当使用递增 1(例如 epochDay(的随机种子为列表生成确定性洗牌时,我观察到对于种子的某些序列,某个条目将始终位于结果的最后一个位置。
这仅适用于大小为 16 的集合。
我已将其缩小到这段代码:
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.Random;
public class Main {
public static void main(String[] args) {
for (int i = 0; i < 50; i++) {
List<String> values = Arrays.asList(
"0", "1", "2", "3", "4", "5", "6", "7", "8", "9",
"A", "B", "C", "D", "E", "F"
);
Random r = new Random(i);
Collections.shuffle(values, r);
System.out.println(values);
}
}
}
这将产生以下输出:
[0, 7, C, 9, 6, 5, A, 4, 3, 1, 2, E, 8, F, D, B]
[A, 5, 0, C, 3, 8, 6, 7, E, 4, F, 2, 9, 1, D, B]
[D, 5, 7, E, A, 2, 3, F, 0, 6, 1, 9, 8, 4, C, B]
[4, F, 8, E, A, 3, D, 1, C, 9, 2, 0, 7, 6, 5, B]
[E, 6, A, 5, F, 2, 8, 0, C, D, 4, 3, 9, 1, 7, B]
[2, 9, 0, 1, C, F, 5, 3, 8, D, E, 6, A, 4, 7, B]
[5, F, 0, D, 3, A, 8, 2, C, 7, 4, 1, 9, E, 6, B]
[3, C, 1, 6, 2, 0, 7, 9, 5, 8, D, 4, A, F, E, B]
[3, 9, 2, A, 5, 6, F, D, 8, E, 7, 0, C, 4, 1, B]
[0, D, 4, 5, E, 2, A, 7, 6, 3, 9, C, F, 8, 1, B]
[2, 5, E, 3, 8, D, 1, 6, 4, C, 7, A, F, 9, 0, B]
[C, 3, 5, A, 6, E, 4, 1, 2, 0, 7, 9, F, D, 8, B]
[F, 7, 3, D, 8, E, C, 9, 5, 0, A, 4, 1, 6, 2, B]
[8, 0, 5, D, E, 6, 4, 9, 2, 3, C, 7, 1, F, A, B]
[5, 4, 7, A, 1, 3, 0, 8, F, 6, E, 2, C, D, 9, B]
[8, C, 6, 4, D, F, 3, 5, 7, 9, A, 1, 0, E, 2, B]
[9, F, A, 4, 0, 6, E, 8, 5, 3, 2, C, 1, D, 7, B]
[8, 5, 9, A, 7, 6, 1, E, 3, F, C, D, 2, 4, 0, B]
[9, D, 5, 6, C, A, 3, 7, 2, 8, 0, F, 1, 4, E, B]
[D, 3, 1, 6, 5, 4, 7, F, 9, C, 2, A, 0, 8, E, B]
[6, 8, 0, 7, A, C, 4, D, 9, 3, F, 5, 2, E, 1, B]
[2, 7, 1, D, C, A, 0, E, F, 4, 5, 8, 3, 6, 9, B]
[D, A, 9, 5, 1, 6, 4, F, E, 7, C, 3, 2, 8, 0, B]
[E, 7, A, C, 2, 9, 1, D, 5, 0, 4, 6, 3, F, 8, B]
[9, E, 8, D, 4, 0, 3, C, 1, F, 7, 2, 5, 6, A, B]
[D, A, C, F, E, 2, 8, 0, 6, 5, 7, 1, 4, 9, 3, B]
[A, 0, C, 2, D, 1, 3, E, 6, 7, 5, 8, 4, F, 9, B]
[D, 5, A, 9, C, 3, 6, 8, E, 0, 7, F, 4, 1, 2, B]
[F, 5, 1, E, D, 3, 9, 0, C, 2, A, 6, 7, 8, 4, B]
[C, 9, 1, 2, 6, 8, 0, 3, E, F, A, 5, 7, D, 4, B]
[1, 0, 9, C, 8, 7, 2, E, F, 6, A, 4, 5, D, 3, B]
[E, 2, 8, C, D, 1, 7, 5, 0, 9, A, 3, 6, 4, F, B]
[4, 2, 6, F, C, 8, 5, A, E, 0, 3, 7, D, 9, 1, B]
[0, 5, E, 7, 4, 2, 1, 6, 8, F, 3, C, A, D, 9, B]
[0, 9, 3, 2, 4, A, E, F, C, 6, 1, 5, 7, D, 8, B]
[D, F, 2, 6, 9, A, 1, 0, 5, 7, 3, C, E, 4, 8, B]
[2, 5, 4, 0, 1, C, D, 7, 8, 9, 6, 3, E, F, A, B]
[7, A, 8, E, 1, 5, 0, 9, 4, C, 6, D, F, 2, 3, B]
[7, 3, D, 0, C, 9, A, F, 5, 6, 4, 1, 8, E, 2, B]
[5, 3, 7, E, C, 8, F, 4, 1, A, D, 0, 9, 6, 2, B]
[C, 5, 1, 2, 7, 0, E, 3, 6, A, 9, 8, F, D, 4, B]
[7, 8, 5, 0, D, 3, 1, 6, A, 2, 9, F, E, 4, C, B]
[F, 4, 0, 1, 7, A, E, 9, 2, 5, 8, D, C, 6, 3, B]
[1, C, 7, 3, 2, 4, 6, 0, 9, A, 8, 5, D, E, F, B]
[1, A, 2, 5, 6, E, 9, 7, 3, 8, F, C, 0, 4, D, B]
[7, F, 5, D, 4, 8, E, 2, A, 1, C, 3, 0, 9, 6, B]
[7, 3, 8, 2, 6, D, 4, 1, F, 5, E, A, 0, 9, C, B]
[2, 9, 7, F, 3, 6, A, 4, E, 8, 0, C, 1, D, 5, B]
[7, 6, 1, D, 8, 5, 0, 4, A, C, 3, 9, 2, E, F, B]
[5, A, 1, F, 6, 2, 9, 7, 8, C, 4, 0, E, D, 3, B]
(在 Oracle JRE 1.8、10 和 12 上测试(
如您所见,条目 B 总是在列表的最后一个位置结束。当我让种子运行到其他范围内的值时,不同的元素将放在最后,但始终是相同的元素,间隔 100-300 个增量种子值。
我可以通过在洗牌之前调用 r.nextInt(( 来解决此问题,但我仍然对这种行为感到困惑,并想知道是否有解释或文档。
源代码shuffle(List<?> list, Random rnd)
public static void shuffle(List<?> list, Random rnd) {
int size = list.size();
if (size < SHUFFLE_THRESHOLD || list instanceof RandomAccess) {
for (int i=size; i>1; i--)
swap(list, i-1, rnd.nextInt(i));
} else {
...
所以问题是random.nextInt(16)
总是返回11
和Random.java
private static final long multiplier = 0x5DEECE66DL;
private static final long addend = 0xBL;
private static final long mask = (1L << 48) - 1;
public Random(long seed) {
if (getClass() == Random.class)
this.seed = new AtomicLong(initialScramble(seed));
else {
// subclass might have overriden setSeed
this.seed = new AtomicLong();
setSeed(seed);
}
}
private static long initialScramble(long seed) {
return (seed ^ multiplier) & mask;
}
protected int next(int bits) {
long oldseed, nextseed;
AtomicLong seed = this.seed;
do {
oldseed = seed.get();
nextseed = (oldseed * multiplier + addend) & mask;
} while (!seed.compareAndSet(oldseed, nextseed));
return (int)(nextseed >>> (48 - bits));
}
public int nextInt(int bound) {
if (bound <= 0)
throw new IllegalArgumentException(BadBound);
int r = next(31);
int m = bound - 1;
if ((bound & m) == 0) // i.e., bound is a power of 2
r = (int)((bound * (long)r) >> 31);
else {
for (int u = r;
u - (r = u % bound) + m < 0;
u = next(31))
;
}
return r;
}
random.nextInt(16)
的第一个电话
int bound = 16;
long seed = (i ^ multiplier) & mask;
long r = (seed * multiplier + addend) & mask;
r = (r >>> (48 - 31));
r = (bound * r) >> 31;
简化后
//j is in [-64, 64]
r = (((multiplier + j) * multiplier + addend) >>> 44) & 0xF
和multiplier < (1 << 35)
,所以
r = ((multiplier * multiplier) >> 44) & 0xF
那是11