Zeckendorf arithmetic

Zeckendorf arithmetic is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Zeckendorf number representation is described Here. It looks like binary. The purpose of this task is to explore why Babbage et al. may have preferred binary to zeckendorf when designing computers. Here I extended the representation task by adding an increment and comparison operator. This task will implement addition; subtraction; multiplication; and division using zeckendorf.

This task is a total immersion zeckendorf task, using decimal numbers will attract serious approbation.

Addition.

Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.

Occurances of 11 must be changed to 100. Ocurances of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;

Subtraction.

10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:

  abcde
  10100 -
   1000
  _____
    100  borrow 1 from a leaves 100
  + 100  add the carry
  _____
   1001

A larger example:

  abcdef
  100100 -
    1000
  ______
  1*0100 borrow 1 from b
   + 100 add the carry
  ______
  1*1001

Sadley we borrowed 1 from b which didn't have it to lend. So now b borrows from a:

    1001
  + 1000 add the carry
    ____
   10100

Multiplication.

Here you teach your computer its zeckendorf tables. eg. 101 * 1001:

  a = 1 * 101 = 101
  b = 10 * 101 = a + a = 10000
  c = 100 * 101 = b + a = 10101
  d = 1000 * 101 = c + b = 101010

  1001 = d + a therefore 101 * 1001 =
 
  101010
   + 101
  ______
 1000100

Division.

Lets try 1000101 divided by 101, so we can use the same table used for addition.

  1000101 -
   101010 subtract d (1000 * 101)
  _______
     1000 -
      101 b and c are too large to subtract, so subtract a
     ____
        1 so 1000101 divided by 101 is d + a (1001) remainder 1

C++

<lang cpp> // For a class N which implements Zeckendorf numbers: // I define an increment operation ++() // I define a comparison operation <=(other N) // I define an addition operation +=(other N) // Nigel Galloway October 28th., 2012

1. include <iostream>

class N { private:

   int dVal = 0, dLen;
   void _a(int i) {
       for (;; i++) {
           if (dLen < i) dLen = i;
           switch ((dVal >> (i*2)) & 3) {
           case 0: case 1: return;
           case 2: if (((dVal >> ((i+1)*2)) & 1) != 1) return;
                   dVal &= ~(3 << (i*2)); dVal += 1 << (i+1)*2; return;
           case 3: dVal &= ~(3 << (i*2)); _b((i+1)*2);
   }}}
   void _b(int pos) {
     if (pos == 0) {++*this; return;}
     if (((dVal >> pos) & 1) == 0) {
       dVal += 1 << pos;
       _a(pos/2);
       if (pos > 1) _a((pos/2)-1);
       }
     else {
       dVal &= ~(1 << pos);
       _b(pos + 1);
       _b(pos - ((pos > 1)? 2:1));
   }}

public:

   N(char const* x = "0") {
       int i = 0, q = 1;
       for (; x[i] > 0; i++);
       for (dLen = --i/2; i >= 0; i--) {
           dVal+=(x[i]-48)*q;
           q*=2;
   }}
   const N& operator++() {
     dVal += 1;
     _a(0);
     return *this;
   }
   const N& operator+=(const N& other) {
     for (int GN = 0; GN < (other.dLen + 1) * 2; GN++) if ((other.dVal >> GN) & 1 == 1) _b(GN);
     return *this;
   }
   const bool operator<=(const N& other) const {
       return dVal <= other.dVal;
   }
   friend std::ostream& operator<<(std::ostream&, const N&);

}; N operator "" N(char const* x) {

   return N(x);

} std::ostream &operator<<(std::ostream &os, const N &G) {

   const static std::string dig[] {"00","01","10"}, dig1[] {"","1","10"};
   if (G.dVal == 0) return os << "0";
   os << dig1[(G.dVal >> (G.dLen*2)) & 3];
   for (int i = G.dLen-1; i >= 0; i--) os << dig[(G.dVal >> (i*2)) & 3];
   return os;

} </lang> The the following test:

int main(void) {
  N G;
  G = 10N;
  G += 10N;
  std::cout << G << std::endl;
  G += 10N;
  std::cout << G << std::endl;
  G += 1001N;
  std::cout << G << std::endl;
  G += 1000N;
  std::cout << G << std::endl;
  G += 10101N;
  std::cout << G << std::endl;
  return 0;
}

Produces this

Output:

101
1001
10101
100101
1010000