heapq.py revision 0cd53a6c37347800f786c4ddaa2e91af30350b5a
1# -*- coding: Latin-1 -*- 2 3"""Heap queue algorithm (a.k.a. priority queue). 4 5Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for 6all k, counting elements from 0. For the sake of comparison, 7non-existing elements are considered to be infinite. The interesting 8property of a heap is that a[0] is always its smallest element. 9 10Usage: 11 12heap = [] # creates an empty heap 13heappush(heap, item) # pushes a new item on the heap 14item = heappop(heap) # pops the smallest item from the heap 15item = heap[0] # smallest item on the heap without popping it 16heapify(x) # transforms list into a heap, in-place, in linear time 17item = heapreplace(heap, item) # pops and returns smallest item, and adds 18 # new item; the heap size is unchanged 19 20Our 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 28These two make it possible to view the heap as a regular Python list 29without surprises: heap[0] is the smallest item, and heap.sort() 30maintains the heap invariant! 31""" 32 33# Original code by Kevin O'Connor, augmented by Tim Peters 34 35__about__ = """Heap queues 36 37[explanation by Fran�ois Pinard] 38 39Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for 40all k, counting elements from 0. For the sake of comparison, 41non-existing elements are considered to be infinite. The interesting 42property of a heap is that a[0] is always its smallest element. 43 44The strange invariant above is meant to be an efficient memory 45representation 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 58In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'. In 59an usual binary tournament we see in sports, each cell is the winner 60over the two cells it tops, and we can trace the winner down the tree 61to see all opponents s/he had. However, in many computer applications 62of such tournaments, we do not need to trace the history of a winner. 63To be more memory efficient, when a winner is promoted, we try to 64replace it by something else at a lower level, and the rule becomes 65that a cell and the two cells it tops contain three different items, 66but the top cell "wins" over the two topped cells. 67 68If this heap invariant is protected at all time, index 0 is clearly 69the overall winner. The simplest algorithmic way to remove it and 70find the "next" winner is to move some loser (let's say cell 30 in the 71diagram above) into the 0 position, and then percolate this new 0 down 72the tree, exchanging values, until the invariant is re-established. 73This is clearly logarithmic on the total number of items in the tree. 74By iterating over all items, you get an O(n ln n) sort. 75 76A nice feature of this sort is that you can efficiently insert new 77items while the sort is going on, provided that the inserted items are 78not "better" than the last 0'th element you extracted. This is 79especially useful in simulation contexts, where the tree holds all 80incoming events, and the "win" condition means the smallest scheduled 81time. When an event schedule other events for execution, they are 82scheduled into the future, so they can easily go into the heap. So, a 83heap is a good structure for implementing schedulers (this is what I 84used for my MIDI sequencer :-). 85 86Various structures for implementing schedulers have been extensively 87studied, and heaps are good for this, as they are reasonably speedy, 88the speed is almost constant, and the worst case is not much different 89than the average case. However, there are other representations which 90are more efficient overall, yet the worst cases might be terrible. 91 92Heaps are also very useful in big disk sorts. You most probably all 93know that a big sort implies producing "runs" (which are pre-sorted 94sequences, which size is usually related to the amount of CPU memory), 95followed by a merging passes for these runs, which merging is often 96very cleverly organised[1]. It is very important that the initial 97sort produces the longest runs possible. Tournaments are a good way 98to that. If, using all the memory available to hold a tournament, you 99replace and percolate items that happen to fit the current run, you'll 100produce runs which are twice the size of the memory for random input, 101and much better for input fuzzily ordered. 102 103Moreover, if you output the 0'th item on disk and get an input which 104may not fit in the current tournament (because the value "wins" over 105the last output value), it cannot fit in the heap, so the size of the 106heap decreases. The freed memory could be cleverly reused immediately 107for progressively building a second heap, which grows at exactly the 108same rate the first heap is melting. When the first heap completely 109vanishes, you switch heaps and start a new run. Clever and quite 110effective! 111 112In a word, heaps are useful memory structures to know. I use them in 113a few applications, and I think it is good to keep a `heap' module 114around. :-) 115 116-------------------- 117[1] The disk balancing algorithms which are current, nowadays, are 118more annoying than clever, and this is a consequence of the seeking 119capabilities of the disks. On devices which cannot seek, like big 120tape drives, the story was quite different, and one had to be very 121clever to ensure (far in advance) that each tape movement will be the 122most effective possible (that is, will best participate at 123"progressing" the merge). Some tapes were even able to read 124backwards, and this was also used to avoid the rewinding time. 125Believe me, real good tape sorts were quite spectacular to watch! 126From all times, sorting has always been a Great Art! :-) 127""" 128 129def heappush(heap, item): 130 """Push item onto heap, maintaining the heap invariant.""" 131 heap.append(item) 132 _siftdown(heap, 0, len(heap)-1) 133 134def 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 140 _siftup(heap, 0) 141 else: 142 returnitem = lastelt 143 return returnitem 144 145def 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 """ 153 154 if heap: 155 returnitem = heap[0] 156 heap[0] = item 157 _siftup(heap, 0) 158 return returnitem 159 heap.pop() # raise IndexError 160 161def heapify(x): 162 """Transform list into a heap, in-place, in O(len(heap)) time.""" 163 n = len(x) 164 # Transform bottom-up. The largest index there's any point to looking at 165 # is the largest with a child index in-range, so must have 2*i + 1 < n, 166 # or i < (n-1)/2. If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so 167 # j-1 is the largest, which is n//2 - 1. If n is odd = 2*j+1, this is 168 # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1. 169 for i in xrange(n//2 - 1, -1, -1): 170 _siftup(x, i) 171 172# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos 173# is the index of a leaf with a possibly out-of-order value. Restore the 174# heap invariant. 175def _siftdown(heap, startpos, pos): 176 newitem = heap[pos] 177 # Follow the path to the root, moving parents down until finding a place 178 # newitem fits. 179 while pos > startpos: 180 parentpos = (pos - 1) >> 1 181 parent = heap[parentpos] 182 if parent <= newitem: 183 break 184 heap[pos] = parent 185 pos = parentpos 186 heap[pos] = newitem 187 188# The child indices of heap index pos are already heaps, and we want to make 189# a heap at index pos too. We do this by bubbling the smaller child of 190# pos up (and so on with that child's children, etc) until hitting a leaf, 191# then using _siftdown to move the oddball originally at index pos into place. 192# 193# We *could* break out of the loop as soon as we find a pos where newitem <= 194# both its children, but turns out that's not a good idea, and despite that 195# many books write the algorithm that way. During a heap pop, the last array 196# element is sifted in, and that tends to be large, so that comparing it 197# against values starting from the root usually doesn't pay (= usually doesn't 198# get us out of the loop early). See Knuth, Volume 3, where this is 199# explained and quantified in an exercise. 200# 201# Cutting the # of comparisons is important, since these routines have no 202# way to extract "the priority" from an array element, so that intelligence 203# is likely to be hiding in custom __cmp__ methods, or in array elements 204# storing (priority, record) tuples. Comparisons are thus potentially 205# expensive. 206# 207# On random arrays of length 1000, making this change cut the number of 208# comparisons made by heapify() a little, and those made by exhaustive 209# heappop() a lot, in accord with theory. Here are typical results from 3 210# runs (3 just to demonstrate how small the variance is): 211# 212# Compares needed by heapify Compares needed by 1000 heapppops 213# -------------------------- --------------------------------- 214# 1837 cut to 1663 14996 cut to 8680 215# 1855 cut to 1659 14966 cut to 8678 216# 1847 cut to 1660 15024 cut to 8703 217# 218# Building the heap by using heappush() 1000 times instead required 219# 2198, 2148, and 2219 compares: heapify() is more efficient, when 220# you can use it. 221# 222# The total compares needed by list.sort() on the same lists were 8627, 223# 8627, and 8632 (this should be compared to the sum of heapify() and 224# heappop() compares): list.sort() is (unsurprisingly!) more efficent 225# for sorting. 226 227def _siftup(heap, pos): 228 endpos = len(heap) 229 startpos = pos 230 newitem = heap[pos] 231 # Bubble up the smaller child until hitting a leaf. 232 childpos = 2*pos + 1 # leftmost child position 233 while childpos < endpos: 234 # Set childpos to index of smaller child. 235 rightpos = childpos + 1 236 if rightpos < endpos and heap[rightpos] < heap[childpos]: 237 childpos = rightpos 238 # Move the smaller child up. 239 heap[pos] = heap[childpos] 240 pos = childpos 241 childpos = 2*pos + 1 242 # The leaf at pos is empty now. Put newitem there, and and bubble it up 243 # to its final resting place (by sifting its parents down). 244 heap[pos] = newitem 245 _siftdown(heap, startpos, pos) 246 247if __name__ == "__main__": 248 # Simple sanity test 249 heap = [] 250 data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0] 251 for item in data: 252 heappush(heap, item) 253 sort = [] 254 while heap: 255 sort.append(heappop(heap)) 256 print sort 257