Thread overview
Array permutations
Sep 11
Vino
Sep 11
jfondren
Sep 11
jfondren
Sep 11
Vino
Sep 11
Vino
Sep 11
Dukc
5 days ago
Elmar
5 days ago
Elmar
September 11

Hi All,

Request your help on the below to print the below array as "Required output", Was able to get these values "[1,2],[2,3],[3,4],[4,5]" by using list.slide(2), need your help to get values "1,3],[1,4],[1,5],[2,4],[2,5],[3,5]"

auto list[] = [1,2,3,4,5]

Required output
[1,2],[2,3],[3,4],[4,5],[1,3],[1,4],[1,5],[2,4],[2,5],[3,5]

From,
Vino

September 11

On Saturday, 11 September 2021 at 19:37:42 UTC, Vino wrote:

>

Hi All,

Request your help on the below to print the below array as "Required output", Was able to get these values "[1,2],[2,3],[3,4],[4,5]" by using list.slide(2), need your help to get values "1,3],[1,4],[1,5],[2,4],[2,5],[3,5]"

auto list[] = [1,2,3,4,5]

Required output
[1,2],[2,3],[3,4],[4,5],[1,3],[1,4],[1,5],[2,4],[2,5],[3,5]

From,
Vino

I don't see the pattern just from the required output. If you have an English description of what you want, it might be easier.

Anyway, take a look at

unittest {
    import std.algorithm : cartesianProduct, filter, map, sort;
    import std.range : drop;
    import std.array : array;

    auto list = [1,2,3,4,5];
    auto required = [[1,2],[2,3],[3,4],[4,5],[1,3],[1,4],[1,5],[2,4],[2,5],[3,5]];
    auto pairs = list.cartesianProduct(list.drop(1))
        .filter!"a[0] < a[1]"
        .map!array
        .array;
    assert(pairs == [[1, 2], [1, 3], [1, 4], [1, 5], [2, 3], [2, 4], [2, 5], [3, 4], [3, 5], [4, 5]]);
    assert(required.sort.array == pairs);
}
September 11

On Saturday, 11 September 2021 at 19:37:42 UTC, Vino wrote:

>

Request your help on the below to print the below array as "Required output", Was able to get these values "[1,2],[2,3],[3,4],[4,5]" by using list.slide(2), need your help to get values "1,3],[1,4],[1,5],[2,4],[2,5],[3,5]"

import std;
auto slideEnds(List)(List list, size_t slideLen)
{	return list.slide(slideLen).map!(window => [window.front, window.back]);
}
void main()
{	auto list = [1,2,3,4,5];

    iota(2, list.length)
    .map!(slideLen => list.slideEnds(slideLen))
    .joiner
    .writeln;
}

This almost does the trick. It leaves out [1, 5] for some reason I didn't bother to check though.

September 11

On Saturday, 11 September 2021 at 19:57:26 UTC, jfondren wrote:

>
auto pairs = list.cartesianProduct(list.drop(1))

This drop isn't necessary.

September 11

On Saturday, 11 September 2021 at 19:57:26 UTC, jfondren wrote:

>

On Saturday, 11 September 2021 at 19:37:42 UTC, Vino wrote:

>

Hi All,

Request your help on the below to print the below array as "Required output", Was able to get these values "[1,2],[2,3],[3,4],[4,5]" by using list.slide(2), need your help to get values "1,3],[1,4],[1,5],[2,4],[2,5],[3,5]"

auto list[] = [1,2,3,4,5]

Required output
[1,2],[2,3],[3,4],[4,5],[1,3],[1,4],[1,5],[2,4],[2,5],[3,5]

From,
Vino

I don't see the pattern just from the required output. If you have an English description of what you want, it might be easier.

Anyway, take a look at

unittest {
    import std.algorithm : cartesianProduct, filter, map, sort;
    import std.range : drop;
    import std.array : array;

    auto list = [1,2,3,4,5];
    auto required = [[1,2],[2,3],[3,4],[4,5],[1,3],[1,4],[1,5],[2,4],[2,5],[3,5]];
    auto pairs = list.cartesianProduct(list.drop(1))
        .filter!"a[0] < a[1]"
        .map!array
        .array;
    assert(pairs == [[1, 2], [1, 3], [1, 4], [1, 5], [2, 3], [2, 4], [2, 5], [3, 4], [3, 5], [4, 5]]);
    assert(required.sort.array == pairs);
}

Hi,

Thank you very much, let me try to explain the actual requirement, the actual requirement is to find all the permutations of a given array, I modified your code and it gives me the exact output , but i need help to remove the array which has the same value as below

Example:
void main() {
import std.algorithm : cartesianProduct, filter, map, sort;
import std.range : drop;
import std.array : array;
import std.stdio : writeln;

auto list = [1,2,3,4,5];
auto pairs = list.cartesianProduct(list)
    .map!array
    .array;
 writeln(pairs);

}

Output
[
[1, 1], /* shodule not print this array /
[1, 2],
[1, 3],
[1, 4],
[1, 5],
[2, 1],
[2, 2], /
should not print this array /
[2, 3],
[2, 4],
[2, 5],
[3, 1],
[3, 2],
[3, 3], /
should not print this array /
[3, 4],
[3, 5],
[4, 1],
[4, 2],
[4, 3],
[4, 4], /
should not print this array /
[4, 5],
[5, 1],
[5, 2],
[5, 3],
[5, 4],
[5, 5] /
should not print this array */
]

From,
Vino

September 11

On Saturday, 11 September 2021 at 23:04:29 UTC, Vino wrote:

>

On Saturday, 11 September 2021 at 19:57:26 UTC, jfondren wrote:

>

[...]

Hi,

Thank you very much, let me try to explain the actual requirement, the actual requirement is to find all the permutations of a given array, I modified your code and it gives me the exact output , but i need help to remove the array which has the same value as below

[...]

Hi,

Was able to resolve the issue by adding the filter ".filter!"a[0] != a[1]"

From,
Vino

5 days ago

On Saturday, 11 September 2021 at 19:37:42 UTC, Vino wrote:

>

Hi All,

Request your help on the below to print the below array as "Required output", Was able to get these values "[1,2],[2,3],[3,4],[4,5]" by using list.slide(2), need your help to get values "1,3],[1,4],[1,5],[2,4],[2,5],[3,5]"

auto list[] = [1,2,3,4,5]

Required output
[1,2],[2,3],[3,4],[4,5],[1,3],[1,4],[1,5],[2,4],[2,5],[3,5]

From,
Vino

Would this be a valid solution to your problem?

pure @safe nothrow
T[][] computeChoices(T)(T[] values, size_t tupleSize = 2)
{
    if (tupleSize == 0) {
     	return [[]];
    }

    T[][] choices = [];
    tupleSize--;
    foreach(i, element; values[0 .. $ - tupleSize])
    {
        import std.algorithm.iteration : map;
        import std.array : array;
        choices ~= computeChoices(values[i+1 .. $], tupleSize)
            		.map!(choice => element ~ choice)
            		.array;
    }

    return choices;
}

unittest
{
    assert(computeChoices([1, 2, 3, 4, 5], 2)
           	== [[1,2], [1,3], [1,4], [1,5], [2,3], [2,4], [2,5], [3,4], [3,5], [4,5]] );
}

You can choose in the 2nd parameter how large the inner arrays should be. It uses GC to allocate the result via the function array. If that is a problem, you could choose an allocator from std.experimental.allocator and use the makeArray function with the allocator instead of the array function. (Manual deallocation could be required then as well.)

5 days ago

I also should discourage its current form with large tupleSizes. The computation is in O(exp(values.length)). Instead of ~= I would suggest an std.array.appender of arrays instead of an 2D-array for the choices, if the choices become large. Most efficient is a preallocated array capacity. In that case ~= has very low overhead.