# TokenPass free response answer 2013

The TokenPass problem from the 2013 AP Computer Science Exam is typical of free response problems that test arrays.

TokenPass is #2 from the from the 2013 AP Computer Science A Free Response problems.

https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap13_frq_comp_sci.pdf

## Part (a) TokenPass constructor

public TokenPass(int playerCount)
{
board = new int[playerCount];

for(int i = 0; i < board.length; i++)
board[i] = (int) (Math.random() * 10) + 1;

currentPlayer = (int) (Math.random() * board.length);
}

A regular for loop through board is necessary for initialization of each element. A foreach loop won’t work since changes will not affect the array itself.

Math.random returns a value in the range of 0.0 <= board[i] < 1.0. The steps to convert this to the desired range of 1 <= board[i] <= 10 are below.

Statement Range
board[i] = Math.random(); 0.0 <= board[i] < 1.0
board[i] = Math.random() * 10; 0.0 <= board[i] < 10.0
board[i] = (int) (Math.random() * 10); 0 <= board[i] <= 9
board[i] = (int) (Math.random() * 10) + 1; 1 <= board[i] <= 10

currentPlayer must be set to a random valid index in board. This is a very common task that can be accomplished by multiplying the return value of Math.random by the length/size of the array/list and casting the result to an int.

For more on working with Math.random() see Generating numbers with Math.random().

## Part (b) distributeCurrentPlayerTokens method

public void distributeCurrentPlayerTokens()
{
int tokens = board[currentPlayer];
board[currentPlayer] = 0;
int targetIndex = currentPlayer + 1;

while(tokens > 0)
{
if(targetIndex == board.length)
targetIndex = 0;

board[targetIndex]++;
tokens--;
targetIndex++;
}
}

This problem requires traversal of an array with modification of each element, which precludes a foreach loop. This solution loops while there is at least one token remaining to distribute. A for loop that runs for the number of tokens is also possible.

There are 2 common approaches to ensuring that targetIndex remains a valid index in board. The solution above explicitly checks for validity and resets targetIndex to 0 when it exceeds the array bounds.

An alternative is to use modular division (%) to constrain targetIndex. See Division operations for more discussion of using % to constrain values.

It is important that the original value of board[currentPlayer] is used as the number of tokens to distribute. board[currentPlayer] can be modified during token distriution.

## Part (b) distributeCurrentPlayerTokens method (alternate solution)

public void distributeCurrentPlayerTokens()
{
int tokens = board[currentPlayer];
board[currentPlayer] = 0;
int targetIndex = currentPlayer + 1;

for(int t = 1; t <= tokens; t++)
{
board[targetIndex % board.length]++;
targetIndex++;
}
}

## Help & comments

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