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# std.algorithm.setops

This is a submodule of std.algorithm. It contains generic algorithms that implement set operations.
The functions multiwayMerge, multiwayUnion, setDifference, setIntersection, setSymmetricDifference expect a range of sorted ranges as input.
All algorithms are generalized to accept as input not only sets but also multisets. Each algorithm documents behaviour in the presence of duplicated inputs.

Cheat Sheet
Function Name | Description
--- | ---
cartesianProduct | Computes Cartesian product of two ranges.
largestPartialIntersection | Copies out the values that occur most frequently in a range of ranges.
largestPartialIntersectionWeighted | Copies out the values that occur most frequently (multiplied by per-value weights) in a range of ranges.
multiwayMerge | Merges a range of sorted ranges.
multiwayUnion | Computes the union of a range of sorted ranges.
setDifference | Lazily computes the set difference of two or more sorted ranges.
setIntersection | Lazily computes the intersection of two or more sorted ranges.
setSymmetricDifference | Lazily computes the symmetric set difference of two or more sorted ranges.

License:
Boost License 1.0.
Authors:
Andrei Alexandrescu

Source std/algorithm/setops.d

auto `cartesianProduct`(R1, R2)(R1 `range1`, R2 `range2`)
if (!allSatisfy!(isForwardRange, R1, R2) || anySatisfy!(isInfinite, R1, R2));
auto `cartesianProduct`(RR...)(RR `ranges`)
if (`ranges`.length >= 2 && allSatisfy!(isForwardRange, RR) && !anySatisfy!(isInfinite, RR));
auto `cartesianProduct`(R1, R2, RR...)(R1 `range1`, R2 `range2`, RR `otherRanges`)
if (!allSatisfy!(isForwardRange, R1, R2, RR) || anySatisfy!(isInfinite, R1, R2, RR));
  Lazily computes the Cartesian product of two or more `ranges`. The product is a range of tuples of elements from each respective range.
  The conditions for the two-range case are as follows:
  If both `ranges` are finite, then one must be (at least) a forward range and the other an input range.
  If one range is infinite and the other finite, then the finite range must be a forward range, and the infinite range can be an input range.
  If both `ranges` are infinite, then both must be forward `ranges`.
  When there are more than two `ranges`, the above conditions apply to each adjacent pair of `ranges`.
  Parameters:

  R1 `range1` | The first range
  R2 `range2` | The second range
  RR `ranges` | Two or more non-infinite forward `ranges`
  RR `otherRanges` | Zero or more non-infinite forward `ranges`

  Returns:
  A forward range of std.typecons.Tuple representing elements of the cartesian product of the given `ranges`.
  Examples:

      import std.algorithm.searching : canFind;
      import std.range;
      import std.typecons : tuple;

      auto N = sequence!"n"(0);         // the range of natural numbers
      auto N2 = cartesianProduct(N, N); // the range of all pairs of natural numbers

      // Various arbitrary number pairs can be found in the range in finite time.
      assert(canFind(N2, tuple(0, 0)));
      assert(canFind(N2, tuple(123, 321)));
      assert(canFind(N2, tuple(11, 35)));
      assert(canFind(N2, tuple(279, 172)));

  Examples:

      import std.algorithm.searching : canFind;
      import std.typecons : tuple;

      auto B = [ 1, 2, 3 ];
      auto C = [ 4, 5, 6 ];
      auto BC = cartesianProduct(B, C);

      foreach (n; [[1, 4], [2, 4], [3, 4], [1, 5], [2, 5], [3, 5], [1, 6],
                   [2, 6], [3, 6]])
      {
          assert(canFind(BC, tuple(n[0], n[1])));
      }

  Examples:

      import std.algorithm.comparison : equal;
      import std.typecons : tuple;

      auto A = [ 1, 2, 3 ];
      auto B = [ 'a', 'b', 'c' ];
      auto C = [ "x", "y", "z" ];
      auto ABC = cartesianProduct(A, B, C);

      assert(ABC.equal([
          tuple(1, 'a', "x"), tuple(1, 'a', "y"), tuple(1, 'a', "z"),
          tuple(1, 'b', "x"), tuple(1, 'b', "y"), tuple(1, 'b', "z"),
          tuple(1, 'c', "x"), tuple(1, 'c', "y"), tuple(1, 'c', "z"),
          tuple(2, 'a', "x"), tuple(2, 'a', "y"), tuple(2, 'a', "z"),
          tuple(2, 'b', "x"), tuple(2, 'b', "y"), tuple(2, 'b', "z"),
          tuple(2, 'c', "x"), tuple(2, 'c', "y"), tuple(2, 'c', "z"),
          tuple(3, 'a', "x"), tuple(3, 'a', "y"), tuple(3, 'a', "z"),
          tuple(3, 'b', "x"), tuple(3, 'b', "y"), tuple(3, 'b', "z"),
          tuple(3, 'c', "x"), tuple(3, 'c', "y"), tuple(3, 'c', "z")
      ]));

void `largestPartialIntersection`(alias less = "a < b", RangeOfRanges, Range)(RangeOfRanges `ror`, Range `tgt`, SortOutput `sorted` = No.sortOutput);
  Given a range of `sorted` forward ranges `ror`, copies to `tgt` the elements that are common to most ranges, along with their number of occurrences. All ranges in `ror` are assumed to be `sorted` by less. Only the most frequent `tgt`.length elements are returned.
  Parameters:

  less | The predicate the ranges are `sorted` by.
  RangeOfRanges `ror` | A range of forward ranges `sorted` by less.
  Range `tgt` | The target range to copy common elements to.
  SortOutput `sorted` | Whether the elements copied should be in `sorted` order. The function `largestPartialIntersection` is useful for e.g. searching an inverted index for the documents most likely to contain some terms of interest. The complexity of the search is Ο(n * log(`tgt`.length)), where n is the sum of lengths of all input ranges. This approach is faster than keeping an associative array of the occurrences and then selecting its top items, and also requires less memory (`largestPartialIntersection` builds its result directly in `tgt` and requires no extra memory). If at least one of the ranges is a multiset, then all occurences of a duplicate element are taken into account. The result is equivalent to merging all ranges and picking the most frequent `tgt`.length elements.

  Warning Because `largestPartialIntersection` does not allocate extra memory, it will leave `ror` modified. Namely, `largestPartialIntersection` assumes ownership of `ror` and discretionarily swaps and advances elements of it. If you want `ror` to preserve its contents after the call, you may want to pass a duplicate to `largestPartialIntersection` (and perhaps cache the duplicate in between calls).

  Examples:

      import std.typecons : tuple, Tuple;

      // Figure which number can be found in most arrays of the set of
      // arrays below.
      double[][] a =
      [
          [ 1, 4, 7, 8 ],
          [ 1, 7 ],
          [ 1, 7, 8],
          [ 4 ],
          [ 7 ],
      ];
      auto b = new Tuple!(double, uint)[1];
      // it will modify the input range, hence we need to create a duplicate
      largestPartialIntersection(a.dup, b);
      // First member is the item, second is the occurrence count
      writeln(b[0]); // tuple(7.0, 4u)
      // 7.0 occurs in 4 out of 5 inputs, more than any other number

      // If more of the top-frequent numbers are needed, just create a larger
      // tgt range
      auto c = new Tuple!(double, uint)[2];
      largestPartialIntersection(a, c);
      writeln(c[0]); // tuple(1.0, 3u)
      // 1.0 occurs in 3 inputs

      // multiset
      double[][] x =
      [
          [1, 1, 1, 1, 4, 7, 8],
          [1, 7],
          [1, 7, 8],
          [4, 7],
          [7]
      ];
      auto y = new Tuple!(double, uint)[2];
      largestPartialIntersection(x.dup, y);
      // 7.0 occurs 5 times
      writeln(y[0]); // tuple(7.0, 5u)
      // 1.0 occurs 6 times
      writeln(y[1]); // tuple(1.0, 6u)

void `largestPartialIntersectionWeighted`(alias less = "a < b", RangeOfRanges, Range, WeightsAA)(RangeOfRanges `ror`, Range `tgt`, WeightsAA `weights`, SortOutput `sorted` = No.sortOutput);
  Similar to largestPartialIntersection, but associates a weight with each distinct element in the intersection.
  If at least one of the ranges is a multiset, then all occurences of a duplicate element are taken into account. The result is equivalent to merging all input ranges and picking the highest `tgt`.length, weight-based ranking elements.
  Parameters:

  less | The predicate the ranges are `sorted` by.
  RangeOfRanges `ror` | A range of forward ranges `sorted` by less.
  Range `tgt` | The target range to copy common elements to.
  WeightsAA `weights` | An associative array mapping elements to `weights`.
  SortOutput `sorted` | Whether the elements copied should be in `sorted` order.

  Examples:

      import std.typecons : tuple, Tuple;

      // Figure which number can be found in most arrays of the set of
      // arrays below, with specific per-element weights
      double[][] a =
      [
          [ 1, 4, 7, 8 ],
          [ 1, 7 ],
          [ 1, 7, 8],
          [ 4 ],
          [ 7 ],
      ];
      auto b = new Tuple!(double, uint)[1];
      double[double] weights = [ 1:1.2, 4:2.3, 7:1.1, 8:1.1 ];
      largestPartialIntersectionWeighted(a, b, weights);
      // First member is the item, second is the occurrence count
      writeln(b[0]); // tuple(4.0, 2u)
      // 4.0 occurs 2 times -> 4.6 (2 * 2.3)
      // 7.0 occurs 3 times -> 4.4 (3 * 1.1)

      // multiset
      double[][] x =
      [
          [ 1, 1, 1, 4, 7, 8 ],
          [ 1, 7 ],
          [ 1, 7, 8],
          [ 4 ],
          [ 7 ],
      ];
      auto y = new Tuple!(double, uint)[1];
      largestPartialIntersectionWeighted(x, y, weights);
      writeln(y[0]); // tuple(1.0, 5u)
      // 1.0 occurs 5 times -> 1.2 * 5 = 6

struct `MultiwayMerge`(alias less, RangeOfRanges);
MultiwayMerge!(less, RangeOfRanges) `multiwayMerge`(alias less = "a < b", RangeOfRanges)(RangeOfRanges `ror`);
  Merges multiple sets. The input sets are passed as a range of ranges and each is assumed to be sorted by less. Computation is done lazily, one union element at a time. The complexity of one popFront operation is Ο(log(`ror`.length)). However, the length of `ror` decreases as ranges in it are exhausted, so the complexity of a full pass through `MultiwayMerge` is dependent on the distribution of the lengths of ranges contained within `ror`. If all ranges have the same length n (worst case scenario), the complexity of a full pass through `MultiwayMerge` is Ο(n * `ror`.length * log(`ror`.length)), i.e., log(`ror`.length) times worse than just spanning all ranges in turn. The output comes sorted (unstably) by less.
  The length of the resulting range is the sum of all lengths of the ranges passed as input. This means that all elements (duplicates included) are transferred to the resulting range.
  For backward compatibility, `multiwayMerge` is available under the name nWayUnion and `MultiwayMerge` under the name of NWayUnion . Future code should use `multiwayMerge` and `MultiwayMerge` as nWayUnion and NWayUnion will be deprecated.
  Parameters:

  less | Predicate the given ranges are sorted by.
  RangeOfRanges `ror` | A range of ranges sorted by less to compute the union for.

  Returns:
  A range of the union of the ranges in `ror`.

  Warning Because `MultiwayMerge` does not allocate extra memory, it will leave `ror` modified. Namely, `MultiwayMerge` assumes ownership of `ror` and discretionarily swaps and advances elements of it. If you want `ror` to preserve its contents after the call, you may want to pass a duplicate to `MultiwayMerge` (and perhaps cache the duplicate in between calls).

  Examples:

      import std.algorithm.comparison : equal;

      double[][] a =
      [
          [ 1, 4, 7, 8 ],
          [ 1, 7 ],
          [ 1, 7, 8],
          [ 4 ],
          [ 7 ],
      ];
      auto witness = [
          1, 1, 1, 4, 4, 7, 7, 7, 7, 8, 8
      ];
      assert(equal(multiwayMerge(a), witness));

      double[][] b =
      [
          // range with duplicates
          [ 1, 1, 4, 7, 8 ],
          [ 7 ],
          [ 1, 7, 8],
          [ 4 ],
          [ 7 ],
      ];
      // duplicates are propagated to the resulting range
      assert(equal(multiwayMerge(b), witness));

  static bool `compFront`(.ElementType!RangeOfRanges `a`, .ElementType!RangeOfRanges `b`);
  this(RangeOfRanges `ror`);
  @property bool `empty`();
  @property ref auto `front`();
  void `popFront`();
auto `multiwayUnion`(alias less = "a < b", RangeOfRanges)(RangeOfRanges `ror`);
  Computes the union of multiple ranges. The input ranges are passed as a range of ranges and each is assumed to be sorted by less. Computation is done lazily, one union element at a time. `multiwayUnion`(`ror`) is functionally equivalent to multiwayMerge(`ror`).uniq.
  "The output of `multiwayUnion` has no duplicates even when its inputs contain duplicates."
  Parameters:

  less | Predicate the given ranges are sorted by.
  RangeOfRanges `ror` | A range of ranges sorted by less to compute the intersection for.

  Returns:
  A range of the union of the ranges in `ror`.
  See also: multiwayMerge
  Examples:

      import std.algorithm.comparison : equal;

      // sets
      double[][] a =
      [
          [ 1, 4, 7, 8 ],
          [ 1, 7 ],
          [ 1, 7, 8],
          [ 4 ],
          [ 7 ],
      ];

      auto witness = [1, 4, 7, 8];
      assert(equal(multiwayUnion(a), witness));

      // multisets
      double[][] b =
      [
          [ 1, 1, 1, 4, 7, 8 ],
          [ 1, 7 ],
          [ 1, 7, 7, 8],
          [ 4 ],
          [ 7 ],
      ];
      assert(equal(multiwayUnion(b), witness));

struct `SetDifference`(alias less = "a < b", R1, R2) if (isInputRange!R1 && isInputRange!R2);
SetDifference!(less, R1, R2) `setDifference`(alias less = "a < b", R1, R2)(R1 `r1`, R2 `r2`);
  Lazily computes the difference of `r1` and `r2`. The two ranges are assumed to be sorted by less. The element types of the two ranges must have a common type.
  In the case of multisets, considering that element a appears x times in `r1` and y times and `r2`, the number of occurences of a in the resulting range is going to be x-y if x > y or 0 othwerise.
  Parameters:

  less | Predicate the given ranges are sorted by.
  R1 `r1` | The first range.
  R2 `r2` | The range to subtract from `r1`.

  Returns:
  A range of the difference of `r1` and `r2`.
  See Also:
  setSymmetricDifference
  Examples:

      import std.algorithm.comparison : equal;
      import std.range.primitives : isForwardRange;

      //sets
      int[] a = [ 1, 2, 4, 5, 7, 9 ];
      int[] b = [ 0, 1, 2, 4, 7, 8 ];
      assert(equal(setDifference(a, b), [5, 9]));
      static assert(isForwardRange!(typeof(setDifference(a, b))));

      // multisets
      int[] x = [1, 1, 1, 2, 3];
      int[] y = [1, 1, 2, 4, 5];
      auto r = setDifference(x, y);
      assert(equal(r, [1, 3]));
      assert(setDifference(r, x).empty);

  this(R1 `r1`, R2 `r2`);
  void `popFront`();
  @property ref auto `front`();
  @property typeof(this) `save`();
  @property bool `empty`();
struct `SetIntersection`(alias less = "a < b", Rs...) if (Rs.length >= 2 && allSatisfy!(isInputRange, Rs) && !is(CommonType!(staticMap!(ElementType, Rs)) == void));
SetIntersection!(less, Rs) `setIntersection`(alias less = "a < b", Rs...)(Rs `ranges`)
if (Rs.length >= 2 && allSatisfy!(isInputRange, Rs) && !is(CommonType!(staticMap!(ElementType, Rs)) == void));
  Lazily computes the intersection of two or more input `ranges` `ranges`. The `ranges` are assumed to be sorted by less. The element types of the `ranges` must have a common type.
  In the case of multisets, the range with the minimum number of occurences of a given element, propagates the number of occurences of this element to the resulting range.
  Parameters:

  less | Predicate the given `ranges` are sorted by.
  Rs `ranges` | The `ranges` to compute the intersection for.

  Returns:
  A range containing the intersection of the given `ranges`.
  Examples:

      import std.algorithm.comparison : equal;

      // sets
      int[] a = [ 1, 2, 4, 5, 7, 9 ];
      int[] b = [ 0, 1, 2, 4, 7, 8 ];
      int[] c = [ 0, 1, 4, 5, 7, 8 ];
      assert(equal(setIntersection(a, a), a));
      assert(equal(setIntersection(a, b), [1, 2, 4, 7]));
      assert(equal(setIntersection(a, b, c), [1, 4, 7]));

      // multisets
      int[] d = [ 1, 1, 2, 2, 7, 7 ];
      int[] e = [ 1, 1, 1, 7];
      assert(equal(setIntersection(a, d), [1, 2, 7]));
      assert(equal(setIntersection(d, e), [1, 1, 7]));

  this(Rs `input`);
  @property bool `empty`();
  void `popFront`();
  @property ElementType `front`();
  @property SetIntersection `save`();
struct `SetSymmetricDifference`(alias less = "a < b", R1, R2) if (isInputRange!R1 && isInputRange!R2);
SetSymmetricDifference!(less, R1, R2) `setSymmetricDifference`(alias less = "a < b", R1, R2)(R1 `r1`, R2 `r2`);
  Lazily computes the symmetric difference of `r1` and `r2`, i.e. the elements that are present in exactly one of `r1` and `r2`. The two ranges are assumed to be sorted by less, and the output is also sorted by less. The element types of the two ranges must have a common type.
  If both ranges are sets (without duplicated elements), the resulting range is going to be a set. If at least one of the ranges is a multiset, the number of occurences of an element x in the resulting range is abs(a-b) where a is the number of occurences of x in `r1`, b is the number of occurences of x in `r2`, and abs is the absolute value.
  If both arguments are ranges of L-values of the same type then `SetSymmetricDifference` will also be a range of L-values of that type.
  Parameters:

  less | Predicate the given ranges are sorted by.
  R1 `r1` | The first range.
  R2 `r2` | The second range.

  Returns:
  A range of the symmetric difference between `r1` and `r2`.
  See Also:
  setDifference
  Examples:

      import std.algorithm.comparison : equal;
      import std.range.primitives : isForwardRange;

      // sets
      int[] a = [ 1, 2, 4, 5, 7, 9 ];
      int[] b = [ 0, 1, 2, 4, 7, 8 ];
      assert(equal(setSymmetricDifference(a, b), [0, 5, 8, 9][]));
      static assert(isForwardRange!(typeof(setSymmetricDifference(a, b))));

      //mutisets
      int[] c = [1, 1, 1, 1, 2, 2, 2, 4, 5, 6];
      int[] d = [1, 1, 2, 2, 2, 2, 4, 7, 9];
      assert(equal(setSymmetricDifference(c, d), setSymmetricDifference(d, c)));
      assert(equal(setSymmetricDifference(c, d), [1, 1, 2, 5, 6, 7, 9]));

  this(R1 `r1`, R2 `r2`);
  void `popFront`();
  @property ref auto `front`();
  @property typeof(this) `save`();
  ref auto `opSlice`();
  @property bool `empty`();

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