Gray codes addition

Question

Is there any known way to compute the addition (and maybe the subtraction) of two Gray codes without having to convert the two Gray codes to regular binary, perform a binary addition then convert the result back to a Gray code? I managed to write increment and decrement functions, but the addition and subtraction seem even less documented and harder to write.

Answer 1

In this document under #6, there is an algorithm for serial Gray code addition (copied directly; note, that ⊕ is xor):

procedure add (n: integer; A,B:word; PA,PB:bit;
               var S:word; var PS:bit; var CE, CF:bit);
var i: integer; E, F, T: bit;
begin
   E := PA; F := PB;
   for i:= 0 to n-1 do begin {in parallel, using previous inputs}
       S[i] := (E and F) ⊕ A[i] ⊕ B[i];
       E := (E and (not F)) ⊕ A[i];
       F := ((not E) and F) ⊕ B[i];
   end;
   CE := E; CF := F;
end;

This adds the Gray code words A and B to form the Gray code word S. The operand parities are PA and PB, the sum parity is PS. This propagates two carry bits internally, E and F, and produces two external carry bits CE and CF

Unfortunately, it doesn't state anything about subtraction, but I assume, when you can encode negative numbers, you can use addition for that.

Answer 2

I recently devised a new algorithm to add two Gray codes. Unfortunately, it is still slower than the naive double-conversion solution and is also slower than Harold Lucal's algorithm (the one in the accepted answer). But any new solution to a problem is welcome, right?

// lhs and rhs are encoded as Gray codes
unsigned add_gray(unsigned lhs, unsigned rhs)
{
    // Highest power of 2 in lhs and rhs
    unsigned lhs_base = hyperfloor(lhs);
    unsigned rhs_base = hyperfloor(rhs);

    if (lhs_base == rhs_base) {
        // If lhs and rhs are equal, return lhs * 2
        if (lhs == rhs) {
            return (lhs << 1u) ^ __builtin_parity(lhs);
        }
        // Else return base*2 + (lhs - base) + (rhs - base)
        return (lhs_base << 1u) ^ add_gray(lhs_base ^ lhs, lhs_base ^ rhs);
    }

    // It's easier to operate from the greatest value
    if (lhs_base < rhs_base) {
        swap(&lhs, &rhs);
        swap(&lhs_base, &rhs_base);
    }

    // Compute lhs + rhs
    if (lhs == lhs_base) {
        return lhs ^ rhs;
    }

    // Compute (lhs - base) + rhs
    unsigned tmp = add_gray(lhs ^ lhs_base, rhs);
    if (hyperfloor(tmp) < lhs_base) {
        // Compute base + (lhs - base) + rhs
        return lhs_base ^ tmp;
    }
    // Here, hyperfloor(lhs) == hyperfloor(tmp)
    // Compute hyperfloor(lhs) * 2 + ((lhs - hyperfloor(lhs)) + rhs) - hyperfloor(lhs)
    return (lhs_base << 1u) ^ (lhs_base ^ tmp);
}

The algorithm uses the following utility functions in order tow work correctly:

// Swap two values
void swap(unsigned* a, unsigned* b)
{
    unsigned temp = *a;
    *a = *b;
    *b = temp;
}

// Isolate the most significant bit
unsigned isomsb(unsigned x)
{
    for (unsigned i = 1u ; i <= CHAR_BIT * sizeof(unsigned) / 2u ; i <<= 1u) {
        x |= x >> i;
    }
    return x & ~(x >> 1u);
}

// Return the greatest power of 2 not higher than
// x where x and the power of 2 are encoded in Gray
// code
unsigned hyperfloor(unsigned x)
{
    unsigned msb = isomsb(x);
    return msb | (msb >> 1u);
}

---

So, how does it work?

I have to admit, it's quite a wall of code for something as « simple » as an addition. It is mostly based on observations about bit patterns in Gray codes; that is, I didn't prove anything formally, but I have yet to find cases where the algorithm doesn't work (if I don't take into account overflow, it does not handle overflow). Here are the main observations used to construct the algorithm, assume that everything is a Gray code:

• 2 * n = (n << 1) ⊕ parity(n)
• If a is a power of 2 and a > b, then a ⊕ b = a + b
• Consequently, i a is a power of 2 and a < b, then a ⊕ b = a - b. This will only work if b < 2 * a though.
• If a and b have the same hyperfloor but are not equal, then a + b = (hyperfloor(a) << 1) ⊕ ((hyperfloor(a) ⊕ a) + (hyperfloor(b) ⊕ b)).

Basically, it means that we know how to multiply by 2, how to add a power of 2 to a smaller Gray code and how to subtract a power of 2 from a Gray code that is bigger than that power of 2 but smaller than the next power of 2. Everything else are tricks so that we can reason in terms of equal values or powers of 2.

If you want more details/information, you can also check this Q&A on Code Review which proposes a modern C++ implementation of the algorithm as well as some optimizations (as a bonus, there are some nice MathJax equations that we can't have here :D).

Answer 3

I accepted @Sebastian Dressler's answer because the suggested algorithm indeed works. For the sake of completeness, I propose here a corresponding C99 implementation of the algorithm (C++ compatible):

// lhs and rhs are encoded as Gray codes
unsigned add_gray(unsigned lhs, unsigned rhs)
{
    // e and f, initialized with the parity of lhs and rhs
    // (0 means even, 1 means odd)
    bool e = __builtin_parity(lhs);
    bool f = __builtin_parity(rhs);

    unsigned res = 0u;
    for (unsigned i = 0u ; i < CHAR_BIT * sizeof(unsigned) ; ++i)
    {
        // Get the ith bit of rhs and  lhs
        bool lhs_i = (lhs >> i) & 1u;
        bool rhs_i = (rhs >> i) & 1u;

        // Copy e and f (see {in parallel} in the original paper)
        bool e_cpy = e;
        bool f_cpy = f;

        // Set the ith bit of res
        unsigned res_i = (e_cpy & f_cpy) ^ lhs_i ^ rhs_i;
        res |= (res_i << i);

        // Update e and f
        e = (e_cpy & (!f_cpy)) ^ lhs_i;
        f = ((!e_cpy) & f_cpy) ^ rhs_i;
    }
    return res;
}

Note: __builtin_parity is a compiler intrinsic (GCC and Clang) that returns the parity of the number of set bits in an integer (if the intrinsic does not exist, there are other methods to compute it by hand). A gray code is even when it has an even number of set bits. The algorithm can still be improved, but this implementation is rather faithful to the original algorithm. If you want details about an optimized implementation, you can have a look at this Q&A on Code Review.