对于简单的循环而言，Clojure的性能与Java相比非常糟糕

Question

Spoiler alert, this is problem 5 of Project Euler.

I am attempting to learn Clojure and solved problem 5, but it is a couple orders of magnitude slower (1515 ms in Java versus 169932 ms in Clojure). I even tried using type hinting, unchecked math operations, and inlining functions all for naught.

Why is my Clojure code so much slower?

Clojure code:

(set! *unchecked-math* true)
(defn divides? [^long number ^long divisor] (zero? (mod number divisor)))

(defn has-all-divisors [divisors ^long num]
  (if (every? (fn [i] (divides? num i)) divisors) num false))

(time (prn (some (fn [^long i] (has-all-divisors (range 2 20) i)) (iterate inc 1))))

Java code:

public class Problem5 {
  public static void main(String[] args) {
    long start = System.currentTimeMillis();
    int i = 1;
    while(!hasAllDivisors(i, 2, 20)) {
      i++;
    }
    long end = System.currentTimeMillis();
    System.out.println(i);
    System.out.println("Elapsed time " + (end - start));
  }

  public static boolean hasAllDivisors(int num, int startDivisor, int stopDivisor) {
    for(int divisor=startDivisor; divisor<=stopDivisor; divisor++) {
      if(!divides(num, divisor)) return false;
    }
    return true;
  }

  public static boolean divides(int num, int divisor) {
    return num % divisor == 0;
  }
}

Answer 1

Some performance problems:

• The (range 2 20) call is creating a new lazy list of numbers for every increment of i . This is expensive, and is causing lots of unnecessary GC.
• You are doing a lot of boxing by passing through function calls. Even the (iterate inc 1) is doing boxing / unboxing at every increment.
• You are traversing a sequence of divisors. This is slower than a straight iterative loop
• mod is actually not a very well optimised function in Clojure at present. You are much better off using rem

You can solve the first problem by using a let statement to define the range just once:

(time (let [rng (range 2 20)]
  (prn (some (fn [^long i] (has-all-divisors rng i)) (iterate inc 1)))))
=> "Elapsed time: 48863.801522 msecs"

You can solve the second problem with loop/recur:

(time (let [rng (range 2 20)
           f (fn [^long i] (has-all-divisors rng i))]
       (prn (loop [i 1] 
              (if (f i)
                i
                (recur (inc i)))))))
=> "Elapsed time: 32757.594957 msecs"

You can solve the third problem by using an iterative loop over the possible divisors:

(defn has-all-divisors [^long num]
  (loop [d (long 2)]
    (if (zero? (mod num d))
      (if (>= d 20) true (recur (inc d)))
      false)))

 (time (prn (loop [i (long 1)] (if (has-all-divisors i) i (recur (inc i))))))
 => "Elapsed time: 13369.525651 msecs"

You can solve the final problem using rem

(defn has-all-divisors [^long num]
  (loop [d (long 2)]
    (if (== 0 (rem num d))
      (if (>= d 20) true (recur (inc d)))
      false)))

 (time (prn (loop [i (long 1)] (if (has-all-divisors i) i (recur (inc i))))))
=> "Elapsed time: 2423.195407 msecs"

As you can see, it is now competitive with the Java version.

In general, you can usually make Clojure almost as fast as Java with a bit of effort. The main tricks are usually:

• Avoid lazy functional features. They are nice, but add some overhead which can be problematic in low-level computation-intensive code.
• Use primitive / unchecked maths
• Use loop/recur rather than sequences
• Ensure you are not doing any reflection on Java objects (ie (set! *warn-on-reflection* true) and eliminate all warnings you find)

Answer 2

I have not been able to reproduce the 1500 ms performance. The Clojure code seems actually twice as fast as the Java version after compilation to uberjar.

Now timing Java version
    232792560
"Elapsed time: 4385.205 msecs"

Now timing Clojure version
    232792560
"Elapsed time: 2511.916 msecs"

I put the java class in resources/HasAllDivisors.java

public class HasAllDivisors {

    public static long findMinimumWithAllDivisors() {
        long i = 1;
        while(!hasAllDivisors(i,2,20)) i++;
        return i;
    }

    public static boolean hasAllDivisors(long num, int startDivisor, int stopDivisor) {
        for(int divisor = startDivisor; divisor <= stopDivisor; divisor++) {
            if(num % divisor > 0) return false;
        }
        return true;
    }

    public static void main(String[] args){
        long start = System.currentTimeMillis();
        long i = findMinimumWithAllDivisors();
        long end = System.currentTimeMillis();
        System.out.println(i);
        System.out.println("Elapsed time " + (end - start));
    }

}

And in Clojure

(time (prn (HasAllDivisors/findMinimumWithAllDivisors)))

(println "Now timing Clojure version")
(time
    (prn
        (loop [i (long 1)]
            (if (has-all-divisors i)
                i
                (recur (inc i))))))

Even on the command line the java class is not reproducing the fast speed.

$ time java HasAllDivisors
  232792560
Elapsed time 4398

real   0m4.563s
user   0m4.597s
sys    0m0.029s

Answer 3

I know this is an old question, but I've been running into similar things. It looks like the statement from the OP, that Clojure is much worse than Java on simple loops, is true. I went through the process in this thread, starting with OP's code and then adding the performance improvements. At the end of it all, the java code runs in around 300 ms and the optimized Clojure code runs in 3000 ms. Creating an uberjar with lein gets the Clojure code down to 2500 ms.

Since we know the answer that the given code spits out, I used that to have the Clojure code merely loop the number of times without doing the mod/rem calculations. It simply goes through the loops.

(def target 232792560)

(defn has-all-divisors [divisors ^long num]
    (loop [d (long 2)]
        (if (< d 20) (recur (inc d)))))

(time (let [rng (range 2 20)
            f (fn [^long i] (has-all-divisors (range 2 20) i)) ]
    (prn (loop [i 1] 
            (if (< i target)
                (do (f i) (recur (inc i))))))))

The resulting times are roughly the same as doing the calculations, ie 3000 ms. So, it's taking Clojure that long simply to walk through that many loops.