# Find i^n, given n

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!

• In what format do we return exactly? Through function output or stdout? – proud haskeller Nov 21 '14 at 15:59
• @MartinBüttner but if i choose function output, how should the output should be formatted/stored without native complex numbers in my language? – proud haskeller Nov 21 '14 at 16:02
• @BetaDecay What are floating point integers? o.O – Martin Ender Nov 21 '14 at 18:04
• @MartinBüttner Haha wrong word :/ Floating point number then – Beta Decay Nov 21 '14 at 18:17
• I feel so stupid. I only just realized that python has a built in complex type. – Justin Nov 22 '14 at 6:03

# Groovy: 27-5 = 22

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


## C 105, was 117

char c;
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);
}

• It doesn't even compile because you must parenthesize the assignments after : in ?: statements in plain C. Also, what's the point of using 0==0 when you can use single char 1? And no need in parentheses before ?. Also, the last ?: statement could be shortened to c[j]=i&1?'i':'1';. – Ruslan Nov 23 '14 at 14:20
• @Ruslan In CodeBlocks Mingw there is only 1 warning around i&0. i&0==0 is a test for even, if i is even the result is 0 else it is 1. – bacchusbeale Nov 25 '14 at 6:13
• Yeah, but what's the point of using 0==0 when it's identical to 1? Note that == has higher precedence than &, otherwise your (supposed) test of (i&0)==0 would always be true. – Ruslan Nov 25 '14 at 6:18

# 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.

# Jelly, 2 - 20 = -18 bytes

ı*


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

• "If your language supports complex numbers, don't use any functions or arithmetic that could work this out." I think it's pretty clear... – totallyhuman Dec 20 '17 at 18:17
• @totallyhuman Well, I'm not sure if the 1j literal is banned either though? – Erik the Outgolfer Dec 20 '17 at 18:22
• Yeah, but the arithmetic (*) is. – totallyhuman Dec 20 '17 at 18:22
• @totallyhuman Hm, maybe I'll add another version below, although "don't use any functions or arithmetic that could work this out" seems to suggest that I can't use a built-in to do this exact task...BTW the way he phrases it gets me to think that you're encouraged to use 1j literals. – Erik the Outgolfer Dec 20 '17 at 18:24

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

'i®„-i1)Iè


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)


{["1","i","-1","-i"][$0%4]}  Try it online • Welcome to PPCG! – alephalpha Apr 18 '19 at 14:11 # Perl 6, 19 bytes - 5 = 14 {<1 i -1 -i>[$_%4]}


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

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# C (gcc), 30 bytes

f(x){printf(L"iㄭ椭1"+x%4);}


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• Input is in A1
• B1: A1/2*PI(
• Result: =COMPLEX(COS(B1),SIN(B1

# Husk, (14 bytes - 5) = 9

!%4⁰C2¨¹-1-i 1


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# C# 69-5 = 64

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

• output is in form -1i, 1i, i, or 1 – binderbound Nov 23 '14 at 23:38
• You cannot claim the -15 bonus as this would fail on the input 3.247104. You may only claim the -5 bonus. – Beta Decay Nov 24 '14 at 7:06
• Oh, sorry, yes - I didn't read the bonuses bit properly – binderbound Nov 27 '14 at 3:34

# 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


# C, 61

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


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# Java (OpenJDK 8), 39 bytes

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


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


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# 05AB1E, 12 bytes

®sfgm¹Éi1'i:


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®             # 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.


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

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


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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


# Zsh, 30 bytes, score 25 (-5)

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


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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.

# Tcl, 50 bytes

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


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# Keg, 15 - 5 = 10 bytes

1i0;-i¿4%⊙&ø&


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

• You program supports negative numbers, you should have the -5 bonus! -5 if you can work the value out where n is also negative – user85052 Nov 16 '19 at 6:09

# Vyxal, 12 - 5 = 7 bytes

1\i1Œ-iW?i


## Explained

1\i1Œ-iW?i
1               # Push the number 1
\i             # Push the letter 'i'
1Œ           # Push 1 and then negate it to get -1
-i       # Push the string "-i"
W      # Wrap the entire stack into a list
?i    # Wrapping index the list based on the input