1/* Drop in replacement for heapq.py
2
3C implementation derived directly from heapq.py in Py2.3
4which was written by Kevin O'Connor, augmented by Tim Peters,
5annotated by François Pinard, and converted to C by Raymond Hettinger.
6
7*/
8
9#include "Python.h"
10
11/* Older implementations of heapq used Py_LE for comparisons.  Now, it uses
12   Py_LT so it will match min(), sorted(), and bisect().  Unfortunately, some
13   client code (Twisted for example) relied on Py_LE, so this little function
14   restores compatibility by trying both.
15*/
16static int
17cmp_lt(PyObject *x, PyObject *y)
18{
19    int cmp;
20    static PyObject *lt = NULL;
21
22    if (lt == NULL) {
23        lt = PyString_FromString("__lt__");
24        if (lt == NULL)
25            return -1;
26    }
27    if (PyObject_HasAttr(x, lt))
28        return PyObject_RichCompareBool(x, y, Py_LT);
29    cmp = PyObject_RichCompareBool(y, x, Py_LE);
30    if (cmp != -1)
31        cmp = 1 - cmp;
32    return cmp;
33}
34
35static int
36_siftdown(PyListObject *heap, Py_ssize_t startpos, Py_ssize_t pos)
37{
38    PyObject *newitem, *parent;
39    int cmp;
40    Py_ssize_t parentpos;
41
42    assert(PyList_Check(heap));
43    if (pos >= PyList_GET_SIZE(heap)) {
44        PyErr_SetString(PyExc_IndexError, "index out of range");
45        return -1;
46    }
47
48    newitem = PyList_GET_ITEM(heap, pos);
49    Py_INCREF(newitem);
50    /* Follow the path to the root, moving parents down until finding
51       a place newitem fits. */
52    while (pos > startpos){
53        parentpos = (pos - 1) >> 1;
54        parent = PyList_GET_ITEM(heap, parentpos);
55        cmp = cmp_lt(newitem, parent);
56        if (cmp == -1) {
57            Py_DECREF(newitem);
58            return -1;
59        }
60        if (cmp == 0)
61            break;
62        Py_INCREF(parent);
63        Py_DECREF(PyList_GET_ITEM(heap, pos));
64        PyList_SET_ITEM(heap, pos, parent);
65        pos = parentpos;
66    }
67    Py_DECREF(PyList_GET_ITEM(heap, pos));
68    PyList_SET_ITEM(heap, pos, newitem);
69    return 0;
70}
71
72static int
73_siftup(PyListObject *heap, Py_ssize_t pos)
74{
75    Py_ssize_t startpos, endpos, childpos, rightpos;
76    int cmp;
77    PyObject *newitem, *tmp;
78
79    assert(PyList_Check(heap));
80    endpos = PyList_GET_SIZE(heap);
81    startpos = pos;
82    if (pos >= endpos) {
83        PyErr_SetString(PyExc_IndexError, "index out of range");
84        return -1;
85    }
86    newitem = PyList_GET_ITEM(heap, pos);
87    Py_INCREF(newitem);
88
89    /* Bubble up the smaller child until hitting a leaf. */
90    childpos = 2*pos + 1;    /* leftmost child position  */
91    while (childpos < endpos) {
92        /* Set childpos to index of smaller child.   */
93        rightpos = childpos + 1;
94        if (rightpos < endpos) {
95            cmp = cmp_lt(
96                PyList_GET_ITEM(heap, childpos),
97                PyList_GET_ITEM(heap, rightpos));
98            if (cmp == -1) {
99                Py_DECREF(newitem);
100                return -1;
101            }
102            if (cmp == 0)
103                childpos = rightpos;
104        }
105        /* Move the smaller child up. */
106        tmp = PyList_GET_ITEM(heap, childpos);
107        Py_INCREF(tmp);
108        Py_DECREF(PyList_GET_ITEM(heap, pos));
109        PyList_SET_ITEM(heap, pos, tmp);
110        pos = childpos;
111        childpos = 2*pos + 1;
112    }
113
114    /* The leaf at pos is empty now.  Put newitem there, and and bubble
115       it up to its final resting place (by sifting its parents down). */
116    Py_DECREF(PyList_GET_ITEM(heap, pos));
117    PyList_SET_ITEM(heap, pos, newitem);
118    return _siftdown(heap, startpos, pos);
119}
120
121static PyObject *
122heappush(PyObject *self, PyObject *args)
123{
124    PyObject *heap, *item;
125
126    if (!PyArg_UnpackTuple(args, "heappush", 2, 2, &heap, &item))
127        return NULL;
128
129    if (!PyList_Check(heap)) {
130        PyErr_SetString(PyExc_TypeError, "heap argument must be a list");
131        return NULL;
132    }
133
134    if (PyList_Append(heap, item) == -1)
135        return NULL;
136
137    if (_siftdown((PyListObject *)heap, 0, PyList_GET_SIZE(heap)-1) == -1)
138        return NULL;
139    Py_INCREF(Py_None);
140    return Py_None;
141}
142
143PyDoc_STRVAR(heappush_doc,
144"Push item onto heap, maintaining the heap invariant.");
145
146static PyObject *
147heappop(PyObject *self, PyObject *heap)
148{
149    PyObject *lastelt, *returnitem;
150    Py_ssize_t n;
151
152    if (!PyList_Check(heap)) {
153        PyErr_SetString(PyExc_TypeError, "heap argument must be a list");
154        return NULL;
155    }
156
157    /* # raises appropriate IndexError if heap is empty */
158    n = PyList_GET_SIZE(heap);
159    if (n == 0) {
160        PyErr_SetString(PyExc_IndexError, "index out of range");
161        return NULL;
162    }
163
164    lastelt = PyList_GET_ITEM(heap, n-1) ;
165    Py_INCREF(lastelt);
166    PyList_SetSlice(heap, n-1, n, NULL);
167    n--;
168
169    if (!n)
170        return lastelt;
171    returnitem = PyList_GET_ITEM(heap, 0);
172    PyList_SET_ITEM(heap, 0, lastelt);
173    if (_siftup((PyListObject *)heap, 0) == -1) {
174        Py_DECREF(returnitem);
175        return NULL;
176    }
177    return returnitem;
178}
179
180PyDoc_STRVAR(heappop_doc,
181"Pop the smallest item off the heap, maintaining the heap invariant.");
182
183static PyObject *
184heapreplace(PyObject *self, PyObject *args)
185{
186    PyObject *heap, *item, *returnitem;
187
188    if (!PyArg_UnpackTuple(args, "heapreplace", 2, 2, &heap, &item))
189        return NULL;
190
191    if (!PyList_Check(heap)) {
192        PyErr_SetString(PyExc_TypeError, "heap argument must be a list");
193        return NULL;
194    }
195
196    if (PyList_GET_SIZE(heap) < 1) {
197        PyErr_SetString(PyExc_IndexError, "index out of range");
198        return NULL;
199    }
200
201    returnitem = PyList_GET_ITEM(heap, 0);
202    Py_INCREF(item);
203    PyList_SET_ITEM(heap, 0, item);
204    if (_siftup((PyListObject *)heap, 0) == -1) {
205        Py_DECREF(returnitem);
206        return NULL;
207    }
208    return returnitem;
209}
210
211PyDoc_STRVAR(heapreplace_doc,
212"Pop and return the current smallest value, and add the new item.\n\
213\n\
214This is more efficient than heappop() followed by heappush(), and can be\n\
215more appropriate when using a fixed-size heap.  Note that the value\n\
216returned may be larger than item!  That constrains reasonable uses of\n\
217this routine unless written as part of a conditional replacement:\n\n\
218    if item > heap[0]:\n\
219        item = heapreplace(heap, item)\n");
220
221static PyObject *
222heappushpop(PyObject *self, PyObject *args)
223{
224    PyObject *heap, *item, *returnitem;
225    int cmp;
226
227    if (!PyArg_UnpackTuple(args, "heappushpop", 2, 2, &heap, &item))
228        return NULL;
229
230    if (!PyList_Check(heap)) {
231        PyErr_SetString(PyExc_TypeError, "heap argument must be a list");
232        return NULL;
233    }
234
235    if (PyList_GET_SIZE(heap) < 1) {
236        Py_INCREF(item);
237        return item;
238    }
239
240    cmp = cmp_lt(PyList_GET_ITEM(heap, 0), item);
241    if (cmp == -1)
242        return NULL;
243    if (cmp == 0) {
244        Py_INCREF(item);
245        return item;
246    }
247
248    returnitem = PyList_GET_ITEM(heap, 0);
249    Py_INCREF(item);
250    PyList_SET_ITEM(heap, 0, item);
251    if (_siftup((PyListObject *)heap, 0) == -1) {
252        Py_DECREF(returnitem);
253        return NULL;
254    }
255    return returnitem;
256}
257
258PyDoc_STRVAR(heappushpop_doc,
259"Push item on the heap, then pop and return the smallest item\n\
260from the heap. The combined action runs more efficiently than\n\
261heappush() followed by a separate call to heappop().");
262
263static PyObject *
264heapify(PyObject *self, PyObject *heap)
265{
266    Py_ssize_t i, n;
267
268    if (!PyList_Check(heap)) {
269        PyErr_SetString(PyExc_TypeError, "heap argument must be a list");
270        return NULL;
271    }
272
273    n = PyList_GET_SIZE(heap);
274    /* Transform bottom-up.  The largest index there's any point to
275       looking at is the largest with a child index in-range, so must
276       have 2*i + 1 < n, or i < (n-1)/2.  If n is even = 2*j, this is
277       (2*j-1)/2 = j-1/2 so j-1 is the largest, which is n//2 - 1.  If
278       n is odd = 2*j+1, this is (2*j+1-1)/2 = j so j-1 is the largest,
279       and that's again n//2-1.
280    */
281    for (i=n/2-1 ; i>=0 ; i--)
282        if(_siftup((PyListObject *)heap, i) == -1)
283            return NULL;
284    Py_INCREF(Py_None);
285    return Py_None;
286}
287
288PyDoc_STRVAR(heapify_doc,
289"Transform list into a heap, in-place, in O(len(heap)) time.");
290
291static PyObject *
292nlargest(PyObject *self, PyObject *args)
293{
294    PyObject *heap=NULL, *elem, *iterable, *sol, *it, *oldelem;
295    Py_ssize_t i, n;
296    int cmp;
297
298    if (!PyArg_ParseTuple(args, "nO:nlargest", &n, &iterable))
299        return NULL;
300
301    it = PyObject_GetIter(iterable);
302    if (it == NULL)
303        return NULL;
304
305    heap = PyList_New(0);
306    if (heap == NULL)
307        goto fail;
308
309    for (i=0 ; i<n ; i++ ){
310        elem = PyIter_Next(it);
311        if (elem == NULL) {
312            if (PyErr_Occurred())
313                goto fail;
314            else
315                goto sortit;
316        }
317        if (PyList_Append(heap, elem) == -1) {
318            Py_DECREF(elem);
319            goto fail;
320        }
321        Py_DECREF(elem);
322    }
323    if (PyList_GET_SIZE(heap) == 0)
324        goto sortit;
325
326    for (i=n/2-1 ; i>=0 ; i--)
327        if(_siftup((PyListObject *)heap, i) == -1)
328            goto fail;
329
330    sol = PyList_GET_ITEM(heap, 0);
331    while (1) {
332        elem = PyIter_Next(it);
333        if (elem == NULL) {
334            if (PyErr_Occurred())
335                goto fail;
336            else
337                goto sortit;
338        }
339        cmp = cmp_lt(sol, elem);
340        if (cmp == -1) {
341            Py_DECREF(elem);
342            goto fail;
343        }
344        if (cmp == 0) {
345            Py_DECREF(elem);
346            continue;
347        }
348        oldelem = PyList_GET_ITEM(heap, 0);
349        PyList_SET_ITEM(heap, 0, elem);
350        Py_DECREF(oldelem);
351        if (_siftup((PyListObject *)heap, 0) == -1)
352            goto fail;
353        sol = PyList_GET_ITEM(heap, 0);
354    }
355sortit:
356    if (PyList_Sort(heap) == -1)
357        goto fail;
358    if (PyList_Reverse(heap) == -1)
359        goto fail;
360    Py_DECREF(it);
361    return heap;
362
363fail:
364    Py_DECREF(it);
365    Py_XDECREF(heap);
366    return NULL;
367}
368
369PyDoc_STRVAR(nlargest_doc,
370"Find the n largest elements in a dataset.\n\
371\n\
372Equivalent to:  sorted(iterable, reverse=True)[:n]\n");
373
374static int
375_siftdownmax(PyListObject *heap, Py_ssize_t startpos, Py_ssize_t pos)
376{
377    PyObject *newitem, *parent;
378    int cmp;
379    Py_ssize_t parentpos;
380
381    assert(PyList_Check(heap));
382    if (pos >= PyList_GET_SIZE(heap)) {
383        PyErr_SetString(PyExc_IndexError, "index out of range");
384        return -1;
385    }
386
387    newitem = PyList_GET_ITEM(heap, pos);
388    Py_INCREF(newitem);
389    /* Follow the path to the root, moving parents down until finding
390       a place newitem fits. */
391    while (pos > startpos){
392        parentpos = (pos - 1) >> 1;
393        parent = PyList_GET_ITEM(heap, parentpos);
394        cmp = cmp_lt(parent, newitem);
395        if (cmp == -1) {
396            Py_DECREF(newitem);
397            return -1;
398        }
399        if (cmp == 0)
400            break;
401        Py_INCREF(parent);
402        Py_DECREF(PyList_GET_ITEM(heap, pos));
403        PyList_SET_ITEM(heap, pos, parent);
404        pos = parentpos;
405    }
406    Py_DECREF(PyList_GET_ITEM(heap, pos));
407    PyList_SET_ITEM(heap, pos, newitem);
408    return 0;
409}
410
411static int
412_siftupmax(PyListObject *heap, Py_ssize_t pos)
413{
414    Py_ssize_t startpos, endpos, childpos, rightpos;
415    int cmp;
416    PyObject *newitem, *tmp;
417
418    assert(PyList_Check(heap));
419    endpos = PyList_GET_SIZE(heap);
420    startpos = pos;
421    if (pos >= endpos) {
422        PyErr_SetString(PyExc_IndexError, "index out of range");
423        return -1;
424    }
425    newitem = PyList_GET_ITEM(heap, pos);
426    Py_INCREF(newitem);
427
428    /* Bubble up the smaller child until hitting a leaf. */
429    childpos = 2*pos + 1;    /* leftmost child position  */
430    while (childpos < endpos) {
431        /* Set childpos to index of smaller child.   */
432        rightpos = childpos + 1;
433        if (rightpos < endpos) {
434            cmp = cmp_lt(
435                PyList_GET_ITEM(heap, rightpos),
436                PyList_GET_ITEM(heap, childpos));
437            if (cmp == -1) {
438                Py_DECREF(newitem);
439                return -1;
440            }
441            if (cmp == 0)
442                childpos = rightpos;
443        }
444        /* Move the smaller child up. */
445        tmp = PyList_GET_ITEM(heap, childpos);
446        Py_INCREF(tmp);
447        Py_DECREF(PyList_GET_ITEM(heap, pos));
448        PyList_SET_ITEM(heap, pos, tmp);
449        pos = childpos;
450        childpos = 2*pos + 1;
451    }
452
453    /* The leaf at pos is empty now.  Put newitem there, and and bubble
454       it up to its final resting place (by sifting its parents down). */
455    Py_DECREF(PyList_GET_ITEM(heap, pos));
456    PyList_SET_ITEM(heap, pos, newitem);
457    return _siftdownmax(heap, startpos, pos);
458}
459
460static PyObject *
461nsmallest(PyObject *self, PyObject *args)
462{
463    PyObject *heap=NULL, *elem, *iterable, *los, *it, *oldelem;
464    Py_ssize_t i, n;
465    int cmp;
466
467    if (!PyArg_ParseTuple(args, "nO:nsmallest", &n, &iterable))
468        return NULL;
469
470    it = PyObject_GetIter(iterable);
471    if (it == NULL)
472        return NULL;
473
474    heap = PyList_New(0);
475    if (heap == NULL)
476        goto fail;
477
478    for (i=0 ; i<n ; i++ ){
479        elem = PyIter_Next(it);
480        if (elem == NULL) {
481            if (PyErr_Occurred())
482                goto fail;
483            else
484                goto sortit;
485        }
486        if (PyList_Append(heap, elem) == -1) {
487            Py_DECREF(elem);
488            goto fail;
489        }
490        Py_DECREF(elem);
491    }
492    n = PyList_GET_SIZE(heap);
493    if (n == 0)
494        goto sortit;
495
496    for (i=n/2-1 ; i>=0 ; i--)
497        if(_siftupmax((PyListObject *)heap, i) == -1)
498            goto fail;
499
500    los = PyList_GET_ITEM(heap, 0);
501    while (1) {
502        elem = PyIter_Next(it);
503        if (elem == NULL) {
504            if (PyErr_Occurred())
505                goto fail;
506            else
507                goto sortit;
508        }
509        cmp = cmp_lt(elem, los);
510        if (cmp == -1) {
511            Py_DECREF(elem);
512            goto fail;
513        }
514        if (cmp == 0) {
515            Py_DECREF(elem);
516            continue;
517        }
518
519        oldelem = PyList_GET_ITEM(heap, 0);
520        PyList_SET_ITEM(heap, 0, elem);
521        Py_DECREF(oldelem);
522        if (_siftupmax((PyListObject *)heap, 0) == -1)
523            goto fail;
524        los = PyList_GET_ITEM(heap, 0);
525    }
526
527sortit:
528    if (PyList_Sort(heap) == -1)
529        goto fail;
530    Py_DECREF(it);
531    return heap;
532
533fail:
534    Py_DECREF(it);
535    Py_XDECREF(heap);
536    return NULL;
537}
538
539PyDoc_STRVAR(nsmallest_doc,
540"Find the n smallest elements in a dataset.\n\
541\n\
542Equivalent to:  sorted(iterable)[:n]\n");
543
544static PyMethodDef heapq_methods[] = {
545    {"heappush",        (PyCFunction)heappush,
546        METH_VARARGS,           heappush_doc},
547    {"heappushpop",     (PyCFunction)heappushpop,
548        METH_VARARGS,           heappushpop_doc},
549    {"heappop",         (PyCFunction)heappop,
550        METH_O,                 heappop_doc},
551    {"heapreplace",     (PyCFunction)heapreplace,
552        METH_VARARGS,           heapreplace_doc},
553    {"heapify",         (PyCFunction)heapify,
554        METH_O,                 heapify_doc},
555    {"nlargest",        (PyCFunction)nlargest,
556        METH_VARARGS,           nlargest_doc},
557    {"nsmallest",       (PyCFunction)nsmallest,
558        METH_VARARGS,           nsmallest_doc},
559    {NULL,              NULL}           /* sentinel */
560};
561
562PyDoc_STRVAR(module_doc,
563"Heap queue algorithm (a.k.a. priority queue).\n\
564\n\
565Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for\n\
566all k, counting elements from 0.  For the sake of comparison,\n\
567non-existing elements are considered to be infinite.  The interesting\n\
568property of a heap is that a[0] is always its smallest element.\n\
569\n\
570Usage:\n\
571\n\
572heap = []            # creates an empty heap\n\
573heappush(heap, item) # pushes a new item on the heap\n\
574item = heappop(heap) # pops the smallest item from the heap\n\
575item = heap[0]       # smallest item on the heap without popping it\n\
576heapify(x)           # transforms list into a heap, in-place, in linear time\n\
577item = heapreplace(heap, item) # pops and returns smallest item, and adds\n\
578                               # new item; the heap size is unchanged\n\
579\n\
580Our API differs from textbook heap algorithms as follows:\n\
581\n\
582- We use 0-based indexing.  This makes the relationship between the\n\
583  index for a node and the indexes for its children slightly less\n\
584  obvious, but is more suitable since Python uses 0-based indexing.\n\
585\n\
586- Our heappop() method returns the smallest item, not the largest.\n\
587\n\
588These two make it possible to view the heap as a regular Python list\n\
589without surprises: heap[0] is the smallest item, and heap.sort()\n\
590maintains the heap invariant!\n");
591
592
593PyDoc_STRVAR(__about__,
594"Heap queues\n\
595\n\
596[explanation by Fran�ois Pinard]\n\
597\n\
598Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for\n\
599all k, counting elements from 0.  For the sake of comparison,\n\
600non-existing elements are considered to be infinite.  The interesting\n\
601property of a heap is that a[0] is always its smallest element.\n"
602"\n\
603The strange invariant above is meant to be an efficient memory\n\
604representation for a tournament.  The numbers below are `k', not a[k]:\n\
605\n\
606                                   0\n\
607\n\
608                  1                                 2\n\
609\n\
610          3               4                5               6\n\
611\n\
612      7       8       9       10      11      12      13      14\n\
613\n\
614    15 16   17 18   19 20   21 22   23 24   25 26   27 28   29 30\n\
615\n\
616\n\
617In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'.  In\n\
618an usual binary tournament we see in sports, each cell is the winner\n\
619over the two cells it tops, and we can trace the winner down the tree\n\
620to see all opponents s/he had.  However, in many computer applications\n\
621of such tournaments, we do not need to trace the history of a winner.\n\
622To be more memory efficient, when a winner is promoted, we try to\n\
623replace it by something else at a lower level, and the rule becomes\n\
624that a cell and the two cells it tops contain three different items,\n\
625but the top cell \"wins\" over the two topped cells.\n"
626"\n\
627If this heap invariant is protected at all time, index 0 is clearly\n\
628the overall winner.  The simplest algorithmic way to remove it and\n\
629find the \"next\" winner is to move some loser (let's say cell 30 in the\n\
630diagram above) into the 0 position, and then percolate this new 0 down\n\
631the tree, exchanging values, until the invariant is re-established.\n\
632This is clearly logarithmic on the total number of items in the tree.\n\
633By iterating over all items, you get an O(n ln n) sort.\n"
634"\n\
635A nice feature of this sort is that you can efficiently insert new\n\
636items while the sort is going on, provided that the inserted items are\n\
637not \"better\" than the last 0'th element you extracted.  This is\n\
638especially useful in simulation contexts, where the tree holds all\n\
639incoming events, and the \"win\" condition means the smallest scheduled\n\
640time.  When an event schedule other events for execution, they are\n\
641scheduled into the future, so they can easily go into the heap.  So, a\n\
642heap is a good structure for implementing schedulers (this is what I\n\
643used for my MIDI sequencer :-).\n"
644"\n\
645Various structures for implementing schedulers have been extensively\n\
646studied, and heaps are good for this, as they are reasonably speedy,\n\
647the speed is almost constant, and the worst case is not much different\n\
648than the average case.  However, there are other representations which\n\
649are more efficient overall, yet the worst cases might be terrible.\n"
650"\n\
651Heaps are also very useful in big disk sorts.  You most probably all\n\
652know that a big sort implies producing \"runs\" (which are pre-sorted\n\
653sequences, which size is usually related to the amount of CPU memory),\n\
654followed by a merging passes for these runs, which merging is often\n\
655very cleverly organised[1].  It is very important that the initial\n\
656sort produces the longest runs possible.  Tournaments are a good way\n\
657to that.  If, using all the memory available to hold a tournament, you\n\
658replace and percolate items that happen to fit the current run, you'll\n\
659produce runs which are twice the size of the memory for random input,\n\
660and much better for input fuzzily ordered.\n"
661"\n\
662Moreover, if you output the 0'th item on disk and get an input which\n\
663may not fit in the current tournament (because the value \"wins\" over\n\
664the last output value), it cannot fit in the heap, so the size of the\n\
665heap decreases.  The freed memory could be cleverly reused immediately\n\
666for progressively building a second heap, which grows at exactly the\n\
667same rate the first heap is melting.  When the first heap completely\n\
668vanishes, you switch heaps and start a new run.  Clever and quite\n\
669effective!\n\
670\n\
671In a word, heaps are useful memory structures to know.  I use them in\n\
672a few applications, and I think it is good to keep a `heap' module\n\
673around. :-)\n"
674"\n\
675--------------------\n\
676[1] The disk balancing algorithms which are current, nowadays, are\n\
677more annoying than clever, and this is a consequence of the seeking\n\
678capabilities of the disks.  On devices which cannot seek, like big\n\
679tape drives, the story was quite different, and one had to be very\n\
680clever to ensure (far in advance) that each tape movement will be the\n\
681most effective possible (that is, will best participate at\n\
682\"progressing\" the merge).  Some tapes were even able to read\n\
683backwards, and this was also used to avoid the rewinding time.\n\
684Believe me, real good tape sorts were quite spectacular to watch!\n\
685From all times, sorting has always been a Great Art! :-)\n");
686
687PyMODINIT_FUNC
688init_heapq(void)
689{
690    PyObject *m;
691
692    m = Py_InitModule3("_heapq", heapq_methods, module_doc);
693    if (m == NULL)
694        return;
695    PyModule_AddObject(m, "__about__", PyString_FromString(__about__));
696}
697
698