Path: utzoo!utgpu!watmath!clyde!att!osu-cis!tut.cis.ohio-state.edu!mailrus!cornell!batcomputer!itsgw!imagine!rpics!guilford From: guilford@rpics (Jim Guilford) Newsgroups: comp.sys.amiga.tech Subject: Re: MINTERM_SWAP? Summary: Needs three blits Keywords: MINTERM graphics how Message-ID: <1946@imagine.PAWL.RPI.EDU> Date: 5 Dec 88 15:50:35 GMT References: <1576@nmtsun.nmt.edu> Sender: news@imagine.PAWL.RPI.EDU Reply-To: guilford@turing.cs.rpi.edu (Jim Guilford) Organization: RPI CS Dept. Lines: 42 In article <1576@nmtsun.nmt.edu> wbnsnsr@nmtsun.nmt.edu (William Norris) writes: >Is there a simple way to swap two bitmaps without a third temporary bitmap? > >I've looked over the MINTERMS, but none seem to do the trick. How come >the lower nibble in the MINTERM is always 0? If 0xC0 is a copy, shouldn't >0xCC be a swap? > > >-- >wbnsnsr@nmtsun.nmt.edu | /// Seulement >William B. Norris IV |\\ /// l'Amiga peut >POB #2185 C/S | \\// vous l'offrir. >Socorro, NM 87801 |=-=-=-=-=-=-=-=-=-=-=-=-= I don't have the RKM handy, so I can't give you a detailed answer about 0xC0. Off hand, I would say that the low order nibble is zer because the examples are only using two input sources. In this case there are 2^2 or four minterms (bits) needed. The only time you need to worry about the low order nibble is when you are dealing with three sources (don't quote me on this, this is off the top of my head). In any event, since there is only one destination, it is definately impossible to do a swap in one blit. It is possible to do a swap in three blits without a temp space. If you let your two bit maps be A and B, then the following will swap them: A <- A xor B B <- A xor B A <- A xor B If you check the math: A1 <- A0 + B0 B1 <- A1 + B0 == (A0 + B0) + B0 == A0 + (B0 + B0) == A0 + 0 == A0 A2 <- A1 + B1 == (A0 + B0) + A0 == B0 + (A0 + A0) == B0 Check the RKM, but I would guess that an XOR minterm is $60 ?? --JimG guilford@cs.rpi.edu ...!rutgers!nysernic!rpics!guilford