till now = 5919
At 19 million, duplicates encountered till now = 6611
At 20 million, duplicates encountered till now = 7343
At 21 million, duplicates encountered till now = 8095
At 22 million, duplicates encountered till now = 8858
At 23 million, duplicates encountered till now = 9707
At 24 million, duplicates encountered till now = 10547
At 25 million, duplicates encountered till now = 11452
At 26 million, duplicates encountered till now = 12399
At 27 million, duplicates encountered till now = 13356
At 28 million, duplicates encountered till now = 14393
At 29 million, duplicates encountered till now = 15369
At 30 million, duplicates encountered till now = 16436
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以下是测试类:
import java.util.Set;
import org.apache.commons.lang3.RandomStringUtils;
import com.google.common.collect.Sets;
public class RandomUnivmylocaL {
public static void main(String[] argv) {
final int million = 1_000_000;
final int iterations = 30;
// 31 chars
final char[] charArr = new char[] { '1', '2', '3', '4', '5', '6', '7',
'8', '9', '0', 'B', 'C', 'D', 'F', 'G', 'H', 'J', 'K', 'L',
'M', 'N', 'P', 'Q', 'R', 'S', 'T', 'V', 'W', 'X', 'Y', 'Z' };
// System.out.println(charArr.length);
final Set<String> set = Sets.newHashSetWithExpectedSize(
iterations * million);
for (int i = 0; i < iterations; i++) {
for (int j = 0; j < million; j++) {
final String univCode = RandomStringUtils.random(7, charArr);
set.add(univCode);
}
System.out.println("At " + (i + 1) + " million, " +
"duplicates encountered till now = " +
(((i + 1) * million) - set.size()));
}
System.out.println("done");
}
}
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这就是生日悖论。
sqrt(275亿) = 165831,大M的生日悖论公式是1.177*sqrt(M),所以当你生成了大约200,000个时,有50/50的机会出现问题,当你生成100万个时,你会有大约18个问题等等。
生日问题 - 有多少个可能会发生碰撞 - 大约 200,000。 http://www.wolframalpha.com/input/?i=n%3D200000,+m%3D(31%5E7),+n+-+m(1+-+((m-1)%2Fm)%5En )
设置 n = 23.0 和 m = 365 以查看房间内 23 人的等效值。 http://www.wolframalpha.com/input/?i=n%3D23.0,+m%3D365,+n+-+m(1+-+((m-1)%2Fm)%5En)
您可以看到您的模拟与大量数据的预期答案有多接近。
http://www.wolframalpha.com/input/?i=n%3D30e6,+m%3D(31%5E7),+n+-+m(1+-+((m-1)%2Fm)%5En )
Quora 的文章不错。http://www.quora.com/How-can-I-calculate-the-expected-number-of-cache-hits。
因此,您需要增加允许的字符数或使用更长的字符串。或者 - 要使用 7 位数字,只需增加计数器即可。或者使用随机号码,并检查您以前是否使用过它,并在您厌倦寻找新号码时重置。
还有伪随机生成器可以覆盖空间而无需重新命中。7 个字符并不能让您的解决方案安全。
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