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176 lines
7.2 KiB
Python
176 lines
7.2 KiB
Python
"""Heap queue algorithm (a.k.a. priority queue).
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Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
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all k, counting elements from 0. For the sake of comparison,
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non-existing elements are considered to be infinite. The interesting
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property of a heap is that a[0] is always its smallest element.
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Usage:
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heap = [] # creates an empty heap
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heappush(heap, item) # pushes a new item on the heap
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item = heappop(heap) # pops the smallest item from the heap
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item = heap[0] # smallest item on the heap without popping it
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Our API differs from textbook heap algorithms as follows:
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- We use 0-based indexing. This makes the relationship between the
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index for a node and the indexes for its children slightly less
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obvious, but is more suitable since Python uses 0-based indexing.
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- Our heappop() method returns the smallest item, not the largest.
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These two make it possible to view the heap as a regular Python list
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without surprises: heap[0] is the smallest item, and heap.sort()
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maintains the heap invariant!
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"""
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# Code by Kevin O'Connor
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__about__ = """Heap queues
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[explanation by François Pinard]
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Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
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all k, counting elements from 0. For the sake of comparison,
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non-existing elements are considered to be infinite. The interesting
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property of a heap is that a[0] is always its smallest element.
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The strange invariant above is meant to be an efficient memory
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representation for a tournament. The numbers below are `k', not a[k]:
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0
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1 2
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3 4 5 6
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7 8 9 10 11 12 13 14
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15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'. In
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an usual binary tournament we see in sports, each cell is the winner
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over the two cells it tops, and we can trace the winner down the tree
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to see all opponents s/he had. However, in many computer applications
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of such tournaments, we do not need to trace the history of a winner.
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To be more memory efficient, when a winner is promoted, we try to
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replace it by something else at a lower level, and the rule becomes
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that a cell and the two cells it tops contain three different items,
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but the top cell "wins" over the two topped cells.
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If this heap invariant is protected at all time, index 0 is clearly
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the overall winner. The simplest algorithmic way to remove it and
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find the "next" winner is to move some loser (let's say cell 30 in the
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diagram above) into the 0 position, and then percolate this new 0 down
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the tree, exchanging values, until the invariant is re-established.
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This is clearly logarithmic on the total number of items in the tree.
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By iterating over all items, you get an O(n ln n) sort.
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A nice feature of this sort is that you can efficiently insert new
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items while the sort is going on, provided that the inserted items are
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not "better" than the last 0'th element you extracted. This is
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especially useful in simulation contexts, where the tree holds all
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incoming events, and the "win" condition means the smallest scheduled
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time. When an event schedule other events for execution, they are
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scheduled into the future, so they can easily go into the heap. So, a
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heap is a good structure for implementing schedulers (this is what I
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used for my MIDI sequencer :-).
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Various structures for implementing schedulers have been extensively
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studied, and heaps are good for this, as they are reasonably speedy,
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the speed is almost constant, and the worst case is not much different
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than the average case. However, there are other representations which
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are more efficient overall, yet the worst cases might be terrible.
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Heaps are also very useful in big disk sorts. You most probably all
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know that a big sort implies producing "runs" (which are pre-sorted
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sequences, which size is usually related to the amount of CPU memory),
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followed by a merging passes for these runs, which merging is often
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very cleverly organised[1]. It is very important that the initial
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sort produces the longest runs possible. Tournaments are a good way
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to that. If, using all the memory available to hold a tournament, you
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replace and percolate items that happen to fit the current run, you'll
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produce runs which are twice the size of the memory for random input,
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and much better for input fuzzily ordered.
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Moreover, if you output the 0'th item on disk and get an input which
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may not fit in the current tournament (because the value "wins" over
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the last output value), it cannot fit in the heap, so the size of the
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heap decreases. The freed memory could be cleverly reused immediately
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for progressively building a second heap, which grows at exactly the
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same rate the first heap is melting. When the first heap completely
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vanishes, you switch heaps and start a new run. Clever and quite
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effective!
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In a word, heaps are useful memory structures to know. I use them in
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a few applications, and I think it is good to keep a `heap' module
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around. :-)
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--------------------
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[1] The disk balancing algorithms which are current, nowadays, are
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more annoying than clever, and this is a consequence of the seeking
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capabilities of the disks. On devices which cannot seek, like big
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tape drives, the story was quite different, and one had to be very
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clever to ensure (far in advance) that each tape movement will be the
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most effective possible (that is, will best participate at
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"progressing" the merge). Some tapes were even able to read
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backwards, and this was also used to avoid the rewinding time.
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Believe me, real good tape sorts were quite spectacular to watch!
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From all times, sorting has always been a Great Art! :-)
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"""
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def heappush(heap, item):
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"""Push item onto heap, maintaining the heap invariant."""
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pos = len(heap)
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heap.append(None)
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while pos:
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parentpos = (pos - 1) >> 1
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parent = heap[parentpos]
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if item >= parent:
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break
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heap[pos] = parent
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pos = parentpos
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heap[pos] = item
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def heappop(heap):
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"""Pop the smallest item off the heap, maintaining the heap invariant."""
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endpos = len(heap) - 1
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if endpos <= 0:
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return heap.pop()
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returnitem = heap[0]
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item = heap.pop()
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pos = 0
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# Sift item into position, down from the root, moving the smaller
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# child up, until finding pos such that item <= pos's children.
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childpos = 2*pos + 1 # leftmost child position
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while childpos < endpos:
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# Set childpos and child to reflect smaller child.
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child = heap[childpos]
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rightpos = childpos + 1
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if rightpos < endpos:
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rightchild = heap[rightpos]
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if rightchild < child:
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childpos = rightpos
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child = rightchild
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# If item is no larger than smaller child, we're done, else
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# move the smaller child up.
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if item <= child:
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break
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heap[pos] = child
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pos = childpos
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childpos = 2*pos + 1
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heap[pos] = item
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return returnitem
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if __name__ == "__main__":
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# Simple sanity test
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heap = []
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data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
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for item in data:
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heappush(heap, item)
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sort = []
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while heap:
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sort.append(heappop(heap))
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print sort
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