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<header>
<copyright>
<year>2001</year><year>2009</year>
<holder>Ericsson AB. All Rights Reserved.</holder>
</copyright>
<legalnotice>
The contents of this file are subject to the Erlang Public License,
Version 1.1, (the "License"); you may not use this file except in
compliance with the License. You should have received a copy of the
Erlang Public License along with this software. If not, it can be
retrieved online at http://www.erlang.org/.
Software distributed under the License is distributed on an "AS IS"
basis, WITHOUT WARRANTY OF ANY KIND, either express or implied. See
the License for the specific language governing rights and limitations
under the License.
</legalnotice>
<title>gb_sets</title>
<prepared></prepared>
<docno></docno>
<date></date>
<rev></rev>
</header>
<module>gb_sets</module>
<modulesummary>General Balanced Trees</modulesummary>
<description>
<p>An implementation of ordered sets using Prof. Arne Andersson's
General Balanced Trees. This can be much more efficient than
using ordered lists, for larger sets, but depends on the
application.</p>
</description>
<section>
<title>Complexity note</title>
<p>The complexity on set operations is bounded by either O(|S|) or
O(|T| * log(|S|)), where S is the largest given set, depending
on which is fastest for any particular function call. For
operating on sets of almost equal size, this implementation is
about 3 times slower than using ordered-list sets directly. For
sets of very different sizes, however, this solution can be
arbitrarily much faster; in practical cases, often between 10
and 100 times. This implementation is particularly suited for
accumulating elements a few at a time, building up a large set
(more than 100-200 elements), and repeatedly testing for
membership in the current set.</p>
<p>As with normal tree structures, lookup (membership testing),
insertion and deletion have logarithmic complexity.</p>
</section>
<section>
<title>Compatibility</title>
<p>All of the following functions in this module also exist
and do the same thing in the <c>sets</c> and <c>ordsets</c>
modules. That is, by only changing the module name for each call,
you can try out different set representations.</p>
<p></p>
<list type="bulleted">
<item>
<p><c>add_element/2</c></p>
</item>
<item>
<p><c>del_element/2</c></p>
</item>
<item>
<p><c>filter/2</c></p>
</item>
<item>
<p><c>fold/3</c></p>
</item>
<item>
<p><c>from_list/1</c></p>
</item>
<item>
<p><c>intersection/1</c></p>
</item>
<item>
<p><c>intersection/2</c></p>
</item>
<item>
<p><c>is_element/2</c></p>
</item>
<item>
<p><c>is_set/1</c></p>
</item>
<item>
<p><c>is_subset/2</c></p>
</item>
<item>
<p><c>new/0</c></p>
</item>
<item>
<p><c>size/1</c></p>
</item>
<item>
<p><c>subtract/2</c></p>
</item>
<item>
<p><c>to_list/1</c></p>
</item>
<item>
<p><c>union/1</c></p>
</item>
<item>
<p><c>union/2</c></p>
</item>
</list>
</section>
<section>
<title>DATA TYPES</title>
<code type="none">
gb_set() = a GB set</code>
</section>
<funcs>
<func>
<name>add(Element, Set1) -> Set2</name>
<name>add_element(Element, Set1) -> Set2</name>
<fsummary>Add a (possibly existing) element to a gb_set</fsummary>
<type>
<v>Element = term()</v>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Returns a new gb_set formed from <c>Set1</c> with
<c>Element</c> inserted. If <c>Element</c> is already an
element in <c>Set1</c>, nothing is changed.</p>
</desc>
</func>
<func>
<name>balance(Set1) -> Set2</name>
<fsummary>Rebalance tree representation of a gb_set</fsummary>
<type>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Rebalances the tree representation of <c>Set1</c>. Note that
this is rarely necessary, but may be motivated when a large
number of elements have been deleted from the tree without
further insertions. Rebalancing could then be forced in order
to minimise lookup times, since deletion only does not
rebalance the tree.</p>
</desc>
</func>
<func>
<name>delete(Element, Set1) -> Set2</name>
<fsummary>Remove an element from a gb_set</fsummary>
<type>
<v>Element = term()</v>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Returns a new gb_set formed from <c>Set1</c> with
<c>Element</c> removed. Assumes that <c>Element</c> is present
in <c>Set1</c>.</p>
</desc>
</func>
<func>
<name>delete_any(Element, Set1) -> Set2</name>
<name>del_element(Element, Set1) -> Set2</name>
<fsummary>Remove a (possibly non-existing) element from a gb_set</fsummary>
<type>
<v>Element = term()</v>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Returns a new gb_set formed from <c>Set1</c> with
<c>Element</c> removed. If <c>Element</c> is not an element
in <c>Set1</c>, nothing is changed.</p>
</desc>
</func>
<func>
<name>difference(Set1, Set2) -> Set3</name>
<name>subtract(Set1, Set2) -> Set3</name>
<fsummary>Return the difference of two gb_sets</fsummary>
<type>
<v>Set1 = Set2 = Set3 = gb_set()</v>
</type>
<desc>
<p>Returns only the elements of <c>Set1</c> which are not also
elements of <c>Set2</c>.</p>
</desc>
</func>
<func>
<name>empty() -> Set</name>
<name>new() -> Set</name>
<fsummary>Return an empty gb_set</fsummary>
<type>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns a new empty gb_set.</p>
</desc>
</func>
<func>
<name>filter(Pred, Set1) -> Set2</name>
<fsummary>Filter gb_set elements</fsummary>
<type>
<v>Pred = fun (E) -> bool()</v>
<v> E = term()</v>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Filters elements in <c>Set1</c> using predicate function
<c>Pred</c>.</p>
</desc>
</func>
<func>
<name>fold(Function, Acc0, Set) -> Acc1</name>
<fsummary>Fold over gb_set elements</fsummary>
<type>
<v>Function = fun (E, AccIn) -> AccOut</v>
<v>Acc0 = Acc1 = AccIn = AccOut = term()</v>
<v> E = term()</v>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Folds <c>Function</c> over every element in <c>Set</c>
returning the final value of the accumulator.</p>
</desc>
</func>
<func>
<name>from_list(List) -> Set</name>
<fsummary>Convert a list into a gb_set</fsummary>
<type>
<v>List = [term()]</v>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns a gb_set of the elements in <c>List</c>, where
<c>List</c> may be unordered and contain duplicates.</p>
</desc>
</func>
<func>
<name>from_ordset(List) -> Set</name>
<fsummary>Make a gb_set from an ordset list</fsummary>
<type>
<v>List = [term()]</v>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Turns an ordered-set list <c>List</c> into a gb_set. The list
must not contain duplicates.</p>
</desc>
</func>
<func>
<name>insert(Element, Set1) -> Set2</name>
<fsummary>Add a new element to a gb_set</fsummary>
<type>
<v>Element = term()</v>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Returns a new gb_set formed from <c>Set1</c> with
<c>Element</c> inserted. Assumes that <c>Element</c> is not
present in <c>Set1</c>.</p>
</desc>
</func>
<func>
<name>intersection(Set1, Set2) -> Set3</name>
<fsummary>Return the intersection of two gb_sets</fsummary>
<type>
<v>Set1 = Set2 = Set3 = gb_set()</v>
</type>
<desc>
<p>Returns the intersection of <c>Set1</c> and <c>Set2</c>.</p>
</desc>
</func>
<func>
<name>intersection(SetList) -> Set</name>
<fsummary>Return the intersection of a list of gb_sets</fsummary>
<type>
<v>SetList = [gb_set()]</v>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns the intersection of the non-empty list of gb_sets.</p>
</desc>
</func>
<func>
<name>is_disjoint(Set1, Set2) -> bool()</name>
<fsummary>Check whether two gb_sets are disjoint</fsummary>
<type>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Returns <c>true</c> if <c>Set1</c> and
<c>Set2</c> are disjoint (have no elements in common),
and <c>false</c> otherwise.</p>
</desc>
</func>
<func>
<name>is_empty(Set) -> bool()</name>
<fsummary>Test for empty gb_set</fsummary>
<type>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns <c>true</c> if <c>Set</c> is an empty set, and
<c>false</c> otherwise.</p>
</desc>
</func>
<func>
<name>is_member(Element, Set) -> bool()</name>
<name>is_element(Element, Set) -> bool()</name>
<fsummary>Test for membership of a gb_set</fsummary>
<type>
<v>Element = term()</v>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns <c>true</c> if <c>Element</c> is an element of
<c>Set</c>, otherwise <c>false</c>.</p>
</desc>
</func>
<func>
<name>is_set(Term) -> bool()</name>
<fsummary>Test for a gb_set</fsummary>
<type>
<v>Term = term()</v>
</type>
<desc>
<p>Returns <c>true</c> if <c>Set</c> appears to be a gb_set,
otherwise <c>false</c>.</p>
</desc>
</func>
<func>
<name>is_subset(Set1, Set2) -> bool()</name>
<fsummary>Test for subset</fsummary>
<type>
<v>Set1 = Set2 = gb_set()</v>
</type>
<desc>
<p>Returns <c>true</c> when every element of <c>Set1</c> is
also a member of <c>Set2</c>, otherwise <c>false</c>.</p>
</desc>
</func>
<func>
<name>iterator(Set) -> Iter</name>
<fsummary>Return an iterator for a gb_set</fsummary>
<type>
<v>Set = gb_set()</v>
<v>Iter = term()</v>
</type>
<desc>
<p>Returns an iterator that can be used for traversing the
entries of <c>Set</c>; see <c>next/1</c>. The implementation
of this is very efficient; traversing the whole set using
<c>next/1</c> is only slightly slower than getting the list
of all elements using <c>to_list/1</c> and traversing that.
The main advantage of the iterator approach is that it does
not require the complete list of all elements to be built in
memory at one time.</p>
</desc>
</func>
<func>
<name>largest(Set) -> term()</name>
<fsummary>Return largest element</fsummary>
<type>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns the largest element in <c>Set</c>. Assumes that
<c>Set</c> is nonempty.</p>
</desc>
</func>
<func>
<name>next(Iter1) -> {Element, Iter2} | none</name>
<fsummary>Traverse a gb_set with an iterator</fsummary>
<type>
<v>Iter1 = Iter2 = Element = term()</v>
</type>
<desc>
<p>Returns <c>{Element, Iter2}</c> where <c>Element</c> is the
smallest element referred to by the iterator <c>Iter1</c>,
and <c>Iter2</c> is the new iterator to be used for
traversing the remaining elements, or the atom <c>none</c> if
no elements remain.</p>
</desc>
</func>
<func>
<name>singleton(Element) -> gb_set()</name>
<fsummary>Return a gb_set with one element</fsummary>
<type>
<v>Element = term()</v>
</type>
<desc>
<p>Returns a gb_set containing only the element <c>Element</c>.</p>
</desc>
</func>
<func>
<name>size(Set) -> int()</name>
<fsummary>Return the number of elements in a gb_set</fsummary>
<type>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns the number of elements in <c>Set</c>.</p>
</desc>
</func>
<func>
<name>smallest(Set) -> term()</name>
<fsummary>Return smallest element</fsummary>
<type>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns the smallest element in <c>Set</c>. Assumes that
<c>Set</c> is nonempty.</p>
</desc>
</func>
<func>
<name>take_largest(Set1) -> {Element, Set2}</name>
<fsummary>Extract largest element</fsummary>
<type>
<v>Set1 = Set2 = gb_set()</v>
<v>Element = term()</v>
</type>
<desc>
<p>Returns <c>{Element, Set2}</c>, where <c>Element</c> is the
largest element in <c>Set1</c>, and <c>Set2</c> is this set
with <c>Element</c> deleted. Assumes that <c>Set1</c> is
nonempty.</p>
</desc>
</func>
<func>
<name>take_smallest(Set1) -> {Element, Set2}</name>
<fsummary>Extract smallest element</fsummary>
<type>
<v>Set1 = Set2 = gb_set()</v>
<v>Element = term()</v>
</type>
<desc>
<p>Returns <c>{Element, Set2}</c>, where <c>Element</c> is the
smallest element in <c>Set1</c>, and <c>Set2</c> is this set
with <c>Element</c> deleted. Assumes that <c>Set1</c> is
nonempty.</p>
</desc>
</func>
<func>
<name>to_list(Set) -> List</name>
<fsummary>Convert a gb_set into a list</fsummary>
<type>
<v>Set = gb_set()</v>
<v>List = [term()]</v>
</type>
<desc>
<p>Returns the elements of <c>Set</c> as a list.</p>
</desc>
</func>
<func>
<name>union(Set1, Set2) -> Set3</name>
<fsummary>Return the union of two gb_sets</fsummary>
<type>
<v>Set1 = Set2 = Set3 = gb_set()</v>
</type>
<desc>
<p>Returns the merged (union) gb_set of <c>Set1</c> and
<c>Set2</c>.</p>
</desc>
</func>
<func>
<name>union(SetList) -> Set</name>
<fsummary>Return the union of a list of gb_sets</fsummary>
<type>
<v>SetList = [gb_set()]</v>
<v>Set = gb_set()</v>
</type>
<desc>
<p>Returns the merged (union) gb_set of the list of gb_sets.</p>
</desc>
</func>
</funcs>
<section>
<title>SEE ALSO</title>
<p><seealso marker="gb_trees">gb_trees(3)</seealso>,
<seealso marker="ordsets">ordsets(3)</seealso>,
<seealso marker="sets">sets(3)</seealso></p>
</section>
</erlref>
