Java Math.min/max performance

Question

EDIT: maaartinus gave the answer I was looking for and tmyklebu's data on the problem helped a lot, so thanks both! :)

I've read a bit about how HotSpot has some "intrinsics" that injects in the code, specially for Java standard Math libs (from here)

So I decided to give it a try, to see how much difference HotSpot could make against doing the comparison directly (specially since I've heard min/max can compile to branchless asm).

public class OpsMath {
    public static final int max(final int a, final int b) {
        if (a > b) {
            return a;
        }
        return b;
    }
}

That's my implementation. From another SO question I've read that using the ternary operator uses an extra register, I haven't found significant differences between doing an if block and using a ternary operator (ie, return ( a > b ) ? a : b ).

Allocating a 8Mb int array (ie, 2 million values), and randomizing it, I do the following test:

try ( final Benchmark bench = new Benchmark( "millis to max" ) )
    {
        int max = Integer.MIN_VALUE;

        for ( int i = 0; i < array.length; ++i )
        {
            max = OpsMath.max( max, array[i] );
            // max = Math.max( max, array[i] );
        }
    }

I'm using a Benchmark object in a try-with-resources block. When it finishes, it calls close() on the object and prints the time the block took to complete. The tests are done separately by commenting in/out the max calls in the code above.

'max' is added to a list outside the benchmark block and printed later, so to avoid the JVM optimizing the whole block away.

The array is randomized each time the test runs.

Running the test 6 times, it gives these results:

Java standard Math:

millis to max 9.242167 
millis to max 2.1566199999999998
millis to max 2.046396 
millis to max 2.048616  
millis to max 2.035761
millis to max 2.001044 

So fairly stable after the first run, and running the tests again gives similar results.

OpsMath:

millis to max 8.65418 
millis to max 1.161559  
millis to max 0.955851 
millis to max 0.946642 
millis to max 0.994543 
millis to max 0.9469069999999999 

Again, very stable results after the first run.

The question is: Why? Thats quite a big difference there. And I have no idea why. Even if I implement my max() method exactly like Math.max() (ie, return (a >= b) ? a : b ) I still get better results! It makes no sense.

Specs:

CPU: Intel i5 2500, 3,3Ghz. Java Version: JDK 8 (public march 18 release), x64. Debian Jessie (testing release) x64.

I have yet to try with 32 bit JVM.

EDIT: Self contained test as requested. Added a line to force the JVM to preload Math and OpsMath classes. That eliminates the 18ms cost of the first iteration for OpsMath test.

// Constant nano to millis.
final double TO_MILLIS = 1.0d / 1000000.0d;
// 8Mb alloc.
final int[] array = new int[(8*1024*1024)/4];
// Result and time array.
final ArrayList<Integer> results = new ArrayList<>();
final ArrayList<Double> times = new ArrayList<>();
// Number of tests.
final int itcount = 6;
// Call both Math and OpsMath method so JVM initializes the classes.
System.out.println("initialize classes " + 
OpsMath.max( Math.max( 20.0f, array.length ), array.length / 2.0f ));
    
final Random r = new Random();
for ( int it = 0; it < itcount; ++it )
{
    int max = Integer.MIN_VALUE;
    
    // Randomize the array.
    for ( int i = 0; i < array.length; ++i )
    {
        array[i] = r.nextInt();
    }
    
    final long start = System.nanoTime();
    for ( int i = 0; i < array.length; ++i )
    {
        max = Math.max( array[i], max );
            // OpsMath.max() method implemented as described.
        // max = OpsMath.max( array[i], max );
    }
    // Calc time.
    final double end = (System.nanoTime() - start);
    // Store results.
    times.add( Double.valueOf( end ) );
    results.add( Integer.valueOf(  max ) );
}
// Print everything.
for ( int i = 0; i < itcount; ++i )
{
    System.out.println( "IT" + i + " result: " + results.get( i ) );
    System.out.println( "IT" + i + " millis: " + times.get( i ) * TO_MILLIS );
}

Java Math.max result:

IT0 result: 2147477409
IT0 millis: 9.636998
IT1 result: 2147483098
IT1 millis: 1.901314
IT2 result: 2147482877
IT2 millis: 2.095551
IT3 result: 2147483286
IT3 millis: 1.9232859999999998
IT4 result: 2147482828
IT4 millis: 1.9455179999999999
IT5 result: 2147482475
IT5 millis: 1.882047

OpsMath.max result:

IT0 result: 2147482689
IT0 millis: 9.003616
IT1 result: 2147483480
IT1 millis: 0.882421
IT2 result: 2147483186
IT2 millis: 1.079143
IT3 result: 2147478560
IT3 millis: 0.8861169999999999
IT4 result: 2147477851
IT4 millis: 0.916383
IT5 result: 2147481983
IT5 millis: 0.873984

Still the same overall results. I've tried with randomizing the array only once, and repeating the tests over the same array, I get faster results overall, but the same 2x difference between Java Math.max and OpsMath.max.

Answer 1

It's hard to tell why Math.max is slower than a Ops.max, but it's easy to tell why this benchmark strongly favors branching to conditional moves: On the n-th iteration, the probability of

Math.max( array[i], max );

being not equal to max is the probability that array[n-1] is bigger than all previous elements. Obviously, this probability gets lower and lower with growing n and given

final int[] array = new int[(8*1024*1024)/4];

it's rather negligible most of the time. The conditional move instruction is insensitive to the branching probability, it always take the same amount of time to execute. The conditional move instruction is faster than branch prediction if the branch is very hard to predict. On the other hand, branch prediction is faster if the branch can be predicted well with high probability. Currently, I'm unsure about the speed of conditional move compared to best and worst case of branching.¹

In your case all but first few branches are fairly predictable. From about n == 10 onward, there's no point in using conditional moves as the branch is rather guaranteed to be predicted correctly and can execute in parallel with other instructions (I guess you need exactly one cycle per iteration).

This seems to happen for algorithms computing minimum/maximum or doing some inefficient sorting (good branch predictability means low entropy per step).

---

¹ Both conditional move and predicted branch take one cycle. The problem with the former is that it needs its two operands and this takes additional instruction. In the end the critical path may get longer and/or the ALUs saturated while the branching unit is idle. Often, but not always, branches can be predicted well in practical applications; that's why branch prediction was invented in the first place.

As for the gory details of timing conditional move vs. branch prediction best and worst case, see the discussion below in comments. My my own benchmark shows that conditional move is significantly faster than branch prediction when branch prediction encounters its worst case, but I can't ignore contradictory results. We need some explanation for what exactly makes the difference. Some more benchmarks and/or analysis could help.

Answer 2

When I run your (suitably-modified) code using Math.max on an old (1.6.0_27) JVM, the hot loop looks like this:

0x00007f4b65425c50: mov    %r11d,%edi         ;*getstatic array
                                              ; - foo146::bench@81 (line 40)
0x00007f4b65425c53: mov    0x10(%rax,%rdx,4),%r8d
0x00007f4b65425c58: mov    0x14(%rax,%rdx,4),%r10d
0x00007f4b65425c5d: mov    0x18(%rax,%rdx,4),%ecx
0x00007f4b65425c61: mov    0x2c(%rax,%rdx,4),%r11d
0x00007f4b65425c66: mov    0x28(%rax,%rdx,4),%r9d
0x00007f4b65425c6b: mov    0x24(%rax,%rdx,4),%ebx
0x00007f4b65425c6f: rex mov    0x20(%rax,%rdx,4),%esi
0x00007f4b65425c74: mov    0x1c(%rax,%rdx,4),%r14d  ;*iaload
                                              ; - foo146::bench@86 (line 40)
0x00007f4b65425c79: cmp    %edi,%r8d
0x00007f4b65425c7c: cmovl  %edi,%r8d
0x00007f4b65425c80: cmp    %r8d,%r10d
0x00007f4b65425c83: cmovl  %r8d,%r10d
0x00007f4b65425c87: cmp    %r10d,%ecx
0x00007f4b65425c8a: cmovl  %r10d,%ecx
0x00007f4b65425c8e: cmp    %ecx,%r14d
0x00007f4b65425c91: cmovl  %ecx,%r14d
0x00007f4b65425c95: cmp    %r14d,%esi
0x00007f4b65425c98: cmovl  %r14d,%esi
0x00007f4b65425c9c: cmp    %esi,%ebx
0x00007f4b65425c9e: cmovl  %esi,%ebx
0x00007f4b65425ca1: cmp    %ebx,%r9d
0x00007f4b65425ca4: cmovl  %ebx,%r9d
0x00007f4b65425ca8: cmp    %r9d,%r11d
0x00007f4b65425cab: cmovl  %r9d,%r11d         ;*invokestatic max
                                              ; - foo146::bench@88 (line 40)
0x00007f4b65425caf: add    $0x8,%edx          ;*iinc
                                              ; - foo146::bench@92 (line 39)
0x00007f4b65425cb2: cmp    $0x1ffff9,%edx
0x00007f4b65425cb8: jl     0x00007f4b65425c50

Apart from the weirdly-placed REX prefix (not sure what that's about), here you have a loop that's been unrolled 8 times that does mostly what you'd expect---loads, comparisons, and conditional moves. Interestingly, if you swap the order of the arguments to max, here it outputs the other kind of 8-deep cmovl chain. I guess it doesn't know how to generate a 3-deep tree of cmovls or 8 separate cmovl chains to be merged after the loop is done.

With the explicit OpsMath.max, it turns into a ratsnest of conditional and unconditional branches that's unrolled 8 times. I'm not going to post the loop; it's not pretty. Basically each mov/cmp/cmovl above gets broken into a load, a compare and a conditional jump to where a mov and a jmp happen. Interestingly, if you swap the order of the arguments to max, here it outputs an 8-deep cmovle chain instead. EDIT: As @maaartinus points out, said ratsnest of branches is actually faster on some machines because the branch predictor works its magic on them and these are well-predicted branches.

I would hesitate to draw conclusions from this benchmark. You have benchmark construction issues; you have to run it a lot more times than you are and you have to factor your code differently if you want to time Hotspot's fastest code. Beyond the wrapper code, you aren't measuring how fast your max is, or how well Hotspot understands what you're trying to do, or anything else of value here. Both implementations of max will result in code that's entirely too fast for any sort of direct measurement to be meaningful within the context of a larger program.

Answer 3

Using JDK 8:

java version "1.8.0"
Java(TM) SE Runtime Environment (build 1.8.0-b132)
Java HotSpot(TM) 64-Bit Server VM (build 25.0-b70, mixed mode)

On Ubuntu 13.10

I ran the following:

import java.util.Random;
import java.util.function.BiFunction;

public class MaxPerformance {
  private final BiFunction<Integer, Integer, Integer> max;
  private final int[] array;

  public MaxPerformance(BiFunction<Integer, Integer, Integer> max, int[] array) {
    this.max = max;
    this.array = array;
  }

  public double time() {
    long start = System.nanoTime();

    int m = Integer.MIN_VALUE;
    for (int i = 0; i < array.length; ++i) m = max.apply(m, array[i]);

    m = Integer.MIN_VALUE;
    for (int i = 0; i < array.length; ++i) m = max.apply(array[i], m);

    // total time over number of calls to max
    return ((double) (System.nanoTime() - start)) / (double) array.length / 2.0;
  }

  public double averageTime(int repeats) {
    double cumulativeTime = 0;
    for (int i = 0; i < repeats; i++)
      cumulativeTime += time();
    return (double) cumulativeTime / (double) repeats;
  }

  public static void main(String[] args) {
    int size = 1000000;
    Random random = new Random(123123123L);
    int[] array = new int[size];
    for (int i = 0; i < size; i++) array[i] = random.nextInt();

    double tMath = new MaxPerformance(Math::max, array).averageTime(100);
    double tAlt1 = new MaxPerformance(MaxPerformance::max1, array).averageTime(100);
    double tAlt2 = new MaxPerformance(MaxPerformance::max2, array).averageTime(100);

    System.out.println("Java Math: " + tMath);
    System.out.println("Alt 1:     " + tAlt1);
    System.out.println("Alt 2:     " + tAlt2);
  }

  public static int max1(final int a, final int b) {
    if (a >= b) return a;
    return b;
  }

  public static int max2(final int a, final int b) {
    return (a >= b) ? a : b; // same as JDK implementation
  }
}

And I got the following results (average nanoseconds taken for each call to max):

Java Math: 15.443555810000003
Alt 1:     14.968298919999997
Alt 2:     16.442204045

So on a long run it looks like the second implementation is the fastest, although by a relatively small margin.

In order to have a slightly more scientific test, it makes sense to compute the max of pairs of elements where each call is independent from the previous one. This can be done by using two randomized arrays instead of one as in this benchmark:

import java.util.Random;
import java.util.function.BiFunction;
public class MaxPerformance2 {
  private final BiFunction<Integer, Integer, Integer> max;
  private final int[] array1, array2;

  public MaxPerformance2(BiFunction<Integer, Integer, Integer> max, int[] array1, int[] array2) {
    this.max = max;
    this.array1 = array1;
    this.array2 = array2;
    if (array1.length != array2.length) throw new IllegalArgumentException();
  }

  public double time() {
    long start = System.nanoTime();

    int m = Integer.MIN_VALUE;
    for (int i = 0; i < array1.length; ++i) m = max.apply(array1[i], array2[i]);
    m += m; // to avoid optimizations!

    return ((double) (System.nanoTime() - start)) / (double) array1.length;
  }

  public double averageTime(int repeats) {
    // warm up rounds:
    double tmp = 0;
    for (int i = 0; i < 10; i++) tmp += time();
    tmp *= 2.0;

    double cumulativeTime = 0;
    for (int i = 0; i < repeats; i++)
        cumulativeTime += time();
    return cumulativeTime / (double) repeats;
  }

  public static void main(String[] args) {
    int size = 1000000;
    Random random = new Random(123123123L);
    int[] array1 = new int[size];
    int[] array2 = new int[size];
    for (int i = 0; i < size; i++) {
        array1[i] = random.nextInt();
        array2[i] = random.nextInt();
    }

    double tMath = new MaxPerformance2(Math::max, array1, array2).averageTime(100);
    double tAlt1 = new MaxPerformance2(MaxPerformance2::max1, array1, array2).averageTime(100);
    double tAlt2 = new MaxPerformance2(MaxPerformance2::max2, array1, array2).averageTime(100);

    System.out.println("Java Math: " + tMath);
    System.out.println("Alt 1:     " + tAlt1);
    System.out.println("Alt 2:     " + tAlt2);
  }

  public static int max1(final int a, final int b) {
    if (a >= b) return a;
    return b;
  }

  public static int max2(final int a, final int b) {
    return (a >= b) ? a : b; // same as JDK implementation
  }
}

Which gave me:

Java Math: 15.346468170000005
Alt 1:     16.378737519999998
Alt 2:     20.506475350000006

The way your test is set up makes a huge difference on the results. The JDK version seems to be the fastest in this scenario. This time by a relatively large margin compared to the previous case.

Somebody mentioned Caliper. Well if you read the wiki, one the first things they say about micro-benchmarking is not to do it: this is because it's hard to get accurate results in general. I think this is a clear example of that.

Answer 4

Here's a branchless min operation, max can be implemented by replacing diff=a-b with diff=b-a.

public static final long min(final long a, final long b) {
    final long diff = a - b;
    // All zeroes if a>=b, all ones if a<b because the sign bit is propagated
    final long mask = diff >> 63;
    return (a & mask) | (b & (~mask));
}

It should be as fast as streaming the memory because the CPU operations should be hidden by the sequential memory read latency.