Skip to content

C
cpython
Kaydet (Commit) 9be62833 authored tarafından Guido van Rossum's avatar Guido van Rossum
Dosyalara gözat
Seçenekler

Tim's quicksort on May 25.

üst 16653cb2
Hide whitespace changes
Inline Side-by-side
Showing with 141 additions and 123 deletions
Objects/listobject.c
Dosyayı görüntüle @ 9be62833
......@@ -625,46 +625,11 @@ docompare(x, y, compare)
}
/* MINSIZE is the smallest array we care to partition; smaller arrays
are sorted using a straight insertion sort (above). It must be at
least 3 for the quicksort implementation to work. Assuming that
comparisons are more expensive than everything else (and this is a
good assumption for Python), it should be 10, which is the cutoff
point: quicksort requires more comparisons than insertion sort for
smaller arrays. */
#define MINSIZE 10
/* Straight insertion sort. More efficient for sorting small arrays. */
static int
insertionsort(array, size, compare)
PyObject **array; /* Start of array to sort */
int size; /* Number of elements to sort */
PyObject *compare;/* Comparison function object, or NULL => default */
{
register PyObject **a = array;
register PyObject **end = array+size;
register PyObject **p;
for (p = a+1; p < end; p++) {
register PyObject *key = *p;
register PyObject **q = p;
while (--q >= a) {
register int k = docompare(key, *q, compare);
/* if (p-q >= MINSIZE)
fprintf(stderr, "OUCH! %d\n", p-q); */
if (k == CMPERROR)
return -1;
if (k < 0) {
*(q+1) = *q;
*q = key; /* For consistency */
}
else
break;
}
}
return 0;
}
are sorted using binary insertion. It must be at least 4 for the
quicksort implementation to work. Binary insertion always requires
fewer compares than quicksort, but does O(N**2) data movement. The
more expensive compares, the larger MINSIZE should be. */
#define MINSIZE 49
/* STACKSIZE is the size of our work stack. A rough estimate is that
this allows us to sort arrays of MINSIZE * 2**STACKSIZE, or large
......@@ -673,20 +638,20 @@ insertionsort(array, size, compare)
exactly in two.) */
#define STACKSIZE 64
/* Quicksort algorithm. Return -1 if an exception occurred; in this
/* quicksort algorithm. Return -1 if an exception occurred; in this
case we leave the array partly sorted but otherwise in good health
(i.e. no items have been removed or duplicated). */
static int
quicksort(array, size, compare)
PyObject **array; /* Start of array to sort */
int size; /* Number of elements to sort */
PyObject **array; /* Start of array to sort */
int size; /* Number of elements to sort */
PyObject *compare;/* Comparison function object, or NULL for default */
{
register PyObject *tmp, *pivot;
register PyObject **l, **r, **p;
register PyObject **lo, **hi;
int top, k, n;
PyObject **lo, **hi, **notp;
int top, k, n, lisp, risp;
PyObject **lostack[STACKSIZE];
PyObject **histack[STACKSIZE];
......@@ -699,55 +664,66 @@ quicksort(array, size, compare)
while (--top >= 0) {
lo = lostack[top];
hi = histack[top];
/* If it's a small one, use straight insertion sort */
n = hi - lo;
if (n < MINSIZE)
/* If it's a small one, use binary insertion sort */
if (n < MINSIZE) {
for (notp = lo+1; notp < hi; ++notp) {
/* set l to where *notp belongs */
l = lo;
r = notp;
pivot = *r;
do {
p = l + ((r - l) >> 1);
k = docompare(pivot, *p, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
r = p;
else
l = p + 1;
} while (l < r);
/* Pivot should go at l -- slide over to
make room. Caution: using memmove
is much slower under MSVC 5; we're
not usually moving many slots. */
for (p = notp; p > l; --p)
*p = *(p-1);
*l = pivot;
}
continue;
}
/* Choose median of first, middle and last as pivot;
these 3 are reverse-sorted in the process; the ends
will be swapped on the first do-loop iteration.
*/
l = lo; /* First */
/* Choose median of first, middle and last as pivot */
l = lo; /* First */
p = lo + (n>>1); /* Middle */
r = hi - 1; /* Last */
r = hi - 1; /* Last */
k = docompare(*l, *p, compare);
k = docompare(*p, *l, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
{ tmp = *l; *l = *p; *p = tmp; }
{ tmp = *p; *p = *l; *l = tmp; }
k = docompare(*p, *r, compare);
k = docompare(*r, *p, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
{ tmp = *p; *p = *r; *r = tmp; }
{ tmp = *r; *r = *p; *p = tmp; }
k = docompare(*l, *p, compare);
k = docompare(*p, *l, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
{ tmp = *l; *l = *p; *p = tmp; }
{ tmp = *p; *p = *l; *l = tmp; }
pivot = *p;
l++;
r--;
/* Partition the array */
do {
tmp = *l; *l = *r; *r = tmp;
if (l == p) {
p = r;
l++;
}
else if (r == p) {
p = l;
r--;
}
else {
l++;
r--;
}
for (;;) {
lisp = risp = 1; /* presumed guilty */
/* Move left index to element >= pivot */
while (l < p) {
......@@ -756,8 +732,10 @@ quicksort(array, size, compare)
return -1;
if (k < 0)
l++;
else
else {
lisp = 0;
break;
}
}
/* Move right index to element <= pivot */
while (r > p) {
......@@ -766,79 +744,119 @@ quicksort(array, size, compare)
return -1;
if (k < 0)
r--;
else
else {
risp = 0;
break;
}
}
if (lisp == risp) {
/* assert l < p < r or l == p == r
* This is the most common case, so we
* strive to get back to the top of the
* loop ASAP.
*/
tmp = *l; *l = *r; *r = tmp;
l++; r--;
if (l < r)
continue;
break;
}
} while (l < r);
/* One (exactly) of the pointers is at p */
/* assert (p == l) ^ (p == r) */
notp = lisp ? r : l;
k = (r - l) >> 1;
if (k) {
*p = *notp;
if (lisp) {
p = r - k;
l++;
}
else {
p = l + k;
r--;
}
/* assert l < p < r */
*notp = *p;
*p = pivot; /* for consistency */
continue;
}
/* lo < l == p == r < hi-1
*p == pivot
/* assert l+1 == r */
*p = *notp;
*notp = pivot;
p = notp;
break;
} /* end of partitioning loop */
/* assert *p == pivot
All in [lo,p) are <= pivot
At p == pivot
All in [p+1,hi) are >= pivot
Now extend as far as possible (around p) so that:
All in [lo,r) are <= pivot
All in [r,l) are == pivot
All in [l,hi) are >= pivot
This wastes two compares if no elements are == to the
pivot, but can win big when there are duplicates.
Mildly tricky: continue using only "<" -- we deduce
equality indirectly.
*/
while (r > lo) {
/* because r-1 < p, *(r-1) <= pivot is known */
k = docompare(*(r-1), pivot, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
break;
/* <= and not < implies == */
r--;
}
l++;
while (l < hi) {
/* because l > p, pivot <= *l is known */
k = docompare(pivot, *l, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
break;
/* <= and not < implies == */
r = p;
l = p + 1;
/* Partitions are [lo,r) and [l,hi).
* See whether *l == pivot; we know *l >= pivot, so
* they're equal iff *l <= pivot too, or not pivot < *l.
* This wastes a compare if it fails, but can win big
* when there are runs of duplicates.
*/
k = docompare(pivot, *l, compare);
if (k == CMPERROR)
return -1;
if (!(k < 0)) {
/* Now extend as far as possible (around p) so that:
All in [lo,r) are <= pivot
All in [r,l) are == pivot
All in [l,hi) are >= pivot
Mildly tricky: continue using only "<" -- we
deduce equality indirectly.
*/
while (r > lo) {
/* because r-1 < p, *(r-1) <= pivot is known */
k = docompare(*(r-1), pivot, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
break;
/* <= and not < implies == */
r--;
}
l++;
}
while (l < hi) {
/* because l > p, pivot <= *l is known */
k = docompare(pivot, *l, compare);
if (k == CMPERROR)
return -1;
if (k < 0)
break;
/* <= and not < implies == */
l++;
}
} /* end of checking for duplicates */
/* Push biggest partition first */
if (r - lo >= hi - l) {
/* First one is bigger */
lostack[top] = lo;
lostack[top] = lo;
histack[top++] = r;
lostack[top] = l;
lostack[top] = l;
histack[top++] = hi;
} else {
/* Second one is bigger */
lostack[top] = l;
lostack[top] = l;
histack[top++] = hi;
lostack[top] = lo;
lostack[top] = lo;
histack[top++] = r;
}
/* Should assert top <= STACKSIZE */
}
/*
* Ouch - even if I screwed up the quicksort above, the
* insertionsort below will cover up the problem - just a
* performance hit would be noticable.
*/
/* insertionsort is pretty fast on the partially sorted list */
if (insertionsort(array, size, compare) < 0)
return -1;
/* Success */
return 0;
}
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment