Bits positioning and numbering would be interesting to represent here.. for example...
2 = 110 == (1*(-2)^2 + 1*(-2)^1 + 0*(-2)^0) == (4 + (-2) + 0 ) 3 = 111 == (1 *(-2)^2 + 1*(-2)^1 + 1*(-2)^0) == (4 + (-2) + 1) 4 = 100 == (1 *(-2)^2 + 0*(-2)^1 + 0*(-2)^0) == (4 + 0 + 0) 5 = 101 == (1 *(-2)^2 + 0*(-2)^1 + 1*(-2)^0) == (4 + 0 + 1) 6 = 11010 == (1 *(-2)^4+ 1 *(-2)^3 + 0 *(-2)^2 + 1*(-2)^1 + 0*(-2)^0) == (16 + (-8) + 0 + (-2) + 0) and so on. There would be always odd no. of bits. Also no. of bits would be not as per the base +2 notation. It ends up finding the combination of positive even powers of 2 and negative odd powers of 2 that forms the no. On Mon, Aug 30, 2010 at 2:37 PM, Maria <[email protected]> wrote: > Write some code to convert a positive integer into base minus 2. That > is, whereas base 2 has a 1's place, a 2's place, a 4's place, etc., > base minus 2 has a 1's place, a minus 2's place, a 4's place, a minus > 8's place, ... (-2)^n. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]<algogeeks%[email protected]> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
