Here's a hint for you Scott: with the addition of one instruction to what I suspect is your algorithm, you can leave X untrashed. One caveat: it will only leave the parity in bit 1 for 12c508's, but will leave it in bit 1 and 5 for 16c84's. Brian -----Original Message----- From: Scott Dattalo To: PICLIST@MITVMA.MIT.EDU Date: Friday, March 13, 1998 9:36 PM Subject: Re: Parity Challenge >Hint: My solution leaves the parity bit in two 'convenient' >places: bit1 and bit5 of 'X'. It trashes W too. > >Scott