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For every time step, the doodlebug will firstly try to move to an adjacent cell containing an ant and eat the ant (you can decide if there are several ants in the adjacent cells, how the doodlebug will choose to move). If there are no ants in adjacent cells, the doodlebug moves according to the same rules as the ant. Note that a doodlebug cannot eat other doodlebugs.

• Determine Space to move to.
• Get Space from Board
• Check if Space is Occupied
• Check if Valid Space on Board
• Get Space from Board
• Remove Critter from Spaces
• Switch Spaces on Critter

Initialize the world with 5 doodlebugs and 100 ants. You will randomly place them on the grid. You will prompt the user to enter the number of time steps to run.

For each time step, do the following in your program: after moves, when breeding, eating, and starving are resolved, display the resulting grid. Draw the world using ASCII characters of “O” for an ant, “X” for a doodlebug and blank space for an empty space (the characters should be arranged to look like a grid). The doodlebugs will move before the ants in each time step. When you reach the time steps entered by the user, ask them to enter another number and start to run the simulation again or to exit the program. You must maintain the state of the current grid while creating the next display.

You should use a dynamic array to represent the grid. Each array element will be a pointer to a Critter. Get your program running and tested. You should see a cyclical pattern between the population of predators and prey, although random perturbations may lead to the elimination of one or both species.

For debugging your program, you should save the random placement until you have everything else working properly. In general, “random” is bad for testing and debugging.

Starve: If a doodlebug has not eaten an ant within three time steps, at the end of the third time step it will starve and die. The doodlebug should then be removed from the grid of cells.

If a doodlebug survives for eight time steps, at the end of the time step, it will spawn off a new doodlebug in the same manner as the ant (the Doodlebug will only breed into an empty cell).

• Figure out the location of the new Critter
• Create the Critter using the Board create Critter function. (Critter->GetSpace()->GetBoard()->createCritter())

I think everyone should contribute to the test cases.

• Remove Critter from Space
• Remove Space from Critter
• Delete Critter from Board
• Downsize Critter Array

If an ant survives for three time steps (not been eaten by doodlebugs), at the end of the time step (i.e., after moving) the ant will breed. This is simulated by creating a new ant in an adjacent (up, down, left, or right) cell that is empty randomly. If the cell that is picked is not empty, randomly attempt to pick another one. If there is no empty cell available, no breeding occurs. Once an offspring is produced, an ant cannot produce an offspring again until it has survived three more time steps.

A space should be able to contain a pointer to the Critter instance

For every time step, the ants randomly move up, down, left, or right. If the neighboring cell in the selected direction is occupied or would move the ant off the grid, then the ant stays in the current cell.

Whoever takes this card should also take the main function. It would be nice to contain all user input and validation in the main function