Why SortedSet<T>.GetViewBetween isn't O(log N)?

蓝咒 提交于 2019-11-26 22:17:08

问题


In .NET 4.0+, a class SortedSet<T> has a method called GetViewBetween(l, r), which returns an interface view on a tree part containing all the values between the two specified. Given that SortedSet<T> is implemented as a red-black tree, I naturally expect it to run in O(log N) time. The similar method in C++ is std::set::lower_bound/upper_bound, in Java it's TreeSet.headSet/tailSet, and they are logarithmic.

However, that is not true. The following code runs in 32 sec, whereas the equivalent O(log N) version of GetViewBetween would make this code run in 1-2 sec.

var s = new SortedSet<int>();
int n = 100000;
var rand = new Random(1000000007);
int sum = 0;
for (int i = 0; i < n; ++i) {
    s.Add(rand.Next());
    if (rand.Next() % 2 == 0) {
        int l = rand.Next(int.MaxValue / 2 - 10);
        int r = l + rand.Next(int.MaxValue / 2 - 10);
        var t = s.GetViewBetween(l, r);
        sum += t.Min;
    }
}
Console.WriteLine(sum);

I decompiled System.dll using dotPeek and here's what I got:

public TreeSubSet(SortedSet<T> Underlying, T Min, T Max, bool lowerBoundActive, bool upperBoundActive)
    : base(Underlying.Comparer)
{
    this.underlying = Underlying;
    this.min = Min;
    this.max = Max;
    this.lBoundActive = lowerBoundActive;
    this.uBoundActive = upperBoundActive;
    this.root = this.underlying.FindRange(this.min, this.max, this.lBoundActive, this.uBoundActive);
    this.count = 0;
    this.version = -1;
    this.VersionCheckImpl();
}

internal SortedSet<T>.Node FindRange(T from, T to, bool lowerBoundActive, bool upperBoundActive)
{
  SortedSet<T>.Node node = this.root;
  while (node != null)
  {
    if (lowerBoundActive && this.comparer.Compare(from, node.Item) > 0)
    {
      node = node.Right;
    }
    else
    {
      if (!upperBoundActive || this.comparer.Compare(to, node.Item) >= 0)
        return node;
      node = node.Left;
    }
  }
  return (SortedSet<T>.Node) null;
}

private void VersionCheckImpl()
{
    if (this.version == this.underlying.version)
      return;
    this.root = this.underlying.FindRange(this.min, this.max, this.lBoundActive, this.uBoundActive);
    this.version = this.underlying.version;
    this.count = 0;
    base.InOrderTreeWalk((TreeWalkPredicate<T>) (n =>
    {
      SortedSet<T>.TreeSubSet temp_31 = this;
      int temp_34 = temp_31.count + 1;
      temp_31.count = temp_34;
      return true;
    }));
}

So, FindRange is obviously O(log N), but after that we call VersionCheckImpl... which does a linear-time traversal of the found subtree only for recounting its nodes!

  1. Why would you need to do that traversal all the time?
  2. Why .NET does not contain a O(log N) method for splitting a tree based on key, like C++ or Java? It is really helpful in lots of situations.

回答1:


about the version field

UPDATE1:

In my memory, a lot of(maybe all?) collections in BCL have the field version.

First,about foreach:

according to this msdn link

The foreach statement repeats a group of embedded statements for each element in an array or an object collection. The foreach statement is used to iterate through the collection to get the desired information, but should not be used to change the contents of the collection to avoid unpredictable side effects.

In many other collections, version is protected the data is not modified during the foreach

For example, HashTable's MoveNext():

public virtual bool MoveNext()
{
    if (this.version != this.hashtable.version)
    {
        throw new InvalidOperationException(Environment.GetResourceString("InvalidOperation_EnumFailedVersion"));
    }
    ..........
}

But in the in the SortedSet<T>'s MoveNext() method:

public bool MoveNext()
{
    this.tree.VersionCheck();
    if (this.version != this.tree.version)
    {
        ThrowHelper.ThrowInvalidOperationException(ExceptionResource.InvalidOperation_EnumFailedVersion);
    }       
    ....
}

UPDATE2:

But the O(N) loop maybe not only for version but also for the Count property.

Because the MSDN of GetViewBetween said:

This method returns a view of the range of elements that fall between lowerValue and upperValue, as defined by the comparer .... You can make changes in both the view and in the underlying SortedSet(Of T).

So for every update it should be sync the count field (key and value are already same). To make sure the Count is correct

There were two policies to reach the target:

  1. Microsoft's
  2. Mono's

First.MS's,in their code, they sacrifice the GetViewBetween()'s performance and win the Count Property's performance.

VersionCheckImpl() is one way to sync the Count property.

Second,Mono. In mono's code,GetViewBetween() is Faster, but in their GetCount()method:

internal override int GetCount ()
{
    int count = 0;
    using (var e = set.tree.GetSuffixEnumerator (lower)) {
        while (e.MoveNext () && set.helper.Compare (upper, e.Current) >= 0)
            ++count;
    }
    return count;
}

It is always an O(N) operation!



来源:https://stackoverflow.com/questions/9850975/why-sortedsett-getviewbetween-isnt-olog-n

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