If a word had the letter B, and the letter after that was A, # item (A) of list would be smaller than that of # item (B) of list. The script above compares a letter with the letter after that, to see if they are in alphebatical order. # item (B) of list would give me 2, and so on. So basically, # item (A) of list would give me 1, since item 1 in the alphabet list is the letter a. define check (word) set to set to repeat (length of ((word) - ) change by (1) if (# item (letter (counter) of (word) ) of :: list) > then set to end end if then add (word) to else word not in alphebatical order ::grey end The “# item () of list” block reports the number of the item in a list that matches the given input. And so on …you can then use the script below. MasterofTheBrick wrote:Assuming that the pre-set list with alphabets is as shown below: 1. However if the variable isn't set to n throughout the loop, it will return as y at the end, which will allow the script to classify the word as one which is spelt in alphabetical order. Upon sensing this, the script above will set the true? variable to n.

define check (word) set to set to repeat ((length of (word)) - (1)) change by (1) if (# item (letter (counter) of (word) ) of :: list) > then set to end end if then add (word) to else word not in alphebatical order ::grey end The “# item () of list” block reports the number of the item in a list that matches the given input. Assuming that the pre-set list with alphabets is as shown below: 1.

Word spelled in alphabetical order