Y1 = Q1' Q2 X' Y2 = X + Q1' Q2 Z = Q1 Q2 X
5 = 101 6 = 11010 -7 = 1001 11 = 11111
A four-bit base (-2) system is contiguous:
5 = 0101 4 = 0100 3 = 0111 2 = 0110 1 = 0001 0 = 0000 -1 = 0011 -2 = 0010 -3 = 1101 -4 = 1100 -5 = 1111 -6 = 1110 -7 = 1001 -8 = 1000 -9 = 1011 -10 = 1010The maximum and minimum values are 5 and -10, respectively.
Any n-bit base (-2) system is contiguous. Proof by induction: Suppose an (n-1)-bit base (-2) system is contiguous. Note that such a system covers 2^n consecutive integers. Adding another bit covers twice as many integers; if x is covered by the (n-1)-bit system, then both x and x+((-2)^n) are covered by the new system. These integers are unique, since the (n-1) bit system only covers 2^n consecutive integers, so the new integers covered do not overlap with the existing integers. Also, the new system is contiguous, since adding (-2)^n is equivalent to shifting the covered integers by (-2)^n on a number line.
The base case for the induction is trivial; consider a 1-bit base (-2) system.
111 <- carry 1 011 <- carry 2 001110 (-6) + 001111 (-5) -------- 110101 (-11)
A B CIN | COUT1 COUT2 S --------+-------------- 0 0 0 | 0 0 0 0 0 1 | 0 0 1 0 1 0 | 0 0 1 0 1 1 | 1 1 0 1 0 0 | 0 0 1 1 0 1 | 1 1 0 1 1 0 | 1 1 0 1 1 1 | 1 1 1