Question 2 (25 points)

Complete the definition of method Cellular below. Sample output might be:

1 1 0 1 0 0 1

0 1 1 0 0 0 1

1 1 1 0 1 0 0

1 0 1 1 0 0 0

etc.

// Given the initial state of a 1-D cellular automaton A containing

// just 0's and 1's, simulate A indefinitely according to the rule:

// Each cell remains unchanged unless its two neighbors are equal,

// in which case the cell changes to the opposite binary value (i.e.,

// 0->1, 1->0 ).

// Note that A[ 0 ] and A[ A.length - 1 ] are considered neighbors. Output

// the initial state of A, and all subsequent states.

static void Cellular( int[] A )

{

int n = A.length;

int[] B = new int[ n ];

while ( true )

{

for( int i = 0; i < n; i ++ )

{

System.out.print( A[ i ] + " " );

if ( A[ (n - 1 + i)%n ] == A[ (i + 1)% n ] )

B[ i ] = ( A[ i ] + 1 )%2;

else

B[ i ] = A[ i ];

}

System.out.println();

A = B;

B = new int[ n ];

}

}

Comments

A careful construction can make work easy. The easiest solution comes from using the modulo function to inspect nearest neighbors; without using the function, more care is required to identify and correctly handle the end cases. This approach not only takes up valuable examination time, but is far more prone to error. Students who avoided modular arithmetic should study the solution as it avoids the necessity of having to consider cases. Note too how modular arithmetic was again used to add a briskness to the solution by cleverly "flipping" the value of B without the use of an if-else construct.