| Guido van Rossum | 4b48d6b | 2002-08-02 20:23:56 +0000 | [diff] [blame] | 1 | # -*- coding: Latin-1 -*- | 
 | 2 |  | 
| Guido van Rossum | 0a82438 | 2002-08-02 16:44:32 +0000 | [diff] [blame] | 3 | """Heap queue algorithm (a.k.a. priority queue). | 
 | 4 |  | 
 | 5 | Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for | 
 | 6 | all k, counting elements from 0.  For the sake of comparison, | 
 | 7 | non-existing elements are considered to be infinite.  The interesting | 
 | 8 | property of a heap is that a[0] is always its smallest element. | 
 | 9 |  | 
 | 10 | Usage: | 
 | 11 |  | 
 | 12 | heap = []            # creates an empty heap | 
 | 13 | heappush(heap, item) # pushes a new item on the heap | 
 | 14 | item = heappop(heap) # pops the smallest item from the heap | 
 | 15 | item = heap[0]       # smallest item on the heap without popping it | 
| Tim Peters | aa7d243 | 2002-08-03 02:11:26 +0000 | [diff] [blame] | 16 | heapify(x)           # transforms list into a heap, in-place, in linear time | 
| Tim Peters | 0cd53a6 | 2002-08-03 10:10:10 +0000 | [diff] [blame] | 17 | item = heapreplace(heap, item) # pops and returns smallest item, and adds | 
 | 18 |                                # new item; the heap size is unchanged | 
| Guido van Rossum | 0a82438 | 2002-08-02 16:44:32 +0000 | [diff] [blame] | 19 |  | 
 | 20 | Our API differs from textbook heap algorithms as follows: | 
 | 21 |  | 
 | 22 | - We use 0-based indexing.  This makes the relationship between the | 
 | 23 |   index for a node and the indexes for its children slightly less | 
 | 24 |   obvious, but is more suitable since Python uses 0-based indexing. | 
 | 25 |  | 
 | 26 | - Our heappop() method returns the smallest item, not the largest. | 
 | 27 |  | 
 | 28 | These two make it possible to view the heap as a regular Python list | 
 | 29 | without surprises: heap[0] is the smallest item, and heap.sort() | 
 | 30 | maintains the heap invariant! | 
 | 31 | """ | 
 | 32 |  | 
| Guido van Rossum | fbb2992 | 2002-08-02 22:01:37 +0000 | [diff] [blame] | 33 | # Original code by Kevin O'Connor, augmented by Tim Peters | 
| Guido van Rossum | 37c3b27 | 2002-08-02 16:50:58 +0000 | [diff] [blame] | 34 |  | 
| Guido van Rossum | 0a82438 | 2002-08-02 16:44:32 +0000 | [diff] [blame] | 35 | __about__ = """Heap queues | 
 | 36 |  | 
 | 37 | [explanation by François Pinard] | 
 | 38 |  | 
 | 39 | Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for | 
 | 40 | all k, counting elements from 0.  For the sake of comparison, | 
 | 41 | non-existing elements are considered to be infinite.  The interesting | 
 | 42 | property of a heap is that a[0] is always its smallest element. | 
 | 43 |  | 
 | 44 | The strange invariant above is meant to be an efficient memory | 
 | 45 | representation for a tournament.  The numbers below are `k', not a[k]: | 
 | 46 |  | 
 | 47 |                                    0 | 
 | 48 |  | 
 | 49 |                   1                                 2 | 
 | 50 |  | 
 | 51 |           3               4                5               6 | 
 | 52 |  | 
 | 53 |       7       8       9       10      11      12      13      14 | 
 | 54 |  | 
 | 55 |     15 16   17 18   19 20   21 22   23 24   25 26   27 28   29 30 | 
 | 56 |  | 
 | 57 |  | 
 | 58 | In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'.  In | 
 | 59 | an usual binary tournament we see in sports, each cell is the winner | 
 | 60 | over the two cells it tops, and we can trace the winner down the tree | 
 | 61 | to see all opponents s/he had.  However, in many computer applications | 
 | 62 | of such tournaments, we do not need to trace the history of a winner. | 
 | 63 | To be more memory efficient, when a winner is promoted, we try to | 
 | 64 | replace it by something else at a lower level, and the rule becomes | 
 | 65 | that a cell and the two cells it tops contain three different items, | 
 | 66 | but the top cell "wins" over the two topped cells. | 
 | 67 |  | 
 | 68 | If this heap invariant is protected at all time, index 0 is clearly | 
 | 69 | the overall winner.  The simplest algorithmic way to remove it and | 
 | 70 | find the "next" winner is to move some loser (let's say cell 30 in the | 
 | 71 | diagram above) into the 0 position, and then percolate this new 0 down | 
 | 72 | the tree, exchanging values, until the invariant is re-established. | 
 | 73 | This is clearly logarithmic on the total number of items in the tree. | 
 | 74 | By iterating over all items, you get an O(n ln n) sort. | 
 | 75 |  | 
 | 76 | A nice feature of this sort is that you can efficiently insert new | 
 | 77 | items while the sort is going on, provided that the inserted items are | 
 | 78 | not "better" than the last 0'th element you extracted.  This is | 
 | 79 | especially useful in simulation contexts, where the tree holds all | 
 | 80 | incoming events, and the "win" condition means the smallest scheduled | 
 | 81 | time.  When an event schedule other events for execution, they are | 
 | 82 | scheduled into the future, so they can easily go into the heap.  So, a | 
 | 83 | heap is a good structure for implementing schedulers (this is what I | 
 | 84 | used for my MIDI sequencer :-). | 
 | 85 |  | 
 | 86 | Various structures for implementing schedulers have been extensively | 
 | 87 | studied, and heaps are good for this, as they are reasonably speedy, | 
 | 88 | the speed is almost constant, and the worst case is not much different | 
 | 89 | than the average case.  However, there are other representations which | 
 | 90 | are more efficient overall, yet the worst cases might be terrible. | 
 | 91 |  | 
 | 92 | Heaps are also very useful in big disk sorts.  You most probably all | 
 | 93 | know that a big sort implies producing "runs" (which are pre-sorted | 
 | 94 | sequences, which size is usually related to the amount of CPU memory), | 
 | 95 | followed by a merging passes for these runs, which merging is often | 
 | 96 | very cleverly organised[1].  It is very important that the initial | 
 | 97 | sort produces the longest runs possible.  Tournaments are a good way | 
 | 98 | to that.  If, using all the memory available to hold a tournament, you | 
 | 99 | replace and percolate items that happen to fit the current run, you'll | 
 | 100 | produce runs which are twice the size of the memory for random input, | 
 | 101 | and much better for input fuzzily ordered. | 
 | 102 |  | 
 | 103 | Moreover, if you output the 0'th item on disk and get an input which | 
 | 104 | may not fit in the current tournament (because the value "wins" over | 
 | 105 | the last output value), it cannot fit in the heap, so the size of the | 
 | 106 | heap decreases.  The freed memory could be cleverly reused immediately | 
 | 107 | for progressively building a second heap, which grows at exactly the | 
 | 108 | same rate the first heap is melting.  When the first heap completely | 
 | 109 | vanishes, you switch heaps and start a new run.  Clever and quite | 
 | 110 | effective! | 
 | 111 |  | 
 | 112 | In a word, heaps are useful memory structures to know.  I use them in | 
 | 113 | a few applications, and I think it is good to keep a `heap' module | 
 | 114 | around. :-) | 
 | 115 |  | 
 | 116 | -------------------- | 
 | 117 | [1] The disk balancing algorithms which are current, nowadays, are | 
 | 118 | more annoying than clever, and this is a consequence of the seeking | 
 | 119 | capabilities of the disks.  On devices which cannot seek, like big | 
 | 120 | tape drives, the story was quite different, and one had to be very | 
 | 121 | clever to ensure (far in advance) that each tape movement will be the | 
 | 122 | most effective possible (that is, will best participate at | 
 | 123 | "progressing" the merge).  Some tapes were even able to read | 
 | 124 | backwards, and this was also used to avoid the rewinding time. | 
 | 125 | Believe me, real good tape sorts were quite spectacular to watch! | 
 | 126 | From all times, sorting has always been a Great Art! :-) | 
 | 127 | """ | 
 | 128 |  | 
 | 129 | def heappush(heap, item): | 
 | 130 |     """Push item onto heap, maintaining the heap invariant.""" | 
| Tim Peters | 657fe38 | 2002-08-03 09:56:52 +0000 | [diff] [blame] | 131 |     heap.append(item) | 
 | 132 |     _siftdown(heap, 0, len(heap)-1) | 
| Tim Peters | 28c2552 | 2002-08-02 21:48:06 +0000 | [diff] [blame] | 133 |  | 
 | 134 | def heappop(heap): | 
 | 135 |     """Pop the smallest item off the heap, maintaining the heap invariant.""" | 
 | 136 |     lastelt = heap.pop()    # raises appropriate IndexError if heap is empty | 
 | 137 |     if heap: | 
 | 138 |         returnitem = heap[0] | 
 | 139 |         heap[0] = lastelt | 
| Tim Peters | 657fe38 | 2002-08-03 09:56:52 +0000 | [diff] [blame] | 140 |         _siftup(heap, 0) | 
| Tim Peters | 28c2552 | 2002-08-02 21:48:06 +0000 | [diff] [blame] | 141 |     else: | 
 | 142 |         returnitem = lastelt | 
| Guido van Rossum | 0a82438 | 2002-08-02 16:44:32 +0000 | [diff] [blame] | 143 |     return returnitem | 
 | 144 |  | 
| Tim Peters | 0cd53a6 | 2002-08-03 10:10:10 +0000 | [diff] [blame] | 145 | def heapreplace(heap, item): | 
 | 146 |     """Pop and return the current smallest value, and add the new item. | 
 | 147 |  | 
 | 148 |     This is more efficient than heappop() followed by heappush(), and can be | 
 | 149 |     more appropriate when using a fixed-size heap.  Note that the value | 
 | 150 |     returned may be larger than item!  That constrains reasonable uses of | 
 | 151 |     this routine. | 
 | 152 |     """ | 
| Guido van Rossum | 3c8dd0c | 2002-08-07 18:58:11 +0000 | [diff] [blame] | 153 |     returnitem = heap[0]    # raises appropriate IndexError if heap is empty | 
 | 154 |     heap[0] = item | 
 | 155 |     _siftup(heap, 0) | 
 | 156 |     return returnitem | 
| Tim Peters | 0cd53a6 | 2002-08-03 10:10:10 +0000 | [diff] [blame] | 157 |  | 
| Tim Peters | aa7d243 | 2002-08-03 02:11:26 +0000 | [diff] [blame] | 158 | def heapify(x): | 
 | 159 |     """Transform list into a heap, in-place, in O(len(heap)) time.""" | 
 | 160 |     n = len(x) | 
| Tim Peters | 28c2552 | 2002-08-02 21:48:06 +0000 | [diff] [blame] | 161 |     # Transform bottom-up.  The largest index there's any point to looking at | 
 | 162 |     # is the largest with a child index in-range, so must have 2*i + 1 < n, | 
 | 163 |     # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so | 
 | 164 |     # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is | 
 | 165 |     # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1. | 
 | 166 |     for i in xrange(n//2 - 1, -1, -1): | 
| Tim Peters | 657fe38 | 2002-08-03 09:56:52 +0000 | [diff] [blame] | 167 |         _siftup(x, i) | 
 | 168 |  | 
 | 169 | # 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos | 
 | 170 | # is the index of a leaf with a possibly out-of-order value.  Restore the | 
 | 171 | # heap invariant. | 
 | 172 | def _siftdown(heap, startpos, pos): | 
 | 173 |     newitem = heap[pos] | 
 | 174 |     # Follow the path to the root, moving parents down until finding a place | 
 | 175 |     # newitem fits. | 
 | 176 |     while pos > startpos: | 
 | 177 |         parentpos = (pos - 1) >> 1 | 
 | 178 |         parent = heap[parentpos] | 
 | 179 |         if parent <= newitem: | 
 | 180 |             break | 
 | 181 |         heap[pos] = parent | 
 | 182 |         pos = parentpos | 
 | 183 |     heap[pos] = newitem | 
 | 184 |  | 
 | 185 | # The child indices of heap index pos are already heaps, and we want to make | 
 | 186 | # a heap at index pos too.  We do this by bubbling the smaller child of | 
 | 187 | # pos up (and so on with that child's children, etc) until hitting a leaf, | 
 | 188 | # then using _siftdown to move the oddball originally at index pos into place. | 
 | 189 | # | 
 | 190 | # We *could* break out of the loop as soon as we find a pos where newitem <= | 
 | 191 | # both its children, but turns out that's not a good idea, and despite that | 
 | 192 | # many books write the algorithm that way.  During a heap pop, the last array | 
 | 193 | # element is sifted in, and that tends to be large, so that comparing it | 
 | 194 | # against values starting from the root usually doesn't pay (= usually doesn't | 
 | 195 | # get us out of the loop early).  See Knuth, Volume 3, where this is | 
 | 196 | # explained and quantified in an exercise. | 
 | 197 | # | 
 | 198 | # Cutting the # of comparisons is important, since these routines have no | 
 | 199 | # way to extract "the priority" from an array element, so that intelligence | 
 | 200 | # is likely to be hiding in custom __cmp__ methods, or in array elements | 
 | 201 | # storing (priority, record) tuples.  Comparisons are thus potentially | 
 | 202 | # expensive. | 
 | 203 | # | 
 | 204 | # On random arrays of length 1000, making this change cut the number of | 
 | 205 | # comparisons made by heapify() a little, and those made by exhaustive | 
 | 206 | # heappop() a lot, in accord with theory.  Here are typical results from 3 | 
 | 207 | # runs (3 just to demonstrate how small the variance is): | 
 | 208 | # | 
 | 209 | # Compares needed by heapify     Compares needed by 1000 heapppops | 
 | 210 | # --------------------------     --------------------------------- | 
 | 211 | # 1837 cut to 1663               14996 cut to 8680 | 
 | 212 | # 1855 cut to 1659               14966 cut to 8678 | 
 | 213 | # 1847 cut to 1660               15024 cut to 8703 | 
 | 214 | # | 
 | 215 | # Building the heap by using heappush() 1000 times instead required | 
 | 216 | # 2198, 2148, and 2219 compares:  heapify() is more efficient, when | 
 | 217 | # you can use it. | 
 | 218 | # | 
 | 219 | # The total compares needed by list.sort() on the same lists were 8627, | 
 | 220 | # 8627, and 8632 (this should be compared to the sum of heapify() and | 
 | 221 | # heappop() compares):  list.sort() is (unsurprisingly!) more efficent | 
 | 222 | # for sorting. | 
 | 223 |  | 
 | 224 | def _siftup(heap, pos): | 
 | 225 |     endpos = len(heap) | 
 | 226 |     startpos = pos | 
 | 227 |     newitem = heap[pos] | 
 | 228 |     # Bubble up the smaller child until hitting a leaf. | 
 | 229 |     childpos = 2*pos + 1    # leftmost child position | 
 | 230 |     while childpos < endpos: | 
 | 231 |         # Set childpos to index of smaller child. | 
 | 232 |         rightpos = childpos + 1 | 
| Tim Peters | 6681de2 | 2002-08-03 19:20:16 +0000 | [diff] [blame] | 233 |         if rightpos < endpos and heap[rightpos] <= heap[childpos]: | 
| Tim Peters | 469cdad | 2002-08-08 20:19:19 +0000 | [diff] [blame] | 234 |             childpos = rightpos | 
| Tim Peters | 657fe38 | 2002-08-03 09:56:52 +0000 | [diff] [blame] | 235 |         # Move the smaller child up. | 
 | 236 |         heap[pos] = heap[childpos] | 
 | 237 |         pos = childpos | 
 | 238 |         childpos = 2*pos + 1 | 
 | 239 |     # The leaf at pos is empty now.  Put newitem there, and and bubble it up | 
 | 240 |     # to its final resting place (by sifting its parents down). | 
 | 241 |     heap[pos] = newitem | 
 | 242 |     _siftdown(heap, startpos, pos) | 
| Tim Peters | 28c2552 | 2002-08-02 21:48:06 +0000 | [diff] [blame] | 243 |  | 
| Guido van Rossum | 0a82438 | 2002-08-02 16:44:32 +0000 | [diff] [blame] | 244 | if __name__ == "__main__": | 
 | 245 |     # Simple sanity test | 
 | 246 |     heap = [] | 
 | 247 |     data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0] | 
 | 248 |     for item in data: | 
 | 249 |         heappush(heap, item) | 
 | 250 |     sort = [] | 
 | 251 |     while heap: | 
 | 252 |         sort.append(heappop(heap)) | 
 | 253 |     print sort |