Ali Çehreli:
This is a somewhat common little exercise: Write a function
that takes the size of a diamond and produces a diamond of that
size.
When printed, here is the output for size 11:
*
***
*****
*******
*********
***********
*********
*******
*****
***
*
Some of my solutions (using each() in the last two is easy):
import std.stdio, std.array, std.string, std.range,
std.algorithm, std.math;
void printDiamond1(in uint n) {
immutable k = (n % 2 == 1) ? 1 : 2;
for (int i = k; i <= n; i += 2)
writeln("*".replicate(i).center(n));
for (int i = n - 2; i >= k; i -= 2)
writeln("*".replicate(i).center(n));
}
void printDiamond2(in int n) {
iota(!(n % 2), n)
.map!(i => "*"
.replicate((n % 2) + ((n / 2) - abs(i - (n / 2)))
* 2)
.center(n))
.join("\n")
.writeln;
}
void printDiamond3(in int n) {
writefln("%-(%s\n%)",
iota(!(n % 2), n)
.map!(i => "*"
.replicate((n % 2) + ((n / 2) - abs(i -
(n / 2))) * 2)
.center(n)));
}
void main() {
foreach (immutable i; 0 .. 15) {
printDiamond3(i);
writeln;
}
}
Output:
*
**
*
***
*
**
****
**
*
***
*****
***
*
**
****
******
****
**
*
***
*****
*******
*****
***
*
**
****
******
********
******
****
**
*
***
*****
*******
*********
*******
*****
***
*
**
****
******
********
**********
********
******
****
**
*
***
*****
*******
*********
***********
*********
*******
*****
***
*
**
****
******
********
**********
************
**********
********
******
****
**
*
***
*****
*******
*********
***********
*************
***********
*********
*******
*****
***
*
**
****
******
********
**********
************
**************
************
**********
********
******
****
**
Bye,
bearophile
