Java中128位序列位操作的优化

Question

为了加快我的Java代码解决问题的速度，我一直在专门处理一个类，它通过操作两个longs (参见实现)来对128 bit进行位级操作。实际上，我也只需要100位的数据结构，但我认为没有更好的方法来实现这一点。

public class BitBoard {

//Bit-Masks for all N-Bits from the RIGHT
public final static long[] GET_N_BITS_FROM_RIGHT = {0x0000000000000000L, 0x0000000000000001L, 0x0000000000000003L, 0x0000000000000007L, 0x000000000000000fL, 0x000000000000001fL, 0x000000000000003fL, 0x000000000000007fL, 0x00000000000000ffL, 0x00000000000001ffL, 0x00000000000003ffL, 0x00000000000007ffL, 0x0000000000000fffL, 0x0000000000001fffL, 0x0000000000003fffL, 0x0000000000007fffL, 0x000000000000ffffL, 0x000000000001ffffL, 0x000000000003ffffL, 0x000000000007ffffL, 0x00000000000fffffL, 0x00000000001fffffL, 0x00000000003fffffL, 0x00000000007fffffL, 0x0000000000ffffffL, 0x0000000001ffffffL, 0x0000000003ffffffL, 0x0000000007ffffffL, 0x000000000fffffffL, 0x000000001fffffffL, 0x000000003fffffffL, 0x000000007fffffffL, 0x00000000ffffffffL, 0x00000001ffffffffL, 0x00000003ffffffffL, 0x00000007ffffffffL, 0x0000000fffffffffL, 0x0000001fffffffffL, 0x0000003fffffffffL, 0x0000007fffffffffL, 0x000000ffffffffffL, 0x000001ffffffffffL, 0x000003ffffffffffL, 0x000007ffffffffffL, 0x00000fffffffffffL, 0x00001fffffffffffL, 0x00003fffffffffffL, 0x00007fffffffffffL, 0x0000ffffffffffffL, 0x0001ffffffffffffL, 0x0003ffffffffffffL, 0x0007ffffffffffffL, 0x000fffffffffffffL, 0x001fffffffffffffL, 0x003fffffffffffffL, 0x007fffffffffffffL, 0x00ffffffffffffffL, 0x01ffffffffffffffL, 0x03ffffffffffffffL, 0x07ffffffffffffffL, 0x0fffffffffffffffL, 0x1fffffffffffffffL, 0x3fffffffffffffffL, 0x7fffffffffffffffL, 0xffffffffffffffffL,};

public final static long[] GET_N_BITS_FROM_LEFT = {0x0000000000000000L, 0x8000000000000000L, 0xc000000000000000L, 0xe000000000000000L, 0xf000000000000000L, 0xf800000000000000L, 0xfc00000000000000L, 0xfe00000000000000L, 0xff00000000000000L, 0xff80000000000000L, 0xffc0000000000000L, 0xffe0000000000000L, 0xfff0000000000000L, 0xfff8000000000000L, 0xfffc000000000000L, 0xfffe000000000000L, 0xffff000000000000L, 0xffff800000000000L, 0xffffc00000000000L, 0xffffe00000000000L, 0xfffff00000000000L, 0xfffff80000000000L, 0xfffffc0000000000L, 0xfffffe0000000000L, 0xffffff0000000000L, 0xffffff8000000000L, 0xffffffc000000000L, 0xffffffe000000000L, 0xfffffff000000000L, 0xfffffff800000000L, 0xfffffffc00000000L, 0xfffffffe00000000L, 0xffffffff00000000L, 0xffffffff80000000L, 0xffffffffc0000000L, 0xffffffffe0000000L, 0xfffffffff0000000L, 0xfffffffff8000000L, 0xfffffffffc000000L, 0xfffffffffe000000L, 0xffffffffff000000L, 0xffffffffff800000L, 0xffffffffffc00000L, 0xffffffffffe00000L, 0xfffffffffff00000L, 0xfffffffffff80000L, 0xfffffffffffc0000L, 0xfffffffffffe0000L, 0xffffffffffff0000L, 0xffffffffffff8000L, 0xffffffffffffc000L, 0xffffffffffffe000L, 0xfffffffffffff000L, 0xfffffffffffff800L, 0xfffffffffffffc00L, 0xfffffffffffffe00L, 0xffffffffffffff00L, 0xffffffffffffff80L, 0xffffffffffffffc0L, 0xffffffffffffffe0L, 0xfffffffffffffff0L, 0xfffffffffffffff8L, 0xfffffffffffffffcL, 0xfffffffffffffffeL, 0xffffffffffffffffL,};

//Sequence left
public long l0;
//Sequence right
public long l1;

public BitBoard(long l0, long l1) {
    this.l0 = l0;
    this.l1 = l1;
}

public BitBoard and(BitBoard b) {
    return new BitBoard(l0 & b.l0, l1 & b.l1);
}

public void andEquals(BitBoard b) {
    l0 &= b.l0;
    l1 &= b.l1;
}

public BitBoard or(BitBoard b) {
    return new BitBoard(l0 | b.l0, l1 | b.l1);
}

public void orEquals(BitBoard b) {
    l0 |= b.l0;
    l1 |= b.l1;
}

public BitBoard not() {
    return new BitBoard(~l0, ~l1);
}

public void notEquals() {
    l0 = ~l0;
    l1 = ~l1;
}

public BitBoard rightShift(int amount) {
    if (amount <= 63) {
        return new BitBoard(l0 >>> amount, l1 >>> amount | ((l0 & GET_N_BITS_FROM_RIGHT[amount]) << (64 - amount)));
    } else {
        return new BitBoard(0, l0 >>> (amount - 64));
    }
}

public void rightShiftEquals(int amount) {
    if (amount <= 63) {
        l1 = l1 >>> amount | ((l0 & GET_N_BITS_FROM_RIGHT[amount]) << (64 - amount));
        l0 = l0 >>> amount;
    } else {
        l1 = l0 >>> (amount - 64);
        l0 = 0;
    }
}

public BitBoard leftShift(int amount) {
    if (amount <= 63) {
        return new BitBoard(l0 << amount | ((l1 & GET_N_BITS_FROM_LEFT[amount]) >>> (64 - amount)), l1 << amount);
    } else {
        return new BitBoard(l1 << (amount - 64), 0);
    }
}

public void leftShiftEquals(int amount) {
    if (amount <= 63) {
        l0 = l0 << amount | ((l1 & GET_N_BITS_FROM_LEFT[amount]) >>> (64 - amount));
        l1 = l1 << amount;
    } else {
        l0 = l1 << (amount - 64);
        l1 = 0;
    }
}

public BitBoard xOr(BitBoard b) {
    return new BitBoard(b.l0 ^ l0, b.l1 ^ l1);
}

public void xOrEquals(BitBoard b) {
    l0 ^= b.l0;
    l1 ^= b.l1;
}

public int popCount() {
    return Long.bitCount(l0) + Long.bitCount(l1);
}

public boolean equalsZero() {
    return l1 == 0 && l0 == 0;
}

public int numberOfTrailingZeros() {
    int l1Trail = Long.numberOfTrailingZeros(l1);
    if (l1Trail == 64) {
        return 64 + Long.numberOfTrailingZeros(l0);
    } else {
        return l1Trail;
    }
}

public BitBoard unsetBit(int bit) {
    if (bit <= 63) {
        return new BitBoard(l0, l1 & ~(1L << bit));
    } else {
        return new BitBoard(l0 & ~(1L << (bit - 64)), l1);
    }
}

public void unsetBitEquals(int bit) {
    if (bit <= 63) {
        l1 &= ~(1L << bit);
    } else {
        l0 &= ~(1L << (bit - 64));
    }
}}

值得注意的是，我必须经常使用这些操作，我完全依赖它们的速度。然而，大多数情况下，我无法使用就地方法和简单的操作，如add和shift将生成新的对象。这将导致大约20%的运行时开销，用于初始化此数据结构(见下图)。

因不对称而产生的开销

还有其他方法来优化这个吗？

另外，这是代码片段吗？

BitBoard bb;
BitBoard bb2;
BitBoard bb3;
BitBoard res = bb.and(bb2).not().xOr(bb3)

慢于

BitBoard bb;
BitBoard bb2;
BitBoard bb3;
BitBoard res=bb;
res.andEquals(bb2);
res.notEquals();
res.xOrEquals(bb3);

因为它正在为中间步骤分配新内存？

编辑：

我一直在和JMH比较我的方法。

Benchmark 1在本地测试该方法：

public class MyBenchmark {

@State(Scope.Thread)
public static class Status{
    BitBoard[] arr;
    @Setup(Level.Trial)
    public void init(){
        arr= new BitBoard[1000];
        for(int i=0;i<arr.length;i++){
            arr[i]= new BitBoard((long)(Math.random()*Integer.MAX_VALUE),i);
        }
    }
}
@Benchmark @OutputTimeUnit(TimeUnit.NANOSECONDS) @BenchmarkMode(Mode.AverageTime)
public BitBoard[] testMethod(Status s) {
    BitBoard[] res= new BitBoard[s.arr.length];
    for(int i=0;i<s.arr.length;i++){
        res[i]= new BitBoard(0,0);
        for(int j=i+1;j<s.arr.length-1;j++){
            res[i].andEquals(s.arr[j]);
            res[i].andEquals(s.arr[j-1]);
            res[i].xOrEquals(s.arr[j+1]);
        }
    }
    return res;
}

}

结果：基准1结果

第二个基准测试不使用就地方法。

public class MyBenchmark {

@State(Scope.Thread)
public static class Status{
    BitBoard[] arr;
    @Setup(Level.Trial)
    public void init(){
        arr= new BitBoard[1000];
        for(int i=0;i<arr.length;i++){
            arr[i]= new BitBoard((long)(Math.random()*Integer.MAX_VALUE),i);
        }
    }
}
@Benchmark @OutputTimeUnit(TimeUnit.NANOSECONDS) @BenchmarkMode(Mode.AverageTime)
public BitBoard[] testMethod(Status s) {
    BitBoard[] res= new BitBoard[s.arr.length];
    for(int i=0;i<s.arr.length;i++){
        for(int j=i+1;j<s.arr.length-1;j++){
            res[i]=s.arr[j].and(s.arr[j-1]).xOr(s.arr[j+1]);
        }
    }
    return res;
}

}

基准2结果

看来，就地的方法确实提供了一个加速！

Answer 1

您所做的是分析而不是基准。对于基准测试，有接近完美的JMH。我不确定侧写器，但大多数都是谎话。很多。

如果您真的需要避免分配，您可以在紧循环中重用一些对象。对于这样的小对象，您绝对不应该使用池，因为它有更多的开销，分配和GC在一起。

如何最小化分配

我非常不喜欢你的名字，所以我会用我自己的名字。您可以像这样扩展您的操作集

void assign(BitBoard that) {
    this.high = that.high;
    this.low = that.low;
}

void inplaceAnd(BitBoard that) {
    this.high &= that.high;
    this.low &= that.low;
}

void inplaceAndNot(BitBoard that) {
    this.high &= ~that.high;
    this.low &= ~that.low;
}

然后，您可以将分配从紧循环中移开(代价是使代码变得更加丑陋)。

BitBoard tmp = new BitBoard(0, 0);
BitBoard result = new BitBoard(0, 0);
for (...) {
    // Let's say, you get a, b and c as inputs.
    // You should compute a&b | a&~b
    // Let's assume, none of a, b, c may be overwritten.
    tmp.assign(a);
    tmp.inplaceAnd(b);
    result.assign(a);
    result.inplaceAndNot(c);
    result.inplaceOr(tmp);    
}

为什么你不应该最小化分配

所有这些内嵌操作都使代码更容易出错，并且比使用

BitBoard result = a.and(b).or(a.andNot(c));

还有这段代码片段..。比..。因为它正在为中间步骤分配新内存？

你必须自己回答自己的问题，正如我们所能说的“可能是，但通常是微不足道的”。在您的情况下，这可能很重要，但唯一的方法是对您的情况进行基准测试。忘记分析器，让JMH比较这两个版本。JVM可以在重要的地方优化大部分的分配。