Find i^n, given n

Question

The Challenge

In as few characters as possible, find the value of i^n, given n, a positive integer greater than 0. This should be outputted as a String.

For those that don't know, i is defined such that i^2=-1. So:

• i^1=i
• i^2=-1
• i^3=-i
• i^4=1

This then repeats..

Rules

• If your language supports complex numbers, don't use any functions or arithmetic that could work this out.
• Floating point inaccuracies are fine for answers that would return decimals anyway, but integer inputs should give exact results

Bonus Points

-5 if you can work the value out where n is also negative

-15 if you can work out the value for any real number (this bonus includes the -5 from the above bonus)

Good luck!

Answer 1

Ruby, score -2

(13 bytes, -15 bonus)

->n{[1,90*n]}

Features include: no rounding errors! (if you pass the input as a Rational)

---

posted by the author, Kezz101

If you support any real number you can output in any valid complex form.

Negatives scores make my adrenaline rush forth. Thus the rules get abused are made use of to achieve this noble goal.

Creates an anonymous function and outputs an array with 2 entries representing a complex number in polar form (angular unit: degrees).

Answer 2

CJam, 12 characters - 5 = 7

1'iW"-i"]li=

Test it here.

Supports negative inputs.

1              "Push 1.";
 'i            "Push the character i.";
   W           "Push -1.";
    "-i"       "Push the string -i.";
        ]      "Wrap all four in an array.";
         li    "Read STDIN and convert to integer.";
           =   "Access array. The index is automatically taken module the array length.";

The result is printed automatically at the end of the program.

Mathematica, 22 20 19 characters - 15 = 4

Sin[t=π#/2]i+Cos@t&

This is an anonymous function, which you can use like

Sin[t=π#/2]i+Cos@t&[15]

(Or assign it to f say, and then do f[15].)

Supports reals and gives exact results for integer input.

Note that the i is not Mathematica's complex i (which is I). It's just an undefined variable.

Also, despite the order of the expression, Mathematica will reorder the output into R+Ci form.

Answer 3

Python 2 - (24-5)=19

lambda n:'1i--'[n%4::-2]

Most credit belongs to @user2357112, I just golfed his answer from the comments on this answer a bit more.

Explanation: Starts at the index n%4 in the string '1i--'. Then iterates backwards in steps of two over each letter in the string. So, for example, n=6 would start at index 2, the first -, then skip the i and take the 1, to return -1.

@xnor pointed out a same-length solution:

lambda n:'--i1'[~n%4::2] 

Pyth - (14-5)=9

I can only seem to get 14, no matter how I try to reverse/slice/etc. :'(

%_2<"1i--"h%Q4

Which is essentially the same as the above python answer, but in 2 steps, because pyth doesn't support the full indexing options of python. Try it online.

I'm going to go have a talk with isaacg about Pyth indexing ;)

Answer 4

TI-BASIC (NSpire) - 5 (20 characters-15)

cos(nπ/2)+sin(nπ/2)i

If you want to recieve a complex return value, replace the i at the end with (complex i).

Answer 5

Marbelous, 43 bytes

Not really a winner, but Marbelous is fun. :)

}0
^0
=0&2
&1
}0
^1\/
=0&02D3169
\/\/&0&1&2

This is a program which reads the input as a single integer from the first command-line argument. Note that the input is taken modulo 256, but this doesn't affect validity of the result for inputs greater than 255, because 256 is divisible by 4.

Explanation

Marbelous is a 2D programming language, which simulates "marbles" (byte values) falling through a bunch of devices. The board is made up of 2-character wide cells (the devices), which can process the marbles. Everything that falls off the bottom of a board is printed to STDOUT.

Let's go through the devices in use:

• }0 is where the first command-line argument goes. I've used two instances of this device, so I get two copies of the input value (at the same time).
• ^n checks for the nth bit of the input marble (where n=0 is the least significant bit), and produces 1 or 0 depending on the bit.
• =0 checks for equality with 0. If the input marble is equal, it just drops straight through, if it isn't it is pushed to the right.
• \/ is a trash can, so it just swallows the input marble and never produces anything.
• 2D is the ASCII code of -, 31 is the ASCII code of 1 and 69 is the ASCII code of i.
• The &n are synchronisers. Synchronisers stall a marble until all synchronisers with the same n hold a marble, at which point they'll all let their stored marble fall through.

So in effect, what I do is to hold the three relevant characters in three synchronisers, and release those depending on how the least significant bits are set in the input.

For more information, see the spec draft.

Answer 6

JavaScript (ES6) 29-5 = 24

Supports negative power.

f=n=>(n&2?'-':'')+(n&1?'i':1)

ES5 :

function f(n){return (n&2?'-':'')+(n&1?'i':1)}

Answer 7

Python 28 bytes - 5 = 23

Supports -ve inputs.

Assuming lambda functions are acceptable (Thanks FryAmTheEggman!):

lambda n:n%4/2*'-'+'1i'[n%2]

otherwise 31 bytes - 5 = 26

print[1,"i",-1,"-i"][input()%4]

Answer 8

APL (Dyalog), 8 chars - 15 bonus = score -7

The built-in (and thus prohibited) function is 0J1*⊢, but this uses @blutorange's method.

¯12○.5×○

The challenge author, Kezz101, wrote:

If you support any real number you can output in any valid complex form.

This returns a complex number in the form aJb which is the normal way for APL to display complex numbers.

Try it online!

Explanation

¯12○ find the unit vector which has the angle in radians of

.5× one half times

○ the argument multiplied by (the circle constant)

Answer 9

(Emacs) Lisp – 34

Just for fun, in (Emacs) Lisp:

(lambda(n)(elt[1 i -1 -i](% n 4)))

If you want to use it, use a defun or use funcall:

(funcall
 (lambda (n) (elt [1 i -1 -i] (% n 4)))
 4) => 1

(mapcar
 (lambda(n)(elt[1 i -1 -i](% n 4)))
 [0 1 2 3 4 5 6 7 8])
 => (1 i -1 -i 1 i -1 -i 1)

Answer 10

Pari/GP, 19 bytes - 5 = 14

As a ring, \$\mathbb{C}\$ is isomorphic to \$\mathbb{R}[x]/(x^2+1)\$.

n->Str(i^n%(i^2+1))

The i here is just a symbol, not the imaginary unit (which is I in Pari/GP).

Try it online!

Answer 11

Pure bash, 29 bytes - 5 = 24

Supports -ve inputs.

a=(1 i -1 -i)
echo ${a[$1%4]}

Answer 12

Befunge-98, 41-5 = 36 35-5 = 30 32-5 = 27

&4%:01-`!4*+:2%'8*'1+\2/d*' +,,@

Supports negative integers. Not going to win any awards with this solution, but whatever.

It just accepts a number as input, does some trickery on the modulus (which, frustratingly, doesn't work like usual modulus for negative numbers in the interpreter I used to test it) to make negatives work, and then does some silly conditionals to decide what each character should be.

I'm sure this can be golfed down plenty further. For now, here's another solution that doesn't accept negatives, but makes up for the loss of the bonus by being shorter:

Befunge-98, 32 26 23

&4%:2%'8*'1+\2/d*' +,,@

Edit - Now takes advantage of the fact that "-" is 13 (0xd) characters away from " ".

Edit 2 - Now, again, takes advantage of the fact that "i" is 56 (0x38, or '8) characters away from "1".

Answer 13

MATLAB, 33 bytes - 5 = 28

x={'i','-1','-i','1'};x(mod(n,4))

Even though it's a few bytes more (37-5=32), I actually like this approach better:

x='1i -';x((mod([n/2,n],2)>=1)+[3,1])

Answer 14

Java 8 Score: 72

In Java, the worst golfing language ever! Golfed:

java.util.function.Function s=n->{new String[]{"i","-1","-i","1"}[n%4]};

Expanded:

class Complex{

    java.util.function.Function s = n -> {new String[]{"i","-1","-i","1"}[n%4]};

}

Note: I'm not used to Java 8. I also do not have the runtime for it yet. Please tell me if there are any syntax errors. This is also my first golf.

Edit: Removed import.

Edit: Removed class declaration.

Another answer with score = 87 - 15 = 72

java.util.function.Function s=n->{Math.cos(n*Math.PI/2)+"+"+Math.sin(n*Math.PI/2)+"i"};

Expanded:

class Complex{

    java.util.function.Function s = n -> {Math.cos(n * Math.PI/2) + " + " + Math.sin(n * Math.PI/2) + "i"};

}

Answer 15

C 77

main(){int n;scanf("%d",&n);char c[]={n%4>1?'-':' ',~n%2?'1':'i',0};puts(c);}

Improved thanks to Ruslan

C 74-5=69

Oh and of course the most obvious approach

main(){unsigned n,*c[]={"1","i","-1","-i"};scanf("%d",&n);puts(c[n%4]);}

Answer 16

OCaml 47

let m n=List.nth["1";"i";"-1";"-i"](n mod 4);;

Not an award winning solution, but this is my first time code-golfing, so I'm not exactly sure of what I'm doing. I tried to use pattern matching, but that got me over 58.

Answer 17

R, 29 - 5 = 24 bytes

c(1,"i",-1,"-i")[scan()%%4+1]

Try it online!

Same as most methods above, takes a modulo of 4, and increases that by 1, because R's arrays are 1-indexed. Works for negative integers as well.

I was worried about mixed outputs here, but Giuseppe pointed out that R coerces numeric types to string types when they are mixed.

Answer 18

PowerShell, 28 bytes -5 = 23

('i',-1,'-i',1)["$args"%4-1]

Try it online!

Port of all the cyclic indexing

Answer 19

Ruby 32-5=27

puts(%w[1 i -1 -i][gets.to_i%4])

Works for negative powers!

Answer 20

Perl, 26 - 5 = 21

say qw(1 i -1 -i)[pop()%4]

works as a standalone program (argument on the commandline) or the body of a function.

Answer 21

Java: 151 131-5=126

Golfed:

class A{public static void main(String[]a){int n=Integer.parseInt(a[0]);System.out.print(n%4==0?"1":n%4==1?"i":n%4==2?"-1":"-i");}}

Ungolfed:

class A {
    public static void main(String[] a) {
        int n = Integer.parseInt(a[0]);
        System.out.print(n % 4 == 0 ? "1" : n % 4 == 1 ? "i" : n % 4 == 2 ? "-1" : "-i");
    }
}

As a function: 72-5=67

Golfed:

void f(int n){System.out.print(n%4==0?"1":n%4==1?"i":n%4==2?"-1":"-i");}

Ungolfed:

public void f(int n) {
    System.out.print(n % 4 == 0 ? "1" : n % 4 == 1 ? "i" : n % 4 == 2 ? "-1" : "-i");
}

Yes, yet another Java reply - and golfed even worse than ever. But you work with what you can...

EDIT: added function version.

EDIT 2: so, after a bit of trial and error, here's a version that tries to do it by the book, without exploring the cycle loophole. So…

Java with value calculation: 146-15=131

Golfed:

class B{public static void main(String[]a){float n=Float.parseFloat(a[0]);System.out.print(Math.cos((n*Math.PI)/2)+Math.sin((n*Math.PI)/2)+"i");}}

Ungolfed:

class B {
    public static void main(String[] a) {
        float n = Float.parseFloat(a[0]);
        System.out.print(Math.cos((n * Math.PI) / 2) + Math.sin((n * Math.PI) / 2) + "i");
    }
}

(at least, I think I can claim the top bonus, correct me otherwise)

Answer 22

Python - 31

print[1,'i',-1,'-i'][input()%4]

I have only recently started learning python. Even though I know it's not good, it's the best I can do.

Answer 23

Haskell GHCi, 29 Bytes - 15 = 14

i p=[cos(p*pi/2),sin(p*pi/2)]

Usage:

*Main> i 0
[1.0,0.0]
*Main> i pi
[0.22058404074969779,-0.9753679720836315]
*Main> i (-6.4)
[-0.8090169943749477,0.5877852522924728]

Answer 24

Jelly, 2 - 20 = -18 bytes

ı*

Try it online!

It doesn't use an i ^ x builtin but it uses builtins for 1j and ** so not sure if allowed.

Answer 25

05AB1E, score 5 (10 bytes - 5 bonus)

'i®„-i1)Iè

Try it online or verify some more test cases.

Explanation:

'i         '# Push character "i"
  ®         # Push -1
   „-i      # Push string "-i"
      1     # Push 1
       )    # Wrap everything on the stack into a list: ["i", -1, "-i", 1]
        Iè  # Use the input to index into it (0-based and with automatic wrap-around)
            # (and output the result implicitly)

Answer 26

Swift, 27

{["1","i","-1","-i"][$0%4]}

Try it online

Answer 27

Perl 6, 19 bytes - 5 = 14

{<1 i -1 -i>[$_%4]}

The result of % has the same sign as the divisor in this language.

Try it online!

Answer 28

C (gcc), 30 bytes

f(x){printf(L"iㄭ椭1"+x%4);}

Try it online!

Answer 29

C (gcc), 44 bytes -15=29

#define f(x)cos(x*acos(0))+1i*sin(x*acos(0))

Try it online!

Handles signed and noninteger arguments.

Answer 30

Haskell, 29 bytes - 5 = 24

f n=words"1 i -1 -i"!!mod n 4

Works for negative powers.

I had a pointfree version worked out, but it turns out it's actually longer.

f=(words"1 i -1 -i"!!).(`mod`4)

Answer 31

C 105, was 117

char c[2];
int main()
{
int n,i,j=0;scanf("%d",&n);i=n%4;
i>1?c[j++]='-':c[j+1]='\0';
c[j]=i&1?'i':'1';
puts(c);
}

Answer 32

Clojure (64 54 31 chars)

(defn i2n[n](println(case(mod n 4) 1 "i" 2 "-1" 3 "-i" 0 "1")))

Edit

Per @SeanAllred's suggestion, here's a version which uses a literal vector instead of a case function:

(defn i2n[n](println (["1" "i" "-1" "-i"] (mod n 4))))

Edit 2

By counting on the REPL to print out the resultant collection and coding the function using the #() shortcut we can reduce it to

#(["1" "i" "-1" "-i"](mod % 4))

(Which is actually much more Clojure/Lisp-ish as the function now actually returns the generated result, allowing the function to be used with map, as in

(map #(["1" "i" "-1" "-i"](mod % 4)) [0 1 2 3 4 5 6 7 8])

which prints

("1" "i" "-1" "-i" "1" "i" "-1" "-i" "1")

Share and enjoy.

Answer 33

Groovy: 27-5 = 22

f={n->[1,'i',-1,'-i'][n%4]}

Answer 34

PARI/GP, 26 - 5 = 21

n->Str([1,I,-1,-I][n%4+1])

n->cos(t=Pi*n/2)+I*sin(t) is one character shorter, but doesn't handle exact answers. Of course n->I^n is disallowed, and presumably also PARI's powIs.

Answer 35

Java (OpenJDK 8), 39 bytes

n->new String[]{"1","i","-1","-i"}[n%4]

Try it online!

Answer 36

Haskell, 27 bytes

(cycle(words"1 i -1 -i")!!)

Try it online!

Answer 37

Tcl, 50 bytes

proc I n {expr $n%4==3?"-i":$n%4==2?-1:$n%4?"i":1}

Try it online!

Answer 38

05AB1E, 12 bytes

®sfgm¹Éi1'i:

Try it online!

---

®             # Push -1 onto the stack.
 sfg          # Push the number of prime factors of input.
    m         # -1^(number_prime_factors)
     ¹Éi      # is input even?
        1'i:  # if input was even, replace 1 with i.

Answer 39

Forth (gforth), 53 bytes - 5 bytes (bonus) = 48 bytes

: f 4 mod dup 1 > if ." -"then 2 mod 56 * '1 + emit ;

Try it online!

This works for both positive and negative numbers

Code Explanation

: f             \ start a new word definition
  4 mod dup     \ get result of n % 4 (and make a copy)
  1 > if        \ if result is 2 or 3
    ." -'       \ print a minus sign
  then          \ end the if
  2 mod 56 *    \ get result modulo 2 and multiply by 56 (difference between ascii for '1' and 'i')
  '1 +          \ add to ascii value of '1'
  emit          \ output the character for the given value
;               \ end the word definition

Answer 40

Zsh, 30 bytes, score 25 (-5)

a=({,}{,-}{1,i})
<<<$a[$1%4+5]

Try it online!

Based on the bash answer, but with some changes needed:

• %4 results in a value in the range -3 to 3 (negative inputs give negative outputs), so we shift up with +5.
• Arrays are 1-indexed, hence +5 instead of +4.
• We make our array twice as long by using brace expansion: {,}foo expands to foo foo.

Answer 41

Keg, 15 - 5 = 10 bytes

1i0;`-i`¿4%⊙&ø&

Try it online!

Pushes the four different possibilities and then takes the input modulo 4 and indexes the stack.

Answer 42

C# 69-5 = 64

public string m(int n){return n%2==0?(n%4==0?"1":"-1"):(m(n-1)+"i");}

Answer 43

Ruby, 208 - 15 = 193

p=Math::PI/2;t=gets.to_f*p;l=->n{Math::cos(n).round(2)};k=l.call(t);if k!=0;k=k.to_i if k.abs==1;print k;end;m=l.call(t+p);if m!=0;r="i";r='-'+r if m==-1;r="#{m}"+r if 1>m && -1<m;r=" + "+r if k!=0;puts r end

This implementation completely conforms to the spec! 1.0 is written as 1, -1.0i is written as -i, etc. I don't see any other solution that incorporates this along with supporting floating points, except for maybe the TI-Calculator.

Ungolfed:

p=Math::PI/2
t=gets.to_f*p
l=->n{Math::cos(n).round(2)}

k=l.call(t)
if k!=0
  k=k.to_i if k.abs==1
  print k
end

m=l.call(t+p)
if m!=0
  r="i"
  r='-'+r if m==-1
  r="#{m}"+r if 1>m && -1<m
  r=" + "+r if k!=0
  puts r
end

Answer 44

C, 61

main(n){scanf("%d",&n);printf("%c%c",n%4/2*45,n%4%2?'i':49);}

Try it here.