Roots of unity
You are encouraged to solve this task according to the task description, using any language you may know.
The purpose of this task is to explore working with complex numbers. Given n, find the n-th roots of unity.
Ada
with Ada.Text_Io; use Ada.Text_Io;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
with Ada.Numerics.Generic_Complex_Types;
with Ada.Numerics.Generic_Complex_Elementary_Functions;
with Ada.Text_Io.Complex_Io;
procedure Roots_Of_Unity is
type Real is digits 10;
package Complex_Type is new Ada.Numerics.Generic_Complex_Types(Real);
use Complex_Type;
package Complex_Functions is new Ada.Numerics.Generic_Complex_Elementary_Functions(Complex_Type);
use Complex_Functions;
package Complex_Io is new Ada.Text_Io.Complex_Io(Complex_Type);
Factor : Complex;
begin
for Root in 2..10 loop
Put(Item => Root, Width => 4);
Put(" ");
for K in 1..Root/2 loop
Factor := Exp(Real(K) * 2.0 * Ada.Numerics.Pi / Real(Root));
Complex_Io.Put(Item => Factor, Fore => 2, Aft => 4, Exp => 0);
end loop;
New_Line;
end loop;
end Roots_Of_Unity;
Output:
2 (-1.0000, 0.0000) 3 (-0.5000, 0.8660) 4 ( 0.0000, 1.0000)(-1.0000, 0.0000) 5 ( 0.3090, 0.9511)(-0.8090, 0.5878) 6 ( 0.5000, 0.8660)(-0.5000, 0.8660)(-1.0000, 0.0000) 7 ( 0.6235, 0.7818)(-0.2225, 0.9749)(-0.9010, 0.4339) 8 ( 0.7071, 0.7071)( 0.0000, 1.0000)(-0.7071, 0.7071)(-1.0000, 0.0000) 9 ( 0.7660, 0.6428)( 0.1736, 0.9848)(-0.5000, 0.8660)(-0.9397, 0.3420) 10 ( 0.8090, 0.5878)( 0.3090, 0.9511)(-0.3090, 0.9511)(-0.8090, 0.5878)(-1.0000, 0.0000)
ALGOL 68
FORMAT complex fmt=$g(-6,4)"⊥"g(-6,4)$;
FOR root FROM 2 TO 10 DO
printf(($g(4)$,root));
FOR n TO root OVER 2 DO
printf(($xf(complex fmt)$,complex exp( 0 I 2*pi*n/root)))
OD;
printf($l$)
OD
Output:
+2 -1.000⊥0.0000 +3 -.5000⊥0.8660 +4 0.0000⊥1.0000 -1.000⊥0.0000 +5 0.3090⊥0.9511 -.8090⊥0.5878 +6 0.5000⊥0.8660 -.5000⊥0.8660 -1.000⊥0.0000 +7 0.6235⊥0.7818 -.2225⊥0.9749 -.9010⊥0.4339 +8 0.7071⊥0.7071 0.0000⊥1.0000 -.7071⊥0.7071 -1.000⊥0.0000 +9 0.7660⊥0.6428 0.1736⊥0.9848 -.5000⊥0.8660 -.9397⊥0.3420 +10 0.8090⊥0.5878 0.3090⊥0.9511 -.3090⊥0.9511 -.8090⊥0.5878 -1.000⊥0.0000
BASIC
For high n's, this may repeat the root of 1 + 0*i.
CLS PI = 3.1415926# n = 5 'this can be changed for any desired n angle = 0 'start at angle 0 DO real = COS(angle) 'real axis is the x axis IF (ABS(real) < 10 ^ -5) THEN real = 0 'get rid of annoying sci notation imag = SIN(angle) 'imaginary axis is the y axis IF (ABS(imag) < 10 ^ -5) THEN imag = 0 'get rid of annoying sci notation PRINT real; "+"; imag; "i" 'answer on every line angle = angle + (2 * PI) / n 'all the way around the circle at even intervals LOOP WHILE angle < 2 * PI
C++
<cpp>#include <complex>
- include <cmath>
- include <iostream>
double const pi = 4 * std::atan(1);
int main() {
for (int n = 2; n <= 10; ++n)
{
std::cout << n << ": ";
for (int k = 0; k < n; ++k)
std::cout << std::polar(1, 2*pi*k/n) << " ";
std::cout << std::endl;
}
}</cpp>
Common Lisp
(defun roots-of-unity (n)
(loop for i below n
collect (cis (* pi (/ (* 2 i) n)))))
The expression is slightly more complicated than necessary in order to preserve exact rational arithmetic until multiplying by pi. The author of this example is not a floating point expert and not sure whether this is actually useful; if not, the simpler expression is (cis (/ (* 2 pi i) n)).
D
<d>module nthroots ; import std.stdio, std.math ;
creal[] nthroots(int n) {
creal[] res ; for(int k = 1 ; k <= n ; k++) res ~= expi(PI*2*k/n) ; return res ;
} void main() {
for(int i = 1; i <= 8 ; i++)
writefln("%2dth : %5.2f", i, nthroots(i)) ;
}</d>
Forth
Complex numbers are not a native type in Forth, so we calculate the roots by hand.
: f0. ( f -- )
fdup 0e 0.001e f~ if fdrop 0e then f. ;
: .roots ( n -- )
dup 1 do
pi i 2* 0 d>f f* dup 0 d>f f/ ( F: radians )
fsincos cr ." real " f0. ." imag " f0.
loop drop ;
3 set-precision
5 .roots
Fortran
PROGRAM Roots
COMPLEX :: root
INTEGER :: n
REAL :: angle, pi
pi = 4.0 * ATAN(1.0)
DO n = 2, 7
angle = 0.0
WRITE(*,"(I1,A)", ADVANCE="NO") n,": "
DO WHILE (angle < 6.283)
root = CMPLX(COS(angle), SIN(angle))
WRITE(*,"(SP,2F7.4,A)", ADVANCE="NO") root, "j "
angle = angle + (2.0*pi) / REAL(n)
END DO
WRITE(*,*)
END DO
END PROGRAM Roots
Output
2: +1.0000+0.0000j -1.0000+0.0000j 3: +1.0000+0.0000j -0.5000+0.8660j -0.5000-0.8660j 4: +1.0000+0.0000j +0.0000+1.0000j -1.0000+0.0000j +0.0000-1.0000j 5: +1.0000+0.0000j +0.3090+0.9511j -0.8090+0.5878j -0.8090-0.5878j +0.3090-0.9511j 6: +1.0000+0.0000j +0.5000+0.8660j -0.5000+0.8660j -1.0000+0.0000j -0.5000-0.8660j +0.5000-0.8660j 7: +1.0000+0.0000j +0.6235+0.7818j -0.2225+0.9749j -0.9010+0.4339j -0.9010-0.4339j -0.2225-0.9749j +0.6235-0.7818j
Haskell
import Data.Complex rootsOfUnity n = [mkPolar 1.0 (2*pi*k/n) | k <- [1..n]]
Output:
*Main> rootsOfUnity 3 [(-0.4999999999999998) :+ 0.8660254037844387, (-0.5000000000000004) :+ (-0.8660254037844384), 1.0 :+ (-2.4492127076447545e-16)]
IDL
For some example n:
n = 5 print, exp( dcomplex( 0, 2*!pi/n) ) ^ ( 1 + indgen(n) )
Outputs:
( 0.30901696, 0.95105653)( -0.80901704, 0.58778520)( -0.80901693, -0.58778534)( 0.30901713, -0.95105647)( 1.0000000, 1.7484556e-007)
J
rou=: [: ^ i. * (o. 0j2) % ] rou 4 1 0j1 _1 0j_1 rou 5 1 0.309017j0.951057 _0.809017j0.587785 _0.809017j_0.587785 0.309017j_0.951057
The computation can also be written as a loop, shown here for comparison only.
rou1=: 3 : 0 z=. 0 $ r=. ^ o. 0j2 % y [ e=. 1 for. i.y do. z=. z,e e=. e*r end. z )
Java
Java doesn't have a nice way of dealing with complex numbers, so the real and imaginary parts are calculated separately based on the angle and printed together. There are also checks in this implementation to get rid of extremely small values (< 1.0E-3 where scientific notation sets in for Doubles). Instead, they are simply represented as 0. To remove those checks (for very high n's), remove both if statements. <java>public static void unity(int n){ //all the way around the circle at even intervals for(double angle = 0;angle < 2 * Math.PI;angle += (2 * Math.PI) / n){ double real = Math.cos(angle); //real axis is the x axis if(Math.abs(real) < 1.0E-3) real = 0.0; //get rid of annoying sci notation double imag = Math.sin(angle); //imaginary axis is the y axis if(Math.abs(imag) < 1.0E-3) imag = 0.0; //get rid of annoying sci notation System.out.print(real + " + " + imag + "i\t"); //tab-separated answers } }</java>
OCaml
<ocaml>open Complex
let pi = 4. *. atan 1.
let () =
for n = 1 to 10 do
Printf.printf "%2d " n;
for k = 1 to n do
let ret = polar 1. (2. *. pi *. float_of_int k /. float_of_int n) in
Printf.printf "(%f + %f i)" ret.re ret.im
done;
print_newline ()
done</ocaml>
Perl
<perl>use Math::Complex;
foreach $n (2 .. 10) {
printf "%2d", $n;
foreach $k (0 .. $n-1) {
$ret = cplxe(1, 2 * pi * $k / $n);
$ret->display_format('style' => 'cartesian', 'format' => '%.3f');
print " $ret";
}
print "\n";
}</perl> Output:
2 1.000 -1.000+0.000i 3 1.000 -0.500+0.866i -0.500-0.866i 4 1.000 0.000+1.000i -1.000+0.000i -0.000-1.000i 5 1.000 0.309+0.951i -0.809+0.588i -0.809-0.588i 0.309-0.951i 6 1.000 0.500+0.866i -0.500+0.866i -1.000+0.000i -0.500-0.866i 0.500-0.866i 7 1.000 0.623+0.782i -0.223+0.975i -0.901+0.434i -0.901-0.434i -0.223-0.975i 0.623-0.782i 8 1.000 0.707+0.707i 0.000+1.000i -0.707+0.707i -1.000+0.000i -0.707-0.707i -0.000-1.000i 0.707-0.707i 9 1.000 0.766+0.643i 0.174+0.985i -0.500+0.866i -0.940+0.342i -0.940-0.342i -0.500-0.866i 0.174-0.985i 0.766-0.643i 10 1.000 0.809+0.588i 0.309+0.951i -0.309+0.951i -0.809+0.588i -1.000+0.000i -0.809-0.588i -0.309-0.951i 0.309-0.951i 0.809-0.588i
Python
<python> import cmath class Complex(complex):
def __repr__(self):
rp = '%7.5f'%self.real if not self.pureImag() else
ip = '%7.5fj'%self.imag if not self.pureReal() else
conj = if (self.pureImag() or self.pureReal() or self.imag<0.0) else '+'
return '0.0' if (self.pureImag() and self.pureReal()) else rp+conj+ip
def pureImag(self):
return abs( self.real) < 0.000005
def pureReal(self):
return abs( self.imag) < 0.000005
def croots(n):
if n<=0:
return None
return ((Complex(cmath.exp(2j*k*cmath.pi/n))) for k in range(n))
for nr in range(2,11):
print nr, list(croots(nr))
</python> Output:
2 [1.00000, -1.00000] 3 [1.00000, -0.50000+0.86603j, -0.50000-0.86603j] 4 [1.00000, 1.00000j, -1.00000, -1.00000j] 5 [1.00000, 0.30902+0.95106j, -0.80902+0.58779j, -0.80902-0.58779j, 0.30902-0.95106j] 6 [1.00000, 0.50000+0.86603j, -0.50000+0.86603j, -1.00000, -0.50000-0.86603j, 0.50000-0.86603j] 7 [1.00000, 0.62349+0.78183j, -0.22252+0.97493j, -0.90097+0.43388j, -0.90097-0.43388j, -0.22252-0.97493j, 0.62349-0.78183j] 8 [1.00000, 0.70711+0.70711j, 1.00000j, -0.70711+0.70711j, -1.00000, -0.70711-0.70711j, -1.00000j, 0.70711-0.70711j] 9 [1.00000, 0.76604+0.64279j, 0.17365+0.98481j, -0.50000+0.86603j, -0.93969+0.34202j, -0.93969-0.34202j, -0.50000-0.86603j, 0.17365-0.98481j, 0.76604-0.64279j] 10 [1.00000, 0.80902+0.58779j, 0.30902+0.95106j, -0.30902+0.95106j, -0.80902+0.58779j, -1.00000, -0.80902-0.58779j, -0.30902-0.95106j, 0.30902-0.95106j, 0.80902-0.58779j]
Seed7
$ include "seed7_05.s7i";
include "float.s7i";
include "complex.s7i";
const proc: main is func
local
var integer: n is 0;
var integer: k is 0;
begin
for n range 2 to 10 do
write(n lpad 2 <& ": ");
for k range 0 to pred(n) do
write(polar(1.0, 2.0 * PI * flt(k) / flt(n)) digits 4 lpad 15 <& " ");
end for;
writeln;
end for;
end func;
Output:
2: 1.0000+0.0000i -1.0000+0.0000i 3: 1.0000+0.0000i -0.5000+0.8660i -0.5000-0.8660i 4: 1.0000+0.0000i 0.0000+1.0000i -1.0000+0.0000i 0.0000-1.0000i 5: 1.0000+0.0000i 0.3090+0.9511i -0.8090+0.5878i -0.8090-0.5878i 0.3090-0.9511i 6: 1.0000+0.0000i 0.5000+0.8660i -0.5000+0.8660i -1.0000+0.0000i -0.5000-0.8660i 0.5000-0.8660i 7: 1.0000+0.0000i 0.6235+0.7818i -0.2225+0.9749i -0.9010+0.4339i -0.9010-0.4339i -0.2225-0.9749i 0.6235-0.7818i 8: 1.0000+0.0000i 0.7071+0.7071i 0.0000+1.0000i -0.7071+0.7071i -1.0000+0.0000i -0.7071-0.7071i 0.0000-1.0000i 0.7071-0.7071i 9: 1.0000+0.0000i 0.7660+0.6428i 0.1736+0.9848i -0.5000+0.8660i -0.9397+0.3420i -0.9397-0.3420i -0.5000-0.8660i 0.1736-0.9848i 0.7660-0.6428i 10: 1.0000+0.0000i 0.8090+0.5878i 0.3090+0.9511i -0.3090+0.9511i -0.8090+0.5878i -1.0000+0.0000i -0.8090-0.5878i -0.3090-0.9511i 0.3090-0.9511i 0.8090-0.5878i
Scheme
<scheme>(define pi (* 4 (atan 1)))
(do ((n 2 (+ n 1)))
((> n 10))
(display n)
(do ((k 0 (+ k 1)))
((>= k n))
(display " ")
(display (make-polar 1 (* 2 pi (/ k n)))))
(newline))</scheme>