Gaussian integers are complex numbers of the form a+bi where a and b are both integers. In base -1+i, all Gaussian integers can be uniquely represented using the digits 0 and 1, without the need for a symbol to denote sign.

For instance, 1100 in base -1+i represents the decimal number 2, since

1*(-1+i)^3 + 1*(-1+i)^2 + 0*(-1+i)^1 + 0*(-1+i)^0
= (2+2i) + (-2i) + 0 + 0
= 2


Input will be two Gaussian integers in base -1+i represented using the digits 01. This can take one of the following forms:

• Two separate digit strings,
• Two decimal integers consisting of 01 representing the base -1+i numbers (e.g. 1100 for 2 in base -1+i),
• Two binary integers representing the base -1+i numbers (e.g. decimal 12 or 0b1100 for 2 in base -1+i)
• A single string separating two digit strings/binary integers by a single non-alphanumeric separator (e.g. 1100 1100 or 12,12 for 2+2)

Output the sum of the two Gaussian integers, also in base -1+i and represented using the digits 01 (in one of the formats allowed as input, not necessarily the same choice). The output is allowed to contain a finite number of leading zeroes.

Your function or program must terminate within 2 seconds for inputs of at most 30 digits each.

• You may assume that the input contains no extraneous leading zeroes. For the special case of 0, you may choose either 0 or the empty string as the representation.

## Test cases

0, 0 => 0                                      # 0 + 0 = 0
0, 1 => 1                                      # 0 + 1 = 1
1, 1 => 1100                                   # 1 + 1 = 2
1100, 1100 => 111010000                        # 2 + 2 = 4
1101, 1101 => 111011100                        # 3 + 3 = 6
110111001100, 1110011011100 => 0               # 42 + (-42) = 0
11, 111 => 0                                   # i + (-i) = 0
11, 110 => 11101                               # i + (-1-i) = -1
10101, 11011 => 10010                          # (-3-2i) + (-2+3i) = (-5+i)
1010100101, 111101 => 1110100000100            # (-19+2i) + (3-4i) = (-16-2i)


Longer test cases:

11011011010110101110010001001, 111100010100101001001010010101 => 0
111111111111111111111111111111, 111111111111111111111111111111 => 100100100100100100100100100100
101101110111011101110111011101, 101101110111011101110111011101 => 11101001010001000100010001000100011100
100100010101001101010110101010, 100010011101001011111110101000 => 110000110010101100001100111100010

• No digit lists? Mar 23, 2016 at 14:48
• @CatsAreFluffy No digit lists, sorry. Mar 23, 2016 at 15:09
• You can save one byte by changing -1+i to i-1 in the title. Mar 23, 2016 at 18:11
• Now we need a conversion the other way around. :P Mar 23, 2016 at 18:35
• There are 1100 types of people in the world. Those who understand binary, those who don't, those who confuse it with ternary, those who confuse it with base 4, those who confuse it with base 5, those who confuse it with base -1+i, those who confuse it with base 6, those who confuse it with base 7, those who confuse it with base 8, those who confuse it with base 9... Mar 27, 2016 at 9:51

# Python 2, 989791 84 bytes

s=input();L=1
for _ ins*8:s+=1098*int(str(s).translate('0011'*64));L*=10
print s%L


This does I/O in decimal. The integers have to be separated by the non-alphanumeric character +.

Thanks to @xnor for golfing off 2 bytes!

Try it on Ideone.

### How it works

In Arithmetic in Complex Bases, the author shows how to add and multiply complex numbers in bases of the form -n + i.

For base -1 + i, addition is done similarly to regular, binary addition with carry, with two differences:

• Instead of carrying 1 to the next higher position, we carry 110 to the next three.

• Carry digits can propagate indefinitely. However, without leading zeroes, the sum a + b has at most eight digits more than the maximum of a and b.

We proceed as follows.

1. First, we add a and b as if their digits were decimal digits.

For a = 10101 and b = 11011, this gives 21112.

2. Next, we form a new number by replacing the digits larger than 1 with a 1, others with a 0.1

For the sum 21112, this gives 10001.

3. For each digit larger than 1, we have to subtract 2 from that digit and carry 110 to the next three higher positions. Since 1098 = 10 * 110 - 2, we can achieve this by multiplying the result from step 2 by 1098, then adding that product to the sum.2

For the sum 21112, this gives 21112 + 1098 * 10001 = 21112 + 10981098 = 11002210.

4. We repeat steps 2 and 3 a total of d * 8 times, where d is the number of digits of a + b. 3

For the initial sum 21112, the results are

                      11002210
12210010
1220010010
122000010010
12200000010010
1220000000010010
122000000000010010
12200000000000010010
1220000000000000010010
122000000000000000010010
12200000000000000000010010
1220000000000000000000010010
122000000000000000000000010010
.
.
.

5. We take the final sum modulo 10d + 8, discarding all but the last d + 8 digits.

For the initial sum 21112, the final result is 10010.

1 This is achieved with translate. Repeating the string 0011 64 times makes one repetition line up with the sequence of ASCII characters 0123, achieving the desired replacement.

2 Note that the digits of this sum cannot exceed 3 (initial value 1 plus two 1's from carries).

3 This happens to work for d = 1, and d * 8 > d + 8 otherwise. The code may repeat the steps (d + 1) * 8 times, since s has a trailing L if s is a long integer.

• This is deep magic. What format is input() expecting? (I get 21112 when I input 10101, 11011.) Mar 23, 2016 at 15:20
• Never mind; that was running a version translated (unsuccessfully, it seems) to Python 3. It works fine under Python 2. Mar 23, 2016 at 15:28
– anon
Mar 23, 2016 at 18:50
• @QPaysTaxes I've edited my answer. Mar 24, 2016 at 2:12
• @Dennis Now could you explain why that works? For example, why d+8 and not, say, d+9? How????
– anon
Mar 24, 2016 at 2:15

# Pyth, 34 bytes

_shM.u,%J/eMeN\12-+PMeNm.B6/J2k,kQ


Try it online: Demonstration or Test Suite (takes quite a while). It should satisfy the time restriction though easily, since the online compiler is quite slow in comparison with the normal (offline) compiler.

### Explanation:

My algorithm is basically an implementation of addition with carrying. But instead of carrying 1, I have to carry 110 (1100 in base -1+i is the same as 2 in base -1+i). This works mostly fine, but you can get stuck in an infinite loop printing zeros. For instance if you are adding 1 with 11 and currently have the carry 110. So I basically add until I get stuck in a loop and then stop. I think that a loop that a loop will always print zeros and therefore this should be fine.

_shM.u,%J/eMeN\12-+PMeNm.B6/J2k,kQ   implicit: Q = input list of strings
,kQ   create the pair ["", Q]
.u                               modify the pair N (^) until loop:
,                                replace N with a new pair containing:
eN                           N[1] (the remaining summand)
eM                             take the last digits of each summand
/    \1                         count the ones
J                                store the count in J
%J       2                        J % 2 (this is the first element of the new pair)
PMeN                  remove the last digit of each summand
+    m   /J2           and add J / 2 new summand:
.B6                 with the value "110" (binary of 6)
-            k          remove empty summand
.u                               returns all intermediate results
hM                                 extract the digits
s                                   sum them up to a long string
_                                    reverse


# Python 2, 69 67 bytes

f=lambda a,b:a*a+b*b^58and 2*f(a*b%2*6,f(a/2,b/2))|a+b&1if a else b


I/O is done with base 2 integers.

-2 thanks @Dennis.

• I take it a*a+b*b^58==0 when a and b are inverses? How does that work?
– xnor
Mar 23, 2016 at 23:59
• @xnor No, a*a+b*b==58 when one of them is 3 and the other is 7. Mar 24, 2016 at 0:06
• It's not obvious me to that (3,7) is the only pair that gives a cycle and needs special-casing. If it is true, then you only actually need to check (a,b)==(3,7) in that order, since (7,3) recurses to (3,7), and maybe there's a shorter expression for that.
– xnor
Mar 24, 2016 at 0:09
• Now this is guaranteed to confuse anyone who doesn't know (or forgets) that (a) ^ is XOR, not exponentiation, or (b) XOR has lower precedence than +. Mar 24, 2016 at 8:03

# Jelly, 2928262421 20 bytes

DBḅ1100ḌµDL+8µ¡Dṣ2ṪḌ


This does I/O in decimal. The integers have to be separated by the non-alphanumeric character +.

### Background

In Arithmetic in Complex Bases, the author shows how to add and multiply complex numbers in bases of the form -n + i.

For base -1 + i, addition is done similarly to regular, binary addition with carry, with two differences:

• Instead of carrying 1 to the next higher position, we carry 110 to the next three.

• Carry digits can propagate indefinitely. However, without leading zeroes, the sum a + b has at most eight digits more than the maximum of a and b.

We proceed as follows.

1. First, we add a and b as if their digits were decimal digits.

For a = 10101 and b = 11011, this gives 21112.

2. For each digit larger than 1, we have to subtract 2 from that digit and carry 110 to the next three higher positions. We can achieve this by converting each decimal digit to binary, the resulting binary arrays from base 1100 to integer, and interpret the resulting list of 0's, 1's, 1100's and 1101's as a non-canonical base 10 number.1

For the sum 21112, this gives 21112 + 1098 * 10001 = 21112 + 10981098 = 11002210.

3. We repeat steps 2 a total of d + 8 times, where d is the number of digits of a + b.

For the initial sum 21112, the results are

                      11002210
12210010
1220010010
122000010010
12200000010010
1220000000010010
122000000000010010
12200000000000010010
1220000000000000010010
122000000000000000010010
12200000000000000000010010
1220000000000000000000010010
122000000000000000000000010010

4. We discard all but the last d + 8 digits from the final result. This is achieved by discarding everything after the last 2.2

For the initial sum 21112, the final result is 10010.

### How it works

DBḅ1100ḌµDL+8µ¡Dṣ2ṪḌ  Main link. Argument: a + b (implicit sum)

µ    µ¡       Execute the chain before the first µ n times, where n is
the result of executing the chain before the second µ.
D            Convert a + b to base 10.
L           Length; count the decimal digits.
+8         Add 8 to the number of digits.
D                     Convert the initial/previous sum to base 10.
B                    Convert each digit (0 - 3) to binary.
ḅ1100               Convert each binary array from base 1100 to integer.
Ḍ              Interpret the resulting list as a base 10 number.
D      Convert the final sum to base 10.
ṣ2    Split at occurrences of 2.
Ṫ   Select the last chunk.
Ḍ  Convert from base 10 to integer.


1 Note that the digits of this sum cannot exceed 3 (initial value 1 plus two 1's from carries).

2 This works because the last digit that will cancel out cannot be a 3.

## Retina, 100 bytes

r+(.*)(\d|(?!\4))( .*)(.?)
$2$4:$1$3
T 0
+1:11(1*:1*)11
:$1 ^:* ::: }:(1*:1*:)11 1:1$1
(1)*:
$#1  This takes the input separated with a comma. The output always starts with three leading zeroes. Try it online! I really wonder if there's a shorter solution for the first stage... • No, no, the score is perfect as it is ;) Mar 23, 2016 at 18:29 • Nice score of -2i! – anon Mar 23, 2016 at 18:51 • Wow. I did not see this solution when I posted mine... Much more superior than my solution. Apr 29, 2016 at 14:07 • @KennyLau I was just looking at it and thinking "hm, I guess I should've added an explanation at some point..." Apr 29, 2016 at 14:09 • ...-2i? This is decimal but the program uses a base that not. Dec 28, 2017 at 15:42 ## Python 3, 289 bytes This performs digitwise addition from least to most significant digit (in other words, the exact same algorithm you were taught in primary school). The differences are that (a) it's in binary, not decimal, so you carry whenever a digit is 2 or more, and (b) 1 + 1 = 1100, not 10. Actually, it's also necessary to note that 11 + 111 = 0, or else sums that should become zero will never terminate. from collections import* def a(*s,p=0): r=defaultdict(int,{0:0}) for S in s: n=0 for d in S[::-1]:r[n]+=d=='1';n+=1 while p<=max(r): while r[p]>1: r[p]-=2 if r[p+1]>1<=r[p+2]:r[p+1]-=2;r[p+2]-=1 else:r[p+2]+=1;r[p+3]+=1 p+=1 return str([*map(r.get,sorted(r))])[-2::-3]  More golfing is surely possible. • How certain are you that your "zero detector" is sufficient? – Yakk Mar 23, 2016 at 19:55 • @Yakk: On a scale of one to peer-reviewed-journal, maybe give it a no-counterexamples-yet? Mar 24, 2016 at 7:43 ## JavaScript (ES6), 146 126 bytes r=n=>n&&n%2-r(n>>=1)-i(n) i=n=>n&&r(n>>=1)-i(n) g=(x,y,b=(x^y)&1)=>x|y&&b+2*g(b-x+y>>1,b-x-y>>1) (x,y)=>g(r(x)+r(y),i(x)+i(y))  g converts a Gaussian integer (real and imaginary parts) into base i-1, while r and i converts a base i-1 integer into a Gaussian integer (real and imaginary parts respectively). Once the conversions are in place, I just have to do the arithmetic. Edit: Saved 20 bytes by calculating the real and imaginary parts separately. C++ 416 bytes, plus #include <vector>\n#include <algorithm>\n (another 40) using I=int;using v=std::vector<I>;void r(v&x){v r{rbegin(x),rend(x)};x=r;}v a(v L,v R){r(L);r(R);L.resize(std::max(L.size(),R.size()));for(int&r:R)L[&r-R.data()]+=r;while(1){L.resize(L.size()+3);auto it=find(rbegin(L),rend(L),2);if(it==rend(L))break;I i=-1+it.base()-begin(L);i&&L[i+1]&&L[i-1]/2?L[i+1]=L[i]=L[i-1]=0:(++L[i+2],++L[i+3],L[i]=0);}L.erase( std::find(rbegin(L),rend(L),1).base(),end(L));r(L);return L;}  or, with more whitespace: using I=int; using v=std::vector<I>; void r(v&x){v r{rbegin(x),rend(x)};x=r;} v a(v L,v R) { r(L);r(R); L.resize(std::max(L.size(),R.size())); for(int&r:R) L[&r-R.data()]+=r; while(1) { L.resize(L.size()+3); auto it=find(rbegin(L), rend(L), 2); if(it==rend(L)) break; I i=-1+it.base()-begin(L); i&&L[i+1]&&L[i-1]/2? L[i+1]=L[i]=L[i-1]=0 : (++L[i+2],++L[i+3],L[i]=0); } L.erase( std::find(rbegin(L),rend(L),1).base(), end(L)); r(L); return L; }  Barely golfed. It takes input as a vector of ints, and returns a vector of ints. # Retina, 157151134133124 123 bytes 1 byte off thanks to Martin Büttner. (.+),(.+)$.1$*0$2,$.2$*0$1, 1 0x +(0x*)(,.*)0(x*),$2,$1$3
{,

(^|0x0xx0xx)
000
(0x*)(0x*)(0x*0)xx
$1x$2x\$3
)^0+
0
0x
1


Try it online!

Converts to unary, and then repeat the following replacements (shown here in decimal):

122 -> 000
0002 -> 1100 (this can also be 0012 -> 1110 and 1112 -> 2210 or even 2222 -> 3320 or even 3333 -> 4431)


Basically, when larger than two: take away two, add nothing in the previous digit, add one to the previous digit, then add one to the previous digit.

In pseudocode:

if(a[n]>2):
a[n] -= 2;
a[n-2] += 1;
a[n-3] += 1;


### Unary implementation:

Each digit (e.g. 3) is shown as the number of xs (e.g. xxx) and then prefixed with 0.

For example, 1234 would be expressed as 0x0xx0xxx0xxxx.

This leaves 0 unchanged, as 101 would be expressed by 0x00x.

Since initially and finally, there is only 0 and 1, the conversion could be easily done by 1->0x and 0x->1`.