如何获得用于生成随机数的范围？

Question

我使用 Java 中的种子生成随机数。知道最终输出是235，种子数是532，我如何在java中得到intBound数？例如

int randomNumber

int seed=532;

int intBound=800;

Random rand = Random(seed); 

randomNumber=rand.nextInt(intBound);

System.out.println("The generated Random number using the above seed and int bound is:- "+randomNumber);

//Results is: The generated Random number using the above seed and int bound is: 235

该问题的简化数学版本是：仅知道数学公式的两个值，如何产生第三个值？例如 1+2=3 这也意味着如果我们只知道 2 个值和使用的公式，我们可以很容易地得到第三个值，知道用于获得结果的公式。

Answer 1

这不可能。许多上限可以产生相同的输出。例如，对 Ideone的快速测试显示了 1000000 下的 9 个可能边界，这将产生 235 的输出，种子 532（800 不是其中之一）：237、369、711、3239、9717、29151、505164、155164、15164和 454941。import java.util.*;class Test{&nbsp; &nbsp; public static void main (String[] args) throws java.lang.Exception&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; List<Integer> bounds = new ArrayList<Integer>();&nbsp; &nbsp; &nbsp; &nbsp; for (int i = 1; i < 1000000; i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Random rng = new Random(532);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (rng.nextInt(i) == 235) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; bounds.add(i);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; System.out.println(bounds);&nbsp; &nbsp; }}您能做的最好的事情就是确定可能的界限。的实现nextInt(int)是需要等同于&nbsp;public int nextInt(int bound) {&nbsp; &nbsp;if (bound <= 0)&nbsp; &nbsp; &nbsp;throw new IllegalArgumentException("bound must be positive");&nbsp; &nbsp;if ((bound & -bound) == bound)&nbsp; // i.e., bound is a power of 2&nbsp; &nbsp; &nbsp;return (int)((bound * (long)next(31)) >> 31);&nbsp; &nbsp;int bits, val;&nbsp; &nbsp;do {&nbsp; &nbsp; &nbsp; &nbsp;bits = next(31);&nbsp; &nbsp; &nbsp; &nbsp;val = bits % bound;&nbsp; &nbsp;} while (bits - val + (bound-1) < 0);&nbsp; &nbsp;return val;&nbsp;}该算法给出特定输出的方式可以分为三种可能性：bound&nbsp;是二的幂bound&nbsp;不是 2 的幂，循环在第一次迭代时终止bound&nbsp;不是 2 的幂，循环继续经过第一次迭代二的幂的bound情况很容易 - 只需尝试bound适合int.&nbsp;其中只有31个。你可以优化这个，但没有多大意义。可以通过计算next(31)本来是的值（可以通过播种一个Random实例并调用next(31)）来处理第一次迭代的非二次幂情况，然后查看哪些值bound会给出正确的值val并终止做的时候。要给出 的正确值val，bound必须是一个bits - val大于的因数val。（有时bits - val会是 0，任何大于它的整数val都会通过。）要终止 do-while，bits - val + (bound-1)一定不能溢出。因此，落入这种情况的可能界限bits - val是一定范围内的因子，而不是 2 的幂。至于最后一个案例，我不想经历它，所以这将“留给读者作为练习”。（这是最困难的情况，困难在于弄清楚bound当您不知道时哪些值会导致溢出val，而这比我花费的时间更长。）

Answer 2

这是一种寻找隐藏边界的实验方法。获取随机对象的副本并记录其输出。创建where&nbsp;is max int 的n实例。对于这些实例中的每一个，使用它们的索引作为 的参数。仅存储遵循原始序列的实例。继续淘汰候选人，直到只剩下一个。RandomnRandomnextIntRandom考虑下表。在顶部，我们使用随机值被采样的迭代来命名列。另一方面，我们有具有相同种子的序列，它们应用了不同的边界值。中间单元格中的值表示为给定的 Random 实例和迭代检索到的随机值。在第一次遍历所有可能的整数后，我们剩下 4 个可能的候选者。我们不断与权威进行比较，直到只剩下一个可能的候选人为止。如果您没有提供足够的权威样本，您可能会留下多个匹配的上限值候选。public static void main(String [] args) throws Exception {&nbsp; &nbsp; final Scanner scanner = new Scanner(System.in);&nbsp; &nbsp; System.out.print("Please enter the seed number: ");&nbsp; &nbsp; final long seed = scanner.nextLong();&nbsp; &nbsp; System.out.print("Please enter the hidden bound: ");&nbsp; &nbsp; final int bound = scanner.nextInt();&nbsp; &nbsp; final long start = System.nanoTime();&nbsp; &nbsp; final IntSupplier original = rand(seed, bound);&nbsp; &nbsp; final int firstRandom = original.getAsInt();&nbsp; &nbsp; Map<Integer, IntSupplier> candidates = new HashMap<>();&nbsp; &nbsp; for (int i = 1; i < Integer.MAX_VALUE; i++) {&nbsp; &nbsp; &nbsp; &nbsp; final IntSupplier candidate = rand(seed, i);&nbsp; &nbsp; &nbsp; &nbsp; if (firstRandom == candidate.getAsInt()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; candidates.put(i, candidate);&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; int iterations = 1;&nbsp; &nbsp; while (candidates.size() > 1) {&nbsp; &nbsp; &nbsp; &nbsp; iterations += 1;&nbsp; &nbsp; &nbsp; &nbsp; final int current = original.getAsInt();&nbsp; &nbsp; &nbsp; &nbsp; Map<Integer, IntSupplier> survivors = new HashMap<>();&nbsp; &nbsp; &nbsp; &nbsp; for (Map.Entry<Integer, IntSupplier> entry : candidates.entrySet()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (entry.getValue().getAsInt() == current) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; survivors.put(entry.getKey(), entry.getValue());&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; candidates = survivors;&nbsp; &nbsp; }&nbsp; &nbsp; if (candidates.size() == 1) {&nbsp; &nbsp; &nbsp; &nbsp; System.out.println("Upper bound is " + candidates.keySet().iterator().next());&nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; System.out.println("No upper bound found");&nbsp; &nbsp; }&nbsp; &nbsp; System.out.println("Completed in " + iterations +&nbsp; " iterations");&nbsp; &nbsp; final long end = System.nanoTime();&nbsp; &nbsp; System.out.println("Completed in " + (end - start) / Math.pow(10,9) + "seconds");}static IntSupplier rand(long seed, int bound) {&nbsp; &nbsp; final Random rand = new Random(seed);&nbsp; &nbsp; return () -> rand.nextInt(bound);}这会产生输出：Please enter the seed number: 532Please enter the hidden bound: 800Upper bound is 800Completed in 4 iterationsCompleted in 46.778499624 seconds