This particular function is designed to determine whether a specific cell on a scrabble board qualifies as a double letter bonus spot. With a cyclomatic complexity of 23, it exceeds the recommended threshold of 20. Despite this, I am unsure of an alternative approach and believe this method is the most effective. Below is the implementation of the function:
checkDoubleLetterCell(row: number, column: number): boolean
{
if((row === middle- 1 || row === middle + 1)
&& (column === middle - 1 || column === middle + 1)
|| (row === 0 || row == SCRABBLE_SIZE - 1 || row === middle)
&& (column === middle + Math.round(middle/2) || column === middle - Math.round(middle/2))
|| (column === 0 || column === SCRABBLE_SIZE - 1 || column === middle)
&& (row === middle + Math.round(middle/2) || row === middle - Math.round(middle/2))
|| (column === middle + 1 || column === middle - 1)
&& (row === middle + Math.round(middle/2) + 1 || row === middle - Math.round(middle/2) - 1)
|| (row === middle + 1 || row === middle - 1)
&& (column === middle + Math.round(middle/2) + 1 || column === middle - Math.round(middle/2) - 1))
{
return true;
}
return false;
}