Re: Replacing nested loops foreach using map/each/etc
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 = 1` ? :)
import std.stdio, std.algorithm, std.range;
int c;
const x = 12, y = 65, z = 50, s = 10;
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 = 10;
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
Re: Replacing nested loops foreach using map/each/etc
On Monday, 25 May 2015 at 17:52:09 UTC, Dennis Ritchie wrote:
But why is the solution breaks down when `s = 1` ? :)
import std.stdio, std.algorithm, std.range;
int c;
const x = 12, y = 65, z = 50, s = 10;
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.
Re: Replacing nested loops foreach using map/each/etc
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 = 1` ? :)
import std.stdio, std.algorithm, std.range;
int c;
const x = 12, y = 65, z = 50, s = 10;
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
Re: Replacing nested loops foreach using map/each/etc
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
Re: Replacing nested loops foreach using map/each/etc
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.
Re: Replacing nested loops foreach using map/each/etc
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]);
});
}
Re: Replacing nested loops foreach using map/each/etc
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
Re: Replacing nested loops foreach using map/each/etc
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
Replacing nested loops foreach using map/each/etc
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
