How do you replace Java nested for loops in Clojure?

Question

I have created a very simple nested loop example and am struggling to write the equivalent Clojure code. I've been trying to do it by list comprehensions but cannot get the same answer. Any help appreciated.

public class Toy {

    public static void main(String[] args) {
        int maxMod = 0;
        for (int i=0;i<1000;i++) {
            for (int j=i;j<1000;j++) {
                if ((i * j) % 13 == 0 && i % 7 == 0) maxMod = i * j;
            }
        }
        System.out.println(maxMod);
    }
}

I guess this isn't the point of the exercise, but you can save a bunch of wasted work by incrementing i by seven at a time instead of 1. — amalloy
– amalloy, Commented Feb 4, 2012 at 3:02

dnolen · Accepted Answer · 2012-02-04 19:24:06Z

25

Here's a list comprehension solution:

(last 
  (for [i (range 1000) 
        j (range 1000)
        :let [n (* i j)] 
        :when (and (= (mod n 13) 0) 
                   (= (mod i 7) 0))] 
    n))

edited Feb 4, 2012 at 19:24

answered Feb 4, 2012 at 1:40

dnolen

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4 Comments

BillRobertson42 Over a year ago

What's the difference between that and using doseq?

seh Over a year ago

Don't you really just want the last entry, which we assume will also happen to be the maximum entry? In your expression above, the max function will keep trying to figure out if it's seen a new maximum for every element yielded by the for sequence.

Retief Over a year ago

@Bill for returns a lazy seq, while doseq executes the body howevery many times (presumably for side effects) and returns nil

seh Over a year ago

By the way, nice job hoisting out the computation of n, which the original Java function failed to do.

Retief · Accepted Answer · 2012-02-04 06:00:01Z

In general, you want to use some sort of sequence operation (like dnolen's answer). However, if you need to do something that is not expressible in some combination of sequence functions, using the loop macro works as well. For this precise problem, dnolen's answer is better than anything using loop, but for illustrative purposes, here is how you would write it with loop.

(loop [i 0
       max-mod 0]
  (if (>= i 1000)
    (println max-mod)
    (recur (inc i)
           (loop [j 0
                  max-mod max-mod]
             (if (>= j 1000)
               max-mod
               (recur (inc j)
                      (if (and (= (mod (* i j) 13) 0)
                               (= (mod 1 7) 0))
                        (* i j)
                        max-mod)))))))

This is pretty much an exact translation of your given code. That said, this is obviously ugly, which is why a solution using for (or other similar functions) is preferred whenever possible.

Rörd · Accepted Answer · 2012-02-06 23:17:22Z

2

List comprehensions create lists from other lists, but you want just a single value as result. You can create the input values (i and j) with a list comprehension, and then use reduce to get a single value from the list:

(reduce (fn [max-mod [i j]]
          (if (and (zero? (mod (* i j) 13))
                   (zero? (mod i 7)))
            (* i j)
            max-mod))
        0
        (for [i (range 1000) j (range 1000)]
             [i j]))

answered Feb 6, 2012 at 23:17

Rörd

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