Hi,
Is it possible to write such a construction that could push immediately to a conditional statement without using nested loops? Ie organize search directly in the iterator if provided by a map / each / iota and other support functions.
Ie I want to write this code shorter :)

import std.stdio;

void main() {

    const x = 12, y = 65, z = 50, s = 1435;
	
    foreach (i; 0 .. x + 1)
        foreach (j; 0 .. x + 1)
            foreach (k; 0 .. x + 1)
                if (i * (y + 3 * z) + j * (y + 2 * z) + k * (y + z) == s) {
                    writeln(i + j + k);
                    writeln(i * 3 + j * 2 + k);
                    writeln(i * 3 + j * 2 + k);
                    writeln(i, j, k);
                }
}
-----
http://rextester.com/SJTU87854

On Monday, 25 May 2015 at 14:25:52 UTC, Dennis Ritchie wrote:
> Hi,
> Is it possible to write such a construction that could push immediately to a conditional statement without using nested loops? Ie organize search directly in the iterator if provided by a map / each / iota and other support functions.
> Ie I want to write this code shorter :)
>
> import std.stdio;
>
> void main() {
>
>     const x = 12, y = 65, z = 50, s = 1435;
> 	
>     foreach (i; 0 .. x + 1)
>         foreach (j; 0 .. x + 1)
>             foreach (k; 0 .. x + 1)
>                 if (i * (y + 3 * z) + j * (y + 2 * z) + k * (y + z) == s) {
>                     writeln(i + j + k);
>                     writeln(i * 3 + j * 2 + k);
>                     writeln(i * 3 + j * 2 + k);
>                     writeln(i, j, k);
>                 }
> }
> -----
> http://rextester.com/SJTU87854

Hint: Use `cartesianProduct` [1] with three iota ranges to replace the foreachs, and `filter` to replace the if

[1] http://dlang.org/phobos/std_algorithm_setops.html#.cartesianProduct

On Monday, 25 May 2015 at 15:06:45 UTC, Alex Parrill wrote:
> Hint: Use `cartesianProduct` [1] with three iota ranges to replace the foreachs, and `filter` to replace the if
>
> [1] http://dlang.org/phobos/std_algorithm_setops.html#.cartesianProduct

Thank you. Is it possible to replace the loop `foreach` function `each` or `chunkBy` with the auxiliary functions?

import std.stdio, std.algorithm, std.range;

void main() {

    const x = 12, y = 65, z = 50, s = 1435;

    auto a = iota(0, x + 1);

    foreach (idx; cartesianProduct(a, a, a).filter!(i => i[0] * (y + 3 * z) + i[1] * (y + 2 * z) + i[2] * (y + z) == s)) {
        writeln(idx[0] + idx[1] + idx[2]);
        writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
        writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
        writeln(idx[0], idx[1], idx[2]);
    }
}
-----
http://rextester.com/HZP96719

On Monday, 25 May 2015 at 15:46:54 UTC, Dennis Ritchie wrote:
> On Monday, 25 May 2015 at 15:06:45 UTC, Alex Parrill wrote:
>> Hint: Use `cartesianProduct` [1] with three iota ranges to replace the foreachs, and `filter` to replace the if
>>
>> [1] http://dlang.org/phobos/std_algorithm_setops.html#.cartesianProduct
>
> Thank you. Is it possible to replace the loop `foreach` function `each` or `chunkBy` with the auxiliary functions?
>
> import std.stdio, std.algorithm, std.range;
>
> void main() {
>
>     const x = 12, y = 65, z = 50, s = 1435;
>
>     auto a = iota(0, x + 1);
>
>     foreach (idx; cartesianProduct(a, a, a).filter!(i => i[0] * (y + 3 * z) + i[1] * (y + 2 * z) + i[2] * (y + z) == s)) {
>         writeln(idx[0] + idx[1] + idx[2]);
>         writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
>         writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
>         writeln(idx[0], idx[1], idx[2]);
>     }
> }
> -----
> http://rextester.com/HZP96719

import std.algorithm;
import std.range;
import std.stdio;

void main()
{
    const x = 12, y = 65, z = 50, s = 1435;
    auto a = iota(0, x + 1);
    cartesianProduct(a, a, a)
      	.filter!(i => i[0] * (y + 3 * z)
            + i[1] * (y + 2 * z)
            + i[2] * (y + z)
            == s)
      	.each!((idx)
        {
            writeln(idx[0] + idx[1] + idx[2]);
            writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
            writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
            writeln(idx[0], idx[1], idx[2]);
        });
}

On Monday, 25 May 2015 at 16:41:35 UTC, Meta wrote:
> import std.algorithm;
> import std.range;
> import std.stdio;
>
> void main()
> {
>     const x = 12, y = 65, z = 50, s = 1435;
>     auto a = iota(0, x + 1);
>     cartesianProduct(a, a, a)
>       	.filter!(i => i[0] * (y + 3 * z)
>             + i[1] * (y + 2 * z)
>             + i[2] * (y + z)
>             == s)
>       	.each!((idx)
>               {
>                   writeln(idx[0] + idx[1] + idx[2]);
>                   writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
>                   writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
>                   writeln(idx[0], idx[1], idx[2]);
>               });
> }

Thanks. I do not even know what `each` support braces.

On Monday, 25 May 2015 at 17:05:35 UTC, Dennis Ritchie wrote:
> On Monday, 25 May 2015 at 16:41:35 UTC, Meta wrote:
>> import std.algorithm;
>> import std.range;
>> import std.stdio;
>>
>> void main()
>> {
>>    const x = 12, y = 65, z = 50, s = 1435;
>>    auto a = iota(0, x + 1);
>>    cartesianProduct(a, a, a)
>>      	.filter!(i => i[0] * (y + 3 * z)
>>            + i[1] * (y + 2 * z)
>>            + i[2] * (y + z)
>>            == s)
>>      	.each!((idx)
>>              {
>>                  writeln(idx[0] + idx[1] + idx[2]);
>>                  writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
>>                  writeln(idx[0] * 3 + idx[1] * 2 + idx[2]);
>>                  writeln(idx[0], idx[1], idx[2]);
>>              });
>> }
>
> Thanks. I do not even know what `each` support braces.

`each` doesn't support braces. There are 4 ways to write a function/delegate literal in D (with a few minor variations):

Short form:
function(int i) => i;
(int i) => i
(i) => i
i => i

Long form:
function(int i) { return i; }
(int i) { return i; }
(i) { return i; }
{ return 0; }

http://dlang.org/expression.html#FunctionLiteral
function

On Monday, 25 May 2015 at 17:19:27 UTC, Meta wrote:
> `each` doesn't support braces. There are 4 ways to write a function/delegate literal in D (with a few minor variations):
>
> Short form:
> function(int i) => i;
> (int i) => i
> (i) => i
> i => i
>
> Long form:
> function(int i) { return i; }
> (int i) { return i; }
> (i) { return i; }
> { return 0; }
>
> http://dlang.org/expression.html#FunctionLiteral
> function

Thanks.

But why is the solution breaks down when `s = 10000` ? :)

import std.stdio, std.algorithm, std.range;

int c;
const x = 12, y = 65, z = 50, s = 100000;

void solve(Range)(Range r) {
	cartesianProduct(r, r, r).filter!(i => i[0] * (y + 3 * z) + i[1] * (y + 2 * z) + i[2] * (y + z) == s).each!dout;
}

void main() {

    auto a = iota(0, x + 1).array;

    solve(a);

    writefln(`%s total`, c);
}

void dout(Tuple)(Tuple idx) {
    ++c;
}
-----
http://rextester.com/XGDL26042

On Monday, 25 May 2015 at 17:52:09 UTC, Dennis Ritchie wrote:
> But why is the solution breaks down when `s = 10000` ? :)
>
> import std.stdio, std.algorithm, std.range;
>
> int c;
> const x = 12, y = 65, z = 50, s = 100000;

Which is it, now? 4 or 5 zeros?

> void solve(Range)(Range r) {
> 	cartesianProduct(r, r, r).filter!(i => i[0] * (y + 3 * z) + i[1] * (y + 2 * z) + i[2] * (y + z) == s).each!dout;
> }
>
> void main() {
>
>     auto a = iota(0, x + 1).array;
>
>     solve(a);
>
>     writefln(`%s total`, c);
> }
>
> void dout(Tuple)(Tuple idx) {
>     ++c;
> }
> -----
> http://rextester.com/XGDL26042

What do you mean it "breaks down"? Your original code doesn't print anything for s = 10_000 or s = 100_000, either.

On Monday, 25 May 2015 at 19:16:04 UTC, anonymous wrote:
> On Monday, 25 May 2015 at 17:52:09 UTC, Dennis Ritchie wrote:
>> But why is the solution breaks down when `s = 10000` ? :)
>>
>> import std.stdio, std.algorithm, std.range;
>>
>> int c;
>> const x = 12, y = 65, z = 50, s = 100000;
>
> Which is it, now? 4 or 5 zeros?

No difference!

>> void solve(Range)(Range r) {
>> 	cartesianProduct(r, r, r).filter!(i => i[0] * (y + 3 * z) + i[1] * (y + 2 * z) + i[2] * (y + z) == s).each!dout;
>> }
>>
>> void main() {
>>
>>    auto a = iota(0, x + 1).array;
>>
>>    solve(a);
>>
>>    writefln(`%s total`, c);
>> }
>>
>> void dout(Tuple)(Tuple idx) {
>>    ++c;
>> }
>> -----
>> http://rextester.com/XGDL26042
>
> What do you mean it "breaks down"? Your original code doesn't print anything for s = 10_000 or s = 100_000, either.

Excuse me, this is my blemish! I forgot that the constant `x` depends on `s`. Everything works correctly:

import std.stdio, std.algorithm, std.range;

void solve(Range)(Range r) {
    cartesianProduct(r, r, r).filter!(i => i[0] * (y + 3 * z) + i[1] * (y + 2 * z) + i[2] * (y + z) == s).each!dout;
}

const y = 65, z = 50, s = 100000;
const x = s / (y + z);

void main() {

    auto a = iota(0, x + 1);

    solve(a);
}

auto dout(Tuple)(Tuple idx) {
    writefln(`%s apples`, idx[0] + idx[1] + idx[2]);
    writefln(`%s gingerbread`, idx[0] * 3 + idx[1] * 2 + idx[2]);
    writefln(`%s pharynx tea`, idx[0] * 3 + idx[1] * 2 + idx[2]);
    writefln("Sour: %s; semi-acid: %s; sweet: %s.\n", idx[0], idx[1], idx[2]);
}
-----
http://rextester.com/MMCI9993