//assuming 4 letter word
//"2" because checks one ahead, last element needs no further connection
^
for (x=0; x>=2; x++;) {
if (abs(numarrayHERE[].index'X - (numarrayHERE[+1].index'X).indexX) = 1) {
boolConnected = true;
} else {
boolConnected = false;
};
//find if they are in line, result of zero above
if (so) {
//do the same for y coordinate as was done above;
};
return boolConnected;
};
//where "abs" is absolute value, numarray's x index is taken, then the next in the numarrays value is taken, if the x coordinate is one away-determined by subtraction 1 or -1 is left or right, check if is zero, then see if it is one away in the same way it was checked for the x coord. the for loop should carry through til the end of the word.