fix ya.make

1 files changed, 0 insertions, 601 deletions

diff --git a/contrib/tools/python3/src/Lib/heapq.py b/contrib/tools/python3/src/Lib/heapq.py
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--- a/contrib/tools/python3/src/Lib/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