# Implement an 8 bit adder

## The Challenge

Implement a function which accepts two integers whose values range from 0 - 255 and returns the sum of those integers mod 256. You may only use bitwise negation (~), bitwise or (|), bit shifting operators (>>,<<), and assignment (=).

Things you cannot use include (but are not limited to)

• Addition, subtraction, multiplication, and division
• Loops
• Conditional statements
• Function calls

Fewest uses of binary or, binary negation, and bit shift operations wins. In the event of a tie, the most popular solution wins. As always, standard loopholes apply.

Here is an example of a simple 2-bit adder. It uses 77 binary negations, 28 binary ors, and 2 bit-shifts for a total score of 107 (this can be seen by running the C preprocessor with gcc -E). It could be made much more efficient by removing the #defines and simplifying the resulting expressions, but I've left them in for clarity.

#include <stdio.h>

#define and(a, b) (~((~a)|(~b)))
#define xor(a, b) (and(~a,b) | and(a,~b))

{
int x, carry;
x = xor(and(a, 1), and(b, 1));
carry = and(and(a, 1), and(b, 1));
carry = xor(xor(and(a, 2), and(b, 2)), (carry << 1));
x = x | carry;
return x;
}

int main(int argc, char **argv)
{
int i, j;
for (i = 0; i < 4; i++) {
for (j = 0; j < 4; j++) {
if (adder(i, j) != (i + j) % 4) {
printf("Failed on %d + %d = %d\n", i, j, adder(i, j));
}
}
}
}


Update: Added example and changed scoring critera

• why not bitwise "and"? – rdans Aug 30 '14 at 0:05
• @Ryan Most people are more familiar with NAND gates than NOR gates :) – Orby Aug 30 '14 at 0:08
• does recursion count as a loop? – rdans Aug 30 '14 at 1:21
• @Ryan Recursion does count as a loop, though I'm not sure how you'd implement it without a conditional statement. – Orby Aug 30 '14 at 1:25
• Is overflow defined or can I just output anything if it overflows? – Comintern Aug 30 '14 at 2:51

## Python, 36 operations

A methods that's logarithmic in the parameter "8"!

def add(a,b):
H = a&b   #4 for AND
L = a|b   #1
NX = H | (~L) #2
K = NX

H = H | ~(K | ~(H<<1)) #5
K = K | (K<<1) #2

H = H | ~(K | ~(H<<2)) #5
K = K | (K<<2) #2

H = H | ~(K | ~(H<<4)) #5

carry = H<<1 #1

neg_res = NX ^ carry  #7 for XOR
res_mod_256 = ~(neg_res|-256) #2
return res_mod_256


The idea is to figure out which indices overflow and cause carries. Initially, this is just the places where both a andd b have a 1. But since carried bits can cause further overlows, this needs to be determined iteratively.

Rather than overflowing each index into the next one, we speed up the process by moving 1 index, then 2 indices, then 4 indices, being sure to remember places where an overflow happened (H) and where an overflow cannot happen any more (K).

A simpler iterative solution with 47 operations:

def add(a,b):
H = a&b   #4 for AND
L = a|b   #1
NX = H | (~L) #2

c=H<<1  #1

for _ in range(6): #6*5
d = (~c)|NX
e = ~d
c = c|(e<<1)

res = c ^ NX  #7 for XOR

res_mod_256 = ~(res|-256) #2
return res_mod_256


Test rig, for anyone who wants to copy it.

errors=[]
for a in range(256):
for b in range(256):
if res!=(a+b)%256: errors+=[(a,b,res)]

print(len(errors),errors[:10])


## python, score = 83 80

def g(x,y):
for i in xrange(7):
nx = ~x
ny = ~y
x,y = ~(x|ny)|~(nx|y), (~(nx|ny))<<1
x = ~(x|~y)|~(~x|y)
return ~(~x|256)


Unroll the loop. It's 10 ops per loop times 7 loops, 7 for the last xor, and 3 to squash the 9th bit at the end.

Implements the equation x+y = x^y + 2*(x&y) by repeating it 8 times. Each time there is one more zero bit at the bottom of y.

# C, Score: 77 60

Golfed just for the hell of it, 206 169 131 bytes:

#define F c=((~(~c|~m))|n)<<1;
a(x,y){int m=(~(x|~y))|~(~x|y),n=~(~x|~y),c;F F F F F F F return (unsigned char)(~(m|~c))|~(~m|c);}


Expanded:

int add(x,y)
{
int m=(~(x|~y))|~(~x|y);
int n=~(~x|~y);
int c = 0;
c=((~(~c|~m))|n)<<1;
c=((~(~c|~m))|n)<<1;
c=((~(~c|~m))|n)<<1;
c=((~(~c|~m))|n)<<1;
c=((~(~c|~m))|n)<<1;
c=((~(~c|~m))|n)<<1;
c=((~(~c|~m))|n)<<1;
return (int)((unsigned char)(~(m|~c))|~(~m|c));
}


Essentially the same solution (mathematically) that @KeithRandall @JuanICarrano came up with, but takes advantage of C's ability to play fast and loose with variable types and pointers to wipe everything after the first 8 bits without using any more operators.

Depends on the endian-ness of the machine and the sizeof() an int and a char, but should be able to be ported to most machine specific applications with the proper pointer math.

EDIT: This is a challenge that C (or other low level languages) will have a distinct upper hand at -- unless somebody comes up with an algorithm that doesn't have to carry.

• If you're going to handle the wrap around that way, you could just cast to unsigned char. It's cleaner and more portable. – Orby Aug 30 '14 at 18:57
• @Orby - I guess typing out unsigned doesn't come naturally to me in code golf. You're right of course - updated. – Comintern Aug 30 '14 at 23:41

# C - 0

It does use operators outside of ~, |, >>, <<, and =, but I see solutions using casting and comma operators, so I guess the rule isn't too strict provided it isn't using the forbidden operators.

unsigned char sum(unsigned char x, unsigned char y)
{
static unsigned char z[] = {
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,
32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,
64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,
80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,
96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,
112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,
128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,
144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,
160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,
176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,
192,193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,
208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,
224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,
240,241,242,243,244,245,246,247,248,249,250,251,252,253,254,255,
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,
32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,
64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,
80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,
96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,
112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,
128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,
144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,
160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,
176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,
192,193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,
208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,
224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,
240,241,242,243,244,245,246,247,248,249,250,251,252,253,254
};

return (&z[x])[y];
}

• This is obviously a loophole, but +1 for pointing it out. – Orby Sep 2 '14 at 20:56

# Python - Score 66 64

def xand(a,b):
return ~(~a|~b) #4

def xxor(a,b):
return (~(a|~b))|~(~a|b) #7

def s(a,b):
axb = xxor(a,b)   #7
ayb = xand(a,b)   #4

C = 0
for i in range(1,8):
C = ((xand(C,axb))|ayb)<<1    #(1+1+4)x7=6x7=42

return xxor(axb,xand(C,255))    #7 + 4 = 11
#total: 7+4+42+11 = 64


It is the equation for a ripple adder. C is the carry. It is computed one bit at a time: in each iteration the carry is propagated left. As pointed out by @Orby, the original version did not make a modular addition. I fixed it and also saved a cycle in the iteration, as the first carry-in is always zero.

• Nice job, but your code does not wrap around properly (i.e. s(255,2) returns 257 rather than 1). You can correct this by changing your last line to return ~(~xxor(axb,C)|256)  which adds 3 points. – Orby Aug 30 '14 at 6:59

# C++ - score: 113

#define ands(x, y) ~(~x | ~y) << 1
#define xorm(x, y) ~(y | ~(x | y)) | ~(x | ~(x | y))

{
int x1 = xorm(x, y);
int y1 = ands(x, y);

int x2 = xorm(x1, y1);
int y2 = ands(x1, y1);

int x3 = xorm(x2, y2);
int y3 = ands(x2, y2);

int x4 = xorm(x3, y3);
int y4 = ands(x3, y3);

int x5 = xorm(x4, y4);
int y5 = ands(x4, y4);

int x6 = xorm(x5, y5);
int y6 = ands(x5, y5);

int x7 = xorm(x6, y6);
int y7 = ands(x6, y6);

int x8 = xorm(x7, y7);
int y8 = ands(x7, y7);

return (x8 | y8) % 256;
}

• add(1, 255) is returning 128 for me, @Ryan. – Orby Aug 30 '14 at 2:02
• @Orby its fixed now – rdans Aug 30 '14 at 2:08
• % is not on the list of permitted operators, namely ~, |, >>, and <<. Maybe replace it with ands(x8|y8, 255)>>1? – Orby Aug 30 '14 at 2:11
• Actually, ~(~x8 | y8 | 0xFFFFFF00) would do the trick nicely with only 4+ to your score. – Orby Aug 30 '14 at 2:21
• But wouldn't making the type byte instead of int would make it overflow automatically? – proud haskeller Aug 30 '14 at 10:54