(1)	_____	Which of the following is a type of automatic vehicle location system?
		Dead Reckoning System Contiguous Encapsulation System Discrete J-Bus System None of the above
(2)	_____	In an even parity checksum scheme:
		The parity bit is set to make the number even The parity bit is set to make the data word flat The parity bit is set to make the number of bits containing ones even None of the above
(3)	_____	The RS232-C protocol:
		Uses timing to reduce errors Only works with DB-9 connectors Is only used to connect terminals to modems None of the above
(4)	_____	The worst-case asymptotic (time) efficiency of retrieving an element from an two-dimensional array is:
(5)	_____	GPS satellites transmit:
		The location of each GPS receiver The location of beacons at intersections The date and time None of the above

Evaluate each of the following expressions related to checksums (where ^ is the bitwise exclusive or operator, and % is the modulo operator). Your answer must be in decimal (i.e., base 10).

16 ^ 16
(1)
(7 + 4) % 8
(2)
(211 + 32) % 256
(3)
1 ^ 2 ^ 3 ^ 4
(4)

public class Subdivide
{
    public static Point      i, j, one, startMax, startMin;
    

    public static void main(String[] args)
    {
       double[]        v;


       // Create the unit vector i = (0,1)
       v = new double[2];
       v[0] = 1.0;
       v[1] = 0.0;
       i = new Point(v);
       

       // Create the unit vector j = (1,0)
       v[0] = 0.0;
       v[1] = 1.0;
       j = new Point(v);
       
       // Create the vector one = (1,1);
       one = i.plus(j);       

       // Create the vector lowerLeft = (0,0)
       v[0] = 0.0;
       v[1] = 0.0;
       startMin = new Point(v);

       // Create the vector upperRight = (4,4)
       v[0] = 4.0;
       v[1] = 4.0;
       startMax = new Point(v);
       
       // Quarter the rectangle formed by (0,0) and (4,4)
       quarter(startMin, startMax);
    }
    



    public static void quarter(Point min, Point max)
    {
       Point               delta, rho, tmin, tmax;
       

       System.out.println("Min: "+min+"     Max: "+max);
       

       rho   = max.minus(min);       
       delta = rho.times(0.5);
       
       if (delta.gt(one))
       {
          // NW
          tmin = min.plus(j.ctimes(delta));
          tmax = tmin.plus(delta);
          quarter(tmin, tmax);
          
          // NE
          tmin = min.plus(delta);
          tmax = tmin.plus(delta);
          quarter(tmin, tmax);
          
          // SE
          tmin = min.plus(i.ctimes(delta));
          tmax = tmin.plus(delta);
          quarter(tmin, tmax);
          
          // SW
          tmin = min;
          tmax = tmin.plus(delta);
          quarter(tmin, tmax);
       }
    }
    
}
    

You must find non-negative values of and that solve this system. Specifically, you must use the Golden Section algorithm to find the value of that minimizes:

Show three iterations of the Golden Section algorithm starting with a lower bound of and an upper bound of .