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Function std.algorithm.setops.cartesianProduct
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.
Prototypes
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));
Parameters
| Name | Description |
|---|---|
| range1 | The first range |
| range2 | The second range |
| ranges | Two or more non-infinite forward ranges |
| 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.
Example
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)));
Example
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]))); }
Example
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") ]));