Update prebuilt Clang to r416183b from Android.
https://android.googlesource.com/platform/prebuilts/clang/host/
linux-x86/+/06a71ddac05c22edb2d10b590e1769b3f8619bef
clang 12.0.5 (based on r416183b) from build 7284624.
Change-Id: I277a316abcf47307562d8b748b84870f31a72866
Signed-off-by: Olivier Deprez <olivier.deprez@arm.com>
diff --git a/linux-x64/clang/python3/lib/python3.9/heapq.py b/linux-x64/clang/python3/lib/python3.9/heapq.py
new file mode 100644
index 0000000..fabefd8
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+++ b/linux-x64/clang/python3/lib/python3.9/heapq.py
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+"""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 and Raymond Hettinger
+
+__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
+a 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', 'merge',
+ 'nlargest', 'nsmallest', 'heappushpop']
+
+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)
+ return returnitem
+ return lastelt
+
+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 unless written as part of a conditional replacement:
+
+ if item > heap[0]:
+ item = heapreplace(heap, item)
+ """
+ returnitem = heap[0] # raises appropriate IndexError if heap is empty
+ heap[0] = item
+ _siftup(heap, 0)
+ return returnitem
+
+def heappushpop(heap, item):
+ """Fast version of a heappush followed by a heappop."""
+ if heap and heap[0] < item:
+ item, heap[0] = heap[0], item
+ _siftup(heap, 0)
+ return item
+
+def heapify(x):
+ """Transform list into a heap, in-place, in O(len(x)) 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(range(n//2)):
+ _siftup(x, i)
+
+def _heappop_max(heap):
+ """Maxheap version of a heappop."""
+ lastelt = heap.pop() # raises appropriate IndexError if heap is empty
+ if heap:
+ returnitem = heap[0]
+ heap[0] = lastelt
+ _siftup_max(heap, 0)
+ return returnitem
+ return lastelt
+
+def _heapreplace_max(heap, item):
+ """Maxheap version of a heappop followed by a heappush."""
+ returnitem = heap[0] # raises appropriate IndexError if heap is empty
+ heap[0] = item
+ _siftup_max(heap, 0)
+ return returnitem
+
+def _heapify_max(x):
+ """Transform list into a maxheap, in-place, in O(len(x)) time."""
+ n = len(x)
+ for i in reversed(range(n//2)):
+ _siftup_max(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 newitem < parent:
+ heap[pos] = parent
+ pos = parentpos
+ continue
+ break
+ 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 comparison 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 not heap[childpos] < heap[rightpos]:
+ 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)
+
+def _siftdown_max(heap, startpos, pos):
+ 'Maxheap variant of _siftdown'
+ 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:
+ heap[pos] = parent
+ pos = parentpos
+ continue
+ break
+ heap[pos] = newitem
+
+def _siftup_max(heap, pos):
+ 'Maxheap variant of _siftup'
+ endpos = len(heap)
+ startpos = pos
+ newitem = heap[pos]
+ # Bubble up the larger child until hitting a leaf.
+ childpos = 2*pos + 1 # leftmost child position
+ while childpos < endpos:
+ # Set childpos to index of larger child.
+ rightpos = childpos + 1
+ if rightpos < endpos and not heap[rightpos] < heap[childpos]:
+ childpos = rightpos
+ # Move the larger 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_max(heap, startpos, pos)
+
+def merge(*iterables, key=None, reverse=False):
+ '''Merge multiple sorted inputs into a single sorted output.
+
+ Similar to sorted(itertools.chain(*iterables)) but returns a generator,
+ does not pull the data into memory all at once, and assumes that each of
+ the input streams is already sorted (smallest to largest).
+
+ >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
+ [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
+
+ If *key* is not None, applies a key function to each element to determine
+ its sort order.
+
+ >>> list(merge(['dog', 'horse'], ['cat', 'fish', 'kangaroo'], key=len))
+ ['dog', 'cat', 'fish', 'horse', 'kangaroo']
+
+ '''
+
+ h = []
+ h_append = h.append
+
+ if reverse:
+ _heapify = _heapify_max
+ _heappop = _heappop_max
+ _heapreplace = _heapreplace_max
+ direction = -1
+ else:
+ _heapify = heapify
+ _heappop = heappop
+ _heapreplace = heapreplace
+ direction = 1
+
+ if key is None:
+ for order, it in enumerate(map(iter, iterables)):
+ try:
+ next = it.__next__
+ h_append([next(), order * direction, next])
+ except StopIteration:
+ pass
+ _heapify(h)
+ while len(h) > 1:
+ try:
+ while True:
+ value, order, next = s = h[0]
+ yield value
+ s[0] = next() # raises StopIteration when exhausted
+ _heapreplace(h, s) # restore heap condition
+ except StopIteration:
+ _heappop(h) # remove empty iterator
+ if h:
+ # fast case when only a single iterator remains
+ value, order, next = h[0]
+ yield value
+ yield from next.__self__
+ return
+
+ for order, it in enumerate(map(iter, iterables)):
+ try:
+ next = it.__next__
+ value = next()
+ h_append([key(value), order * direction, value, next])
+ except StopIteration:
+ pass
+ _heapify(h)
+ while len(h) > 1:
+ try:
+ while True:
+ key_value, order, value, next = s = h[0]
+ yield value
+ value = next()
+ s[0] = key(value)
+ s[2] = value
+ _heapreplace(h, s)
+ except StopIteration:
+ _heappop(h)
+ if h:
+ key_value, order, value, next = h[0]
+ yield value
+ yield from next.__self__
+
+
+# Algorithm notes for nlargest() and nsmallest()
+# ==============================================
+#
+# Make a single pass over the data while keeping the k most extreme values
+# in a heap. Memory consumption is limited to keeping k values in a list.
+#
+# Measured performance for random inputs:
+#
+# number of comparisons
+# n inputs k-extreme values (average of 5 trials) % more than min()
+# ------------- ---------------- --------------------- -----------------
+# 1,000 100 3,317 231.7%
+# 10,000 100 14,046 40.5%
+# 100,000 100 105,749 5.7%
+# 1,000,000 100 1,007,751 0.8%
+# 10,000,000 100 10,009,401 0.1%
+#
+# Theoretical number of comparisons for k smallest of n random inputs:
+#
+# Step Comparisons Action
+# ---- -------------------------- ---------------------------
+# 1 1.66 * k heapify the first k-inputs
+# 2 n - k compare remaining elements to top of heap
+# 3 k * (1 + lg2(k)) * ln(n/k) replace the topmost value on the heap
+# 4 k * lg2(k) - (k/2) final sort of the k most extreme values
+#
+# Combining and simplifying for a rough estimate gives:
+#
+# comparisons = n + k * (log(k, 2) * log(n/k) + log(k, 2) + log(n/k))
+#
+# Computing the number of comparisons for step 3:
+# -----------------------------------------------
+# * For the i-th new value from the iterable, the probability of being in the
+# k most extreme values is k/i. For example, the probability of the 101st
+# value seen being in the 100 most extreme values is 100/101.
+# * If the value is a new extreme value, the cost of inserting it into the
+# heap is 1 + log(k, 2).
+# * The probability times the cost gives:
+# (k/i) * (1 + log(k, 2))
+# * Summing across the remaining n-k elements gives:
+# sum((k/i) * (1 + log(k, 2)) for i in range(k+1, n+1))
+# * This reduces to:
+# (H(n) - H(k)) * k * (1 + log(k, 2))
+# * Where H(n) is the n-th harmonic number estimated by:
+# gamma = 0.5772156649
+# H(n) = log(n, e) + gamma + 1 / (2 * n)
+# http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence
+# * Substituting the H(n) formula:
+# comparisons = k * (1 + log(k, 2)) * (log(n/k, e) + (1/n - 1/k) / 2)
+#
+# Worst-case for step 3:
+# ----------------------
+# In the worst case, the input data is reversed sorted so that every new element
+# must be inserted in the heap:
+#
+# comparisons = 1.66 * k + log(k, 2) * (n - k)
+#
+# Alternative Algorithms
+# ----------------------
+# Other algorithms were not used because they:
+# 1) Took much more auxiliary memory,
+# 2) Made multiple passes over the data.
+# 3) Made more comparisons in common cases (small k, large n, semi-random input).
+# See the more detailed comparison of approach at:
+# http://code.activestate.com/recipes/577573-compare-algorithms-for-heapqsmallest
+
+def nsmallest(n, iterable, key=None):
+ """Find the n smallest elements in a dataset.
+
+ Equivalent to: sorted(iterable, key=key)[:n]
+ """
+
+ # Short-cut for n==1 is to use min()
+ if n == 1:
+ it = iter(iterable)
+ sentinel = object()
+ result = min(it, default=sentinel, key=key)
+ return [] if result is sentinel else [result]
+
+ # When n>=size, it's faster to use sorted()
+ try:
+ size = len(iterable)
+ except (TypeError, AttributeError):
+ pass
+ else:
+ if n >= size:
+ return sorted(iterable, key=key)[:n]
+
+ # When key is none, use simpler decoration
+ if key is None:
+ it = iter(iterable)
+ # put the range(n) first so that zip() doesn't
+ # consume one too many elements from the iterator
+ result = [(elem, i) for i, elem in zip(range(n), it)]
+ if not result:
+ return result
+ _heapify_max(result)
+ top = result[0][0]
+ order = n
+ _heapreplace = _heapreplace_max
+ for elem in it:
+ if elem < top:
+ _heapreplace(result, (elem, order))
+ top, _order = result[0]
+ order += 1
+ result.sort()
+ return [elem for (elem, order) in result]
+
+ # General case, slowest method
+ it = iter(iterable)
+ result = [(key(elem), i, elem) for i, elem in zip(range(n), it)]
+ if not result:
+ return result
+ _heapify_max(result)
+ top = result[0][0]
+ order = n
+ _heapreplace = _heapreplace_max
+ for elem in it:
+ k = key(elem)
+ if k < top:
+ _heapreplace(result, (k, order, elem))
+ top, _order, _elem = result[0]
+ order += 1
+ result.sort()
+ return [elem for (k, order, elem) in result]
+
+def nlargest(n, iterable, key=None):
+ """Find the n largest elements in a dataset.
+
+ Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
+ """
+
+ # Short-cut for n==1 is to use max()
+ if n == 1:
+ it = iter(iterable)
+ sentinel = object()
+ result = max(it, default=sentinel, key=key)
+ return [] if result is sentinel else [result]
+
+ # When n>=size, it's faster to use sorted()
+ try:
+ size = len(iterable)
+ except (TypeError, AttributeError):
+ pass
+ else:
+ if n >= size:
+ return sorted(iterable, key=key, reverse=True)[:n]
+
+ # When key is none, use simpler decoration
+ if key is None:
+ it = iter(iterable)
+ result = [(elem, i) for i, elem in zip(range(0, -n, -1), it)]
+ if not result:
+ return result
+ heapify(result)
+ top = result[0][0]
+ order = -n
+ _heapreplace = heapreplace
+ for elem in it:
+ if top < elem:
+ _heapreplace(result, (elem, order))
+ top, _order = result[0]
+ order -= 1
+ result.sort(reverse=True)
+ return [elem for (elem, order) in result]
+
+ # General case, slowest method
+ it = iter(iterable)
+ result = [(key(elem), i, elem) for i, elem in zip(range(0, -n, -1), it)]
+ if not result:
+ return result
+ heapify(result)
+ top = result[0][0]
+ order = -n
+ _heapreplace = heapreplace
+ for elem in it:
+ k = key(elem)
+ if top < k:
+ _heapreplace(result, (k, order, elem))
+ top, _order, _elem = result[0]
+ order -= 1
+ result.sort(reverse=True)
+ return [elem for (k, order, elem) in result]
+
+# If available, use C implementation
+try:
+ from _heapq import *
+except ImportError:
+ pass
+try:
+ from _heapq import _heapreplace_max
+except ImportError:
+ pass
+try:
+ from _heapq import _heapify_max
+except ImportError:
+ pass
+try:
+ from _heapq import _heappop_max
+except ImportError:
+ pass
+
+
+if __name__ == "__main__":
+
+ import doctest # pragma: no cover
+ print(doctest.testmod()) # pragma: no cover