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Commit a64dc245 authored by Tim Peters's avatar Tim Peters
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Replaced samplesort with a stable, adaptive mergesort.

parent 92f81f2e
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Objects/listobject.c
View file @ a64dc245
......@@ -755,15 +755,16 @@ reverse_slice(PyObject **lo, PyObject **hi)
}
}
/* New quicksort implementation for arrays of object pointers.
Thanks to discussions with Tim Peters. */
/* Lots of code for an adaptive, stable, natural mergesort. There are many
* pieces to this algorithm; read listsort.txt for overviews and details.
*/
/* Comparison function. Takes care of calling a user-supplied
comparison function (any callable Python object). Calls the
standard comparison function, PyObject_RichCompareBool(), if the user-
supplied function is NULL.
Returns <0 on error, >0 if x < y, 0 if x >= y. */
* comparison function (any callable Python object). Calls the
* standard comparison function, PyObject_RichCompareBool(), if the user-
* supplied function is NULL.
* Returns -1 on error, 1 if x < y, 0 if x >= y.
*/
static int
islt(PyObject *x, PyObject *y, PyObject *compare)
{
......@@ -799,42 +800,6 @@ islt(PyObject *x, PyObject *y, PyObject *compare)
return i < 0;
}
/* MINSIZE is the smallest array that will get a full-blown samplesort
treatment; smaller arrays are sorted using binary insertion. It must
be at least 7 for the samplesort implementation to work. Binary
insertion does fewer compares, but can suffer O(N**2) data movement.
The more expensive compares, the larger MINSIZE should be. */
#define MINSIZE 100
/* MINPARTITIONSIZE is the smallest array slice samplesort will bother to
partition; smaller slices are passed to binarysort. It must be at
least 2, and no larger than MINSIZE. Setting it higher reduces the #
of compares slowly, but increases the amount of data movement quickly.
The value here was chosen assuming a compare costs ~25x more than
swapping a pair of memory-resident pointers -- but under that assumption,
changing the value by a few dozen more or less has aggregate effect
under 1%. So the value is crucial, but not touchy <wink>. */
#define MINPARTITIONSIZE 40
/* MAXMERGE is the largest number of elements we'll always merge into
a known-to-be sorted chunk via binary insertion, regardless of the
size of that chunk. Given a chunk of N sorted elements, and a group
of K unknowns, the largest K for which it's better to do insertion
(than a full-blown sort) is a complicated function of N and K mostly
involving the expected number of compares and data moves under each
approach, and the relative cost of those operations on a specific
architecure. The fixed value here is conservative, and should be a
clear win regardless of architecture or N. */
#define MAXMERGE 15
/* STACKSIZE is the size of our work stack. A rough estimate is that
this allows us to sort arrays of size N where
N / ln(N) = MINPARTITIONSIZE * 2**STACKSIZE, so 60 is more than enough
for arrays of size 2**64. Because we push the biggest partition
first, the worst case occurs when all subarrays are always partitioned
exactly in two. */
#define STACKSIZE 60
/* Compare X to Y via islt(). Goto "fail" if the comparison raises an
error. Else "k" is set to true iff X<Y, and an "if (k)" block is
started. It makes more sense in context <wink>. X and Y are PyObject*s.
......@@ -853,7 +818,6 @@ islt(PyObject *x, PyObject *y, PyObject *compare)
Even in case of error, the output slice will be some permutation of
the input (nothing is lost or duplicated).
*/
static int
binarysort(PyObject **lo, PyObject **hi, PyObject **start, PyObject *compare)
/* compare -- comparison function object, or NULL for default */
......@@ -902,420 +866,822 @@ binarysort(PyObject **lo, PyObject **hi, PyObject **start, PyObject *compare)
return -1;
}
/* samplesortslice is the sorting workhorse.
[lo, hi) is a contiguous slice of a list, to be sorted in place.
On entry, must have lo <= hi,
If islt() complains return -1, else 0.
Even in case of error, the output slice will be some permutation of
the input (nothing is lost or duplicated).
/*
Return the length of the run beginning at lo, in the slice [lo, hi). lo < hi
is required on entry. "A run" is the longest ascending sequence, with
samplesort is basically quicksort on steroids: a power of 2 close
to n/ln(n) is computed, and that many elements (less 1) are picked at
random from the array and sorted. These 2**k - 1 elements are then
used as preselected pivots for an equal number of quicksort
partitioning steps, partitioning the slice into 2**k chunks each of
size about ln(n). These small final chunks are then usually handled
by binarysort. Note that when k=1, this is roughly the same as an
ordinary quicksort using a random pivot, and when k=2 this is roughly
a median-of-3 quicksort. From that view, using k ~= lg(n/ln(n)) makes
this a "median of n/ln(n)" quicksort. You can also view it as a kind
of bucket sort, where 2**k-1 bucket boundaries are picked dynamically.
The large number of samples makes a quadratic-time case almost
impossible, and asymptotically drives the average-case number of
compares from quicksort's 2 N ln N (or 12/7 N ln N for the median-of-
3 variant) down to N lg N.
We also play lots of low-level tricks to cut the number of compares.
Very obscure: To avoid using extra memory, the PPs are stored in the
array and shuffled around as partitioning proceeds. At the start of a
partitioning step, we'll have 2**m-1 (for some m) PPs in sorted order,
adjacent (either on the left or the right!) to a chunk of X elements
that are to be partitioned: P X or X P. In either case we need to
shuffle things *in place* so that the 2**(m-1) smaller PPs are on the
left, followed by the PP to be used for this step (that's the middle
of the PPs), followed by X, followed by the 2**(m-1) larger PPs:
P X or X P -> Psmall pivot X Plarge
and the order of the PPs must not be altered. It can take a while
to realize this isn't trivial! It can take even longer <wink> to
understand why the simple code below works, using only 2**(m-1) swaps.
The key is that the order of the X elements isn't necessarily
preserved: X can end up as some cyclic permutation of its original
order. That's OK, because X is unsorted anyway. If the order of X
had to be preserved too, the simplest method I know of using O(1)
scratch storage requires len(X) + 2**(m-1) swaps, spread over 2 passes.
Since len(X) is typically several times larger than 2**(m-1), that
would slow things down.
*/
lo[0] <= lo[1] <= lo[2] <= ...
struct SamplesortStackNode {
/* Represents a slice of the array, from (& including) lo up
to (but excluding) hi. "extra" additional & adjacent elements
are pre-selected pivots (PPs), spanning [lo-extra, lo) if
extra > 0, or [hi, hi-extra) if extra < 0. The PPs are
already sorted, but nothing is known about the other elements
in [lo, hi). |extra| is always one less than a power of 2.
When extra is 0, we're out of PPs, and the slice must be
sorted by some other means. */
PyObject **lo;
PyObject **hi;
int extra;
};
or the longest descending sequence, with
/* The number of PPs we want is 2**k - 1, where 2**k is as close to
N / ln(N) as possible. So k ~= lg(N / ln(N)). Calling libm routines
is undesirable, so cutoff values are canned in the "cutoff" table
below: cutoff[i] is the smallest N such that k == CUTOFFBASE + i. */
#define CUTOFFBASE 4
static long cutoff[] = {
43, /* smallest N such that k == 4 */
106, /* etc */
250,
576,
1298,
2885,
6339,
13805,
29843,
64116,
137030,
291554,
617916,
1305130,
2748295,
5771662,
12091672,
25276798,
52734615,
109820537,
228324027,
473977813,
982548444, /* smallest N such that k == 26 */
2034159050 /* largest N that fits in signed 32-bit; k == 27 */
};
lo[0] > lo[1] > lo[2] > ...
Boolean *descending is set to 0 in the former case, or to 1 in the latter.
For its intended use in a stable mergesort, the strictness of the defn of
"descending" is needed so that the caller can safely reverse a descending
sequence without violating stability (strict > ensures there are no equal
elements to get out of order).
Returns -1 in case of error.
*/
static int
samplesortslice(PyObject **lo, PyObject **hi, PyObject *compare)
/* compare -- comparison function object, or NULL for default */
count_run(PyObject **lo, PyObject **hi, PyObject *compare, int *descending)
{
register PyObject **l, **r;
register PyObject *tmp, *pivot;
register int k;
int n, extra, top, extraOnRight;
struct SamplesortStackNode stack[STACKSIZE];
assert(lo <= hi);
n = hi - lo;
if (n < MINSIZE)
return binarysort(lo, hi, lo, compare);
/* ----------------------------------------------------------
* Normal case setup: a large array without obvious pattern.
* --------------------------------------------------------*/
/* extra := a power of 2 ~= n/ln(n), less 1.
First find the smallest extra s.t. n < cutoff[extra] */
for (extra = 0;
extra < sizeof(cutoff) / sizeof(cutoff[0]);
++extra) {
if (n < cutoff[extra])
break;
/* note that if we fall out of the loop, the value of
extra still makes *sense*, but may be smaller than
we would like (but the array has more than ~= 2**31
elements in this case!) */
}
/* Now k == extra - 1 + CUTOFFBASE. The smallest value k can
have is CUTOFFBASE-1, so
assert MINSIZE >= 2**(CUTOFFBASE-1) - 1 */
extra = (1 << (extra - 1 + CUTOFFBASE)) - 1;
/* assert extra > 0 and n >= extra */
/* Swap that many values to the start of the array. The
selection of elements is pseudo-random, but the same on
every run (this is intentional! timing algorithm changes is
a pain if timing varies across runs). */
{
unsigned int seed = n / extra; /* arbitrary */
unsigned int i;
for (i = 0; i < (unsigned)extra; ++i) {
/* j := random int in [i, n) */
unsigned int j;
seed = seed * 69069 + 7;
j = i + seed % (n - i);
tmp = lo[i]; lo[i] = lo[j]; lo[j] = tmp;
int k;
int n;
assert(lo < hi);
*descending = 0;
++lo;
if (lo == hi)
return 1;
n = 2;
IFLT(*lo, *(lo-1)) {
*descending = 1;
for (lo = lo+1; lo < hi; ++lo, ++n) {
IFLT(*lo, *(lo-1))
;
else
break;
}
}
else {
for (lo = lo+1; lo < hi; ++lo, ++n) {
IFLT(*lo, *(lo-1))
break;
}
}
/* Recursively sort the preselected pivots. */
if (samplesortslice(lo, lo + extra, compare) < 0)
goto fail;
return n;
fail:
return -1;
}
top = 0; /* index of available stack slot */
lo += extra; /* point to first unknown */
extraOnRight = 0; /* the PPs are at the left end */
/*
Locate the proper position of key in a sorted vector; if the vector contains
an element equal to key, return the position immediately to the left of
the leftmost equal element. [gallop_right() does the same except returns
the position to the right of the rightmost equal element (if any).]
/* ----------------------------------------------------------
* Partition [lo, hi), and repeat until out of work.
* --------------------------------------------------------*/
for (;;) {
assert(lo <= hi);
n = hi - lo;
"a" is a sorted vector with n elements, starting at a[0]. n must be > 0.
/* We may not want, or may not be able, to partition:
If n is small, it's quicker to insert.
If extra is 0, we're out of pivots, and *must* use
another method.
*/
if (n < MINPARTITIONSIZE || extra == 0) {
if (n >= MINSIZE) {
assert(extra == 0);
/* This is rare, since the average size
of a final block is only about
ln(original n). */
if (samplesortslice(lo, hi, compare) < 0)
goto fail;
}
else {
/* Binary insertion should be quicker,
and we can take advantage of the PPs
already being sorted. */
if (extraOnRight && extra) {
/* swap the PPs to the left end */
k = extra;
do {
tmp = *lo;
*lo = *hi;
*hi = tmp;
++lo; ++hi;
} while (--k);
}
if (binarysort(lo - extra, hi, lo,
compare) < 0)
goto fail;
}
"hint" is an index at which to begin the search, 0 <= hint < n. The closer
hint is to the final result, the faster this runs.
The return value is the int k in 0..n such that
a[k-1] < key <= a[k]
pretending that *(a-1) is minus infinity and a[n] is plus infinity. IOW,
key belongs at index k; or, IOW, the first k elements of a should precede
key, and the last n-k should follow key.
Returns -1 on error. See listsort.txt for info on the method.
*/
static int
gallop_left(PyObject *key, PyObject **a, int n, int hint, PyObject *compare)
{
int ofs;
int lastofs;
int k;
assert(key && a && n > 0 && hint >= 0 && hint < n);
/* Find another slice to work on. */
if (--top < 0)
break; /* no more -- done! */
lo = stack[top].lo;
hi = stack[top].hi;
extra = stack[top].extra;
extraOnRight = 0;
if (extra < 0) {
extraOnRight = 1;
extra = -extra;
a += hint;
lastofs = 0;
ofs = 1;
IFLT(*a, key) {
/* a[hint] < key -- gallop right, until
* a[hint + lastofs] < key <= a[hint + ofs]
*/
const int maxofs = n - hint; /* &a[n-1] is highest */
while (ofs < maxofs) {
IFLT(a[ofs], key) {
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) /* int overflow */
ofs = maxofs;
}
continue;
else /* key <= a[hint + ofs] */
break;
}
if (ofs > maxofs)
ofs = maxofs;
/* Translate back to offsets relative to &a[0]. */
lastofs += hint;
ofs += hint;
}
else {
/* key <= a[hint] -- gallop left, until
* a[hint - ofs] < key <= a[hint - lastofs]
*/
const int maxofs = hint + 1; /* &a[0] is lowest */
while (ofs < maxofs) {
IFLT(*(a-ofs), key)
break;
/* key <= a[hint - ofs] */
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) /* int overflow */
ofs = maxofs;
}
if (ofs > maxofs)
ofs = maxofs;
/* Translate back to positive offsets relative to &a[0]. */
k = lastofs;
lastofs = hint - ofs;
ofs = hint - k;
}
a -= hint;
assert(-1 <= lastofs && lastofs < ofs && ofs <= n);
/* Now a[lastofs] < key <= a[ofs], so key belongs somewhere to the
* right of lastofs but no farther right than ofs. Do a binary
* search, with invariant a[lastofs-1] < key <= a[ofs].
*/
++lastofs;
while (lastofs < ofs) {
int m = lastofs + ((ofs - lastofs) >> 1);
IFLT(a[m], key)
lastofs = m+1; /* a[m] < key */
else
ofs = m; /* key <= a[m] */
}
assert(lastofs == ofs); /* so a[ofs-1] < key <= a[ofs] */
return ofs;
/* Pretend the PPs are indexed 0, 1, ..., extra-1.
Then our preselected pivot is at (extra-1)/2, and we
want to move the PPs before that to the left end of
the slice, and the PPs after that to the right end.
The following section changes extra, lo, hi, and the
slice such that:
[lo-extra, lo) contains the smaller PPs.
*lo == our PP.
(lo, hi) contains the unknown elements.
[hi, hi+extra) contains the larger PPs.
fail:
return -1;
}
/*
Exactly like gallop_left(), except that if key already exists in a[0:n],
finds the position immediately to the right of the rightmost equal value.
The return value is the int k in 0..n such that
a[k-1] <= key < a[k]
or -1 if error.
The code duplication is massive, but this is enough different given that
we're sticking to "<" comparisons that it's much harder to follow if
written as one routine with yet another "left or right?" flag.
*/
static int
gallop_right(PyObject *key, PyObject **a, int n, int hint, PyObject *compare)
{
int ofs;
int lastofs;
int k;
assert(key && a && n > 0 && hint >= 0 && hint < n);
a += hint;
lastofs = 0;
ofs = 1;
IFLT(key, *a) {
/* key < a[hint] -- gallop left, until
* a[hint - ofs] <= key < a[hint - lastofs]
*/
const int maxofs = hint + 1; /* &a[0] is lowest */
while (ofs < maxofs) {
IFLT(key, *(a-ofs)) {
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) /* int overflow */
ofs = maxofs;
}
else /* a[hint - ofs] <= key */
break;
}
if (ofs > maxofs)
ofs = maxofs;
/* Translate back to positive offsets relative to &a[0]. */
k = lastofs;
lastofs = hint - ofs;
ofs = hint - k;
}
else {
/* a[hint] <= key -- gallop right, until
* a[hint + lastofs] <= key < a[hint + ofs]
*/
k = extra >>= 1; /* num PPs to move */
if (extraOnRight) {
/* Swap the smaller PPs to the left end.
Note that this loop actually moves k+1 items:
the last is our PP */
do {
tmp = *lo; *lo = *hi; *hi = tmp;
++lo; ++hi;
} while (k--);
const int maxofs = n - hint; /* &a[n-1] is highest */
while (ofs < maxofs) {
IFLT(key, a[ofs])
break;
/* a[hint + ofs] <= key */
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) /* int overflow */
ofs = maxofs;
}
else {
/* Swap the larger PPs to the right end. */
while (k--) {
--lo; --hi;
tmp = *lo; *lo = *hi; *hi = tmp;
if (ofs > maxofs)
ofs = maxofs;
/* Translate back to offsets relative to &a[0]. */
lastofs += hint;
ofs += hint;
}
a -= hint;
assert(-1 <= lastofs && lastofs < ofs && ofs <= n);
/* Now a[lastofs] <= key < a[ofs], so key belongs somewhere to the
* right of lastofs but no farther right than ofs. Do a binary
* search, with invariant a[lastofs-1] <= key < a[ofs].
*/
++lastofs;
while (lastofs < ofs) {
int m = lastofs + ((ofs - lastofs) >> 1);
IFLT(key, a[m])
ofs = m; /* key < a[m] */
else
lastofs = m+1; /* a[m] <= key */
}
assert(lastofs == ofs); /* so a[ofs-1] <= key < a[ofs] */
return ofs;
fail:
return -1;
}
/* The maximum number of entries in a MergeState's pending-runs stack.
* This is enough to sort arrays of size up to about
* 32 * phi ** MAX_MERGE_PENDING
* where phi ~= 1.618. 85 is ridiculouslylarge enough, good for an array
* with 2**64 elements.
*/
#define MAX_MERGE_PENDING 85
/* If a run wins MIN_GALLOP times in a row, we switch to galloping mode,
* and stay there until both runs win less often than MIN_GALLOP
* consecutive times. See listsort.txt for more info.
*/
#define MIN_GALLOP 8
/* Avoid malloc for small temp arrays. */
#define MERGESTATE_TEMP_SIZE 256
/* One MergeState exists on the stack per invocation of mergesort. It's just
* a convenient way to pass state around among the helper functions.
*/
typedef struct s_MergeState {
/* The user-supplied comparison function. or NULL if none given. */
PyObject *compare;
/* 'a' is temp storage to help with merges. It contains room for
* alloced entries.
*/
PyObject **a; /* may point to temparray below */
int alloced;
/* A stack of n pending runs yet to be merged. Run #i starts at
* address base[i] and extends for len[i] elements. It's always
* true (so long as the indices are in bounds) that
*
* base[i] + len[i] == base[i+1]
*
* so we could cut the storage for this, but it's a minor amount,
* and keeping all the info explicit simplifies the code.
*/
int n;
PyObject **base[MAX_MERGE_PENDING];
int len[MAX_MERGE_PENDING];
/* 'a' points to this when possible, rather than muck with malloc. */
PyObject *temparray[MERGESTATE_TEMP_SIZE];
} MergeState;
/* Conceptually a MergeState's constructor. */
static void
merge_init(MergeState *ms, PyObject *compare)
{
assert(ms != NULL);
ms->compare = compare;
ms->a = ms->temparray;
ms->alloced = MERGESTATE_TEMP_SIZE;
ms->n = 0;
}
/* Free all the temp memory owned by the MergeState. This must be called
* when you're done with a MergeState, and may be called before then if
* you want to free the temp memory early.
*/
static void
merge_freemem(MergeState *ms)
{
assert(ms != NULL);
if (ms->a != ms->temparray)
PyMem_Free(ms->a);
ms->a = ms->temparray;
ms->alloced = MERGESTATE_TEMP_SIZE;
}
/* Ensure enough temp memory for 'need' array slots is available.
* Returns 0 on success and -1 if the memory can't be gotten.
*/
static int
merge_getmem(MergeState *ms, int need)
{
assert(ms != NULL);
if (need <= ms->alloced)
return 0;
/* Don't realloc! That can cost cycles to copy the old data, but
* we don't care what's in the block.
*/
merge_freemem(ms);
ms->a = (PyObject **)PyMem_Malloc(need * sizeof(PyObject*));
if (ms->a) {
ms->alloced = need;
return 0;
}
PyErr_NoMemory();
merge_freemem(ms); /* reset to sane state */
return -1;
}
#define MERGE_GETMEM(MS, NEED) ((NEED) <= (MS)->alloced ? 0 : \
merge_getmem(MS, NEED))
/* Merge the na elements starting at pa with the nb elements starting at pb
* in a stable way, in-place. na and nb must be > 0, and pa + na == pb.
* Must also have that *pb < *pa, that pa[na-1] belongs at the end of the
* merge, and should have na <= nb. See listsort.txt for more info.
* Return 0 if successful, -1 if error.
*/
static int
merge_lo(MergeState *ms, PyObject **pa, int na, PyObject **pb, int nb)
{
int k;
PyObject *compare;
PyObject **dest;
int result = -1; /* guilty until proved innocent */
assert(ms && pa && pb && na > 0 && nb > 0 && pa + na == pb);
if (MERGE_GETMEM(ms, na) < 0)
return -1;
memcpy(ms->a, pa, na * sizeof(PyObject*));
dest = pa;
pa = ms->a;
*dest++ = *pb++;
--nb;
if (nb == 0)
goto Succeed;
if (na == 1)
goto CopyB;
compare = ms->compare;
for (;;) {
int acount = 0; /* # of times A won in a row */
int bcount = 0; /* # of times B won in a row */
/* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
for (;;) {
k = islt(*pb, *pa, compare);
if (k) {
if (k < 0)
goto Fail;
*dest++ = *pb++;
++bcount;
acount = 0;
--nb;
if (nb == 0)
goto Succeed;
if (bcount >= MIN_GALLOP)
break;
}
}
--lo; /* *lo is now our PP */
pivot = *lo;
/* Now an almost-ordinary quicksort partition step.
Note that most of the time is spent here!
Only odd thing is that we partition into < and >=,
instead of the usual <= and >=. This helps when
there are lots of duplicates of different values,
because it eventually tends to make subfiles
"pure" (all duplicates), and we special-case for
duplicates later. */
l = lo + 1;
r = hi - 1;
assert(lo < l && l < r && r < hi);
else {
*dest++ = *pa++;
++acount;
bcount = 0;
--na;
if (na == 1)
goto CopyB;
if (acount >= MIN_GALLOP)
break;
}
}
/* One run is winning so consistently that galloping may
* be a huge win. So try that, and continue galloping until
* (if ever) neither run appears to be winning consistently
* anymore.
*/
do {
/* slide l right, looking for key >= pivot */
do {
IFLT(*l, pivot)
++l;
else
k = gallop_right(*pb, pa, na, 0, compare);
acount = k;
if (k) {
if (k < 0)
goto Fail;
memcpy(dest, pa, k * sizeof(PyObject *));
dest += k;
pa += k;
na -= k;
if (na == 1)
goto CopyB;
/* na==0 is impossible now if the comparison
* function is consistent, but we can't assume
* that it is.
*/
if (na == 0)
goto Succeed;
}
*dest++ = *pb++;
--nb;
if (nb == 0)
goto Succeed;
k = gallop_left(*pa, pb, nb, 0, compare);
bcount = k;
if (k) {
if (k < 0)
goto Fail;
memmove(dest, pb, k * sizeof(PyObject *));
dest += k;
pb += k;
nb -= k;
if (nb == 0)
goto Succeed;
}
*dest++ = *pa++;
--na;
if (na == 1)
goto CopyB;
} while (acount >= MIN_GALLOP || bcount >= MIN_GALLOP);
}
Succeed:
result = 0;
Fail:
if (na)
memcpy(dest, pa, na * sizeof(PyObject*));
return result;
CopyB:
assert(na == 1 && nb > 0);
/* The last element of pa belongs at the end of the merge. */
memmove(dest, pb, nb * sizeof(PyObject *));
dest[nb] = *pa;
return 0;
}
/* Merge the na elements starting at pa with the nb elements starting at pb
* in a stable way, in-place. na and nb must be > 0, and pa + na == pb.
* Must also have that *pb < *pa, that pa[na-1] belongs at the end of the
* merge, and should have na >= nb. See listsort.txt for more info.
* Return 0 if successful, -1 if error.
*/
static int
merge_hi(MergeState *ms, PyObject **pa, int na, PyObject **pb, int nb)
{
int k;
PyObject *compare;
PyObject **dest;
int result = -1; /* guilty until proved innocent */
PyObject **basea;
PyObject **baseb;
assert(ms && pa && pb && na > 0 && nb > 0 && pa + na == pb);
if (MERGE_GETMEM(ms, nb) < 0)
return -1;
dest = pb + nb - 1;
memcpy(ms->a, pb, nb * sizeof(PyObject*));
basea = pa;
baseb = ms->a;
pb = ms->a + nb - 1;
pa += na - 1;
*dest-- = *pa--;
--na;
if (na == 0)
goto Succeed;
if (nb == 1)
goto CopyA;
compare = ms->compare;
for (;;) {
int acount = 0; /* # of times A won in a row */
int bcount = 0; /* # of times B won in a row */
/* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
for (;;) {
k = islt(*pb, *pa, compare);
if (k) {
if (k < 0)
goto Fail;
*dest-- = *pa--;
++acount;
bcount = 0;
--na;
if (na == 0)
goto Succeed;
if (acount >= MIN_GALLOP)
break;
} while (l < r);
/* slide r left, looking for key < pivot */
while (l < r) {
register PyObject *rval = *r--;
IFLT(rval, pivot) {
/* swap and advance */
r[1] = *l;
*l++ = rval;
}
else {
*dest-- = *pb--;
++bcount;
acount = 0;
--nb;
if (nb == 1)
goto CopyA;
if (bcount >= MIN_GALLOP)
break;
}
}
}
} while (l < r);
/* One run is winning so consistently that galloping may
* be a huge win. So try that, and continue galloping until
* (if ever) neither run appears to be winning consistently
* anymore.
*/
do {
k = gallop_right(*pb, basea, na, na-1, compare);
if (k < 0)
goto Fail;
k = na - k;
acount = k;
if (k) {
dest -= k;
pa -= k;
memmove(dest+1, pa+1, k * sizeof(PyObject *));
na -= k;
if (na == 0)
goto Succeed;
}
*dest-- = *pb--;
--nb;
if (nb == 1)
goto CopyA;
k = gallop_left(*pa, baseb, nb, nb-1, compare);
if (k < 0)
goto Fail;
k = nb - k;
bcount = k;
if (k) {
dest -= k;
pb -= k;
memcpy(dest+1, pb+1, k * sizeof(PyObject *));
nb -= k;
if (nb == 1)
goto CopyA;
/* nb==0 is impossible now if the comparison
* function is consistent, but we can't assume
* that it is.
*/
if (nb == 0)
goto Succeed;
}
*dest-- = *pa--;
--na;
if (na == 0)
goto Succeed;
} while (acount >= MIN_GALLOP || bcount >= MIN_GALLOP);
}
Succeed:
result = 0;
Fail:
if (nb)
memcpy(dest-(nb-1), baseb, nb * sizeof(PyObject*));
return result;
CopyA:
assert(nb == 1 && na > 0);
/* The first element of pb belongs at the front of the merge. */
dest -= na;
pa -= na;
memmove(dest+1, pa+1, na * sizeof(PyObject *));
*dest = *pb;
return 0;
}
assert(lo < r && r <= l && l < hi);
/* everything to the left of l is < pivot, and
everything to the right of r is >= pivot */
/* Merge the two runs at stack indices i and i+1.
* Returns 0 on success, -1 on error.
*/
static int
merge_at(MergeState *ms, int i)
{
PyObject **pa, **pb;
int na, nb;
int k;
PyObject *compare;
assert(ms != NULL);
assert(ms->n >= 2);
assert(i >= 0);
assert(i == ms->n - 2 || i == ms->n - 3);
pa = ms->base[i];
pb = ms->base[i+1];
na = ms->len[i];
nb = ms->len[i+1];
assert(na > 0 && nb > 0);
assert(pa + na == pb);
/* Record the length of the combined runs; if i is the 3rd-last
* run now, also slide over the last run (which isn't involved
* in this merge). The current run i+1 goes away in any case.
*/
if (i == ms->n - 3) {
ms->len[i+1] = ms->len[i+2];
ms->base[i+1] = ms->base[i+2];
}
ms->len[i] = na + nb;
--ms->n;
if (l == r) {
IFLT(*r, pivot)
++l;
else
--r;
}
assert(lo <= r && r+1 == l && l <= hi);
/* assert r == lo or a[r] < pivot */
assert(*lo == pivot);
/* assert l == hi or a[l] >= pivot */
/* Swap the pivot into "the middle", so we can henceforth
ignore it. */
*lo = *r;
*r = pivot;
/* The following is true now, & will be preserved:
All in [lo,r) are < pivot
All in [r,l) == pivot (& so can be ignored)
All in [l,hi) are >= pivot */
/* Check for duplicates of the pivot. One compare is
wasted if there are no duplicates, but can win big
when there are.
Tricky: we're sticking to "<" compares, so deduce
equality indirectly. We know pivot <= *l, so they're
equal iff not pivot < *l.
*/
while (l < hi) {
/* pivot <= *l known */
IFLT(pivot, *l)
break;
else
/* <= and not < implies == */
++l;
}
/* Where does b start in a? Elements in a before that can be
* ignored (already in place).
*/
compare = ms->compare;
k = gallop_right(*pb, pa, na, 0, compare);
if (k < 0)
return -1;
pa += k;
na -= k;
if (na == 0)
return 0;
/* Where does a end in b? Elements in b after that can be
* ignored (already in place).
*/
nb = gallop_left(pa[na-1], pb, nb, nb-1, compare);
if (nb <= 0)
return nb;
assert(lo <= r && r < l && l <= hi);
/* Partitions are [lo, r) and [l, hi)
:ush fattest first; remember we still have extra PPs
to the left of the left chunk and to the right of
the right chunk! */
assert(top < STACKSIZE);
if (r - lo <= hi - l) {
/* second is bigger */
stack[top].lo = l;
stack[top].hi = hi;
stack[top].extra = -extra;
hi = r;
extraOnRight = 0;
/* Merge what remains of the runs, using a temp array with
* min(na, nb) elements.
*/
if (na <= nb)
return merge_lo(ms, pa, na, pb, nb);
else
return merge_hi(ms, pa, na, pb, nb);
}
/* Examine the stack of runs waiting to be merged, merging adjacent runs
* until the stack invariants are re-established:
*
* 1. len[-3] > len[-2] + len[-1]
* 2. len[-2] > len[-1]
*
* See listsort.txt for more info.
*
* Returns 0 on success, -1 on error.
*/
static int
merge_collapse(MergeState *ms)
{
int *len = ms->len;
assert(ms);
while (ms->n > 1) {
int n = ms->n - 2;
if (n > 0 && len[n-1] <= len[n] + len[n+1]) {
if (len[n-1] < len[n+1])
--n;
if (merge_at(ms, n) < 0)
return -1;
}
else {
/* first is bigger */
stack[top].lo = lo;
stack[top].hi = r;
stack[top].extra = extra;
lo = l;
extraOnRight = 1;
else if (len[n] <= len[n+1]) {
if (merge_at(ms, n) < 0)
return -1;
}
++top;
} /* end of partitioning loop */
else
break;
}
return 0;
}
/* Regardless of invariants, merge all runs on the stack until only one
* remains. This is used at the end of the mergesort.
*
* Returns 0 on success, -1 on error.
*/
static int
merge_force_collapse(MergeState *ms)
{
int *len = ms->len;
assert(ms);
while (ms->n > 1) {
int n = ms->n - 2;
if (n > 0 && len[n-1] < len[n+1])
--n;
if (merge_at(ms, n) < 0)
return -1;
}
return 0;
}
fail:
return -1;
/* Compute a good value for the minimum run length; natural runs shorter
* than this are boosted artificially via binary insertion.
*
* If n < 64, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return 32.
* Else return an int k, 32 <= k <= 64, such that n/k is close to, but
* strictly less than, an exact power of 2.
*
* See listsort.txt for more info.
*/
static int
merge_compute_minrun(int n)
{
int r = 0; /* becomes 1 if any 1 bits are shifted off */
assert(n >= 0);
while (n >= 64) {
r |= n & 1;
n >>= 1;
}
return n + r;
}
static PyTypeObject immutable_list_type;
/* An adaptive, stable, natural mergesort. See listsort.txt.
* Returns Py_None on success, NULL on error. Even in case of error, the
* list will be some permutation of its input state (nothing is lost or
* duplicated).
*/
static PyObject *
listsort(PyListObject *self, PyObject *args)
{
int k;
PyObject *compare = NULL;
PyObject **hi, **p;
MergeState ms;
PyObject **lo, **hi;
int nremaining;
int minrun;
PyTypeObject *savetype;
PyObject *compare = NULL;
PyObject *result = NULL; /* guilty until proved innocent */
assert(self != NULL);
if (args != NULL) {
if (!PyArg_ParseTuple(args, "|O:sort", &compare))
if (!PyArg_ParseTuple(args, "|O:msort", &compare))
return NULL;
}
merge_init(&ms, compare);
savetype = self->ob_type;
if (self->ob_size < 2) {
k = 0;
goto done;
}
self->ob_type = &immutable_list_type;
hi = self->ob_item + self->ob_size;
/* Set p to the largest value such that [lo, p) is sorted.
This catches the already-sorted case, the all-the-same
case, and the appended-a-few-elements-to-a-sorted-list case.
If the array is unsorted, we're very likely to get out of
the loop fast, so the test is cheap if it doesn't pay off.
*/
for (p = self->ob_item + 1; p < hi; ++p) {
IFLT(*p, *(p-1))
break;
}
/* [lo, p) is sorted, [p, hi) unknown. Get out cheap if there are
few unknowns, or few elements in total. */
if (hi - p <= MAXMERGE || self->ob_size < MINSIZE) {
k = binarysort(self->ob_item, hi, p, compare);
goto done;
}
/* Check for the array already being reverse-sorted, or that with
a few elements tacked on to the end. */
for (p = self->ob_item + 1; p < hi; ++p) {
IFLT(*(p-1), *p)
break;
}
/* [lo, p) is reverse-sorted, [p, hi) unknown. */
if (hi - p <= MAXMERGE) {
/* Reverse the reversed prefix, then insert the tail */
reverse_slice(self->ob_item, p);
k = binarysort(self->ob_item, hi, p, compare);
goto done;
}
nremaining = self->ob_size;
if (nremaining < 2)
goto succeed;
/* March over the array once, left to right, finding natural runs,
* and extending short natural runs to minrun elements.
*/
lo = self->ob_item;
hi = lo + nremaining;
minrun = merge_compute_minrun(nremaining);
do {
int descending;
int n;
/* A large array without obvious pattern. */
k = samplesortslice(self->ob_item, hi, compare);
/* Identify next run. */
n = count_run(lo, hi, compare, &descending);
if (n < 0)
goto fail;
if (descending)
reverse_slice(lo, lo + n);
/* If short, extend to min(minrun, nremaining). */
if (n < minrun) {
const int force = nremaining <= minrun ?
nremaining : minrun;
if (binarysort(lo, lo + force, lo + n, compare) < 0)
goto fail;
n = force;
}
/* Push run onto pending-runs stack, and maybe merge. */
assert(ms.n < MAX_MERGE_PENDING);
ms.base[ms.n] = lo;
ms.len[ms.n] = n;
++ms.n;
if (merge_collapse(&ms) < 0)
goto fail;
/* Advance to find next run. */
lo += n;
nremaining -= n;
} while (nremaining);
assert(lo == hi);
if (merge_force_collapse(&ms) < 0)
goto fail;
assert(ms.n == 1);
assert(ms.base[0] == self->ob_item);
assert(ms.len[0] == self->ob_size);
done: /* The IFLT macro requires a label named "fail". */;
succeed:
result = Py_None;
fail:
self->ob_type = savetype;
if (k >= 0) {
Py_INCREF(Py_None);
return Py_None;
}
else
return NULL;
merge_freemem(&ms);
Py_XINCREF(result);
return result;
}
#undef IFLT
int
......@@ -1645,7 +2011,7 @@ PyDoc_STRVAR(count_doc,
PyDoc_STRVAR(reverse_doc,
"L.reverse() -- reverse *IN PLACE*");
PyDoc_STRVAR(sort_doc,
"L.sort([cmpfunc]) -- sort *IN PLACE*; if given, cmpfunc(x, y) -> -1, 0, 1");
"L.sort([cmpfunc]) -- stable sort *IN PLACE*; cmpfunc(x, y) -> -1, 0, 1");
static PyMethodDef list_methods[] = {
{"append", (PyCFunction)listappend, METH_O, append_doc},
......@@ -1657,7 +2023,7 @@ static PyMethodDef list_methods[] = {
{"count", (PyCFunction)listcount, METH_O, count_doc},
{"reverse", (PyCFunction)listreverse, METH_NOARGS, reverse_doc},
{"sort", (PyCFunction)listsort, METH_VARARGS, sort_doc},
{NULL, NULL} /* sentinel */
{NULL, NULL} /* sentinel */
};
static PySequenceMethods list_as_sequence = {
......
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