A Blog.

I wrote some simple utilities for measuring time and memory usage of Clojure code, and running Clojure code within preset time and memory limits. The methods used for handling memory are a bit hacky, but they seem to work OK for my current purposes. Example snippets and code are posted over at the Google group for now, and if there’s enough interest, maybe this will eventually find its way into clojure.contrib.

If you’re using Clojure’s built-in set operations, clojure.set/union/intersection/difference, you should be aware that their speed currently depends (sometimes dramatically) on the order of the arguments given. And, since clojure.set/difference is not commutative, it is simply needlessly slow if the second set is larger than the first.

… [truncated] …

Update : This post is no longer relevant, since my faster (+ multi-argument) versions are now included in clojure.set.

For some applications, it could be desirable to compute java.util.List hashCode()s backwards, recursively from back to front, rather than front-to-back as defined in the API [1]. For instance, in Clojure this would enable computing the hash values for all of the rests of a seq for free while computing its hash value. This requires just a tiny bit of math and understanding of modular arithmetic. Clojure code showing how this can be done follows.

[1] http://java.sun.com/j2se/1.4.2/docs/api/java/util/List.html#hashCode()

(defn forward-hash "How hashCode() is defined in java.util.List"  [s]
  (loop [s (seq s) h (int 1)]
    (if s 
        (recur (rest s) (int (unchecked-add (unchecked-multiply h 31) (hash (first s)))))
      h)))
 
 
(def int-pow 
     (memoize 
      (fn [x n]
	(let [x (int x) n (int n)]
	  (cond (zero? n) 1
		(even? n) (let [h (int-pow x (/ n 2))] (unchecked-multiply h h))
		:else     (unchecked-multiply x (int-pow x (dec n))))))))
 
(defn backward-hash "Compute the same value as forward-hash, but back-to-front."  [s]
  (let [s (seq s) p (int-pow 31 (count (rest s)))]
    (if s 
        (unchecked-add (unchecked-multiply 30 p)
		       (unchecked-add (int (backward-hash (rest s)))
				      (unchecked-multiply (int (hash (first s))) p)))
	1)))
 
 
user> (map #(% (vec (range 1000))) [forward-hash backward-hash #(.hashCode %)])
(133786869 133786869 133786869)

Update: I posted this on the Clojure issue page here. Note a small caveat there: it is not allowed by Java that (.equals [] (seq [])), so Clojure collections and seqs cannot be comparable in general. However, everything carries through for non-empty seqs and collections.