Homework Policy
All the work for Part 6A must be entirely your own. You may consult books and the course instructors, but nothing else. You may work with other students on Part 6B. At the top of your program, put the names of any others that you have worked with. You must completely understand all work that you submit.

The Student Honor Code applies to this assignment:

"On my honor as a University of Colorado at Boulder student, I have neither received nor given unauthorized assistance on this work."

Part 6A of the Foxes and Geese Program

Due at 9:30am, Tuesday, Oct 8. No late submissions accepted for credit.

There is an island which is shared by foxes and geese. Each year the populations of these animals varies according to these equations:

          New fox population   =  (1 - d + bG)F
          New goose population =  (1 + r - rG/k - aF)G

          F = last year's fox population (try about 100)
          G = last year's goose population (try about 10000)
          d = death rate of foxes (try about 0.1)
          b = conversion rate (try about 0.00001)
          r = growth rate of geese (try about 0.4)
          a = ability rate (try about 0.0005)
          k = carrying capacity of island (try about 30000)

These fancy phrases probably don't mean much to you, but the equations should be enough to program the simulation. If you want to find out more about this mathematical model, see the book Population Biology by P.W. Hendrick.

This program declares a group of constants and then uses a function to compute the fox and goose populations over a period of 100 years. The constants and prototype for the function must be exactly these (so that Dora can check your work):

const int INITIAL_FOX = 100;
const int INITIAL_GEESE = 10000;
const double DEATH_RATE = 0.1;
const double CONVERSION_RATE = 0.00001;
const double GROWTH_RATE = 0.4;
const double FOX_ABILITY = 0.0005;
const int CAPACITY = 30000;
const int YEARS= 100;

void population(int foxes[], int geese[]);
// This function uses the declared constants to compute the fox and
// goose population for future years, putting these values in
// foxes[0]...foxes[YEARS] and geese[0]...geese[YEARS].
// THIS FUNCTION DOES NOT WRITE ANY OUTPUT TO cout.
// IT JUST PUTS THE NUMBERS IN THE ARRAY.

Notice that the size of the two arrays must be at least YEARS+1 since the function will compute values for foxes[0]...foxes[YEARS] and geese[0]...geese[YEARS].

For Part 6A of this assignment (due at 9:30 Oct 8 with no late submissions), your program must produce an output of exactkt this form:

---------- Year ---------- Foxes ---------- Geese
              0              100            10000
             10              215            23238
             20              542            14541
             30              607             8693
             40              518             9005
             50              502            10676
             60              535            10476
             70              539             9913
             80              529             9975
             90              527            10168
            100              531            10148
The minimum population for foxes was 100 in year 0.
The maximum population for foxes was 622 in year 26.
The minimum population for geese was 8338 in year 34.
The maximum population for geese was 23404 in year 9.

    cout << setw(15) << x;

const int YEARS= 100;           // Number of years in the simulation
const int TABLE_INCREMENT = 10; // Size of the steps in the output table

void computeMinimum(int data[], int& minValue, int& i);
// The function examines data[0]...data[YEARS] to find the minimum entry.  
// The value of this entry is placed in the parameter minValue.
// The index of this entry is placed in the parameter i.
// If there are several equally small minimum elements, then i is set
// to the index of the first such item.
// THIS FUNCTION DOES NOT WRITE ANY OUTPUT TO cout.
// IT JUST PUTS THE NUMBERS IN THE TWO REFERENCE PARAMETERS.

void computeMaximum(int data[], int& maxValue, int& i);
// The function examines data[0]...data[YEARS] to find the maximum entry.  
// The value of this entry is placed in the parameter maxValue.
// The index of this entry is placed in the parameter i.
// If there are several equally large maximum elements, then i is set
// to the index of the first such item.
// THIS FUNCTION DOES NOT WRITE ANY OUTPUT TO cout.
// IT JUST PUTS THE NUMBERS IN THE TWO REFERENCE PARAMETERS.

Hint: The formulas produce a double number for each year's population. You should store this value in a double number and then convert it to an integer before assigning it to an element of the integer array. For example, if p is a double number, then you may write the assignment:

    foxes[i] = (int)p;

Part 6B of the Foxes and Geese Program

Due at 9:30am, Tuesday, Oct 15. No late submissions accepted for credit.

Modify your program so that it also draws a graph of these populations, using the x-axis for time and the y-axis for populations. There should be two different lines on the graph for the fox and geese populations. The two lines should be drawn in different colors and using different scales so that the maximum populations are near the top of the screen in both cases. Label the x axis every 20 years; put two different labels on the y-axis to show the two different scales. The y-axis scales must have at least three numbers each (one at the top, one at the bottom and one in the middle). There are notes at www.cs.colorado.edu/~main/intro/info/outtext.html that talk about printing labels on axes in graphics mode.