;;; The SRFI-32 sort package -- quick sort -*- Scheme -*- ;;; Copyright (c) 2002 by Olin Shivers. ;;; This code is open-source; see the end of the file for porting and ;;; more copyright information. ;;; Olin Shivers 2002/7. ;;; (quick-sort < v [start end]) -> vector ;;; (quick-sort! < v [start end]) -> unspecific ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;;; The algorithm is a standard quicksort, but the partition loop is fancier, ;;; arranging the vector into a left part that is <, a middle region that is ;;; =, and a right part that is > the pivot. Here's how it is done: ;;; The partition loop divides the range being partitioned into five ;;; subranges: ;;; =======<<<<<<<<>>>>>>======= ;;; where = marks a value that is equal the pivot, < marks a value that ;;; is less than the pivot, ? marks a value that hasn't been scanned, and ;;; > marks a value that is greater than the pivot. Let's consider the ;;; left-to-right scan. If it checks a ? value that is <, it keeps scanning. ;;; If the ? value is >, we stop the scan -- we are ready to start the ;;; right-to-left scan and then do a swap. But if the rightward scan checks ;;; a ? value that is =, we swap it *down* to the end of the initial chunk ;;; of ====='s -- we exchange it with the leftmost < value -- and then ;;; continue our rightward scan. The leftwards scan works in a similar ;;; fashion, scanning past > elements, stopping on a < element, and swapping ;;; up = elements. When we are done, we have a picture like this ;;; ========<<<<<<<<<<<<>>>>>>>>>>========= ;;; Then swap the = elements up into the middle of the vector to get ;;; this: ;;; <<<<<<<<<<<<=================>>>>>>>>>> ;;; Then recurse on the <'s and >'s. Work out all the tricky little ;;; boundary cases, and you're done. ;;; ;;; Other tricks: ;;; - This quicksort also makes some effort to pick the pivot well -- it uses ;;; the median of three elements as the partition pivot, so pathological n^2 ;;; run time is much rarer (but not eliminated completely). If you really ;;; wanted to get fancy, you could use a random number generator to choose ;;; pivots. The key to this trick is that you only need to pick one random ;;; number for each *level* of recursion -- i.e. you only need (lg n) random ;;; numbers. ;;; - After the partition, we *recurse* on the smaller of the two pending ;;; regions, then *tail-recurse* (iterate) on the larger one. This guarantees ;;; we use no more than lg(n) stack frames, worst case. ;;; - There are two ways to finish off the sort. ;;; A Recurse down to regions of size 10, then sort each such region using ;;; insertion sort. ;;; B Recurse down to regions of size 10, then sort *the entire vector* ;;; using insertion sort. ;;; We do A. Each choice has a cost. Choice A has more overhead to invoke ;;; all the separate insertion sorts -- choice B only calls insertion sort ;;; once. But choice B will call the comparison function *more times* -- ;;; it will unnecessarily compare elt 9 of one segment to elt 0 of the ;;; following segment. The overhead of choice A is linear in the length ;;; of the vector, but *otherwise independent of the algorithm's parameters*. ;;; I.e., it's a *fixed*, *small* constant factor. The cost of the extra ;;; comparisons made by choice B, however, is dependent on an externality: ;;; the comparison function passed in by the client. This can be made ;;; arbitrarily bad -- that is, the constant factor *isn't* fixed by the ;;; sort algorithm; instead, it's determined by the comparison function. ;;; If your comparison function is very, very slow, you want to eliminate ;;; every single one that you can. Choice A limits the potential badness, ;;; so that is what we do. (define (vector-quick-sort! < v . maybe-start+end) (call-with-values (lambda () (vector-start+end v maybe-start+end)) (lambda (start end) (%quick-sort! < v start end)))) (define (vector-quick-sort < v . maybe-start+end) (call-with-values (lambda () (vector-start+end v maybe-start+end)) (lambda (start end) (let ((ans (make-vector (- end start)))) (vector-portion-copy! ans v start end) (%quick-sort! < ans 0 (- end start)) ans)))) ;;; %QUICK-SORT is not exported. ;;; Preconditions: ;;; V vector ;;; START END fixnums ;;; 0 <= START, END <= (vector-length V) ;;; If these preconditions are ensured by the cover functions, you ;;; can safely change this code to use unsafe fixnum arithmetic and vector ;;; indexing ops, for *huge* speedup. ;;; ;;; We bail out to insertion sort for small ranges; feel free to tune the ;;; crossover -- it's just a random guess. If you don't have the insertion ;;; sort routine, just kill that branch of the IF and change the recursion ;;; test to (< 1 (- r l)) -- the code is set up to work that way. (define (%quick-sort! elt< v start end) ;; Swap the N outer pairs of the range [l,r). (define (swap l r n) (if (> n 0) (let ((x (vector-ref v l)) (r-1 (- r 1))) (vector-set! v l (vector-ref v r-1)) (vector-set! v r-1 x) (swap (+ l 1) r-1 (- n 1))))) ;; Choose the median of V[l], V[r], and V[middle] for the pivot. (define (median v1 v2 v3) (call-with-values (lambda () (if (elt< v1 v2) (values v1 v2) (values v2 v1))) (lambda (little big) (if (elt< big v3) big (if (elt< little v3) v3 little))))) (let recur ((l start) (r end)) ; Sort the range [l,r). (if (< 10 (- r l)) ; Ten: the gospel according to Sedgewick. (let ((pivot (median (vector-ref v l) (vector-ref v (quotient (+ l r) 2)) (vector-ref v (- r 1))))) ;; Everything in these loops is driven by the invariants expressed ;; in the little pictures & the corresponding l,i,j,k,m,r indices ;; and the associated ranges. ;; =======<<<<<<<<>>>>>>======= ;; l i j k m r ;; [l,i) [i,j) [j,k] (k,m] (m,r) (letrec ((lscan (lambda (i j k m) ; left-to-right scan (let lp ((i i) (j j)) (if (> j k) (done i j m) (let ((x (vector-ref v j))) (cond ((elt< x pivot) (lp i (+ j 1))) ((elt< pivot x) (rscan i j k m)) (else ; Equal (if (< i j) (begin (vector-set! v j (vector-ref v i)) (vector-set! v i x))) (lp (+ i 1) (+ j 1))))))))) ;; =======<<<<<<<<<>????????>>>>>>>======= ;; l i j k m r ;; [l,i) [i,j) j (j,k] (k,m] (m,r) (rscan (lambda (i j k m) ; right-to-left scan (let lp ((k k) (m m)) (if (<= k j) (done i j m) (let* ((x (vector-ref v k))) (cond ((elt< pivot x) (lp (- k 1) m)) ((elt< x pivot) ; Swap j & k & lscan. (vector-set! v k (vector-ref v j)) (vector-set! v j x) (lscan i (+ j 1) (- k 1) m)) (else ; x=pivot (if (< k m) (begin (vector-set! v k (vector-ref v m)) (vector-set! v m x))) (lp (- k 1) (- m 1))))))))) ;; =======<<<<<<<<<<<<<>>>>>>>>>>>======= ;; l i j m r ;; [l,i) [i,j) [j,m] (m,r) (done (lambda (i j m) (let ((num< (- j i)) (num> (+ 1 (- m j))) (num=l (- i l)) (num=r (- (- r m) 1))) (swap l j (min num< num=l)) ; Swap ='s into (swap j r (min num> num=r)) ; the middle. ;; Recur on the <'s and >'s. Recurring on the ;; smaller range and iterating on the bigger ;; range ensures O(lg n) stack frames, worst case. (cond ((<= num< num>) (recur l (+ l num<)) (recur (- r num>) r)) (else (recur (- r num>) r) (recur l (+ l num<)))))))) (let ((r-1 (- r 1))) (lscan l l r-1 r-1)))) ;; Small segment => punt to insert sort. ;; Use the dangerous subprimitive. (%vector-insert-sort! elt< v l r))))