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+# -*- coding: Latin-1 -*-
+
+"""Heap queue algorithm (a.k.a. priority queue).
+
+Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
+all k, counting elements from 0.  For the sake of comparison,
+non-existing elements are considered to be infinite.  The interesting
+property of a heap is that a[0] is always its smallest element.
+
+Usage:
+
+heap = []            # creates an empty heap
+heappush(heap, item) # pushes a new item on the heap
+item = heappop(heap) # pops the smallest item from the heap
+item = heap[0]       # smallest item on the heap without popping it
+heapify(x)           # transforms list into a heap, in-place, in linear time
+item = heapreplace(heap, item) # pops and returns smallest item, and adds
+                               # new item; the heap size is unchanged
+
+Our API differs from textbook heap algorithms as follows:
+
+- We use 0-based indexing.  This makes the relationship between the
+  index for a node and the indexes for its children slightly less
+  obvious, but is more suitable since Python uses 0-based indexing.
+
+- Our heappop() method returns the smallest item, not the largest.
+
+These two make it possible to view the heap as a regular Python list
+without surprises: heap[0] is the smallest item, and heap.sort()
+maintains the heap invariant!
+"""
+
+# Original code by Kevin O'Connor, augmented by Tim Peters
+
+__about__ = """Heap queues
+
+[explanation by François Pinard]
+
+Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
+all k, counting elements from 0.  For the sake of comparison,
+non-existing elements are considered to be infinite.  The interesting
+property of a heap is that a[0] is always its smallest element.
+
+The strange invariant above is meant to be an efficient memory
+representation for a tournament.  The numbers below are `k', not a[k]:
+
+                                   0
+
+                  1                                 2
+
+          3               4                5               6
+
+      7       8       9       10      11      12      13      14
+
+    15 16   17 18   19 20   21 22   23 24   25 26   27 28   29 30
+
+
+In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'.  In
+an usual binary tournament we see in sports, each cell is the winner
+over the two cells it tops, and we can trace the winner down the tree
+to see all opponents s/he had.  However, in many computer applications
+of such tournaments, we do not need to trace the history of a winner.
+To be more memory efficient, when a winner is promoted, we try to
+replace it by something else at a lower level, and the rule becomes
+that a cell and the two cells it tops contain three different items,
+but the top cell "wins" over the two topped cells.
+
+If this heap invariant is protected at all time, index 0 is clearly
+the overall winner.  The simplest algorithmic way to remove it and
+find the "next" winner is to move some loser (let's say cell 30 in the
+diagram above) into the 0 position, and then percolate this new 0 down
+the tree, exchanging values, until the invariant is re-established.
+This is clearly logarithmic on the total number of items in the tree.
+By iterating over all items, you get an O(n ln n) sort.
+
+A nice feature of this sort is that you can efficiently insert new
+items while the sort is going on, provided that the inserted items are
+not "better" than the last 0'th element you extracted.  This is
+especially useful in simulation contexts, where the tree holds all
+incoming events, and the "win" condition means the smallest scheduled
+time.  When an event schedule other events for execution, they are
+scheduled into the future, so they can easily go into the heap.  So, a
+heap is a good structure for implementing schedulers (this is what I
+used for my MIDI sequencer :-).
+
+Various structures for implementing schedulers have been extensively
+studied, and heaps are good for this, as they are reasonably speedy,
+the speed is almost constant, and the worst case is not much different
+than the average case.  However, there are other representations which
+are more efficient overall, yet the worst cases might be terrible.
+
+Heaps are also very useful in big disk sorts.  You most probably all
+know that a big sort implies producing "runs" (which are pre-sorted
+sequences, which size is usually related to the amount of CPU memory),
+followed by a merging passes for these runs, which merging is often
+very cleverly organised[1].  It is very important that the initial
+sort produces the longest runs possible.  Tournaments are a good way
+to that.  If, using all the memory available to hold a tournament, you
+replace and percolate items that happen to fit the current run, you'll
+produce runs which are twice the size of the memory for random input,
+and much better for input fuzzily ordered.
+
+Moreover, if you output the 0'th item on disk and get an input which
+may not fit in the current tournament (because the value "wins" over
+the last output value), it cannot fit in the heap, so the size of the
+heap decreases.  The freed memory could be cleverly reused immediately
+for progressively building a second heap, which grows at exactly the
+same rate the first heap is melting.  When the first heap completely
+vanishes, you switch heaps and start a new run.  Clever and quite
+effective!
+
+In a word, heaps are useful memory structures to know.  I use them in
+a few applications, and I think it is good to keep a `heap' module
+around. :-)
+
+--------------------
+[1] The disk balancing algorithms which are current, nowadays, are
+more annoying than clever, and this is a consequence of the seeking
+capabilities of the disks.  On devices which cannot seek, like big
+tape drives, the story was quite different, and one had to be very
+clever to ensure (far in advance) that each tape movement will be the
+most effective possible (that is, will best participate at
+"progressing" the merge).  Some tapes were even able to read
+backwards, and this was also used to avoid the rewinding time.
+Believe me, real good tape sorts were quite spectacular to watch!
+From all times, sorting has always been a Great Art! :-)
+"""
+
+__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace']
+
+def heappush(heap, item):
+    """Push item onto heap, maintaining the heap invariant."""
+    heap.append(item)
+    _siftdown(heap, 0, len(heap)-1)
+
+def heappop(heap):
+    """Pop the smallest item off the heap, maintaining the heap invariant."""
+    lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
+    if heap:
+        returnitem = heap[0]
+        heap[0] = lastelt
+        _siftup(heap, 0)
+    else:
+        returnitem = lastelt
+    return returnitem
+
+def heapreplace(heap, item):
+    """Pop and return the current smallest value, and add the new item.
+
+    This is more efficient than heappop() followed by heappush(), and can be
+    more appropriate when using a fixed-size heap.  Note that the value
+    returned may be larger than item!  That constrains reasonable uses of
+    this routine.
+    """
+    returnitem = heap[0]    # raises appropriate IndexError if heap is empty
+    heap[0] = item
+    _siftup(heap, 0)
+    return returnitem
+
+def heapify(x):
+    """Transform list into a heap, in-place, in O(len(heap)) time."""
+    n = len(x)
+    # Transform bottom-up.  The largest index there's any point to looking at
+    # is the largest with a child index in-range, so must have 2*i + 1 < n,
+    # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
+    # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
+    # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
+    for i in reversed(xrange(n//2)):
+        _siftup(x, i)
+
+# 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
+# is the index of a leaf with a possibly out-of-order value.  Restore the
+# heap invariant.
+def _siftdown(heap, startpos, pos):
+    newitem = heap[pos]
+    # Follow the path to the root, moving parents down until finding a place
+    # newitem fits.
+    while pos > startpos:
+        parentpos = (pos - 1) >> 1
+        parent = heap[parentpos]
+        if parent <= newitem:
+            break
+        heap[pos] = parent
+        pos = parentpos
+    heap[pos] = newitem
+
+# The child indices of heap index pos are already heaps, and we want to make
+# a heap at index pos too.  We do this by bubbling the smaller child of
+# pos up (and so on with that child's children, etc) until hitting a leaf,
+# then using _siftdown to move the oddball originally at index pos into place.
+#
+# We *could* break out of the loop as soon as we find a pos where newitem <=
+# both its children, but turns out that's not a good idea, and despite that
+# many books write the algorithm that way.  During a heap pop, the last array
+# element is sifted in, and that tends to be large, so that comparing it
+# against values starting from the root usually doesn't pay (= usually doesn't
+# get us out of the loop early).  See Knuth, Volume 3, where this is
+# explained and quantified in an exercise.
+#
+# Cutting the # of comparisons is important, since these routines have no
+# way to extract "the priority" from an array element, so that intelligence
+# is likely to be hiding in custom __cmp__ methods, or in array elements
+# storing (priority, record) tuples.  Comparisons are thus potentially
+# expensive.
+#
+# On random arrays of length 1000, making this change cut the number of
+# comparisons made by heapify() a little, and those made by exhaustive
+# heappop() a lot, in accord with theory.  Here are typical results from 3
+# runs (3 just to demonstrate how small the variance is):
+#
+# Compares needed by heapify     Compares needed by 1000 heappops
+# --------------------------     --------------------------------
+# 1837 cut to 1663               14996 cut to 8680
+# 1855 cut to 1659               14966 cut to 8678
+# 1847 cut to 1660               15024 cut to 8703
+#
+# Building the heap by using heappush() 1000 times instead required
+# 2198, 2148, and 2219 compares:  heapify() is more efficient, when
+# you can use it.
+#
+# The total compares needed by list.sort() on the same lists were 8627,
+# 8627, and 8632 (this should be compared to the sum of heapify() and
+# heappop() compares):  list.sort() is (unsurprisingly!) more efficient
+# for sorting.
+
+def _siftup(heap, pos):
+    endpos = len(heap)
+    startpos = pos
+    newitem = heap[pos]
+    # Bubble up the smaller child until hitting a leaf.
+    childpos = 2*pos + 1    # leftmost child position
+    while childpos < endpos:
+        # Set childpos to index of smaller child.
+        rightpos = childpos + 1
+        if rightpos < endpos and heap[rightpos] <= heap[childpos]:
+            childpos = rightpos
+        # Move the smaller child up.
+        heap[pos] = heap[childpos]
+        pos = childpos
+        childpos = 2*pos + 1
+    # The leaf at pos is empty now.  Put newitem there, and bubble it up
+    # to its final resting place (by sifting its parents down).
+    heap[pos] = newitem
+    _siftdown(heap, startpos, pos)
+
+# If available, use C implementation
+try:
+    from _heapq import heappush, heappop, heapify, heapreplace
+except ImportError:
+    pass
+
+if __name__ == "__main__":
+    # Simple sanity test
+    heap = []
+    data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
+    for item in data:
+        heappush(heap, item)
+    sort = []
+    while heap:
+        sort.append(heappop(heap))
+    print sort