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######## ################## ###### ###### ##### ##### #### #### ## ##### #### #### #### #### #### ##### ##### ## ## #### ## ## ## ### ## #### ## ## ## ##### ######## ## ## ## ##### ## ## ## ## ## ##### ## ## ######## ## ## ## ### ## ## #### ## ## ##### #### #### #### #### ##### #### #### #### #### #### ###### ##### ## ###### ###### ╔SSUE #10 ################## ╩UNE 30, 1995 ######## ---------------------------------------------------------------------(V1.5)--- ┼DITOR'S ╬OTES BY ├RAIG ╘AYLOR ╘HIS IS MY LAST ISSUE OF ├OMMODORE ╚ACKING (HAVING FINALLY GOTTEN OUT THE DOOR, BUT ╔ COULDN'T BREAK TRADITION AND GET IT OUT ON TIME :-) ). ╔'M HAVING TO GIVE IT UP BECAUSE ╔'VE GRADUALLY LOST INTEREST IN ├OMMODORE COMPUTERS OVER THE YEARS AND WITH THE SEARCH FOR A JOB (ANYONE WANNA HIRE A CSC GRADUATE?) AND AS ╔ GET OLDER ╔ SEEM TO HAVE LESS AND LESS TIME. ╔'M GONNA BE HANDING THE REIGNS OF ├OMMODORE ╚ACKING OVER TO ╩IM ┬RAIN, WHO IS A VERY ACTIVE MEMBER OF THE ├OMMODORE ╔NTERNET COMMUNITY. ╚E WILL ALSO BE RUNNING A MAILSERVER THAT WILL TAKE THE PLACE OF MINE (═INE WILL BECOME UNAVAILABLE AFTER ╩ULY 1ST AND WILL SEND POINTERS TO ╩IM ┬RAIN'S MAILSERVER). ╔T'S BEEN INTERESTING TO WATCH THE ├OMMODORE COMPUTERS EVOLVE, TAKE OFF LIKE A ROCKET AND THEN HAVE ├OMMODORE GO INTO LIQUIDATION. ├OMMODORE COMPUTERS HAVE BEEN AND STILL ARE, (WITH SOME EXCEPTIONS - 1541 HEAD-BANGING COMES TO MIND), TECHNOLOGICALLY SOUND. ╞OR A "HACKING" MACHINE THEY'RE WONDERFUL. ═Y EMAIL ADDRESS HAS CHANGED TO DUCK@NANDO.NET. ╔ PERIODICALLY STILL CHECK MAIL AT DUCK@PEMBVAX1.PEMBROKE.EDU BUT ONLY EVERY 2 WEEKS OR SO. ╔ AM STILL GOING TO TRY TO BE IN THE ├OMMODORE COMMUNITY BUT TIME WILL GOVERN MY ABILITY TO DO THAT. ╔'M GOING TO MISS EDITING THIS RAG.... ┴ND HERE IS ╩IM ┬RAIN: ═AIL ╙ERVER ├HANGES: ╫ITH ╔SSUE 10, THE ADDRESS FOR THE ├OMMODORE ╚ACKING MAIL SERVER HAS CHANGED. ╘HE NEW ADDRESS IS BRAIN@MAIL.MSEN.COM ╘HE COMMANDS ARE THE SAME AS BEFORE. ╬OT ALL OF THE FILES HAVE BEEN MOVED YET, SO PLEASE EMAIL THE ADMINISTRATOR (╩IM ┬RAIN, BRAIN@MAIL.MSEN.COM) IF A FILE YOU NEED IS NOT ON THE NEW SITE. ╚OWDY: ╚OWDY, MY NAME IS ╩IM ┬RAIN, AND ╔ WILL BE TAKING OVER THE POSITION OF EDITOR FOR ├OMMODORE ╚ACKING STARTING WITH ╔SSUE 11. ╙OME OF YOU MAY KNOW ME AS THE ├OMMODORE ╘RIVIA ├ONTEST ADMINISTRATOR, THE ╒╙┼╬┼╘ NEWSGROUP COMP.SYS.CBM ╞┴╤ ═ANITAINER, OR THE KEEPER OF A ├OMMODORE ╔NFORMATION ╫╫╫ ╙ITE AT HTTP://WWW.MSEN.COM/▐BRAIN/CBMHOME.HTML. ╫HEREVER YOU HAVE HEARD OF ME FROM, OR EVEN IF YOU HAVEN'T, ╔ WILL STATE THAT ╔ PLAN ON HANDLING ├OMMODORE ╚ACKING IN THE FOLLOWING WAY. ╘HE NEXT ISSUE WILL POSSIBLY LOOK DIFFERENT COSMETICALLY, AS ╔ EDIT SOMEWHAT DIFFERENTLY THAN ├RAIG, BUT THE CONTENT AND BASIC LAYOUT WILL REMAIN THE SAME. ╘HE TYPES OF MATERIAL WILL NOT CHANGE, AND THE STRUCTURE FOR SUBMITTING ARTICLES WILL CHANGE ONLY IN THE ADDRESS TO MAIL THEM TO: BRAIN@MAIL.MSEN.COM. ╚OWEVER, ╔ DO HAVE A FEW CHANGES IN MIND: 1) ╘RY TO STABILIZE THE ISSUE GENERATION SO THAT ├OMMODORE ╚ACKING WILL BECOME A QUARTERLY PUBLICATION. 2) ┴TTEMPT A FULLY "╚╘═╠"IZED VERSION OF THE MAGAZINE, WHILE STILL PROVIDING A TEXT VERSION. 3) ╨URSUE THE POSSIBILITY OF PROVIDING A PRINTED VERSION OF THESE ISSUES FOR THOSE WHO HAVE NO ONLINE ACCESS TO THEM. 4) ┼NCOURAGE ╒SER ╟ROUPS AND OTHER ├┬═ RELATED ORGANIZATIONS TO CARRY THIS MAGAZINE FOR THEIR MEMBERS. ╙O, AGAIN ╔ SAY HOWDY. ┴S ALWAYS, ├OMMODORE ╚ACKING WILL ACCEPT YOUR ARTICLES AT ANY TIME, SO PLEASE HELP US KEEP THIS QUALITY MAGAZINE RUNNING. ╔F YOU HAVE ANY QUESTIONS OR COMMENTS ABOUT THE CHANGE IN EDITORSHIP, THE POSSIBLE CHANGES, OR OTHER MATTERS, PLEASE FEEL FREE TO DROP ME A NOTE. ╔ LOOK FORWARD TO HEARING FROM YOU AND PUBLISHING YOUR ARTICLES. ╩IM ┬RAIN BRAIN@MAIL.MSEN.COM =========================================================================== ╠EGAL ═UMBO-╩UMBO ╨ERMISSION IS GRANTED TO RE-DISTRIBUTE THIS "NET-MAGAZINE", IN WHOLE, FREELY FOR NON-PROFIT USE. ╚OWEVER, PLEASE CONTACT INDIVIDUAL AUTHORS FOR PERMISSION TO PUBLISH OR RE-DISTRIBUTE ARTICLES SEPARATELY. ┴ CHARGE OF NO GREATER THAN 5 ╒╙ DOLLARS OR EQUIVLENT MAY BE CHARGED FOR LIBRARY SERVICE / DISKETTTE COSTS FOR THIS "NET-MAGAZINE". ------------------------------------------------------------------------------- ╨LEASE NOTE THAT THIS ISSUE AND PRIOR ONES ARE AVAILABLE VIA ANONYMOUS ╞╘╨ FROM CCNGA.UWATERLOO.CA (AMONG OTHERS) UNDER /PUB/CBM/HACKING.MAG AND VIA A MAILSERVER WHICH DOCUMENTATION CAN BE OBTAINED BY SENDING MAIL TO "BRAIN@MAIL.MSEN.COM" WITH A SUBJECT LINE OF "MAILSERVER" AND THE LINES OF "HELP" AND "CATALOG" IN THE BODY OF THE MESSAGE. ------------------------------------------------------------------------------- ╔N ╘HIS ╔SSUE: ├OMMODORE ╘RIVIA ╘RIVIA ┼DITION #13-18 ARE IN THIS ARTICLE. ┴S YOU MAY KNOW, THESE QUESTIONS FORM PART OF A CONTEST IN WHICH THE MONTHLY WINNER GETS A PRIZE (╘HANKS TO MY VARIOUS PRIZE DONATORS). ╘HE WHOLE THING IS MAINLY JUST FOR FUN, SO PLEASE ENJOY. ╘RY YOUR HAND AT ├OMMODORE TRIVIA!! ┬╞╠╔ - ╬EW GRAPHICS MODES 2 ╞╠╔ GAVE US MORE COLOR TO THE SCREEN, ┴╞╠╔ INCREASED THE HORIZONTAL RESOLUTION AND COLOR SELECTION BY USING THE HIRES MODE. ┬╞╠╔ STANDS FOR '┬IG ╞╠╔' AND GIVES US 400 LINES INSTEAD OF THE USUAL TWO HUNDRED. ┴╞╠╔ AND ┬╞╠╔ CAN BE COMBINED, BUT WE ARE NOT GOING INTO THAT. ═AKING STABLE RASTER ROUTINES (├64 AND ╓╔├-20) ╔N THIS ARTICLE, ╔ DOCUMENT TWO METHODS OF CREATING STABLE RASTER ROUTINES ON ├OMMODORE COMPUTERS. ╘HE PRINCIPLES APPLY FOR MOST 8-BIT COMPUTERS, NOT ONLY ├OMMODORES, BUT RASTER EFFECTS ARE VERY RARELY SEEN ON OTHER COMPUTERS. ┴ ─IFFERANT ╨ERSPECTIVE - ╨ART ╔╔╔. ┘ES!!! ╔T'S YET ANOTHER ARTICLE ON 3─ GRAPHICS! ┼VEN IF YOU HAVEN'T BEEN FOLLOWING THIS SERIES, YOU CAN USE THIS PROGRAM. ╘HIS TIME AROUND WE WILL WRITE A COMPLETELY GENERAL POLYGON PLOTTER -- IF YOU CAN TYPE BASIC DATA STATEMENTS, YOU CAN CREATE A THREE-DIMENSIONAL OBJECT OUT OF POLYGONS AND ROTATE AND PROJECT IT TO YOUR HEART'S CONTENT. ╞OR THE MORE TECHNICALLY INCLINED WE WILL LOOK AT OPTIMIZATIONS TO THE LINE ROUTINE, ┼╧╥-BUFFER FILLING, AND MORE! ┘OW! ╙ECOND ╙╔─ ├HIP ╔NSTALLATION ╘HIS ARTICLE DESCRIBES HOW TO ADD A SECOND SID CHIP FOR USE IN ╙ID╨LAYER AND OTHER PROGRAMS. ┴S ALWAYS, BE EXTRA CAREFUL WHEN MAKING MODIFICATIONS TO YOUR COMPUTER. ╙OLVING ╠ARGE ╙YSTEMS OF ╠INEAR ┼QUATIONS ON A ├64 ╫ITHOUT ═EMORY ╧╦, NOW THAT ╔ HAVE YOUR ATTENTION, ╔ LIED. ┘OU CAN'T SOLVE DENSE LINEAR SYSTEMS OF EQUATIONS BY DIRECT METHODS WITHOUT USING MEMORY TO STORE THE PROBLEM DATA. ╚OWEVER, ╔'LL COME BACK TO THIS MEMORY FREE ASSERTION LATER. ╘HE MAIN PURPOSE OF THIS ARTICLE IS TO RESCUE A USEFULL NUMERICAL ALGORITHM, "╤UARTERSOLVE", AND ALSO TO PROVIDE A BRIEF LOOK AT THE ├╧═┴╠ PROGRAMMING LANGUAGE AND ┬╠┴╙ ROUTINES. ╘HE ╫ORLD OF ╔╥├ - ┴ ╬EW ╠IFE FOR THE ├64/128 ╔'VE HEARD PEOPLE TALKING ABOUT ╔╥├. ╫HAT IS IT? ╫HY IS IT USEFUL TO ME AS A ├OMMODORE USER? ┬ILL "├OOLHAND" ╠UECK EXPLAINS THE HOWS AND WHYS IN THIS ARTICLE. ╙WIFT╠INK-232 ┴PPLICATION ╬OTES (VERSION 1.0B) ╘HIS INFORMATION IS MADE AVAILABLE FROM A PAPER DOCUMENT PUBLISHED BY ├═─, WITH ├═─'S PERMISSION. ─ESIGN AND ╔MPLEMENTATION OF A ╙IMPLE/┼FFICIENT ╒PLOAD/─OWNLOAD ╨ROTOCOL ╘HIS ARTICLE DETAILS HOW TO IMPLEMENT A CUSTOM UPLOAD/DOWNLOAD PROTOCOL THAT IS FASTER THAN MOST OF THE ONES COMMON TO THE ├64/128 COMPUTERS. ─ESIGN AND ╔MPLEMENTATION OF A '╥EAL' ╧PERATING ╙YSTEM FOR THE 128: ╨ART ╔╔ ╘HERE HAS BEEN A SLIGHT CHANGE IN PLANS. ╔ ORIGINALLY INTENDED THIS ARTICLE TO GIVE THE DESIGN OF A THEORETICAL DISTRIBUTED MULTITASKING MICROKERNEL OPERATING SYSTEMFOR THE ├128. ╔ HAVE DECIDED TO GO A DIFFERENT ROUTE: TO TAKE OUT THE DISTRIBUTED COMPONENT FOR NOW AND IMPLEMENT A REAL MULTITASKING MICROKERNEL ╧╙ FOR A SINGLE MACHINE AND EXTEND THE SYSTEM TO BE DISTRIBUTED LATER. ╘HE IMPLEMENTATION SO FAR IS, OF COURSE, ONLY IN THE PROTOTYPE STAGE AND THE APPLICATION FOR IT IS ONLY A DEMO. ╨ART ╔╔╔ OF THIS SERIES WILL EXTEND THIS DEMO SYSTEM INTO, PERHAPS, A USABLE DISTRIBUTED OPERATING SYSTEM. ======================================================================== ╘RIVIA BY ╩IM ┬RAIN (BRAIN@MAIL.MSEN.COM) ╫ELL, SUMMER IS UPON THE ┬RAIN HOUSEHOLD, AND THINGS ARE MOVING AT A FAST CLIP AT THE HOUSE. ╚OWEVER, THE TRIVIA STILL KEEPS COMING. ╔ APPRECIATE ALL THE PEOPLE WHO CONTRIBUTE TO THE TRIVIA AND ALL THE PEOPLE WHO TAKE PART IN THE MONTHLY CONTEST. ╔ HAVE COLLECTED ╘RIVIA ┼DITION #13-18 IN THIS ARTICLE. ┴S YOU MAY KNOW, THESE QUESTIONS FORM PART OF A CONTEST IN WHICH THE MONTHLY WINNER GETS A PRIZE (╘HANKS TO MY VARIOUS PRIZE DONATORS). ╘HE WHOLE THING IS MAINLY JUST FOR FUN, SO PLEASE ENJOY. ┴S THE SUMMER MONTHS START UP, NEWS ON THE TRIVIA INCLUDES: 1) ╔ NOW HAVE ACCESS TO SOME MORE ORPHAN MACHINES (├65, ├116), SO EXPECT SOME TRIVIA QUESTIONS ON THOSE MODELS. 2) ╘HE NEW HOME NOW HAS A NUMBER OF MACHINES SET UP, SO TESTING ANSWERS TO THE TRIVIA IS EVEN EASIER. ╔ AM STILL TRYING TO GET THE OLD ╨┼╘ MACHINES IN HOUSE, BUT THE OTHERS ARE HERE. 3) ╘HE ├OMMODORE ╫ORLD ╫IDE ╫EB ╨AGES (HTTP://WWW.MSEN.COM/▐BRAIN/CBMHOME.HTML) THAT ╔ MAINTAIN AND PLACE THE TRIVIA ON CAUGHT THE EYE OF ╒╙┴ ╘ODAY AND THE ╨HEONIX ╟AZETTE. ╔ WAS INTERVIEWED FOR BOTH ARTICLES. ╠OOK IN THE ╩UNE 20TH EDITION OF ╒╙┴ ╘ODAY FOR THE SEGMENT, AND POSSIBLY A PICTURE OF ╩IM ┬RAIN AND THE MACHINES HE USES TO CREATE THE TRIVIA. ┴S ALWAYS, ╔ WELCOME ANY QUESTIONS (WITH ANSWERS), AND ENCOURAGE PEOPLE TO ENTER THEIR RESPONSES TO THE TRIVIA, NOW AT #18. ╩IM. ╘HE FOLLOWING ARTICLE CONTAINS THE ANSWERS TO THE ─ECEMBER EDITION OF TRIVIA ($0├0 - $0├╞), THE QUESTIONS AND ANSWERS FOR ╩ANUARY ($0─0 - $0─╞), ╞EBRUARY ($0┼0 - $0┼╞), ═ARCH ($0╞0 - $0╞╞), ┴PRIL ($100 - $10╞), AND THE QUESTIONS FOR THE ═AY EDITION ($110 - $11╞). ┼NJOY THEM! ╚ERE ARE THE ANSWERS TO ├OMMODORE ╘RIVIA ┼DITION #13 FOR ─ECEMBER, 1994 ╤ $0├0) ╘HE EARLY 1541 DRIVES USED A MECHANISM DEVELOPED BY ______. ╬AME THE COMPANY. ┴ $0├0) ┴LPS. ╤ $0├1) ╧N LATER MODELS, ├OMMODORE SUBSEQUENTLY CHANGED MANUFACTURERS FOR THE 1541 DRIVE MECHANISM. ╬AME THE NEW MANUFACTURER. ┴ $0├1) ╬EWTRONICS. ╤ $0├2) ╫HAT IS THE MOST OBVIOUS DIFFERENCE(S). (╧NLY ONE DIFFERENCE IS NECESSARY) ┴ $0├2) ┴LPS: PUSH-TYPE LATCH, ROUND ╠┼─. ╬EWTRONICS: LEVER-TYPE LATCH, RECTANGULAR ╠┼─. ╤ $0├3) ╧N ├OMMODORE ┬┴╙╔├ ╓2.0, WHAT ANSWER DOES THE FOLLOWING GIVE: ╨╥╔╬╘ (╙╤╥(9)=3) ┴ $0├3) 0. ┴CCORDING TO ├OMMODORE ┬┴╙╔├, THE ANSWER SHOULD BBY -1, WHICH IS THE ┬┴╙╔├ VALUE OF ╘╥╒┼. ╚OWEVER, THE ABOVE EQUATION IS ╬╧╘ TRUE. ─OING ╨╥╔╬╘ ╙╤╥(9) YIELDS 3, BUT DOING ╨╥╔╬╘ (╙╤╥(9)-3) YIELDS 9.31322575┼-10 (├64). ╘HIS ANOMALY CAN BE ATTRIBUTED TO ROUNDOFF ERRORS IN THE FLOATING POINT MATH ROUTINES IN ├OMMODORE ┬┴╙╔├. ╤ $0├4) ╔N ├OMMODORE ┬┴╙╔├ (┴NY VERSION) WHAT DOES ┬ EQUAL AFTER THE FOLLOWING RUNS: ├=0:┬=├=0 ┴ $0├4) ┬ = -1. ╘HE SECOND STATEMENT IS THE ONE TO LOOK AT. ╘HE SECOND EQUALS SIGN IS TREATED AS A COMPARISON, WHILE THE FIRST IS TREATED AS A ASSIGNMENT. ┬ GETS SET TO THE OUTCOME OF THE COMPARISON, WHICH IS ╘╥╒┼ (-1). ╤ $0├5) ╘HE FIRST ╨┼╘ CASSETTE DECKS WERE ACTUALLY _______ BRAND CASSETTE PLAYERS, MODIFIED FOR THE ╨┼╘ COMPUTERS. ╬AME THE COMAPNY. ┴ $0├5) ╙ANYO. ╙PECIFICALLY, ═ODEL ═1540┴. ╫HAT A MODEL NUMBER! ╤ $0├6) ╔N ├OMMODORE ┬┴╙╔├ (┴NY VERSION), WHAT HAPPENS IF THE FOLLOWING PROGRAM IS RUN: 10 ╩=0 20 ╔╞ ╩=0 ╟╧ ╘╧ 40 30 ╨╥╔╬╘ "╩<>0" 40 ╨╥╔╬╘ "╩=0" ┴ $0├6) ╧N ┬┴╙╔├ 2.0 OR GREATER: ?╙┘╬╘┴╪ ┼╥╥╧╥ ╔╬ 20 ╥┼┴─┘. ╧N ┬┴╙╔├ 1.0: (FOUND ON THE ╨┼╘ 2001 SERIES) ╩=0 ╥┼┴─┘. ┬┴╙╔├ 1.0 TOTALLY IGNORED SPACES, SO LINE 20 BECAME "╔╞╩=0╟╧╘╧40". ╘HAT STATEMENT WOULD BE CORRECTLY PARSED, SICNE IT CONTAINS THE "╟╧╘╧" KEYWORD. ╚OWEVER, ON ┬┴╙╔├ 2.0 OR GREATER, SPACES WEREN'T IGNORED SO COMPLETELY, AND THE "╘╧" IN "╟╧ ╘╧" WOULD BE TOKENIZED SEPARATELY, SO SOME CODE WAS ADDED TO ┬┴╙╔├ TO CHECK TO "╟╧". ┴S THE CODE THAT ACCEPTS ╟╧╘╧ AS A SPECIAL CASE FOR ╘╚┼╬ AFTER AN ╔╞ STATEMENT WASN'T PATCHED THIS WAY, THE ABOVE FAILS, BECAUSE ╟╧ IS NOT A VALID KEYWORD AFTER ╔╞. ╘HE STATEMENT ╙╚╧╒╠─ WORK CORRECTLY, BUT DOES NOT BECAUSE OF THIS FAILURE TO FIX THE ╔╞ COMMAND PARSING. ╧N ┬┴╙╔├ 2.0 OR GREATER, SUBSTITUTING THE FOLLOWING LINE FOR LINE 20 WILL CAUSE THE PROGRAM TO WORK: 20 ╔╞ ╩=0 ╘╚┼╬ ╟╧ ╘╧ 40 ╤ $0├7) ╔N QUESTION $068, WE LEARNED HOW ╩ACK ╘RAMIEL FIRST HAPPENED UPON THE NAME "├╧══╧─╧╥┼". ┴CCORDING TO THE STORY, THOUGH, IN WHAT COUNTRY WAS HE IN WHEN HE FIRST SAW IT? ┴ $0├7) ╟ERMANY. ╤ $0├8) ╧N THE ├OMMODORE USER PORT CONNECTOR, HOW MANY EDGE CONTACTS ARE THERE? ┴ $0├8) 24. ╘WO ROWS OF 12 CONTACTS EACH. ╤ $0├9) ╧N MOST ├OMMODORE COMPUTERS, A LOGICAL ┬┴╙╔├ SCREEN LINE CAN CONTAIN UP TO 80 CHARACTERS. ╧N WHAT ├OMMODORE COMPUTER(S) IS THIS NOT TRUE? ┴ $0├9) ┴CCORDING TO ├OMMODORE DOCUMENTATION, A _PHYSICAL_ SCREEN LINE IS DEFINED AS ONE SCREEN LINE OF CHARACTERS. ┴ _LOGICAL_ SCREEN LINE IS DEFINED AS HOW MANY _PHYSICAL_ LINES CAN BE CHAINED TOGETHER TO CREATE A VALID ┬┴╙╔├ PROGRAM LINE. ╫ITH THAT IN MIND, MOST ├OMMODORE COMPUTERS CHOSE A _LOGICAL_ SCREEN LINE THAT WAS A MULTIPLE OF THE SCREEN WIDTH. ╘HIS WORKS FINE FOR 40 AND 80 COLUMN SCREENS, BUT WHAT DO WE DO WITH THE ╓╔├-20, WITH ITS 22 COLUMN SCREEN. ╙OLUTION: MAKE THE _LOGICAL_ LINE LENGTH EQUAL TO 4 _PHYSICAL_ LINES, OR 88 COLUMNS. ╫HEN THE ├OMMDORE 128 WAS INTRODUCED, THE NUMBER ROSE TO 160 CHARACTERS, WHICH IS 4 _PHYSICAL_ LINES IN 40 COLUMN MODE, OR 2 _PHYSICAL_ LINES IN 80 COLUMN MODE. ╚OWEVER, YOU CAN ONLY TAKE ADVANTAGE OF THIS IN 128 MODE. 64 MODE IS LIMITED TO 80 CHARACTERS. ╘O ADD TO ALL THIS CONFUSION, A VALID ┬┴╙╔├ PROGRAM LINE (IN MEMORY) CAN ACTUALLY BE 255 (TOKENIZED) CHARACTERS LONG, BUT CREATING SUCH A LONG LINE CANNOT BE DONE FROM THE BUILT-IN EDITOR IN DIRECT MODE. ╘HE ┴MIGA┬┴╙╔├, AVAILABLE ON THE ┴MIGA, ALSO DOES NOT HAVE THE 80 COLUMN LINE LIMIT. ╚OWEVER, THAT ┬┴╙╔├ IS ╙╧╧╧ MUCH DIFFERENT THAT ╔ AM NOT SURPRISED. ╘HE OLDER ├┬═ ┬┴╙╔├S, ON THE OTHER HAND, WERE ALL DERIVATIVES OF THE ORIGINAL ╠EVEL 1 ┬┴╙╔├ FOR THE ╨┼╘. ╤ $0├┴) ╔F A FILE IS SAVED TO A ├OMMODORE ─ISK ─RIVE WITH THE FOLLOWING CHARACTERS: CHR$(65);CHR$(160);CHR$(66), WHAT WILL THE DIRECTORY ENTRY LOOK LIKE? ┴ $0├┴) ╘HE FILENAME WILL SHOW UP AS "┴"┬, WITH THE '┬' SHOWING UP TO THE RIGHT OF THE '"' MARK. ╘HIS COULD BE USED TO MAKE PROGRAM LOADING EASIER. ┴ FILE THAT SHOWED UP AS "FILENAME",8,1 COULD BE LOADED BY SIMPLY HITTING SHIFT-RUN/STOP ON THAT LINE. ╤ $0├┬) ╫HAT IS THE MAXIMUM LENGTH (IN CHARACTERS) OF A ├┬═ DATASETTE FILENAME? ┴ $0├┬) ╥EFERENCES ╔ HAVE ON HAND SAY 128 CHARACTERS. ╚OWEVER, THE ACTUAL CODE ON THE 8032 AND THE ├64 ACTS AS THOUGH 187 CHARACTERS CAN ACTUALLY BE SENT (TAPE BUFFER-5 CONTROL BYTES = 192-5=187). ╘HE REFERENCES THAT CLAIM 128 CHARACTERS ARE ╬ICK ╚AMPSHIRE'S _╘HE ╓╔├ ╥EVEALED_ AND _╘HE ╨┼╘ ╥EVEALED_. ┴╬YONE CARE TO LAY THIS ONE TO REST? ╤ $0├├) ╚OW MANY KEYS ARE ON A STOCK ├OMMODORE 64 KEYBOARD? ┴ $0├├) 66 KEYS. ╘HIS IS THE SAME NUMBER AS FOUND ON THE ╓╔├-20 AND THE ├OMMODORE 16. ╤ $0├─) ├OMMODORE ┬┴╙╔├ USES KEYWORD "TOKENS" TO SAVE PROGRAM SPACE. ╘OKEN 129 BECOMES "╞╧╥". ╫HAT TWO TOKENS EXPAND TO INCLUDE A LEFT PARENTHESIS AS WELL AS A ┬┴╙╔├ KEYWORD? ┴ $0├─) ╘┴┬( (163) AND ╙╨├( (166). ╤ $0├┼) ╘HERE ARE 6 WIRES IN THE ├OMMODORE SERIAL BUS. ╬AME THE 6 WIRES. ┴ $0├┼) 1) ╙ERIAL /╙╥╤╔╬ 2) ╟╬─ 3) ╙ERIAL ┴╘╬ ╔╬/╧╒╘ 4) ╙ERIAL ├╠╦ ╔╬/╧╒╘ 5) ╙ERIAL ─┴╘┴ ╔╬/╧╒╘ 6) /╥┼╙┼╘ ╤ $0├╞) ╧N THE ├OMMODORE DATASETTE CONNECTOR, HOW MANY LOGICAL CONNECTIONS ARE THERE? ┴ $0├╞) 6. ╧PPOSING PINS ON THE CONNECTOR ARE HOOKED TOGETHER ELECTRICALLY. ╚ERE ARE THE ANSWERS TO ├OMMODORE ╘RIVIA ┼DITION #14 FOR ╩ANUARY, 1995 ╤ $0─0) ╚OW MANY KEYS WERE THERE ON THE "ORIGINAL" ╨┼╘ AND WHAT WAS SPECIAL ABOUT THEM? ┴ $0─0) THE ORIGINAL ╨┼╘ HAD 73 CALCULATOR-STYLE KEYS THAT WERE LAID OUT IN A RECTANGULAR MATRIX, NOT TYPEWRITER-STYLE. ╤ $0─1) ╚OW DO YOU PRODUCE THE "HIDDEN" MESSAGE(S) ON THE ├OMMODORE 128? ┴ $0─1) ╙┘╙ 32800,123,45,6. ╘HE SCREEN WILL CLEAR, AND THE SOFTWARE AND HARDWARE DEVELOPERS ON THE 128 PROJECT WILL BE NAMED. ╘HE EXACT TEXT IS AS FOLLOWS: [╥╓╙] ┬ROUGHT TO YOU BY... ╙OFTWARE: ╞RED ┬OWEN ╘ERRY ╥YAN ╓ON ┼RTWINE ╚ERDWARE: ┬IL ╚ERD ─AVE ╚AYNIE ╞RANK ╨ALAIA [╥╓╙]╠INK ARMS,DON'T MAKE THEM. ╤ $0─2) ╚OW MUCH MEMORY DID THE "ORIGINAL" ╨┼╘ SHOW ON BOOTUP? ┴ $0─2) ╘HE "ORIGINAL" ╨┼╘ CAME IN TWO CONFIGURATIONS, 4╦ AND 8╦, SO: ╘HE ╨┼╘ 2001-4 HAD 3071 BYTES. ╘HE ╨┼╘ 2001-8 HAD 7167 BYTES. ╤ $0─3) ╫E ALL KNOW THE "REBOOT" SYS FOR THE 64 IS SYS 64738, BUT WHO KNOWS THE SAME SYS LOCATION TO REBOOT THE ├┬═ 8032? ┴ $0─3) SYS 64790 ╤ $0─4) ╫HICH COMPUTER(S) BEEPED AT BOOTUP? (═AY BE MORE THAN ONE, BUT ONLY ONE REQUIRED) ┴ $0─4) ╔ KNOW SOME OF THESE ARE CORECT, BUT THE SHEER SIZE OF THE LIST PREVENTS ME FROM CHECKING THEM ┴╠╠ OUT. ╞┴╘ 40╪╪ SERIES 80╪╪ SERIES ╨├-10 (╔ SUSPECT A NUMBER OF ╔┬═ CLONES DID, AND THESE THINGS HAVE NO CONSISTENT NAMING CONVENTION ACROSS COUNTRY BOUNDARIES.) ╨├-20 ┴MIGA 1000 ╙╨9000 (╙UPER╨┼╘) ╤ $0─5) ╚OW MUCH MEMORY DID THE ├┬═ 8032 SHOW ON BOOTUP? ┴ $0─5) 31743 BYTES. ╤ $0─6) ├ERTAIN ├OMMODORE COMPUTERS PROVIDED EMTPY ┼╨╥╧═ SOCKETS ON THE MOTHERBOARD. ╟IVE ME THE NUMBER OF EMPTY SOCKETS ON THE FOLLOWING MACHINES: A) ├┬═ 30╪╪. B) ├┬═ 8╪╪╪. C) ├┬═ ├128. D) ╨LUS/4. ┴ $0─6) A) 3 SOCKETS. B) 2 SOCKETS. C) 1 SOCKET. D) 1 SOCKET. ╤ $0─7) ╔N ╟ERMANY, THE ├┬═ 8032 CAME WITH A 4K┬ ┼╨╥╧═ FOR THE ┼╪╪╪ AREA, WHILE THE ╒╙ VERSION ONLY HAD A 2K┬ ┼╨╥╧═. ╫HY? ┴ $0─7) ╘HE ╟ERMAN VERSION HAD ADDITIONAL KEYBAORD DRIVERS FOR UMLAUT CHARACTERS AND DEAD KEYS. ╤ $0─8) ╫HO PUBLISHED THE FIRST ╨┼╘ MEMORY MAP IN THE "╨┼╘ ╟AZETTE"? ┴ $0─8) ╬ONE OTHER THAN THE INFAMOUS ╩IM ┬UTTERFIELD. ╤ $0─9) ╫HICH IS FASTER TO MOVE THE SURSOR ON A ╨┼╘/├┬═ OR ├64: ╙┘╙ OR ╨╥╔╬╘? ┴ $0─9) ╨╥╔╬╘ IS FASTER, SINCE THE SYS APPROACH MUST PROCESS THE POKES BEFORE THE SYS, WHICH ARE VERY SLOW. ╤ $0─┴) ╧N THE ┴MIGA 1000, WHERE ARE THE SIGNATURES OF THE FIRST ┴MIGA DEVELOPERS LOCATED? ┴ $0─┴) ╔NSIDE THE TOP CASE OF THE ┴MIGA (1000). ╘HERE IS AN INTERESTING FOOTNOTE TO THIS QUESTION. ╔T SEEMS THAT AT LEAST SOME ORIGINAL ┴MIGA MACHINES WERE LABELED AS ┴MIGA (WITH NU NUMBER). ╘HEN, AT SOME LATER POINT, THE NUMBER WAS ADDED. ╔N ADDITION, ├OMMODORE PRODUCED SOME ┴MIGA 1000 MACHINES WITHOUT THE SIGNATURES, BUT MOST HAD THE TELLTALE HANDWRITING ON THE INSIDE OF THE CASE. ╤ $0─┬) ╧N THE 6502, WHAT DOES THE ACCUMULATOR CONTAIN AFTER THE FOLLOWING IS EXECUTED: LDA #$AA SED ADC #01 ┴ $0─┬) ┴SSUME CARRY WAS CLEAR. ╔F SO, THEN $11 IS THE CORRECT ANSWER. ╤ $0─├) ╫HAT IS THE MODEL NUMBER OF THE ╒╙ ╬╘╙├ ╓╔├-╔╔ CHIP? ┴ $0─├) ╔TS FIRST NUMBER WAS 6567, AND THAT IS THE NUMBER MOST PEOPLE KNOW IT BY, BUT ├OMMODORE PRODUCED A ╓╔├-╔╔ USING A NEW MANUFACTURING PROCESS THAT WAS NUMBERED THE 8562. ╤ $0──) ╫HAT IS THE ┼UROPEAN ╨┴╠ ╓╔├-╔╔ CHIP'S MODEL NUMBER? (╬OT SURE IF THAT'S ITS RIGHTFUL TERM, BUT ╔ HOPE YOU UNDERSTAND). ┴ $0──) ╙AME HERE. ╘HE PART NUMBER 6569 IS THE MOST REMEMBERED NUMBER, BUT AN 8565 WILL WORK AS WELL. ╤ $0─┼) ┴SSUME YOU HAVE TWO COMPUTERS, ONE WITH EACH OF THE ABOVE CHIPS INSIDE. ╫HICH CHIP DRAWS MORE PIXELS ON THE SCREEN PER SECOND? ┴ $0─┼) ╬OTE, FOR THE PURPOSES OF THE CALCULATION ╔ AM PERFORMING, "PIXELS" REFERS TO PICTURE ELEMENTS THAT CAN BE ADDDRESS AND MODIFIED USING NORMAL ╓╔├ MODES, SO THERE ARE 320*200 "PIXELS" ON BOTH THE ╨┴╠ AND ╬╘╙├ SCREENS. (╔ PROBABLY SHOULD HAVE STATED THIS, BUT IT IS TOO LATE NOW.) ┴LSO, THE SCREEN REFRESH RATES USED IN THE CALCULATIONS ARE THOSE DEFINED BY THE RESPECTIVE TELEVISION STANDARDS (60╚Z ╒.╙., 50╚Z ┼UROPEAN), EVEN THOUGH THE ACTUAL FREQUENCIES ARE OFF BY A SMALL PERCENTAGE. (FOR EXAMPLE, THE ACTUAL 50╚Z REFRESH RATE ON ┼UROPEAN ╓╔├-╔╔ CHIPS WAS CALCULATES AS 50.124567╚Z BY ┴NDREAS ┬OOSE) ╙O, THE ╨┴╠ DRAWS 320*200*50 PIXELS PER SECOND = 3200000 PIXELS/S ╬╘╙├ DRAWS 320*200*60 PIXELS PER SECOND = 3840000 PIXLES/S ╬OW, SOME PEOPLE THOUGHT ╔ MEANT THE WHOLE SCREEN, NOT JUST THE DISPLAY AREA PROVIDED BY THE ╓╔├-╔╔ CHIP. ╫ELL, ╔ AM NOT SURE EXACTLY YOU CALCULATE PIXELS ON A SCREEN, SINCE THE NUMBERS COULD VARY FROM DISPLAY TO DISPLAY, BUT IF WE MEASURE IN SCANLINES: ╨┴╠ = 312 SCANLINES * 50 = 15600 SCANLINES/S ╬╘╙├ = 262 SCANLINES * 60 = 15720 SCANLINES/S ╘HE ╬╘╙├ MACHINES WINS BOTH WAYS. ╤ $0─╞) ╔N ├OMMODORE ┬┴╙╔├, WHICH STATEMENT EXECUTES FASTER: A = 2--2 OR A = 2+2 ┴ $0─╞) B IS THE CORRECT ANSWER, AND THERE ARE A COUPLE OF REASONS WHY: 1) 2--2 TAKES LONGER TO PARSE IN THE ┬┴╙╔├ INTERPRETER. 2) ├OMMODORE ┬┴╙╔├ SUBTRACTS BY COMPLEMENTING THE SIGN OF THE SECOND NUMBER AND ADDING. ╘HIS INCURS EXTRA TIME. ╘HERE ARE EVEN MORE SUBTLE ONES, BUT ╔ LEAVE THEM AS AN EXERCISE FOR THE READER. ╙END ME YOUR REASON WHY. ╚ERE ARE THE ANSWERS TO ├OMMODORE ╘RIVIA ┼DITION #15 FOR ╞EBRUARY, 1995 ╤ $0┼0) ╫HAT IS THE DIFFERENCE(S) BETWEEN THE ╬EWTRONICS 1541 AND THE 1541├? (ONLY ONE DIFFERENCE IS NEEDED) ┴ $0┼0) (╟EORGE ╨AGE, A NOTED AUTHORITY ON ├┬═ ─RIVES, INDICATED THAT ├OMMODORE MADE THIS A TOUGH QUESTION TO ANSWER.) ┬Y THE TIME THE 1541├ WAS INTRODUCED, ├OMMODORE THREW A NUMBER OF DRIVES TOGETHER AND CALLED THEM 1541├S. ╘HE THEORETICAL 1541├ EXHIBITED THE FOLLOWING FEATURES: ╬O HEAD BANGING, AND OTHER PROBLEMS FIXED BY MODIFIED ╥╧═S. ├ASE COLOR MATCHES ├64├ AND ├128 COMPUTERS. ╤ $0┼1) ╫HAT HAPPENS WHEN YOU TYPE 35072121 IN DIRECT MODE ON THE ├64 AND HIT RETURN? ┴ $0┼1) ╙IMPLE ANSWER: ═OST LIKELY, THE SCREEN CLEARS AND THE WORD ╥┼┴─┘. IS PRINTED AT SCREEN TOP. ╘HIS IS THE BEHAVIOR SEEN WHEN PRESSING ╥╒╬-╙╘╧╨/╥┼╙╘╧╥┼. ┴LTERNATELY, NOTHING COULD HAPPEN, OR THE COMPUTER COULD LOCK UP. ╔NVOLVED ANSWER: ╘HERE IS A BUG IN ┬┴╙╔├ 2.0. ┼ASILY FIXED, BUT DESTINED TO LIVE LIFE IMMORTAL. (LONG) ╘HE BUG IS IN THE ╨┼╘╙├╔╔ NUMBER TO BINARY CONVERSION ROUTINE AT $A69B (╠╔╬╟┼╘). ╘HE ROUTINE BASICALLY READS IN A CHARACTER FROM THE LINE, MULTIPLIES A PARTIAL RESULT BY 10 AND ADDS THE NEW CHARACTER TO THE PARTIAL RESULT. ╚ERE IS A CODE SNIPPET: A96A RTS A96B LDX #$00 ; ZERO OUT PARTIAL RESULT A96D STX $14 A96F STX $15 A971 BCS $A96A ; NOT A NUMBER, RETURN A973 SBC #$2F ; ╨┼╘╙├╔╔ TO BINARY A975 STA $07 A977 LDA $15 ; GET HI BYTE OR PARTIAL RESULT A979 STA $22 A97B CMP #$19 ; PARTIAL > 6399 A97D BCS $A953 ; YES, GOTO ERROR A97F LDA $14 ; LOAD LO BYTE OF RESULT A981 ASL ; LO*2 A982 ROL $22 ; HI*2 + C A984 ASL ; LO*2 A985 ROL $22 ; HI*2 + C A987 ADC $14 ; COMPLETE LO*5 A989 STA $14 A98B LDA $22 A98D ADC $15 ; COMPLETE HI*5 A98F STA $15 A991 ASL $14 ; LO*2 COMPLETE LO*10 A993 ROL $15 ; HI*2 COMPLETE HI*10 A995 LDA $14 A997 ADC $07 ; ADD NEW CHAR A999 STA $14 A99B BCC $A99F ; DID LO OVERFLOW? A99D INC $15 ; YES, INC HI A99F JSR $0073 ; GET NEXT CHAR A9A2 JMP $A971 ; GO THROUGH IT AGAIN. ╘HE PROBLEM IS AT $A97D. WHEN THE PARTIAL RESULT IS GREATER THAN 6399, (IF PARTIAL > 6399, THEN NEW PARTIAL RESULT WILL BE OVER 63999) THE ROUTINE NEEDS TO GET TO $AF08 TO PRINT AN ERROR, BUT CAN'T DUE TO BRANCH RESTRICTIONS. ╚OWEVER, A BRANCH THAT WILL GET THERE IS IN THE PRECEDING FUNCTION, WHICH HANDLES THE ╧╬ ╟╧╘╧/╟╧╙╒┬ KEYWORDS ($A94B, ╧╬╟╧╘╧). ╙O, THE ┬┴╙╔├ WRITERS JUST BRANCHED TO THE CODE IN ╧╬╟╧╘╧; SPECIFICALLY $A953: A94B JSR $B79E A94E PHA A94F CMP #$8D ; IS THE KEYWORD ╟╧╙╒┬ ($8D) A951 BEQ $A957 ; YES A953 CMP #$89 ; IS THE KEYWORD ╟╧╘╧ ($89) A955 BNE $A8E8 ; NO, PRINT ╙┘╬╘┴╪ ┼╥╥╧╥. A957 ... ; HANDLE ╧╬ ╟╧╘╧/╟╧╙╒┬ ╘HIS CODE IS CHECKING TO MAKE SURE THE ╧╬ (VAR) IS FOLLOWED WITH A ╟╧╘╧ OR ╟╧╙╒┬ KEYWORD. ╘HE ╠╔╬╟┼╘ ERROR HANDLER BRANCHES TO $A953, WHICH COMPARES .┴ (WHICH HOLDS HI BYTE OF PARTIAL RESULT) TO $89. ╬ORMALLY, THIS FAILS, AND THE NORMAL ╙┘╬╘┴╪ ┼╥╥╧╥ CODE IS REACHED THROUGH THE BRANCH TO $A8E8. ╚OWEVER, FOR PARTIAL RESULTS OF THE FORM $89╪╪, THE CHECK SUCCEEDS, AND ┬┴╙╔├ TRIES TO EXECUTE AN ╧╬ ╟╧╘╧/╟╧╙╒┬ CALL. ┬Y THE WAY, IT IS NO COINCIDENCE THAT THIS ERROR OCCURS ON 35072121, SINCE ONE OF THE PARTIAL RESULTS IS $8900 (HI BYTE IS $89). ╔N FACT, 350721 WILL ACHIEVE THE SAME RESULT. ╔F THE CHECK SUCCEEDS, THE CODE LIMPS ALONG UNTIL $A96A: A969 PLA ; COMPLEMENT TO $A94E A96A RTS ; RETURN ┬UT WE NEVER EXECUTED $A94E, THE PUSH, SO THE STACK IS NOW MESSED UP. ╙INCE THE STACK HELD $9E, $79, $A5 BEFORE THE ╨╠┴, (╘HE STACK COULD HOLD OTHER VALUES, BUT ╔ ALWAYS SAW THESE) THE ╥╘╙ GETS ADDRESS $A579 TO RETURN TO, WHICH USUALLY HOLDS A ┬╥╦ OPCODE. ╘HE BREAK HANDLER IS INVOKED, AND THE SCREEN CLEARS WITH THE ╥┼┴─┘. AT THE TOP. ╬OW, THE ┬┴╙╔├ 2.0 AUTHORS WERE JUSTIFIED IN REUSING THE ERROR HANDLER CODE IN ╧╬╟╧╘╧ FOR ╠╔╬╟┼╘, BUT THEY CALCULATED THE BRANCH OFFSET WRONG, ACCORDING TO MY TESTS. ╔F YOU HAVE THE ╠╔╬╟┼╘ ERROR HANDLER BRANCH TO $A955, ALL THESE TROUBLES DISAPPEAR. ┘OU CAN VERIFY THIS PROCEDURE WITH THE FOLLOWING ┬┴╙╔├ PROGRAM ON A 64: 10 FOR T=57344 TO 65535:POKE T,PEEK(T):NEXT 20 FOR T=40960 TO 49151:POKE T,PEEK(T):NEXT 30 POKE 43390, 214 40 POKE 1, PEEK(1) AND 254 ╩UST TO BE COMPLETE, THIS ERROR OCCURS WHEN A 6 DIGIT OR GREATER LINE NUMBER IS ENTERED AND THE FIRST 6 DIGITS INDICATE A NUMBER IN THE RANGE 35072-35327 ($8900-$89FF). ┴LSO, IT APPEARS THE ERROR OCCURS ON THE ╓╔├-20, BUT ╔ DIDN'T COMPLETELY VERIFY IT. ╔T WOULD BE INTERESTING TO NOTE IF THE ERROR IS FOUND ON ALL VERSION OF ├┬═ ┬┴╙╔├. ╫HEW, WHAT A MOUTHFUL. ╤ $0┼2) ╔F A ╙╔─ CHIP IS PRODUCING A "SAWTOOTH WAVEFORM", DOES THE WAVEFORM LOOK LIKE: A) "/▄/▄/▄/▄" OR B) "▄\▄\▄\▄\" ? ┴ $0┼2) A IS THE CORRECT ANSWER. ╤ $0┼3) ╧N ┬┴╙╔├ 2.0, WHAT SPECIAL PRECAUTION(S) MUST ONE TAKE WHEN WORKING WITH RELATIVE FILES? (ONLY ONE IS NEEDED) ┴ $0┼3) ┬ECAUSE ┬┴╙╔├ 2.0 DOESN'T HANDLE POSITIONING IN RELATIVE FILES QUITE RIGHT, ONE MUST POSITION THE RELATIVE FILE POINTER BEFORE ┴╬─ ┴╞╘┼╥ A READ OR WRITE TO A RELATIVE FILE. ╤ $0┼4) ╫HAT INCOMPATIBILITY EXISTED BETWEEN ├128 ╥EV. 0 ╥╧═╙ AND THE ╥┼╒? ┴ $0┼4) ╧╦, ╔ ADMIT IT. ╔ PLACED THIS ANSWER AND ITS DISCUSSION SOMEWHERE IN MY STORE OF INFORMATION, AND IT MUST HAVE FALLEN BEHIND THE CABINET, BECAUSE ╔ CANNOT FIND IT. ╔ WILL POST AN ANSWER TO THIS AS SOON AS ╔ CAN FIND IT, BUT THE ANSWERS REALLY MUST GO OUT, AS THEY HAVE BEEN HELD UP LONG ENOUGH. ╤ $0┼5) ╫HAT CAN TRIGGER AN ╬═╔ INTERRUPT? (COUNT ALL SOURCES ON ONE CHIP AS ONE) ┴ $0┼5) ╘HE FOLLOWING SOURCES CAN TRIGGER AN ╬═╔ INTERRUPT: 1) ╘HE EXPANSION PORT. 2) ├╔┴ #2. 3) ╘HE ╥┼╙╘╧╥┼ KEY. ╤ $0┼6) ╫HAT CAN TRIGGER AN ╔╥╤ INTERRUPT? (COUNT ALL SOURCES ON ONE CHIP AS ONE) ┴ $0┼6) ╘HE FOLLOWING SOURCES CAN TRIGGER AN ╔╥╤ INTERRUPT: 1) ╘HE ╓╔├-╔╔ CHIP. 2) ├╔┴ #1. 3) ╘HE EXPANSION PORT. ╤ $0┼7) ╫HERE IS THE ╥╧═ IN A 1541 LOCATED IN THE 64╦ MEMORY MAP? ┴ $0┼7) ╘HE ╥╧═ IS LOCATED FROM $├000 TO $╞╞╞╞, YET THE ╥╧═ CODE DOES NOT BEGIN UNTIL $├100. ╤ $0┼8) ╫HICH ╓╔┴ ON THE 1541 IS HOOKED TO THE READ/WRITE HEAD? ┴ $0┼8) ╓╔┴ #2, LOCATED IN MEMORY FROM $1├00 TO $1├0┼. ╤ $0┼9) ╔N THE ├OMMODORE ─╧╙, WHAT BIT IN THE FILE TYPE BYTE DENOTES A "LOCKED" FILE? ┴ $0┼9) BIT 6. ╤ $0┼┴) ╔F FILES ARE "LOCKED" UNDER ├OMMODORE ─╧╙, UNDER WHAT CONDITION(S) MAY THE FILE BE CHANGED? ┴ $0┼┴) ─EPENDING ON THE FILE, THE FOLLOWING OPERATIONS CAN BE DONE ON A LOCKED FILE: 1) ╥ENAME WILL CHANGE FILE NAME, ALTHOUGH NOT CONTENTS OF FILE. 2) ╥ANDOM ACCESS CAN BE USED TO ALTER FILE. 3) ╞ORMATTING THE DISK WILL ALTER THE FILE. (DUH!) 4) ╙AVE-WITH-REPLACE (@0:) WILL REPLACE FILE AND UNLOCK IT. 5) ╧PENING FILE IN APPEND MODE WILL ALLOW IT TO BE CHANGED, AND UNLOCK IT. 6) ╧PENING A RELATIVE FILE AND ADDING OR CHANGING A RECORD WILL SUCCEED AND UNLOCK FILE. ╤ $0┼┬) ╚OW BIG CAN A PROGRAM FILE BE ON A 1541 OR SIMILAR? ┴ $0┼┬) ╘HE FILE CAN BE AS LARGE AS A SEQUENTIAL FILE, SINCE BOTH ARE STORED IN THE SAME WAY: 168656 BYTES. ╚OWEVER, SINCE A PROGRAM CONTAINS ITS LOAD ADDRESS AS BYTES 0 AND 1, THE LARGEST PROGRAM SIZE IS 168654 BYTES. ╤ $0┼├) ╒NDER ┬┴╙╔├ 2.0, HOW DOES ONE OPEN A RANDOM ACCESS FILE ON A DISK DRIVE? ┴ $0┼├) ╥ANDOM ACCESS (OR DIRECT ACCESS) FILES ARE A MISNOMER. ╫HAT YOU REALLY DOING IS OPENING THE DISK FOR READING AND WRITING. ┘OU NEED TWO OPEN COMMAND TO ACCESS A RANDOM FILE: (ASSUME DRIVE 8) OPEN 15,8,15 AND OPEN 1,8,4,"#1" WILL OPEN A RANDOM ACCESS FILE USING BUFFER 1. OPEN 1,8,4,"#" WILL OPEN A RANDOM ACCESS FILE USING THE FIRST AVAILABLE BUFFER ╬OW, BY USING ┬-╥, ┬-╫, ┬-┴ OR THEIR REPLACEMENTS, YOU CAN WRITE DATA TO SECTORS ON THE DISK. ╬OTE THAT ╥ANDOM ACCESS FILES ARE DIFFERENT FROM RELATIVE FILES. ╤ $0┼─) ┴ FILE THAT HAS A '*' IMMEDIATELY BEFORE THE FILETYPE IS CALLED A _________ FILE. ┴ $0┼─) A SPLAT FILE. ╘HIS IS ITS CORRECT TERM, BELIEVE IT OR NOT. ╤ $0┼┼) ╫E KNOW THE 1541 AND SIMILAR DRIVES HAVE 5 INTERNAL BUFFER AREAS, BUT HOW MANY DOES AN 8050 DRIVE HAVE? ┴ $0┼┼) ╙INCE THE 8050 HAS TWICE THE ON-BOARD ╥┴═ (4K┬), IT HAS 16 BUFFERS, BUT ONLY 13 ARE AVAILABLE. (┴LL ├┬═ DRIVES USE ONE BUFFER FOR ZERO-PAGE MEMORY, ONE FOR STACK MEMORY, AND ONE FOR TEMPORARY VARIABLES.) ╤ $0┼╞) ╧N A "SAVE-WITH-REPLACE", WHERE IS THE LOCATION OF THE FIRST TRACK AND SECTOR OF THE NEW COPY OF THE PROGRAM SAVED IN THE DIRECTORY ENTRY FOR THE OLD COPY? ┴ $0┼╞) ╘HE NEW FIRST TRACK IS STORED AT LOCATION 26, AND THE NEW FIRST SECTOR IS STORED AT LOCATION 27. ╘HESE VALUES ARE COPIED TO THEIR CORRECT LOCATIONS AFTER THE SAVE IS COMPLETED. ╚ERE ARE THE ANSWERS TO ├OMMODORE ╘RIVIA ┼DITION #16 FOR ═ARCH, 1995 ╤ $0╞0) ╫HAT SIZE MATRIX OF PIXELS COMPRISES A CHARACTER ON A ╨┼╘ 2001 COMPUTER? ┴ $0╞0) ╘HE MATRIX WAS 8 BY 8. ╤ $0╞1) ╚OW MANY BYTES DID THE OPENING SCREEN ON A ├┬═ 4016 SHOW AS AVAILABLE FOR USE BY ┬┴╙╔├? ┴ $0╞1) 15359 BYTES FREE. ╤ $0╞2) ╘HE CHARACTER SET THAT PRODUCES UPPERCASE LETTERS ON UNSHIFTED KEYS IS THE ________________ CHARACTER SET. ┴ $0╞2) "STANDARD MODE". ╤ $0╞3) ╘HE CHARACTER SET THAT PRODUCES LOWERCASE LETTERS ON UNSHIFTED KEYS IS THE ________________ CHARACTER SET. ┴ $0╞3) "ALTERNATE MODE" ╤ $0╞4) ╘O GET TO THE SET MENTIONED IN $╞2, WHAT CHARACTER CODE WOULD BE PRINTED TO THE SCREEN? ┴ $0╞4) CHR$(142) ╤ $0╞5) ╫HAT CHARACTER CODE WOULD ONE PRINT TO THE SCREEN TO INVOKE THE CHARARACTER SET IN $╞3? ┴ $0╞5) CHR$(14) ╤ $0╞6) ╔F ONE DOES ╠╔╙╘ 60-100, WILL LINE 100 GET "LISTED"? ┴ $0╞6) ┘ES. ╘HE ABOVE TRANSLATES AS: ╠╔╙╘ 60 THROUGH TO AND INCLUDING 100. ╤ $0╞7) ╘HE ABBREVIATION FOR THE ┬┴╙╔├ 4.0 COMMAND "├╧╠╠┼├╘" IS ________. ┴ $0╞7) CO╠. "├" "╧" "╙╚╔╞╘-╠". ╞OR THOSE WHO ARE INTERESTED, THE ├╧╠╠┼├╘ COMMAND IS ANALOGOUS TO THE ╓┴╠╔─┴╘┼ OPERATION. ╤ $0╞8) ╫HEN YOU USE A SUBSCRIPTED VARIABLE IN ┬┴╙╔├, HOW MANY ELEMENTS ARE CREATED BY DEFAULT IF NO ─╔═ STATEMENT IS ISSUED? ┴ $0╞8) 11 ELEMENTS. ┴(0) - ┴(10). ┴LMOST EVERYONE WHO HAS EVER PROGRAMMED IN ├OMMODORE ┬┴╙╔├ HAS SEEN THE "┬┴─ ╙╒┬╙├╥╔╨╘" ERROR WHEN THEY TRY TO USE THE 12TH ELEMENT IN A UN-─╔═ENSIONED ARRAY. ╤ $0╞9) ╚OW LARGE IS THE KEYBOARD BUFFER IN ├┬═ COMPUTERS? ┴ $0╞9) 10 BYTES. ╙INCE THIS AREA COULD BE ╨╧╦┼D TO, MANY BOOT PROGRAMS WOULD POKE CHARACTERS INTO THIS BUFFER TO SIMULATE KEYPRESSES. ╤ $0╞┴) ╧N THE ├OMMODORE 1581, HOW LARGE IS A PHYSICAL SECTOR IN BYTES? ┴ $0╞┴) ┴ PHYSICAL SECTOR IS 512 BYTES IN LENGTH. ╔NTERNALLY, THE 1581 CREATES 2 256 "LOGICAL" SECTORS IN A PHYSICAL SECTOR, TO MAINTAIN COMPATIBILITY WITH OLDER ├OMMODORE DRIVES. ╤ $0╞┬) ┘OU'LL FIND ┬┴╙╔├ 3.5 ON THE _____________ LINE OF ├┬═ COMPUTERS. ┴ $0╞┬) ╘HE ╪64 SERIES. ╘HAT INCLUDES THE ├OMMODORE 16, THE ├OMMODORE 116, AND THE ├OMMODORE ╨LUS/4. ╤ $0╞├) ╧N THE ├OMMODORE 1351 MOUSE, WHAT REGISTERS IN THE ├OMMODORE COMPUTER WOULD THE ╪ AND ┘ PROPORTIONAL INFORMATION BE READ FROM? ┴ $0╞├) ┼VEN THOUGH YOU ARE LOOKING FOR DIGITAL INFORMATION (HOW FAR THE MOUSE HAS TRAVELED SINCE THE LAST MOVEMENT IN A PARTICULAR AXIS), THE INFORMATION IS READ FROM THE "PADDLE" OR POTENTIOMETER (╨╧╘) REGISTERS. ╧N THE ├64, THE ╨╧╘ REGISTERS ARE PART OF THE ╙╔─ CHIP, AND ARE AT 54297 ($─419) FOR ╨╧╘╪, AND 54298 ($─41┴) FOR ╨╧╘┘. ╤ $0╞─) ╫HAT IS THE MAXIMUM SIZE OF A SEQUENTIAL FILE ON A 1581 DRIVE? ┴ $0╞─) 802640 BYTES. ╤ $0╞┼) ╫HAT FLAW EXISTS IN THE EARLY ├OMMODORE 1670 MODEMS? ┴ $0╞┼) ╫HEN THE 1670 MODEM WAS FIRST INTRODUCED, IT POWERED UP IN AUTO- ANSWER MODE, WHICH MEANS IT WOULD ANSWER INCOMING CALLS AFTER THE PHONG RANG. ┘OU COULD TURN THIS FEATURE OFF THROUGH SOFTWARE CONTROL, BUT IF THE POWER WAS RESET, THE MODEM WOULD ANSWER THE PHONE. ╙O MANY PEOPLE COMPLAINED TO ├OMMODORE THAT ├┬═ REVISED THE 1670 TO INCLUDE AN EXTRA ─╔╨ SWITCH THAT TURNED THIS FEATURE OFF. ╤ $0╞╞) ╫HAT IS THE MODEL NUMBER OF THE FIRST MODEM FOR THE ╓╔├ AND ├64? ┴ $0╞╞) ╘HE 1600 MANUAL DIAL/MANUAL ANSWER 0-300 BPS MODEM. ╘HE AUTHOR OWNS ONE, AND USED IT FOR MANY YEARS. ╘O OPERATE, YOU MUST USE A PHONE WITH A DETACHABLE HANDSET CORD. ┘OU DIALED THE NUMBER ON THE PHONE, WAITED FOR THE ANSWER, UNPLUGGED THE HANDSET, AND PLUGGED THE CORD INTO THE 1600. ┴ SWITCH TOGGLED BETWEEN USING ORIGINATE OR ANSWER FREQUENCIES. ╘HE 1600 WAS MANUFACTURED BY ┴NCHOR ┴UTOMATION FOR ├OMMODORE. (┴S AN ASIDE, THIS UNIT CLAIMED 300 BPS, BUT ╔ NEVER COULD GET 300 TO WORK WELL. ═OST OF MY TELECOMMUNICATIONS HAPPENED AT 150 BPS.) -------├OMMODORE ╘RIVIA ┼DITION #17 ╤UESTIONS AND ┴NSWERS (┬┼╟╔╬)-------- ╤ $100) ╧N THE ═╧╙ ╘ECHNOLOGY'S ╦╔═-1, HOW MANY KEYS WERE ON THE KEYPAD? ┴ $100) 23 KEYS. ╘HE KEYPAD HAS ROOM FOR 24, BUT ONE SPOT IS TAKEN BY A SWITCH THAT PUTS THE SYSTEM INTO SINGLE-STEP MODE. ╔NTERESTINGLY, SOME PICTURES HAVE THE SWITCH ON THE UPPER LEFT, SOME ON THE UPPER RIGHT. ╤ $101) ╘HE ╦╔═-1 KEYPAD HAD THE COMMON 0-9┴-╞ KEYS ON THE KEYPAD, BUT ALSO HAD SOME SPECIAL KEYS. ╬AME THEM. ┴ $101) ╟╧ (╟O) ┼XECUTES AN INSTRUCTION AND DISPLAYS THE ADDRESS OF NEXT, ╙╘ (╙TOP) ╙TOPS EXECUTION OF PROGRAM AND RETURN CONTROL TO MONITOR, ╥╙ (╥ESET), ┴─ (┴DDRESS) ┴DDRESS ENTRY MODE, ─┴ (─ATA) ─ATA ENTRY MODE, ╨├ (╨ROGRAM ├OUNTER) ─ISPLAYS AND RESTORES PROGRAM COUNTER TO VALUES IN ╨├╠ AND ╨├╚, + (╔NCREMENT) ╔NCREMENTS THE ADDRESS WITHOUT CHANGING THE ENTRY MODE. ╤ $102) ╘HE ╦╔═-1 WAS A SET OF MODULES THAT COULD BE PLUGGED TOGETHER TO EXPAND THE SYSTEM. ┼ACH MODULE HAD A MODEL NUMBER. ╫HAT WAS THE MODEL NUMBER OF THE ╦╔═-1 MOTHERBOARD? ┴ $102) ╘HE ╦╔═-4. ╤ $103) ╧N THE 1525 LINE OF PRINTERS, IF YOU WANTED TO CREATE THE FOLLOWING GRAPHIC, WHAT BYTES WOULD YOU SEND TO THE PRINTER AFTER TURNING ON GRAPHICS MODE? **** * * * * * * * * * * **** ┴ $103) ╔ GUESS ╔ SHOULD HAVE STIPULATED THAT THIS IS A BITMAP. ┴╙├╔╔ JUST HAS A FEW LIMITATIONS. ┴NYWAY, THE CORRECT BYTES TO SEND ARE: 255, 193, 193, 255. ┘OU GOT THESE BY ASSIGNING EACH BIT IN A COLUMN A VALUE, AND ADDING 128 TO THE RESULT FOR EACH COLUMN. ╤ $104) ╫HAT IS THE HORIZONTAL RESOLUTION OF THE 1525 LINE OF PRINTERS? ┴ $104) ├HARACTER RESOLUTION: 80 CHARS, OR 10 CHARS/INCH (CPI). ╟RAPHICS RESOLUTION: 480 DOTS, OR 60 DOTS/INCH (DPI). ╤ $105) ╧N ├OMMODORE DRIVES, EXPLAIN THE DIFFERENCE BETWEEN THE ┬-╥ COMMAND AND THE ╒1 COMMAND. ┴ $105) ╘HE TWO COMMANDS READ IN DATA FROM A DISK SECTOR. ╚OWEVER, THE ╒1 COMMAND ALWAYS READS A FULL SECTOR (255 BYTES). ╘HE ┬-╥ COMMAND READS THE NUMBER OF BYTES SPECIFIED IN THE FIRST BYTE OF THE SECTOR. ╔F THE FIRST BYTE IS A 15, ┬-╥ WILL READ 15 BYTES FROM THE SECTOR. (╞ROM THE 1581 MANUAL) ╤ $106) ╧N THE ├OMMODORE 1541 DRIVE, WHAT DOES THE ╒: COMMAND DO? ┴ $106) ╘HIS COMMAND HAS BEEN TRADITIONALLY USED TO RESET ├OMMODORE DRIVES, INCLUDING THE ├┬═ 1541. ╚OWEVER, SOME EARLY VERSIONS OF THE ─RIVE ─╧╙ DID NOT CORRECTLY HANDLE THIS COMMAND. ╔N THESE VERSIONS, THE DRIVE AND COMPUTER FAILED TO COMPLETE THE COMMAND TRANSACTION SUCCESSFULLY, AND WHAT LOOKED LIKE A HUNG MACHINE RESULTED. ├OMMODORE LATER FIXED THIS PROBLEM. ╔F ╒: SEEMS TO NOT WORK ON YOUR DRIVE, TRY ╒; INSTEAD. ╤ $107) ╫HAT DOES THE FIRST ROUTINE IN THE 1541 DRIVE ╥╧═ ACTUALLY DO? ┴ $107) ╘HE FUNCTION, CALLED ╙┼╘╠─┴ AND RESIDING AT $├100, TURNS ON THE DRIVE ACTIVE ╠┼─ FOR THE CURRENT DRIVE. ╘HE ROUTINE LOADS THE CURRENT DRIVE FROM $7╞ AND SETS BIT 3 OF ─╙╦├╬╘ ($1├00). ╤ $108) ╚OW MANY FILES WILL A 1581 DISK DRIVE HOLD? ┴ $108) 296 FILES. ╬OTE THAT IT IS NOT A MULTIPLE OF 144. ╤ $109) ├OMMODORE 1581 DRIVES HAVE A SPECIAL "AUTOBOOT" FEATURE THAT ENABLES THE DRIVE TO LOAD AND RUN A PROGRAM OFF A DISK UPON DRIVE BOOTUP. ╫HAT IS THE REQUIRED NAME OF THE FILE? ┴ $109) ├╧╨┘╥╔╟╚╘ ├┬═ 86 ╤ $10┴) ╫HAT FILETYPE MUST THE FILE MENTIONED IN $109 BE? ┴ $10┴) ╒╙╥. ╤ $10┬) ╘O POWER UP A 1351 MOUSE IN "JOYSTICK MODE", WHAT MUST THE USER DO? ┴ $10┬) ╔F ONE DEPRESSES THE RIGHT MOUSE BUTTON DURING POWER-UP, THE 1351 WILL BEHAVE JUST LIKE A JOYSTICK. ╤ $10├) ─ESCRIBE THE CONTENTS OF THE ╨╧╘╪ OR ╨╧╘┘ REGISTERS WHEN USING A 1351 MOUSE. ┴ $10├) ┼ACH REGISTER HOLDS THE SAME TYPE OF INFORMATION, JUST FOR A SEPARATE AXIS, SO WE WILL DESCRIBE JUST ONE REGISTER: ┬IT: ╞UNCTION 7 ─ON'T CARE 6-1 ═OUSE AXIS POSITION MOD 64. 0 ╬OISE ┬IT. (CHECK THIS BIT TO SEE WHETHER MOUSE HAS MOVED) ╤ $10─) ├OMMODORE COMPUTERS TYPICALLY USE MOST OF ZERO PAGE FOR TEMPORARY VARIABLES AND OTHER ITEMS. ╚OWEVER, BOTH THE ╓╔├-20 AND THE 64 RESERVE 4 BYTES FOR USER PROGRAMS THAT NEED ZERO PAGE MEMORY. ╫HERE ARE THESE LOCATIONS? ┴ $10─) $╞┬-$╞┼ (251-254). ╔ AM NOT SURE THESE WERE "RESERVED" FOR PROGRAMMERS AS MUCH AS THEY WERE JUST NOT UTILIZED BY THE ├┬═ PROGRAMMERS. ╤ $10┼) ╬AME THE 16 COLORS AVAILABLE ON THE 64. ┴ $10┼) ┬LACK ╫HITE ╥ED ├YAN (╠IGHT ┬LUE-╟REEN) ╨URPLE ╟REEN ┬LUE ┘ELLOW ╧RANGE ┬ROWN ╠IGHT ╥ED ─ARK ╟RAY (╟RAY 1) ═EDIUM ╟REY (╟RAY 2) ╠IGHT ╟REEN ╠IGHT ┬LUE ╠IGHT ╟RAY (╟RAY 3) ╤ $10╞) ┬OTH THE ╓╔├-20 AND THE ├64 EMULATE THE OPERATION OF THE 6551 ╒┴╥╘. ╚OW MANY "MOCK 6551" REGISTERS ARE MAPPED INTO THE MEMORY MAP? ┴ $10╞) 5, FROM $293-$297 (659-663). ╘HE REGISTER CONTENTS: $293 6551 ├ONTROL ╥EGISTER $294 6551 ├OMMAND ╥EGISTER $295-6 6551 ╒SER ─EFINED ┬AUD ╥ATE VALUE. $297 6551 ╙TATUS ╥EGISTER ------------├OMMODORE ╘RIVIA ┼DITION #18 ╤UESTIONS (┬┼╟╔╬)-------------- ╤ $110) ╫HAT IS THE NAME OF THE COMPANY THAT RECENTLY PURCHASED THE LIQUIDATED ├OMMODORE ASSETS? ╤ $111) ┴T ONE TIME, ├OMMODORE ATTEMPTED TO MANUFACTURE A DUAL DRIVE VERSION OF THE 1571 CALLED THE 1572. ╞OR WHAT TECHNICAL REASON DID IT UTIMATELY FAIL? ╤ $112) ╧VER WHAT COMPUTER SYSTEM DID A ╒SER ╟ROUP SUE ├OMMODORE AND WIN? ╤ $113) ╔N $103, THE QUESTION ASKED HOW TO CREATE A GRAPHIC OF A SMALL BOX ON THE 1525. ╔N THIS QUESRTION, WE HAVE MADE A DIFFERENT DESIGN. ╔F YOU WANTED TO CREATE THE FOLLOWING GRAPHIC USING INDIVIDUAL DOTS ON THE PRINTER, WHAT BYTES WOULD YOU SEND TO THE PRINTER AFTER TURNING ON GRAPHICS MODE? ** * * * *** * ** *** * * * * * ** ** * * * * ** ╤ $114) (╙OME ├65 QUESTIONS) ╚OW MANY ╙╔─ CHIPS DOES THE THE DEVELOPMENT ├OMMODORE 65 MACHINE CONTAIN? ╤ $115) ╫HAT ├╨╒ DOES THE ├OMMODORE 65 USE? ╤ $116) ╫HAT IS THE ALTERNATE NAME FOR THE ├OMMODORE 65? ╤ $117) ╚OW MANY PROCESSORS DOES THE INTERNAL 1581-COMPATIBLE DRIVE ON THE ├65 CONTAIN? ╤ $118) ╔N THE TRADITION OF NAMING CERTIAN ╔├S AFTER FAMOUS CARTOON CHARACTERS, ONE OF THE ╔├S IN THE ├65 IS NAMED AFTER A ╫ARNER ┬ROTHERS CARTOON CHARACTER. ╫HICH ONE? ╤ $119) ╫HAT VERSION OF ┬┴╙╔├ IS INCLUDED ON THE ├OMMODORE 65 IN ├65 MODE? ╤ $11┴) ╚OW MANY ╔/╧ PORTS DOES A ├OMMODORE 65 CONTAIN? ╤ $11┬) ╫HAT COMMON ├OMMODORE 64 ╔/╧ PORT DOES THE ├65 ╬╧╘ HAVE? ╤ $11├) ╚OW MANY FUNCTION KEYS ARE ON A ├OMMODORE 65? ╤ $11─) ╫HAT ├┬═ DISK DRIVE ─╧╙ WAS USED AS THE TEMPLATE FOR THE INTERNAL ├65 DRIVE ─╧╙? ╤ $11┼) ╫HAT RESOLUTION OF TEXT SCREEN DOES THE ├65 POWER UP IN? (╨LEASE GIVE ANSWERS IN CHARACTERS). ╤ $11╞) ╫HAT DISTINGUISHING NON-TEXTUAL CHARACTERISTIC IN THE ├65 IS NOT PRESENT IN OTHE ├OMMODORE 8-BIT COMPUTERS? ╘HE INFORMATION IN THIS BETWEEN THE LINES MARKED BY (┬┼╟╔╬) AND (┼╬─) IS COPYRIGHT 1995 BY ╩IM ┬RAIN. ╨ROVIDED THAT THE INFORMATION BETWEEN THE (┬┼╟╔╬) AND (┼╬─) LINES IS NOT CHANGED EXCEPT TO CORRECT TYPOGRAPHICAL ERRORS, THE SO MARKED COPYRIGHTED INFORMATION MAY BE REPRODUCED IN ITS ENTIRETY ON OTHER NETWORKS OR IN OTHER MEDIUMS. ╞OR MORE INFORMATION ABOUT USING THIS FILE, PLEASE CONTACT THE ADDRESS SHOWN BELOW. ╩IM ┬RAIN BRAIN@MAIL.MSEN.COM 602 ╬ORTH ╠EMEN ╞ENTON, ═╔ 48430 (810) 737-7300 X8528 ╙OME ARE EASY, SOME ARE HARD, TRY YOUR HAND AT: ├OMMODORE ╘RIVIA #18! ======================================================================== ┬╞╠╔ - ╬EW GRAPHICS MODES 2 BY ╨ASI '┴LBERT' ╧JALA <ALBERT@CS.TUT.FI> ╧NE DAY ╔ WAS WATCHING SOME DEMOS THAT USED LINECRUNCH ROUTINES FOR WHOLE-SCREEN MULTICOLOR-GRAPHICS UPSCROLLERS. ╔ ALREADY HAD MY THEORIES ABOUT HOW AND WHY LINECRUNCH WORKED, BUT BECAUSE ╔ HAD NOT USED IT ANYWHERE, THE DETAILS WERE A BIT VAGUE. ╔N FACT, ╔ HAVE MANY TIMES ACCIDENTALLY CREATED LINECRUNCH EFFECTS WHEN TRYING TO DO SOMETHING ELSE WITH $─011. ╨ROBABLY EVERY DEMO CODER HAS. ┬UT YOU LEARN BY DOING. ╔ HAD THE IDEA OF USING LINECRUNCH FOR ╞╠╔ INSTEAD OF A SIMPLE MULTICOLOR PICTURE AS IT ALWAYS SEEMED TO BE USED. ╚OWEVER, THIS HAS PROBABLY BEEN DONE BEFORE AND BECAUSE ╔ DON'T LIKE TO DO THINGS THAT HAVE BEEN DONE BEFORE, ╔ DECIDED TO USE LINECRUNCH TO SHOW A TWO-SCREEN-TALL ╞╠╔ PICTURE. _╠INECRUNCH ┬ASICS_ ╞OR THOSE NOT FAMILIAR WITH LINECRUNCH ROUTINES: LINECRUNCH IS USED TO SCROLL THE SCREEN ╒╨╫┴╥─╙ BY CONVINCING ╓╔├-╔╔ THAT IT HAS ALREADY SHOWED MORE CHARACTER ROWS THAN IT IN REALITY HAS SHOWN. ╙URPRISINGLY (OR THEN, MAYBE NOT :) THIS CONSISTS OF FIDDLING WITH $─011. ╘HE TIMING IS CRITICAL AS ALWAYS. ╠INECRUNCH WORKS BY SETTING $─011 EQUAL THE LINE BEFORE THE CURRENT LINE AND ╓╔├-╔╔ WILL HAPPILY THINK THAT IT IS TIME TO MOVE ON TO THE NEXT CHARACTER ROW - ADD 40 TO THE VIDEO MATRIX COUNTER, 320 TO THE GRAPHICS MEMORY COUNTER AND BE READY TO START A BAD LINE. ╧R, MAYBE '╬╧╘ TO GO BACK TO THE CURRENT ROW' WOULD BE A MORE SUITABLE DESCRIPTION. (╨ROGRAMMING ╓╔├-╔╔ IS SLOWLY BECOMING A SCIENCE.) ╘HE REQUIRED TIMING ALSO DOES NOT CAUSE BAD LINES SO THAT YOU CAN SKIP ANOTHER LINE IMMEDIATELY ON THE SUCCESSIVE LINE. ╔N ADDITION, LINES CAN BE SKIPPED ONLY AFTER THE FIRST CHARACTER ROW AND HALF OF THE SECOND CHARACTER ROW HAVE BEEN DISPLAYED. ╘HIS HAS SOMETHING TO DO WITH THE WAY ╓╔├-╔╔ DECIDES WHEN THERE IS A BAD LINE. ┬ECAUSE LINECRUNCH CAUSES ╓╔├-╔╔ TO SKIP ROWS, IT WILL RUN OUT OF VIDEO MATRIX AND COLOR MEMORY (AND GRAPHICS MEMORY) BEFORE REACHING THE END OF THE SCREEN. ╚OWEVER, ╓╔├-╔╔ DOES NOT STOP DISPLAYING THE GRAPHICS NOR DOES IT RESET THE INTERNAL COUNTERS. ╘HE COUNTERS KEEP ON RUNNING AND WRAP AROUND INSTEAD. ╬ORMALLY, WHEN ╓╔├-╔╔ IS DISPLAYING THE LAST CHARACTER ROW, IT IS SHOWING THE MEMORY FROM OFFSETS $3C0 TO $3E7. ╔F ╓╔├-╔╔ HAS SKIPPED ONE CHARACTER ROW, IT IS DISPLAYING FROM $3E8 TO $40F INSTEAD. ┬UT, THERE ARE ONLY 10 BITS FOR THE VIDEO MATRIX COUNTER (0..1023), SO IT WRAPS AROUND TO ZERO AFTER $3FF. ╘HIS MEANS THAT THE BEGINNING OF THE VIDEO MATRIX IS DISPLAYED AT THE BOTTOM OF THE SCREEN. ╘HE CHARACTER ROWS BECOME SHIFTED BY 24 CHARACTER POSITIONS TO THE RIGHT BECAUSE THERE WERE ORIGINALLY 24 UNUSED MEMORY LOCATIONS AT THE END OF THE MEMORY (1000..1023). (╘O BE HONEST, SPRITE IMAGE POINTERS ARE NOT UNUSED MEMORY, BUT THEY ARE NOT USED WITH NORMAL ╞╠╔.) ____________________ ____________________ ▄ABCDEFGHIJKLMNOPQRST▄ ▄ABCDEFGHIJKLMNOPQRST▄ ▄ ▄ ▄--------------------▄ <- ╙KIPPED ROW : : : : : : : : : : : : ▄ ▄ ▄NORMALLY LAST LINE ▄ ▄NORMALLY LAST LINE ▄ ▄╪╪╪╪╪╪╪╪┌┌┌┌ABCDEFGH▄ └--------------------' └--------------------' ╪ = UNUSED MEM (1000..1015) ┌ = SPRITE POINTERS (1016..1023) ╞IGURE 1: ╠INECRUNCH ╘HE SAME THING HAPPENS FOR COLOR MEMORY BECAUSE IT USES THE SAME COUNTER FOR ADDRESSING THE MEMORY (IN FACT, COLOR MEMORY ACCESS AND CHARACTER DATA ACCESS ARE PERFORMED SIMULTANEOSLY, 12 BITS AT A TIME). ╘HE GRAPHICS MEMORY BEHAVES THE SAME WAY, EXCEPT THAT THE COUNTER HAS THREE BITS MORE AND IT COUNTS AT EIGHT TIMES THE SPEED, SO THAT IT WRAPS AT THE EXACT SAME TIME AS THE OTHER COUNTER. ╘HE FIRST CHARACTER ROW CAN'T BE USED FOR LINECRUNCH AND THE SECOND ONE IS ALSO LOST IN THE PROCESS. ╘HE FIRST USABLE LINE TO DISPLAY IS THE THIRD CHARACTER ROW. ╚OWEVER, THOSE TWO LOST ROWS CAN STILL BE USED AS AN EXTENSION AT THE END OF THE FIRST SCREEN. ┘OU MUST NOTICE, HOWEVER, THAT THE ALIGNMENT HAS BEEN CHANGED. ┴FTER THESE TWO ROWS HAVE BEEN DISPLAYED, THE VIDEO BANK IS SWITCHED TO GET NEW FRESH DATA ON THE SCREEN. _┬ACK TO ┬╞╠╔_ ╫RAPPED DATA IS NOTHING DIFFICULT TO WORK WITH. ╔T IS JUST THE MATTER OF WRITING THE RIGHT CONVERSION PROGRAM. ┴LSO, THE NORMAL ╞╠╔ ROUTINE CAN BE USED, WE JUST HAVE TO MAKE SURE ╓╔├ ALWAYS HAS THE RIGHT BANK VISIBLE - SIMPLE ╠─┴ BANK,X:STA $──00 CAN ACCOMPLISH THAT. ╘HE MORE DIFFICULT ASPECT IS TO MAKE THE DISPLAY FREELY LOCATABLE. ╫E HAVE 32 KILOBYTES OF GRAPHICS DATA, THIS IS THE MAIN REASON WE CAN'T EVEN THINK ABOUT USING COPYING. ╠INECRUNCH COMBINED WITH THE BAD LINE DELAYING TECHNIQUE WILL DO THE JOB MUCH MORE NICELY. ╞IGURE 2 SHOWS THE PRINCIPLES. ╘O MAKE THINGS SIMPLER ╔ HAVE CHOSEN LOCATION 0 TO MEAN THAT THE TOP OF THE PICTURE IS VISIBLE, 1 MEANS THAT THE PICTURE IS SCROLLED ONE LINE UPWARDS AND SO ON. ╫E CAN SEE THAT LINECRUNCH IS NOT USED AT ALL FOR THE LOCATION 0. ╘O MAKE THE PICTURE START AT THE SAME POINT WHETHER LINECRUNCH HAS CRUNCHED LINES OR NOT WE COMPENSATE THE NON-LOST RASTER LINES BY DELAYING THE NEXT BAD LINE. ╫HEN THE LOCATION IS N*8 (N=0,1,2..), THE SUM OF THE LINECRUNCHED AND DELAYED LINES IS CONSTANT - THE GRAPHICS DISPLAY ALWAYS STARTS AT THE SAME POINT. ╘HEN HOW DO WE DEAL WITH THE LOCATION VALUES THAT ARE NOT EVENLY DIVIDABLE BY EIGHT ? ╬OW, LETS ASSUME THAT THE LOCATION IS ╠, AND WE HAVE ├, WHICH IS THE LOCATION DIVIDED BY EIGHT (├ = ╠/8), AND ╥, WHICH IS THE REMAINDER (╥ = ╠%8). ╘O MAKE THE PICTURE SCROLL TO THE RIGHT POSITION, WE NEED TO DELAY THE BAD LINE LESS THAN BEFORE - ╥ LINES LESS FOR LOCATION ╠ THAN FOR LOCATION ├*8. ┼.G. FOR LOCATION 2 WE DELAY THE BAD LINE TWO LINES LESS THAN FOR LOCATION 0. ╘HIS ALSO SHOWS THAT WE NEED 7 LINES MORE THAN IS NEEDED FOR TO COMPENSATE FOR THE LINECRUNCH. ─ETERMINING THE NUMBER OF LINECRUNCH LINES IS A RECURSIVE PROCESS, BECAUSE WHEN YOU USE MORE LINECRUNCH LINES, THAT DECREASES THE NUMBER OF LINES YOU HAVE AVAILABLE FOR THE DISPLAY AND YOU NEED BIGGER RANGE FOR THE LOCATION VALUE. ╘HE LINECRUNCH CAN BE STARTED AFTER 12 LINES, AND WE NEED AT LEAST 7 LINES TO USE THE SOFT Y-SCROLL. ╘HIS MAKES 181 LINES AVAILABLE FOR THE DISPLAY ORIGINALLY. ┬ECAUSE WE NEED TO SHOW 400 LINES OF GRAPHICS, WE WOULD NEED (400-181)/8=28 LINECRUNCH LINES. ╚OWEVER, THIS IN TURN REDUCES THE NUMBER OF LINES WE HAVE FOR GRAPHICS TO 181-28=153 AND WE NEED (400-153)/8=31 LINECRUNCH LINES. ┴GAIN, 181-31 IS 150. ╫E GET (400-150)/8=32 AND THERE IT FINALLY CONVERGES AND WE HAVE 149 LINES FOR GRAPHICS, WHICH MAKES LOCATION VALUES 0..251 VALID. ╠OCATION 0 1 2 .. 8 9 .. 251 ___________________.. ___________.. ________ ___________________.. ___________.. ________ ╠INECRUNCH -------------------.. ___________.. ^ ^ ^ ▄ ▄ ▄ ^ ^ ▄ ▄ ▄ ▄ ▄ ┬AD LINE DELAYED▄ ▄ ▄ ▄ ▄ ▄ ▄ ▄ ▄ ▄ ======== ▄ ▄ V ▄ ▄ 244 ▄ V ___.. ▄ V : V ________0 V ___.. : ╟FX ┼NABLED ________0_______1__.. ________8__.. 250_____ 0 1 2 8 9 251 1 2 3 9 10 252 2 3 4 10 11 253 3 4 5 11 12 254 4 5 6 12 13 255 5 6 7 13 14 256 6 7 8 14 15 257 7 8 9 15 16 258 : : : : : : : : : : : : 148 149 150.. 156 157.. 399 ╞IGURE 2: ╠INECRUNCH AND ─═┴ DELAY IN ┬╞╠╔ (╟RAPHICS LINES NOT IN SCALE) _├LIPPING ADDED_ ╬OW WE CAN SCROLL THE PICTURE TO ANY LOCATION WE WANT, BUT THE TOP OF THE PICTURE IS NOT CLIPPED AND IT IS VERY ANNOYING TO WATCH. ╫E NEED TO ENABLE THE GRAPHICS AT THE SAME POINT REGARDLESS OF THE Y-SCROLL VALUE. ╘HE ANSWER IS IN THE EXTENDED COLOR MODE (┼├═). ╫HEN BOTH ┼├═ AND MULTICOLOR MODE (═├═) ARE SELECTED, ╓╔├-╔╔ WILL TURN THE DISPLAY TO BLACK. ╘HIS IS BECAUSE THERE IS A CONFLICTING SITUATION AND IT JUST CAN'T DECIDE WHICH COLOR SCHEME TO USE. ╘HE VIDEO ACCESSES WILL CONTINUE TO HAPPEN JUST LIKE BEFORE, THE DATA IS JUST NOT DISPLAYED. ╫HEN THE ┼├═ BIT IS CLEARED AGAIN, THE NORMAL MULTICOLOR GRAPHICS IS SHOWN. ╙O, WE SET THE ┼├═ BIT AND START TO DISPLAY THE FIRST EIGHT LINES OF THE ╞╠╔. ┬ECAUSE THE ╞╠╔ ROUTINE ALREADY WRITES TO $─011, WE JUST MAKE SURE THE ┼├═ BIT IS SET IN THE FIRST ╥ NUMBER OF WRITES TO $─011 AND ZERO IN ALL OTHER. ╘HE VIEWER IS NOW 'COMPLETE'. ┘OU CAN TAKE A LOOK AT THE CODE BELOW OR YOU CAN GET ├64╟FX1_4.LHA AND SEE IT IN ACTION YOURSELF AND NOT JUST RELY ON MY WORD. ╘HE PACKAGE INCLUDES CONVERTER PROGRAMS FOR ┬╞╠╔, ╞╠╔ AND ╦OALA (┴╬╙╔-├), COUPLE OF EXAMPLE PICTURES AND VIEWERS FOR ╨┴╠ AND ╬╘╙├ MACHINES. -╨ASI '┴LBERT' ╧JALA ALBERT@CS.TUT.FI -------------------------------------------------------------------------- ┬╞╠╔ VIEWER PROGRAM FOR ╨┴╠ MACHINES ╒╨╧╙ = $├00 ; TEMPORARY AREA FOR TABLES ┬┴╬╦ = $─00 ; ╒╨╧╙ FOR LINECRUNCH, ┬┴╬╦ FOR ╞╠╔ BANK SELECT ╥┴╙╘┼╥ = 29 ; WHERE TO POSITION THE SPRITE -> ╔╥╤ 20 LINES LATER ─╒══┘ = $╞╞╞ ; DUMMY LOCATION FOR TIMING PURPOSES ╞╠╔╙┌ = 19-1 ; VISIBLE ╞╠╔ SIZE IN CHARACTER ROWS - 1 *= $810 ╙┼╔ ╠─┴ #$7╞:╙╘┴ $─├0─ ; ╔╥╤ SETUP ╠─┴ #1:╙╘┴ $─01┴ ╙╘┴ $─015:╙╘┴ ╦┼┘╫+1 ╠─┴ #<╔╥╤:╙╘┴ $314 ╠─┴ #>╔╥╤:╙╘┴ $315 ╠─┴ #╥┴╙╘┼╥:╙╘┴ $─001:├╠├:┴─├ #20:╙╘┴ $─012 ╠─┴ #0:╙╘┴ $─017 ╠─┴ #0:╙╘┴ 2 ╩╙╥ ╬┼╫╨╧╙ ; ╔NIT THE ╞╠╔ ROUTINES ╠─┴ #$┴:╙╘┴ $─011 ; ┬LANK SCREEN ╠─╪ #23 ; ╔NIT TABLES ┬╠╧╧╨ ╠─┴ #$94:╙╘┴ ┬┴╬╦,╪ ╠─┴ #$96:╙╘┴ ┬┴╬╦+24,╪ ─┼╪:┬╨╠ ┬╠╧╧╨ ╠─╪ #15 ╠╧╧╨0 ╠─┴ ┘╔╬╔╘,╪:┴╬─ #$77 ; ├HANGE TO $37 TO BETTER SEE THE ╙╘┴ ╒╨╧╙,╪ ; WORKINGS OF THE ROUTINES ╙╘┴ ╒╨╧╙+16,╪ ╙╘┴ ╒╨╧╙+32,╪ ─┼╪:┬╨╠ ╠╧╧╨0 ╠─┴ #$34:╙╘┴ 1 ; ├OPY TO THE LAST VIDEO BANK ╠─┴ #$80:╙╘┴ ╙╥├+2 ; FROM $8000-$┬╞╞╞ TO $├000-$╞╞╞╞ ╠─┴ #$├0:╙╘┴ ─╙╘+2 ╠─╪ #0:╠─┘ #$40 ╙╥├ ╠─┴ $8000,╪ ─╙╘ ╙╘┴ $├000,╪ ╔╬╪:┬╬┼ ╙╥├ ╔╬├ ╙╥├+2:╔╬├ ─╙╘+2 ─┼┘:┬╬┼ ╙╥├ ╠─┴ #$37:╙╘┴ 1 ╠─╪ #0 ; ╔NIT COLOR MEMORY ╠╨ ╠─┴ $3├00,╪:╙╘┴ $─800,╪ ; ┴LL 1024 BYTES ARE USED ╠─┴ $3─00,╪:╙╘┴ $─900,╪ ; - SOME EVEN TWICE! ╠─┴ $3┼00,╪:╙╘┴ $─┴00,╪ ╠─┴ $3╞00,╪:╙╘┴ $─┬00,╪ ╔╬╪:┬╬┼ ╠╨ ╠─┴ $─├0─:├╠╔ ╦┼┘╫ ╠─╪ #0:┬╬┼ ╦┼┘╫ ; ╫AIT FOR SPACE TO BE PRESSED ╙┼╔ ; ╙YSTEM TO NORMAL ╠─┴ #$37:╙╘┴ 1 ╩╙╥ $╞─┴3 ╠─┴ #$97:╙╘┴ $──00 ╩╙╥ $┼5┴0 ╠─┘ #3 ╔╥╤╠ ╠─┴ $╞─30,┘:╙╘┴ $314,┘ ─┼┘:┬╨╠ ╔╥╤╠ ╠─╪ #0:╠─┴ #1 ; ├LEAR COLOR MEMORY ├╠╠ ╙╘┴ $─800,╪:╙╘┴ $─900,╪ ╙╘┴ $─┴00,╪:╙╘┴ $─┬00,╪ ╔╬╪:┬╬┼ ├╠╠ ├╠╔:╥╘╙ ┘╔╬╔╘ ┬┘╘ $78,$79,$7┴,$7┬,$7├,$7─,$7┼,$7╞ ┬┘╘ $78,$79,$7┴,$7┬,$7├,$7─,$7┼,$7╞ *=*-<*+256 ╔╥╤ ╠─┴ #$18:╙╘┴ $─016:╠─╪ #0:╠─┴ #$5┴ ╔╬├ ─╒══┘:─┼├ ─╒══┘ ; ╙YNCHRONIZATION ╙╘╪ $─020:╙╘╪ $─021:╙╘┴ $─011 ╠─┴ #$15:╙╘┴ $─018 ╠─┴ #$97:╙╘┴ $──00 ╠─╪ #44 ; ╫AIT FOR THE 4TH LINE ╠╠ ─┼╪:┬╨╠ ╠╠:╬╧╨ ╠─╪ #0 ╠╧╧╨3 ╬╧╨ ; ╠INECRUNCH-PART ROUTINE ╠─┴ ╒╨╧╙+6,╪:╔╬├ ─╒══┘:╙╘┴ $─011 ╬╧╨:╬╧╨:╔╬├ ─╒══┘ ╬╧╨:╬╧╨:╬╧╨:╬╧╨:╬╧╨ ╬╧╨:╬╧╨:╬╧╨:╬╧╨:╬╧╨ ╬╧╨:╬╧╨:╬╧╨:╬╧╨:╬╧╨ ╔╬╪ ┼1 ├╨╪ #$10:┬╬┼ ╠╧╧╨3 ; ╙KIP THAT MANY CHARACTER ROWS-4 ┬╔╘ $┼┴ ╠╧╧╨4 ╠─┴ ╒╨╧╙,╪:╔╬├ ─╒══┘:╙╘┴ $─011 ╬╧╨:╬╧╨:╬╧╨:╔╬├ ─╒══┘ ╬╧╨:╬╧╨:╬╧╨:╬╧╨:╬╧╨ ╬╧╨:╬╧╨:╬╧╨:╬╧╨:╬╧╨ ╬╧╨:╬╧╨:╬╧╨:╬╧╨:╠─┴ #0 ╔╬╪ ┼2 ├╨╪ #$1╞:┬╬┼ ╠╧╧╨4 ; ─ELAY ─═┴ UNTIL WE ARE AT THE ; 'SAME PLACE' EACH TIME ╠─┴ #0:╙╘┴ $─020 ; ╬OW WAIT FOR THE BAD LINE AND START ╞╠╔ ┬╔╘ $┼┴:╬╧╨ ╬╧╨:╬╧╨:╬╧╨:╬╧╨ ╬╧╨:╬╧╨:╬╧╨:╬╧╨ ╬╧╨:╬╧╨:╬╧╨:╬╧╨ ┬0 ╠─┴ #$92:╙╘┴ $──00:╬╧╨ ; ╘HE RIGHT VIDEO BANK ; ╫AIT FOR 0-7 LINES TO SET THE ┼├═ MODE OFF ; (MAKES THE GRAPHICS VISIBLE) ╞0 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$08:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞1 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$18:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞2 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$28:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞3 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$38:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞4 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$48:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞5 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$58:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞6 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$68:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞7 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$78:╙╘┴ $─018 ╠─╪ #╞╠╔╙┌:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ; ─O ╞╠╔ 18 MORE CHARACTER ROWS ╞8 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$08:╙╘┴ $─018 ┬1 ╠─┴ ┬┴╬╦,╪:╙╘┴ $──00:┬╔╘ $┼┴ ╞9 ╠─┴ #0:╙╘┴ $─011:╠─┴ #$18:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞┴ ╠─┴ #0:╙╘┴ $─011:╠─┴ #$28:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞┬ ╠─┴ #0:╙╘┴ $─011:╠─┴ #$38:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞├ ╠─┴ #0:╙╘┴ $─011:╠─┴ #$48:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞─ ╠─┴ #0:╙╘┴ $─011:╠─┴ #$58:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞┼ ╠─┴ #0:╙╘┴ $─011:╠─┴ #$68:╙╘┴ $─018:╬╧╨:╬╧╨:╬╧╨:╬╧╨:┬╔╘ $┼┴ ╞╞ ╠─┴ #0:╙╘┴ $─011:╠─┴ #$78:╙╘┴ $─018:╬╧╨:╬╧╨:─┼╪:┬═╔ ┼╞╠╔:╩═╨ ╞8 ┼╞╠╔ ╬╧╨ ╠─┴ #$╞├ ╫╠ ├═╨ $─012:┬╬┼ ╫╠ ╔╬├ ─╒══┘:╔╬├ ─╒══┘:╔╬├ ─╒══┘:╔╬├ ─╒══┘ ╔╬├ ─╒══┘:╔╬├ ─╒══┘:╔╬├ ─╒══┘:╔╬├ ─╒══┘ ╔╬├ ─╒══┘:╔╬├ ─╒══┘:╙╘┴ $─020 ╩╙╥ ╬┼╫╨╧╙ ; ╒PDATE THE LOCATION ╩╙╥ ├╚╨╧╙ ; ├HANGE TO A NEW LOCATION ╠─┴ $─├01:┴╬─ #$10:┬╬┼ ╧╓3 ; ├HECK FOR THE SPACE BAR ╠─┴ #0:╙╘┴ ╦┼┘╫+1 ╧╓3 ╠─╪ #$53:╙╘╪ $─011:╔╬├ $─019:╩═╨ $┼┴81 ╬┼╫╨╧╙ ╠─┴ #0 ; ╔NIT THE ╔╥╤ ROUTINE FOR THIS POSITION ╠╙╥:╠╙╥:╠╙╥:├╠├:┴─├ #4:╙╘┴ ┼1+1 ╠─┴ #7:╙┼├:╙┬├ ╬┼╫╨╧╙+1:┴╬─ #7:╘┴╪:╘┴┘:├╠├:┴─├ #35:╙╘┴ ┼2+1 ╠─┴ ╒╨╧╙+3+7,┘:─┼╪:┬═╔ ╩0:┴╬─ #$3╞ ╩0 ╙╘┴ ╞7+1:┴╬─ #$3╞:╙╘┴ ╞╞+1 ╠─┴ ╒╨╧╙+3+6,┘:─┼╪:┬═╔ ╩1:┴╬─ #$3╞ ╩1 ╙╘┴ ╞6+1:┴╬─ #$3╞:╙╘┴ ╞┼+1 ╠─┴ ╒╨╧╙+3+5,┘:─┼╪:┬═╔ ╩2:┴╬─ #$3╞ ╩2 ╙╘┴ ╞5+1:┴╬─ #$3╞:╙╘┴ ╞─+1 ╠─┴ ╒╨╧╙+3+4,┘:─┼╪:┬═╔ ╩3:┴╬─ #$3╞ ╩3 ╙╘┴ ╞4+1:┴╬─ #$3╞:╙╘┴ ╞├+1 ╠─┴ ╒╨╧╙+3+3,┘:─┼╪:┬═╔ ╩4:┴╬─ #$3╞ ╩4 ╙╘┴ ╞3+1:┴╬─ #$3╞:╙╘┴ ╞┬+1 ╠─┴ ╒╨╧╙+3+2,┘:─┼╪:┬═╔ ╩5:┴╬─ #$3╞ ╩5 ╙╘┴ ╞2+1:┴╬─ #$3╞:╙╘┴ ╞┴+1 ╠─┴ ╒╨╧╙+3+1,┘:─┼╪:┬═╔ ╩6:┴╬─ #$3╞ ╩6 ╙╘┴ ╞1+1:┴╬─ #$3╞:╙╘┴ ╞9+1 ╠─┴ ╒╨╧╙+3+0,┘:─┼╪:┬═╔ ╩7:┴╬─ #$3╞ ╩7 ╙╘┴ ╞0+1:┴╬─ #$3╞:╙╘┴ ╞8+1 ╠─┴ #$96:╙╘┴ ┬0+1:╠─┴ #199:╙┼├:╙┬├ ╬┼╫╨╧╙+1:┬├├ ╧╓2 ╠╙╥:╠╙╥:╠╙╥:├╠├:┴─├ #5:╙╘┴ ┬1+1 ╥╘╙ ╧╓2 ╠─┴ #0:╙╘┴ ┬1+1:╠─╪ #$94:╙╘╪ ┬0+1:╥╘╙ ├╚╨╧╙ ╠─╪ ╬┼╫╨╧╙+1 ╠─┴ $─├00:╘┴┘ ; ╟ET JOYSTICK ┴╬─ #$10:┬╬┼ ─╔╥ ; ╔F NO BUTTON PRESSED ╘┘┴:┴╬─ #1:┬┼╤ ╒╨ ; ╔F JOY UP ╘┘┴:┴╬─ #2:┬┼╤ ─╧╫╬ ; ╔F JOY DOWN ╥╘╙ ─╔╥ ╠─┴ #0:┬┼╤ ╒╨ ─╧╫╬ ─┼╪:├╨╪ #$╞╞:┬╬┼ ─╧╦ ╠─╪ #0:╙╘╪ ─╔╥+1 ; ├HANGE DIRECTION ─╧╦ ╙╘╪ ╬┼╫╨╧╙+1:╥╘╙ ╒╨ ╔╬╪:├╨╪ #$╞─:┬├├ ╒╧╦ ; 251(LOCATIONS)+149(VISIBLE)=400 ╠─╪ #$╞├:╙╘╪ ─╔╥+1 ; ├HANGE DIRECTION ╒╧╦ ╙╘╪ ╬┼╫╨╧╙+1:╥╘╙ -------------------------------------------------------------------------- ╘HE ┬╞╠╔ FILE FORMAT: ╞ILE ┬╞╠╔ ─ISPLAY ╠INES ╧FFSET ╧FFSET ╠INES ╙IZE ├OLORS 0-1.3 0..55 944..999 22.7-24 56 ╔ 1.3-2 56..79 - 24 2-24 80..999 0..919 0-22 920 24-24.7 1000..1023 920..943 22-22.7 24 ╔╔ 0-1.3 0..55 1968..2024 49.3-50.6 56 1.3-24.7 56..1023 1000..1967 24-49.3 968 ╟FX 0-1.3 0..447 7552..7999 22.7-24 448 ╔ 1.3-2 448..639 - 192 2-24 640..7999 0..7359 0-22 7360 24-24.7 8000..8191 7360..7551 22-22.7 192 ╔╔ 0-1.3 0..447 15744..16192 49.3-50.6 448 1.3-24.7 448..8191 8000..15743 24-49.3 7744 ======================================================================== ═AKING STABLE RASTER ROUTINES (├64 AND ╓╔├-20) BY ═ARKO ═AKELA (═ARKO.═AKELA@╚╒╘.╞╔) ╨REFACE ╘OO MANY GRAPHICAL EFFECTS, ALSO CALLED RASTER EFFECTS, HAVE BEEN CODED IN A VERY SLOPPY WAY. ╞OR INSTANCE, IF THERE ARE ANY COLOR BARS ON THE SCREEN IN A GAME OR DEMO, THE COLORS OFTEN JITTER A BIT, E.G. THEY ARE NOT STABLE. ┴ND ALSO, IT IS FAR TOO EASY TO MAKE VIRTUALLY ANY DEMO CRASH BY HITTING THE ╥ESTORE KEY, OR AT LEAST CAUSE VISUAL DISTORTIONS ON THE SCREEN. ┴S LATE AS A YEAR AGO ╔ STILL HADN'T CODED A STABLE RASTER INTERRUPT ROUTINE MYSELF. ┬UT THEN ╔ HAD TO DO IT, SINCE ╔ WAS RESEARCHING THE VIDEO CHIP TIMING DETAILS TOGETHER WITH MY ╟ERMAN FRIEND ┴NDREAS ┬OOSE. ╔T WAS ASHAMING THAT WE HAD THE SAME LEVEL OF KNOWLEDGE WHEN IT CAME TO THE HARDWARE, BUT HE WAS THE ONLY OF US WHO HAD WRITTEN A STABLE RASTER ROUTINE. ╫ELL, FINALLY ╔ MADE ME TO START CODING. ╔ USED THE SAME DOUBLE-INTERRUPT IDEA AS ┴NDREAS USED IN HIS ROUTINE. ┴FTER A COUPLE OF ERRORS MY ROUTINE WORKED, AND ╔ UNDERSTOOD HOW IT WORKS EXACTLY. (╘HIS IS SOMETHING THAT SEPARATES US NORMAL CODERS FROM DEMO PEOPLE: ╘HEY OFTEN CODE BY INSTINCT; BY PATCHING THE ROUTINE UNTIL IT WORKS, WITHOUT KNOWING EXACTLY WHAT IS HAPPENING. ╘HAT'S WHY DEMOS OFTEN RELY ON WEIRD THINGS, LIKE CRASH IF THE MEMORY IS NOT INITIALIZED PROPERLY.) ╔N THIS ARTICLE, ╔ DOCUMENT TWO METHODS OF CREATING STABLE RASTER ROUTINES ON ├OMMODORE COMPUTERS. ╘HE PRINCIPLES APPLY FOR MOST 8-BIT COMPUTERS, NOT ONLY ├OMMODORES, BUT RASTER EFFECTS ARE VERY RARELY SEEN ON OTHER COMPUTERS. ┬ACKGROUND ╫HAT ARE RASTER EFFECTS? ╘HEY ARE EFFECTS, WHERE YOU CHANGE THE SCREEN APPEARANCE WHILE IT IS BEING DRAWN. ╞OR INSTANCE, YOU CAN SET THE SCREEN COLOR TO WHITE IN THE TOP OF THE SCREEN, AND TO BLACK IN THE MIDDLE OF THE SCREEN. ╔N THAT WAY, YOU WILL GET A PICTURE WHOSE TOP HALF IS WHITE AND BOTTOM HALF BLACK. ╬ORMALLY SUCH EFFECTS ARE IMPLEMENTED WITH INTERRUPT ROUTINES THAT ARE EXECUTED SYNCHRONIZED WITH THE SCREEN REFRESH. ╘HE VIDEO CHIP ON THE ├OMMODORE 64 AND MANY OTHER VIDEOCHIPS HAVE A SPECIAL INTERRUPT FEATURE CALLED THE ╥ASTER INTERRUPT. ╔T WILL GENERATE AN ╔╥╤ IN THE BEGINNING OF A SPECIFIED RASTER LINE. ╧N OTHER COMPUTERS, LIKE THE ╓╔├-20, THERE IS NO ╥ASTER INTERRUPT, BUT YOU CAN GENERATE THE INTERRUPTS WITH A TIMER, PROVIDED THAT THE TIMER AND THE VIDEOCHIP ARE CLOCKED FROM THE SAME SOURCE. ┼VEN IF THE PROCESSOR GETS AN INTERRUPT SIGNAL AT THE SAME POSITION ON EACH VIDEO FRAME, IT WON'T ALWAYS BE EXECUTING THE FIRST INSTRUCTION OF THE INTERRUPT ROUTINE AT THE SAME SCREEN POSITION. ╘HE ╬═╧╙ 6502 MACHINE INSTRUCTIONS CAN TAKE 2 TO 9 MACHINE CYCLES TO EXECUTE, AND IF THE MAIN PROGRAM CONTAINS INSTRUCTIONS OF VERY VARYING LENGTHS, THE BEGINNING POSITION OF THE INTERRUPT CAN JUMP BETWEEN 7 DIFFERENT POSITIONS. ╘HIS IS WHY YOU NEED TO SYNCHRONIZE THE RASTER ROUTINE WHEN DOING SERIOUS EFFECTS. ┴LSO, EXECUTING THE INTERRUPT SEQUENCE WILL TAKE 7 ADDITIONAL CYCLES, AND THE INTERRUPT SEQUENCE WILL ONLY START AFTER THE CURRENT INSTRUCTION IF THE INTERRUPT ARRIVED AT LEAST TWO CYCLES BEFORE THE END OF THE CURRENT INSTRUCTION. ╔T IS EVEN POSSIBLE THAT AN INTERRUPT ARRIVES WHILE INTERRUPTS ARE DISABLED AND THE PROCESSOR IS JUST STARTING TO EXECUTE A ├╠╔ INSTRUCTION. ┴LAS, THE PROCESSOR WILL NOT JUMP TO THE INTERRUPT RIGHT AFTER THE ├╠╔, BUT IT WILL EXECUTE THE NEXT INSTRUCTION BEFORE JUMPING TO IT. ╘HIS IS NATURAL, SINCE THE ├╠╔ TAKES ONLY TWO CYCLES. ┬UT ANYWAY, THIS IS ONLY A CONSTANT IN OUR EQUATION, AND ACTUALLY OUT OF THE SCOPE OF THIS ARTICLE. ╚OW TO SYNCHRONIZE A RASTER INTERRUPT ROUTINE? ╘HE ONLY WAY IS TO CHECK THE CURRENT SCREEN POSITION AND DELAY APPROPRIATELY MANY CYCLES. ╘HERE ARE SEVERAL WAYS OF DOING THIS, SOME OF WHICH ARE VERY AWFUL AND INEFFICIENT. ╘HE UGLIEST WAYS OF DOING THIS ON THE ├OMMODORE 64 ╔ KNOW ARE BUSY-WAITING SEVERAL RASTER LINES AND POLLING THE RASTER LINE VALUE, OR USING THE ╠IGHT PEN FEATURE, WHICH WILL FAIL IF THE USER PRESSES THE FIRE BUTTON ON ╩OYSTICK PORT 1. ╚ERE ╔ WILL PRESENT TWO WAYS, BOTH VERY ELEGANT IN MY OPINION. ╒SING AN AUXILIARY TIMER ╧N THE ╓╔├-20, THERE IS NO ╥ASTER INTERRUPT FEATURE IN THE VIDEO CHIP. ┴LL YOU CAN DO IS TO USE A TIMER FOR GENERATING RASTER INTERRUPTS. ┴ND IF YOU USE TWO TIMERS RUNNING AT A CONSTANT PHASE DIFFERENCE, YOU CAN GET FULL SYNCHRONIZATION. ╘HE FIRST TIMER GENERATES THE RASTER INTERRUPT, AND THE SECOND TIMER, THE AUXILIARY TIMER, TELLS THE RASTER ROUTINE WHERE IT IS RUNNING. ┴CTUALLY YOU COULD EVEN USE THE FIRST TIMER ALSO FOR THE CHECKING, BUT THE CODE WILL LOOK NICER IN THE WAY ╔ WILL BE PRESENTING NOW. ┬ESIDES, YOU CAN USE THE AUXILIARY TIMER IDEA EVEN WHEN REAL RASTER INTERRUPTS ARE AVAILABLE. ╘HE MAJOR DRAWBACK OF USING AN AUXILIARY TIMER IS INITIALIZING IT. ╘HE INITIALIZATION ROUTINE MUST SYNCHRONIZE WITH THE SCREEN, THAT IS, WAIT FOR THE BEGINNING OF THE WANTED RASTER LINE. ╘O ACCOMPLISH THIS, THE ROUTINE MUST FIRST WAIT FOR A RASTER LINE THAT OCCURS A BIT EARLIER. ┴BOUT THE ONLY WAY TO DO THIS IS WITH A LOOP LIKE ╠─┴ #VALUE LOOP ├═╨ RASTER ┬╬┼ LOOP ╧NE ROUND OF THIS LOOP WILL TAKE 4+3=7 CYCLES TO EXECUTE, ASSUMING THAT ABSOLUTE ADDRESSING IS BEING USED. ╘HE LOOP WILL BE FINISHED IF THE RASTER REGISTER CONTAINS THE WANTED VALUE WHILE THE PROCESSOR READS IT ON THE LAST CYCLE OF THE ├═╨ INSTRUCTION. ╘HE RASTER REGISTER CAN ACTUALLY HAVE CHANGED ALREADY ON THE FIRST CYCLE OF THE ┬╬┼ INSTRUCTION ON THE PREVIOUS RUN OF THE LOOP, THAT IS 7 CYCLES EARLIER! ┬ECAUSE OF THIS, THE ROUTINE MUST POLL THE RASTER REGISTER FOR SEVERAL RASTER LINES, ALWAYS CONSUMING ONE CYCLE MORE IF THE RASTER REGISTER CHANGED TOO EARLY. ┴S THE SYNCHRONIZATION CAN BE OFF AT MOST BY 7 CYCLES, A LOOP OF 7 RASTER REGISTER VALUE CHANGES WOULD DO, BUT ╔ MADE THE LOOP A BIT LONGER IN MY ╓╔├-20 ROUTINE. (╫ELL, ╔ HAVE TO ADMIT IT, ╔ WAS TOO LAZY TO MAKE IT WORK ONLY WITH 7 ROUNDS.) ┴FTER THE INITIALIZATION ROUTINE IS FULLY SYNCHRONIZED THE SCREEN, IT CAN SET UP THE TIMER(S) AND INTERRUPTS AND EXIT. ╘HE AUXILIARY TIMER IN MY ╓╔├-20 DEMO ROUTINE IS SEVERAL DOZENS OF CYCLES AFTER THE PRIMARY TIMER, SEE THE SOURCE CODE FOR COMMENTS. ╔T IS ARRANGED SO THAT THE AUXILIARY TIMER WILL BE AT LEAST 0 WHEN IT IS BEING READ IN THE RASTER ROUTINE. ╘HE RASTER ROUTINE WILL WAIT AS MANY EXTRA CYCLES AS THE AUXILIARY TIMER READS, HOWEVER AT MOST 15 CYCLES. ╒SING DOUBLE RASTER INTERRUPT ╧N THE ├OMMODORE 64, ╔ HAVE NEVER SEEN THE AUXILIARY TIMER SCHEME BEING USED. ┴CTUALLY ╔ HAVEN'T SEEN IT BEING USED ANYWHERE, ╔ WAS PROBABLY THE FIRST ONE WHO MADE A STABLE RASTER INTERRUPT ROUTINE ON THE ╓╔├-20. ╔NSTEAD, THE DOUBLE INTERRUPT METHOD IS BECOMING THE STANDARD ON THE ├64 SIDE. ╘HE DOUBLE INTERRUPT METHOD IS BASED ENTIRELY ON THE ╥ASTER INTERRUPT FEATURE OF THE VIDEO CHIP. ╔N THE FIRST RASTER INTERRUPT ROUTINE, THE PROGRAM SETS UP ANOTHER RASTER INTERRUPT ON A FURTHER LINE, CHANGES THE INTERRUPT VECTOR AND ENABLES INTERRUPTS. ╔N THE PLACE WHERE THE SECOND RASTER INTERRUPT WILL OCCUR, THERE WILL BE 2-BYTE INSTRUCTIONS IN THE FIRST INTERRUPT ROUTINE. ╔N THIS WAY, THE BEGINNING OF THE NEXT RASTER INTERRUPT WILL BE OFF AT MOST BY ONE CYCLE. ╙OME CODERS MIGHT NOT CARE ABOUT THIS ONE CYCLE, BUT IF YOU CAN DO IT RIGHT, WHY WOULDN'T YOU DO IT RIGHT UNTIL THE END? ┴T THE BEGINNING OF THE SECOND RASTER INTERRUPT ROUTINE, YOU WILL READ THE RASTER LINE COUNTER REGISTER AT THE POINT WHERE IT IS ABOUT TO CHANGE. ╫HEN THE RASTER ROUTINE IS BEING EXECUTED, THERE ARE TWO POSSIBILITIES: ┼ITHER THE RASTER COUNTER HAS JUST CHANGED, OR IT WILL CHANGE ON THE NEXT CYCLE. ╙O, YOU JUST NEED TO COMPARE IF THE REGISTER CHANGED ONE CYCLE TOO EARLY OR NOT, AND DELAY A CYCLE WHEN NEEDED. ╘HIS IS EASILY ACCOMPLISHED WITH A BRANCH TO THE NEXT ADDRESS. ╧F COURSE, SOMEWHERE IN YOUR SECOND RASTER INTERRUPT ROUTINE YOU MUST RESTORE THE ORIGINAL RASTER INTERRUPT POSITION AND SET THE INTERRUPT VECTOR TO POINT TO THE FIRST INTERRUPT ROUTINE. ┴PPLYING IN PRACTICE ╔ ALMOST FORGOT MY COMPLAINTS ABOUT DEMOS CRASHING WHEN YOU ACTIVELY HIT THE ╥ESTORE KEY. ╧N THE ╓╔├-20, YOU CAN DISABLE ╬═╔ INTERRUPTS GENERATED BY THE ╥ESTORE KEY, AND ON THE ├64, YOU CAN GENERATE AN ╬═╔ INTERRUPT WITH THE ├╔┴2 TIMER AND LEAVE THE ╬═╔-LINE LOW, SO THAT NO FURTHER HIGH-TO-LOW TRANSITIONS WILL BE RECOGNIZED ON THE LINE. ╘HE EXAMPLE PROGRAMS DEMONSTRATE HOW TO DO THIS. ╙O FAR, THIS ARTICLE HAS BEEN PRETTY THEORETICAL. ╘O APPLY THESE RESULTS IN PRACTICE, YOU MUST DEFINITELY KNOW HOW MANY ├╨╒ CLOCK CYCLES THE VIDEO CHIP CONSUMES WHILE DRAWING A SCAN LINE. ╘HIS IS FAIRLY EASY TO MEASURE WITH A TIMER INTERRUPT, IF YOU PATCH THE INTERRUPT HANDLER SO THAT IT CHANGES THE SCREEN COLOR ON EACH RUN. ╙ET THE TIMER INTERVAL TO ╠╔╬┼╙*├╧╠╒═╬╙ CYCLES, WHERE ╠╔╬┼╙ IS THE AMOUNT OF RASTER LINES AND ├╧╠╒═╬╙ IS YOUR GUESS FOR THE AMOUNT OF CLOCK CYCLES SPENT IN A RASTER LINE. ╔F YOUR GUESS IS RIGHT, THE COLOR WILL ALWAYS BE CHANGED IN THE SAME SCREEN POSITION (NEGLECTING THE 7-CYCLE JITTER). ╫HEN ADJUSTING THE TIMER, REMEMBER THAT THE TIMERS ON THE 6522 ╓╔┴ REQUIRE 2 CYCLES FOR RE-LOADING, AND THE ONES ON THE 6526 ├╔┴ NEED ONE EXTRA CYCLE. ╦EEP TRYING DIFFERENT TIMER VALUES UNTIL YOU THE SCREEN COLOR CHANGES AT ONE FIXED POSITION. ├OMMODORE USED SEVERAL DIFFERENT VALUES FOR ╠╔╬┼╙ AND ├╧╠╒═╬╙ ON ITS VIDEOCHIPS. ╘HEY NEVER MANAGED TO MAKE THE SCREEN REFRESH RATE EXACTLY 50 OR 60 ╚ERTZ, BUT THEY DIDN'T HESITATE TO CLAIM THAT THEIR COMPUTERS COMPLY WITH THE ╨┴╠-┬ OR ╬╘╙├-═ STANDARDS. ╔N THE FOLLOWING TABLES ╔ HAVE GATHERED SOME INFORMATION OF SOME ├OMMODORE VIDEO CHIPS. ╬╘╙├-═ SYSTEMS: ├HIP ├RYSTAL ─OT ╨ROCESSOR ├YCLES/ ╠INES/ ╚OST ╔─ FREQ/╚Z CLOCK/╚Z CLOCK/╚Z LINE FRAME ------ -------- -------- -------- --------- ------- ------ ╓╔├-20 6560-101 14318181 4090909 1022727 65 261 ├64 6567╥56┴ 14318181 8181818 1022727 64 262 ├64 6567╥8 14318181 8181818 1022727 65 263 ╠ATER ╬╘╙├-═ VIDEO CHIPS WERE MOST PROBABLY LIKE THE 6567╥8. ╬OTE THAT THE PROCESSOR CLOCK IS A 14TH OF THE CRYSTAL FREQUENCY ON ALL ╬╘╙├-═ SYSTEMS. ╨┴╠-┬ SYSTEMS: ├HIP ├RYSTAL ─OT ╨ROCESSOR ├YCLES/ ╠INES/ ╚OST ╔─ FREQ/╚Z CLOCK/╚Z CLOCK/╚Z LINE FRAME ------ -------- -------- -------- --------- ------- ------ ╓╔├-20 6561-101 4433618 4433618 1108405 71 312 ├64 6569 17734472 7881988 985248 63 312 ╧N THE ╨┴╠-┬ ╓╔├-20, THE CRYSTAL FREQUENCY IS SIMULTANEOUSLY THE DOT CLOCK, WHICH IS ┬╘╫ A 4TH OF THE CRYSTAL FREQUENCY USED ON THE ├64. ╧N THE ├64, THE CRYSTAL FREQUENCY IS DIVIDED BY 18 TO GENERATE THE PROCESSOR CLOCK, WHICH IN TURN IS MULTIPLIED BY 8 TO GENERATE THE DOT CLOCK. ╘HE BASIC TIMINGS ARE THE SAME ON ALL 6569 REVISIONS, AND ALSO ON ANY LATER ├64 AND ├128 VIDEO CHIPS. ╔F ╔ REMEMBER CORRECTLY, THESE VALUES WERE THE SAME ON THE ├16 VIDEOCHIP ╘┼─ AS WELL. ╬OTE THAT THE DOT CLOCK IS 4 TIMES THE PROCESSOR CLOCK ON THE ╓╔├-20, AND 8 TIMES THAT ON THE ├64. ╘HAT IS, ONE PROCESSOR CYCLE IS HALF A CHARACTER WIDE ON THE ╓╔├-20, AND A FULL CHARACTER ON A ├64. ╔ DON'T HAVE EXACT MEASUREMENTS OF THE ╓╔├-20 TIMING, BUT IT SEEMS THAT WHILE THE ╓╔├-20 VIDEOCHIPS DRAW THE CHARACTERS ON THE SCREEN, IT FIRST READS THE CHARACTER CODE, AND THEN, ON THE FOLLOWING VIDEO CYCLE, THE APPEARANCE ON THE CURRENT CHARACTER LINE. ╘HERE ARE NO BAD LINES, LIKE ON THE ├64, WHERE THE CHARACTER CODES (AND COLORS) ARE FETCHED ON EVERY 8TH RASTER LINE. ╘HOSE ONES WHO GOT UPSET WHEN ╔ SAID THAT ├OMMODORE HAS NEVER MANAGED TO MAKE A FULLY ╨┴╠-┬ OR ╬╘╙├-═ COMPLIANT 8-BIT COMPUTER SHOULD TAKE A CLOSER LOOK AT THE "╠INES/FRAME" COLUMNS. ╔F THAT DOES NOT CONVINCE YOU, CALCULATE THE RASTER LINE RATE AND THE SCREEN REFRESH RATE FROM THE VALUES IN THE TABLE AND SEE THAT THEY DON'T COMPLY WITH THE STANDARDS. ╘O CALCULATE THE LINE RATE, DIVIDE THE PROCESSOR CLOCK FREQUENCY BY THE AMOUNT OF CYCLES PER LINE. ╘O GET THE SCREEN REFRESH RATE, DIVIDE THAT FREQUENCY BY THE AMOUNT OF RASTER LINES. ╘HE ├ODE ╧╦, ENOUGH THEORY AND BACKGROUND. ╚ERE ARE THE TWO EXAMPLE PROGRAMS, ONE FOR THE ╓╔├-20 AND ONE FOR THE ├64. ╔N ORDER TO FULLY UNDERSTAND THEM, YOU NEED TO KNOW THE EXACT EXECUTION TIMES OF ╬═╧╙ 6502 INSTRUCTIONS. (┴LL 8-BIT ├OMMODORE COMPUTERS USE THE ╬═╧╙ 6502 PROCESSOR CORE, EXCEPT THE ├65 PROTOTYPE, WHICH USED A INFERIOR ├═╧╙ VERSION WITH ALL NICE POORLY-DOCUMENTED FEATURES REMOVED.) ┘OU SHOULD CHECK THE 64DOC DOCUMENT, AVAILABLE ON MY ╫╫╫ PAGES AT HTTP://WWW.HUT.FI/▐MSMAKELA/CBM/EMUL/X64/64DOC.HTML, OR VIA ╞╘╨ AT FTP.FUNET.FI:/PUB/CBM/DOCUMENTS/64DOC. ╔ CAN ALSO E-MAIL IT TO YOU ON REQUEST. ┴LSO, ╔ HAVE WRITTEN A COMPLETE DESCRIPTION OF THE VIDEO TIMING ON THE 6567╥56┴, 6567╥8 AND 6569 VIDEO CHIPS, WHICH COULD MAYBE BE TURNED INTO ANOTHER ├=╚ACKING ARTICLE. ╘HE DOCUMENT IS CURRENTLY PARTIALLY IN ┼NGLISH AND PARTIALLY IN ╟ERMAN. ╘HE ┼NGLISH PART IS AVAILABLE FROM FTP.FUNET.FI AS /PUB/CBM/DOCUMENTS/PAL.TIMING, AND ╔ CAN SEND COPIES OF THE ╟ERMAN PART (SCREEN RESOLUTION, SPRITE DISTURBANCE MEASUREMENTS, AND MORE PRECISE TIMING INFORMATION) VIA E-MAIL. ╘HE CODE IS WRITTEN FOR THE ─┴╙═ ASSEMBLER, OR MORE PRECISELY FOR A EXTENDED ┴╬╙╔ ├ PORT OF IT MADE BY ╧LAF ╙EIBERT. ╘HIS EXCELLENT CROSS-ASSEMBLER IS AVAILABLE AT FTP.FUNET.FI IN /PUB/CBM/PROGRAMMING. ╞IRST THE RASTER DEMO FOR THE ╓╔├-20. ╬OTE THAT ON THE ╓╔├-20, THE $9004 REGISTER CONTAINS THE UPPER 8 BITS OF THE RASTER COUNTER. ╙O, THIS REGISTER CHANGES ONLY ON EVERY SECOND LINE. ╔ HAVE TESTED THE PROGRAM ON MY 6561-101-BASED ╓╔├-20, BUT NOT ON AN ╬╘╙├-═ SYSTEM. ╔T WAS HARD TO GET IN CONTACT WITH ╬╘╙├-═ ╓╔├-20 OWNERS. ─ANIEL ─ALLMANN, WHO HAS A ╬╘╙├-═ ╓╔├-20, ALTHOUGH HE LIVES IN ╟ERMANY, RAN MY TEST TO DETERMINE THE AMOUNT OF CYCLES PER LINE AND LINES PER FRAME ON THE 6560-101. ╒NFORTUNATELY, THE SECOND ╓╔┴ OF HIS ╓╔├-20 IS PARTIALLY BROKEN, AND BECAUSE OF THIS, THIS PROGRAM DID NOT WORK ON HIS COMPUTER. ├RAIG ┬RUCE RAN THE PROGRAM ONCE, AND HE REPORTED THAT IT ALMOST WORKED. ╔ CORRECTED A LITTLE BUG IN THE CODE, SO THAT NOW THE DISPLAY SHOULD BE STABLE ON AN ╬╘╙├-═ SYSTEM, TOO. ┬UT THE ACTUAL RASTER EFFECT, SIX 16*16-PIXEL BOXES CENTERED AT THE TOP BORDER, ARE VERY LIKELY TO BE OFF THEIR POSITION. PROCESSOR 6502 ╬╘╙├ = 1 ╨┴╠ = 2 ;╙┘╙╘┼═ = ╬╘╙├ ; 6560-101: 65 CYCLES PER RASTER LINE, 261 LINES ╙┘╙╘┼═ = ╨┴╠ ; 6561-101: 71 CYCLES PER RASTER LINE, 312 LINES #IF ╙┘╙╘┼═ & ╨┴╠ ╠╔╬┼╙ = 312 ├┘├╠┼╙_╨┼╥_╠╔╬┼ = 71 #ENDIF #IF ╙┘╙╘┼═ & ╬╘╙├ ╠╔╬┼╙ = 261 ├┘├╠┼╙_╨┼╥_╠╔╬┼ = 65 #ENDIF ╘╔═┼╥_╓┴╠╒┼ = ╠╔╬┼╙ * ├┘├╠┼╙_╨┼╥_╠╔╬┼ - 2 .ORG $1001 ; FOR THE UNEXPANDED ╓IC-20 ; ╘HE ┬┴╙╔├ LINE BASIC: .WORD 0$ ; LINK TO NEXT LINE .WORD 1995 ; LINE NUMBER .BYTE $9┼ ; ╙┘╙ TOKEN ; ╙┘╙ DIGITS .IF (* + 8) / 10000 .BYTE $30 + (* + 8) / 10000 .ENDIF .IF (* + 7) / 1000 .BYTE $30 + (* + 7) % 10000 / 1000 .ENDIF .IF (* + 6) / 100 .BYTE $30 + (* + 6) % 1000 / 100 .ENDIF .IF (* + 5) / 10 .BYTE $30 + (* + 5) % 100 / 10 .ENDIF .BYTE $30 + (* + 4) % 10 0$: .BYTE 0,0,0 ; END OF ┬┴╙╔├ PROGRAM START: LDA #$7F STA $912E ; DISABLE AND ACKNOWLEDGE INTERRUPTS STA $912D STA $911E ; DISABLE ╬═╔S (╥ESTORE KEY) ;SYNCHRONIZE WITH THE SCREEN SYNC: LDX #28 ; WAIT FOR THIS RASTER LINE (TIMES 2) 0$: CPX $9004 BNE 0$ ; AT THIS STAGE, THE INACCURACY IS 7 CLOCK CYCLES ; THE PROCESSOR IS IN THIS PLACE 2 TO 9 CYCLES ; AFTER $9004 HAS CHANGED LDY #9 BIT $24 1$: LDX $9004 TXA BIT $24 #IF ╙┘╙╘┼═ & ╨┴╠ LDX #24 #ENDIF #IF ╙┘╙╘┼═ & ╬╘╙├ BIT $24 LDX #21 #ENDIF DEX BNE *-1 ; FIRST SPEND SOME TIME (SO THAT THE WHOLE CMP $9004 ; LOOP WILL BE 2 RASTER LINES) BCS *+2 ; SAVE ONE CYCLE IF $9004 CHANGED TOO LATE DEY BNE 1$ ; NOW IT IS FULLY SYNCHRONIZED ; 6 CYCLES HAVE PASSED SINCE LAST $9004 CHANGE ; AND WE ARE ON LINE 2(28+9)=74 ;INITIALIZE THE TIMERS TIMERS: LDA #$40 ; ENABLE ╘IMER ┴ FREE RUN OF BOTH ╓╔┴S STA $911B STA $912B LDA #<╘╔═┼╥_╓┴╠╒┼ LDX #>╘╔═┼╥_╓┴╠╒┼ STA $9116 ; LOAD THE TIMER LOW BYTE LATCHES STA $9126 #IF ╙┘╙╘┼═ & ╨┴╠ LDY #7 ; MAKE A LITTLE DELAY TO GET THE RASTER EFFECT TO THE DEY ; RIGHT PLACE BNE *-1 NOP NOP #ENDIF #IF ╙┘╙╘┼═ & ╬╘╙├ LDY #6 DEY BNE *-1 BIT $24 #ENDIF STX $9125 ; START THE ╔╥╤ TIMER ┴ ; 6560-101: 65 CYCLES FROM $9004 CHANGE ; 6561-101: 77 CYCLES FROM $9004 CHANGE LDY #10 ; SPEND SOME TIME (1+5*9+4=55 CYCLES) DEY ; BEFORE STARTING THE REFERENCE TIMER BNE *-1 STX $9115 ; START THE REFERENCE TIMER POINTERS: LDA #<IRQ ; SET THE RASTER ╔╥╤ ROUTINE POINTER STA $314 LDA #>IRQ STA $315 LDA #$C0 STA $912E ; ENABLE ╘IMER ┴ UNDERFLOW INTERRUPTS RTS ; RETURN IRQ: ; IRQ (EVENT) ; > 7 + AT LEAST 2 CYCLES OF LAST INSTRUCTION (9 TO 16 TOTAL) ; PHA ; 3 ; TXA ; 2 ; PHA ; 3 ; TYA ; 2 ; PHA ; 3 ; TSX ; 2 ; LDA $0104,X ; 4 ; AND #XX ; 2 ; BEQ ; 3 ; JMP ($314) ; 5 ; --- ; 38 TO 45 CYCLES DELAY AT THIS STAGE LDA $9114 ; GET THE ╬═╔ TIMER ┴ VALUE ; (42 TO 49 CYCLES DELAY AT THIS STAGE) ; STA $1E00 ; UNCOMMENT THESE IF YOU WANT TO MONITOR ; LDY $9115 ; THE REFERENCE TIMER ON THE SCREEN ; STY $1E01 CMP #8 ; ARE WE MORE THAN 7 CYCLES AHEAD OF TIME? BCC 0$ PHA ; YES, SPEND 8 EXTRA CYCLES PLA AND #7 ; AND RESET THE HIGH BIT 0$: CMP #4 BCC 1$ BIT $24 ; WASTE 4 CYCLES AND #3 1$: CMP #2 ; SPEND THE REST OF THE CYCLES BCS *+2 BCS *+2 LSR BCS *+2 ; NOW IT HAS TAKEN 82 CYCLES FROM THE BEGINNING OF THE ╔╥╤ EFFECT: LDY #16 ; PERFORM AMAZING VIDEO EFFECT LDA $900F TAX EOR #$F7 0$: STA $900F STX $900F STA $900F STX $900F STA $900F STX $900F STA $900F STX $900F STA $900F STX $900F STA $900F STX $900F PHA PLA #IF ╙┘╙╘┼═ & ╨┴╠ PHA PLA NOP #ENDIF #IF ╙┘╙╘┼═ & ╬╘╙├ BIT $24 #ENDIF NOP DEY BNE 0$ ; END OF AMAZING VIDEO EFFECT JMP $EABF ; RETURN TO NORMAL ╔╥╤ ┴ND AFTER YOU HAVE RECOVERED FROM THE SCHOCK OF SEEING A ╓╔├-20 PROGRAM, HERE IS AN EXAMPLE FOR THE ├64. ╔T DOES ALSO SOMETHING NOTEWORTHY; IT REMOVES THE SIDE BORDERS ON A NORMAL SCREEN WHILE DISPLAYING ALL EIGHT SPRITES. ╫ELL, IT CANNOT REMOVE THE BORDERS ON BAD LINES, AND THE BAD LINES LOOK PRETTY BAD. ┬UT ╔ COULD USE THE PROGRAM FOR WHAT ╔ WANTED: ╔ MEASURED THE SPRITE DISTORTIONS ON ALL VIDEOCHIP TYPES ╔ HAD AT HAND. (╞┘╔: THE SPRITES 0-2 GET DISTORTED AT THE VERY RIGHT OF THE SCREEN, AND THE SPRITES 6 AND 7 ARE INVISIBLE AT THE VERY LEFT OF THE SCREEN. ┘OU WILL NEED A MONITOR WITH HORIZONTAL SIZE CONTROLS TO WITNESS THESE EFFECTS.) ╘HIS PROGRAM IS REALLY ROBUST, IT INSTALLS ITSELF NICELY TO THE INTERRUPT ROUTINE CHAIN. ╔T EVEN HAS AN ENTRY POINT FOR DEINSTALLING ITSELF. ┬UT IN ITS ROBUSTNESS IT USES SELF-MODIFYING CODE TO STORE THE ORIGINAL INTERRUPT ROUTINE ADDRESS. :-) ╘HE CODE ALSO RELIES ON THE PAGE BOUNDARIES IN BEING WHERE THEY ARE. ╘HE CYCLES ARE COUNTED SO THAT THE BRANCHES "IRQLOOP" MUST TAKE 4 CYCLES. ╔F THE "IRQLOOP" COMES TO THE SAME ├╨╒ PAGE WITH THE BRANCH INSTRUCTIONS, YOU MUST ADD ONE CYCLE TO THE LOOP IN A WAY OR ANOTHER. ╫HEN CODING THE ROUTINE, ╔ NOTICED AGAIN HOW STUPID ASSEMBLY CODING CAN BE, ESPECIALLY CONDITIONAL ASSEMBLING. ╔N A MACHINE LANGUAGE MONITOR YOU HAVE FAR BETTER CONTROL ON PAGE BOUNDARIES. ┬╘╫, YOU MIGHT WONDER WHY ╔ DISABLE THE ╥ESTORE KEY IN A SUBROUTINE AT THE END AND NOT IN THE BEGINNING OF THE PROGRAM. ╫ELL, THE ROUTINE WAS SO LONG THAT IT WOULD HAVE AFFECTED THE "IRQLOOP" PAGE BOUNDARIES. ┴ND ╔ DIDN'T WANT TO RISK THE MODIFIED PROGRAMS WORKING ON ALL THREE DIFFERENT VIDEOCHIP TYPES ON THE FIRST TRY. ╔N THE CODE, THERE ARE SOME COMMENTS THAT DOCUMENT THE VIDEO TIMING, LIKE THIS ONE: ;3S4S5S6S7SRRRRRGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG--▄▄0S1S2S ╨HI-1 ╓╔├-╔╔ ;SSSSSSSSSS ▄▄SSSSSS ╨HI-2 ╓╔├-╔╔ ;==========XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX▄▄╪╪╪====== ╨HI-2 6510 ; ^ NOW WE ARE HERE ╘HE TWO VERTICAL BARS "▄" DENOTE OPTIONAL CYCLES. ╧N ╨┴╠-┬ SYSTEMS (63 CYCLES PER LINE), THEY ARE NOT PRESENT. ╧N 6567╥56┴, WHICH HAS 64 CYCLES PER LINE, THERE IS ONE ADDITIONAL CYCLE ON THIS POSITION, AND THE 6567╥8 HAS TWO ADDITIONAL CYCLES THERE. ╘HE NUMBERS 0 THROUGH 7 ARE SPRITE POINTER FETCHES (FROM THE END OF THE CHARACTER MATRIX, E.G. THE TEXT SCREEN), THE "S" CHARACTERS DENOTE SPRITE IMAGE FETCHES, THE "R"S ARE MEMORY REFRESH, AND THE "G" ARE GRAPHICS FETCHES. ╘HE TWO IDLE VIDEO CHIP CYCLES ARE MARKED WITH "-". ╧N THE PROCESSOR TIMING LINE, THE "=" SIGNS SHOW HALTED ├╨╒, "X" MEANS FREE BUS, AND "╪" MEANS THAT THE PROCESSOR WILL BE HALTED AT ONCE, UNLESS IT IS PERFORMING WRITE CYCLES. PROCESSOR 6502 ; ╙ELECT THE VIDEO TIMING (PROCESSOR CLOCK CYCLES PER RASTER LINE) ├┘├╠┼╙ = 65 ; 6567╥8 AND ABOVE, ╬╘╙├-═ ;├┘├╠┼╙ = 64 ; 6567╥5 6┴, ╬╘╙├-═ ;├┘├╠┼╙ = 63 ; 6569 (ALL REVISIONS), ╨┴╠-┬ CINV = $314 CNMI = $318 RASTER = 52 ; START OF RASTER INTERRUPT M = $FB ; ZERO PAGE VARIABLE .ORG $801 BASIC: .WORD 0$ ; LINK TO NEXT LINE .WORD 1995 ; LINE NUMBER .BYTE $9┼ ; ╙┘╙ TOKEN ; ╙┘╙ DIGITS .IF (* + 8) / 10000 .BYTE $30 + (* + 8) / 10000 .ENDIF .IF (* + 7) / 1000 .BYTE $30 + (* + 7) % 10000 / 1000 .ENDIF .IF (* + 6) / 100 .BYTE $30 + (* + 6) % 1000 / 100 .ENDIF .IF (* + 5) / 10 .BYTE $30 + (* + 5) % 100 / 10 .ENDIF .BYTE $30 + (* + 4) % 10 0$: .BYTE 0,0,0 ; END OF ┬┴╙╔├ PROGRAM START: JMP INSTALL JMP DEINSTALL INSTALL: ; INSTALL THE RASTER ROUTINE JSR RESTORE ; ─ISABLE THE ╥ESTORE KEY (DISABLE ╬═╔ INTERRUPTS) CHECKIRQ: LDA CINV ; CHECK THE ORIGINAL ╔╥╤ VECTOR LDX CINV+1 ; (TO AVOID MULTIPLE INSTALLATION) CMP #<IRQ1 BNE IRQINIT CPX #>IRQ1 BEQ SKIPINIT IRQINIT: SEI STA OLDIRQ ; STORE THE OLD ╔╥╤ VECTOR STX OLDIRQ+1 LDA #<IRQ1 LDX #>IRQ1 STA CINV ; SET THE NEW INTERRUPT VECTOR STX CINV+1 SKIPINIT: LDA #$1B STA $D011 ; SET THE RASTER INTERRUPT LOCATION LDA #RASTER STA $D012 LDX #$E CLC ADC #3 TAY LDA #0 STA M 0$: LDA M STA $D000,X ; SET THE SPRITE ╪ ADC #24 STA M TYA STA $D001,X ; AND ┘ COORDINATES DEX DEX BPL 0$ LDA #$7F STA $DC0D ; DISABLE TIMER INTERRUPTS STA $DD0D LDX #1 STX $D01A ; ENABLE RASTER INTERRUPT LDA $DC0D ; ACKNOWLEDGE ├╔┴ INTERRUPTS LSR $D019 ; AND VIDEO INTERRUPTS LDY #$FF STY $D015 ; TURN ON ALL SPRITES CLI RTS DEINSTALL: SEI ; DISABLE INTERRUPTS LDA #$1B STA $D011 ; RESTORE TEXT SCREEN MODE LDA #$81 STA $DC0D ; ENABLE ╘IMER ┴ INTERRUPTS ON ├╔┴ 1 LDA #0 STA $D01A ; DISABLE VIDEO INTERRUPTS LDA OLDIRQ STA CINV ; RESTORE OLD ╔╥╤ VECTOR LDA OLDIRQ+1 STA CINV+1 BIT $DD0D ; RE-ENABLE ╬═╔ INTERRUPTS CLI RTS ; ┴UXILIARY RASTER INTERRUPT (FOR SYNCRONIZATION) IRQ1: ; IRQ (EVENT) ; > 7 + AT LEAST 2 CYCLES OF LAST INSTRUCTION (9 TO 16 TOTAL) ; PHA ; 3 ; TXA ; 2 ; PHA ; 3 ; TYA ; 2 ; PHA ; 3 ; TSX ; 2 ; LDA $0104,X ; 4 ; AND #XX ; 2 ; BEQ ; 3 ; JMP ($314) ; 5 ; --- ; 38 TO 45 CYCLES DELAY AT THIS STAGE LDA #<IRQ2 STA CINV LDA #>IRQ2 STA CINV+1 NOP ; WASTE AT LEAST 12 CYCLES NOP ; (UP TO 64 CYCLES DELAY ALLOWED HERE) NOP NOP NOP NOP INC $D012 ; ┴T THIS STAGE, $D012 HAS ALREADY BEEN INCREMENTED BY ONE. LDA #1 STA $D019 ; ACKNOWLEDGE THE FIRST RASTER INTERRUPT CLI ; ENABLE INTERRUPTS (THE SECOND INTERRUPT CAN NOW OCCUR) LDY #9 DEY BNE *-1 ; DELAY NOP ; ╘HE SECOND INTERRUPT WILL OCCUR WHILE EXECUTING THESE NOP ; TWO-CYCLE INSTRUCTIONS. NOP NOP NOP OLDIRQ = * + 1 ; ╨LACEHOLDER FOR SELF-MODIFYING CODE JMP * ; ╥ETURN TO THE ORIGINAL INTERRUPT ; ═AIN RASTER INTERRUPT IRQ2: ; IRQ (EVENT) ; 7 + 2 OR 3 CYCLES OF LAST INSTRUCTION (9 OR 10 TOTAL) ; PHA ; 3 ; TXA ; 2 ; PHA ; 3 ; TYA ; 2 ; PHA ; 3 ; TSX ; 2 ; LDA $0104,X ; 4 ; AND #XX ; 2 ; BEQ ; 3 ; JMP (CINV) ; 5 ; --- ; 38 OR 39 CYCLES DELAY AT THIS STAGE LDA #<IRQ1 STA CINV LDA #>IRQ1 STA CINV+1 LDX $D012 NOP #IF ├┘├╠┼╙ - 63 #IF ├┘├╠┼╙ - 64 NOP ; 6567╥8, 65 CYCLES/LINE BIT $24 #ELSE NOP ; 6567╥56┴, 64 CYCLES/LINE NOP #ENDIF #ELSE BIT $24 ; 6569, 63 CYCLES/LINE #ENDIF CPX $D012 ; ╘HE COMPARISON CYCLE IS EXECUTED ├┘├╠┼╙ OR ├┘├╠┼╙+1 CYCLES ; AFTER THE INTERRUPT HAS OCCURRED. BEQ *+2 ; ─ELAY BY ONE CYCLE IF $D012 HADN'T CHANGED. ; ╬OW EXACTLY ├┘├╠┼╙+3 CYCLES HAVE PASSED SINCE THE INTERRUPT. DEX DEX STX $D012 ; RESTORE ORIGINAL RASTER INTERRUPT POSITION LDX #1 STX $D019 ; ACKNOWLEDGE THE RASTER INTERRUPT LDX #2 DEX BNE *-1 NOP NOP LDA #20 ; SET THE AMOUNT OF RASTER LINES-1 FOR THE LOOP STA M LDX #$C8 IRQLOOP: LDY #2 DEY BNE *-1 ; DELAY DEC $D016 ; NARROW THE SCREEN (EXACT TIMING REQUIRED) ;3S4S5S6S7SRRRRRGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG--▄▄0S1S2S ╨HI-1 ╓╔├-╔╔ ;SSSSSSSSSS ▄▄SSSSSS ╨HI-2 ╓╔├-╔╔ ;==========XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX▄▄╪╪╪====== ╨HI-2 6510 ; ^ NOW WE ARE HERE STX $D016 ; EXPAND THE SCREEN #IF ├┘├╠┼╙ - 63 #IF ├┘├╠┼╙ - 64 BIT $24 ; 6567╥8 #ELSE NOP ; 6567╥56┴ #ENDIF #ELSE NOP ; 6569 #ENDIF DEC M BMI ENDIRQ CLC LDA $D011 SBC $D012 AND #7 BNE IRQLOOP ; ╘HIS INSTRUCTION TAKES 4 CYCLES INSTEAD OF 3, ; BECAUSE THE PAGE BOUNDARY IS CROSSED. BADLINE: DEC M NOP NOP NOP NOP DEC $D016 ;3S4S5S6S7SRRRRRGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG--▄▄0S1S2S ╨HI-1 ╓╔├-╔╔ ;SSSSSSSSSS CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC ▄▄SSSSSS ╨HI-2 ╓╔├-╔╔ ;==========X╪╪╪========================================▄▄***====== ╨HI-2 6510 ; ^ WE ARE HERE STX $D016 ;3S4S5S6S7SRRRRRGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG--▄▄0S1S2S ╨HI-1 ╓╔├-╔╔ ;SSSSSSSSSS ▄▄SSSSSS ╨HI-2 ╓╔├-╔╔ ;==========XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX▄▄╪╪╪====== ╨HI-2 6510 ; ^ ^^- WE ARE HERE (6569) ; ▄ \- OR HERE (6567╥56┴) ; \- OR HERE (6567╥8) LDY #2 DEY BNE *-1 NOP NOP #IF ├┘├╠┼╙ - 63 #IF ├┘├╠┼╙ - 64 NOP ; 6567╥8, 65 CYCLES/LINE NOP NOP #ELSE BIT $24 ; 6567╥56┴, 64 CYCLES/LINE #ENDIF #ELSE NOP ; 6569, 63 CYCLES/LINE #ENDIF DEC M BPL IRQLOOP ; ╘HIS IS A 4-CYCLE BRANCH (PAGE BOUNDARY CROSSED) ENDIRQ: JMP $EA81 ; RETURN TO THE AUXILIARY RASTER INTERRUPT RESTORE: ; DISABLE THE ╥ESTORE KEY LDA CNMI LDY CNMI+1 PHA LDA #<NMI ; ╙ET THE ╬═╔ VECTOR STA CNMI LDA #>NMI STA CNMI+1 LDX #$81 STX $DD0D ; ┼NABLE ├╔┴ 2 ╘IMER ┴ INTERRUPT LDX #0 STX $DD05 INX STX $DD04 ; ╨REPARE ╘IMER ┴ TO COUNT FROM 1 TO 0. LDX #$DD STX $DD0E ; ├AUSE AN INTERRUPT. NMI = * + 1 LDA #$40 ; ╥╘╔ PLACEHOLDER PLA STA CNMI STY CNMI+1 ; RESTORE ORIGINAL ╬═╔ VECTOR (ALTHOUGH IT WON'T BE USED) RTS ┬INARIES ╚ERE ARE THE PROGRAMS IN UUENCODED FORMAT. ╞IRST THE ╓╔├-20 PROGRAMS: ├OLOR BOXES FOR THE ╓╔├-20, ╬╘╙├-═ VERSION (PROBABLY DISTORTED DISPLAY): BEGIN 644 COPPER.6560 ═└1└*$,╠'╟├0╤,#─└└└"╔?╪╘╬─8╘═─8╘>─:(<[└20╘/╬@"20─╦@20┬┬0─)"2┬ ═%<╦0_<╘$─+└└┬-#╦╩4"-&┘&-*┘&╔0┌)"├1:1├2:1╚└:(╘/╘─)(╪┼─:└*┬-#] ═├┴61╩6╥-%└.╔$(╘5└┌╟└├2┌18*╘4─<─(─└1(:"─'╥020!"0─*0/)└╦└└╠└!* ═╠└"@$*╘/─*╔)]╪╘/─(╪/─(╘/─(╪/─(╘/─(╪/─(╘/─(╪/─(╘/─(╪/─(╘/─(╪/ ,─$┴╚)"3╩┬-#43+_╩ └ END ├OLOR BOXES FOR THE ╓╔├-20, ╨┴╠-┬ VERSION: BEGIN 644 COPPER.6561 ═└1└*$,╠'╟├0╤,#─└└└"╔?╪╘╬─8╘═─8╘>─:(<[└20╘/╬@"20─╦@20┬┬0─╚┴├* ═╘/╫-!)"╨└(├0[:┼└├1╬1├2╬1╩8:┬5╚╘6─8╘╞─:└'┬-#]┌╬╩.)9&@"╚├0_8╪5 ═─:┼╩├10#╩1"-%0.╔╨(╘╬─6"═%)')")└$2&@╔!\─$─└0─)"─#╥0*╨└+└└2╦└└ ═╚!"═#┘"╩2?>-#┘".#┘"-#┘".#┘"-#┘".#┘"-#┘".#┘"-#┘".#┘"-#┘".#┘!( +:$┴╚┌╬╩(╘--,╧^╩. └ END ╥EMOVED SIDEBORDERS WITH 8 SPRITES AND BAD LINES, ╨┴╠-┬ VERSION: BEGIN 644 RASTER.63 ═└0@*",╠'╟├(╨-├$└└└!,$╨┴,=└@@'0╞═%└.╬%0/)┼=└$╪└├╨$7┬-╬0┬.╬@┬╔ ═┼:((├10#├┴4#╩1╬-$="╔-(╘2╘*(.&&─#╩*─└┴?╬┼^┘╘└╘&─8┴?╬8╟0'0╥╠╚0 ═[┌┼_├0╫<├0╫=╚@&.&═"═#=╤.&="@_╪╨5╘%┴@>*─;├1'0╩8&-#=╥╔└(╘:╘*╓┘ ═"(╘4└┌╓┌"(╘5└╥╨-╫5┴@╩;╬-%└.╔"(╘5└^╦╩┌╬╦╩┌╬╪2╘*─!├1╟06*└)┬-#] ═┌╬╦╩┌╬╔,╬└┬╔┼8╘4└┌─(├14#╦┴+0┌┬0─[!+0\└#*╥╚╪2╘*(!├┴╟0╚@+*╘/╫╩ ═┌╩─4┴?╬┬╥*└"┬-#]╙┴;0├┴;0┌╠;[,",8╦1'0[1+0*0?0┘<;[┌╬╦╩┌╠╪6╘(╪6 ═╘*└"┬-#]┌╬╦╩╤╧╠0╙4╥!┌╩╘8└┌╨9└╘┬╔0╚╘8└┌─)├1─#╚╚&.#=╓┬└(╪%╫>┬. 1!-╓┬╫8╪.╫:┼└:(╘8└╪╨9└╓└8 └ END ╥EMOVED SIDEBORDERS WITH 8 SPRITES AND BAD LINES, 6567╥56┴ VERSION (VERY OLD ╬╘╙├-═ ├64S): BEGIN 644 RASTER.64 ═└0@*",╠'╟├(╨-├$└└└!,$╨┴,=└@@'@╞═%└.╬%0/)┼=└$╪└├╨$7┬-╬0┬.╬@┬╔ ═┼:((├10#├┴4#╩1╬-$="╔-(╘2╘*(.&&─#╩*─└┴?╬┼^┘╘└╘&─8┴?╬8╟0'0╥╠╚0 ═[┌┼_├0╫<├0╫=╚@&.&═"═#=╤.&="@_╪╨5╘%┴@>*─;├1'0╩8&-#=╥╔└(╘:╘*╓┘ ═"(╘4└┌╓┌"(╘5└╥╨-╫5┴@╩;╬-%└.╔"(╘5└^╦╩┌╬╦╩┌╬╪2╘*─!├1╟06*└)┬-#] ═┌╬╦╩┌╬╔,╬└┬╔┼8╘4└┌─(├14#╦┴+0┌╬╦╩[!+0\└#*╥╚╪2╘*(!├┴╟0╚@+*╘/╫╩ ═┌╩─4┴?╬┬╥*└"┬-#]╙┴;0├┴;0┌╠;[,"08╦1'0[1+0*0?0┘<;[┌╬╦╩┌╠╪6╘(╪6 ═╘*└"┬-#]┌╬╚─),;[$,╤,@>╩═&└.╠&0-(╩4.-&└.╔"8╘9└┌*!├@╫=╚@".!=╫╚ 2├@3=╚═╓.#═╓╔0&┬-&└.,&0-@ └ END ╥EMOVED SIDEBORDERS WITH 8 SPRITES AND BAD LINES, 6567╥8 AND ABOVE (NOT TOO OLD ╬╘╙├-═ ├64S AND ALL ├128S) BEGIN 644 RASTER.65 ═└0@*",╠'╟├(╨-├$└└└!,$╨┴,=└@@(0╞═%└.╬%0/)┼=└$╪└├╨$7┬-╬0┬.╬@┬╔ ═┼:((├10#├┴4#╩1╬-$="╔-(╘2╘*(.&&─#╩*─└┴?╬┼^┘╘└╘&─8┴?╬8╟0'0╥╠╚0 ═[┌┼_├0╫<├0╫=╚@&.&═"═#=╤.&="@_╪╨5╘%┴@>*─;├1'0╩8&-#=╥╔└(╘:╘*╓┘ ═"(╘4└┌╓┌"(╘5└╥╨-╫5┴@╩;╬-%└.╔"(╘5└^╦╩┌╬╦╩┌╬╪2╘*─!├1╟06*└)┬-#] ═┌╬╦╩┌╬╔,╬└┬╔┼8╘4└┌─(├14#╦┴+0┌╬╚─).╨2╘/└└╥╠╩.$═"┬└8╪9╘*("╥═#] ═┌╬╩╔%(7[╚╠┬@└╚├0_<╪6╘(╪6╘"0─╤╧╠╨)1┬═$=#═$═└╔!]#─╤╧╧╩┌╬╦╩╙┴;0 ═├┴;0╚└*(╘/╫╩┌╬╦╩┌╠;[$,╔,@>╩═&└.╠&0-(╩4:-&└.╔"8╘9└┌*!├@╫=╚@". 5!=╫╚├@3=╚═╓.#═╓╔0&┬-&└.,&0-@ └ END ╘HAT WAS ALL, FOLKS! ╔ HOPE YOU LEARNED SOMETHING FROM THIS ARTICLE. ╞EEL FREE TO E-MAIL ME AT ═ARKO.═AKELA@╚╒╘.╞╔, SHOULD ANYTHING REMAIN UNCLEAR. ======================================================================== ┴ ─IFFERENT ╨ERSPECTIVE, PART ╔╔╔ BY ╙TEPHEN ╩UDD --- SJUDD@NWU.EDU ╟EORGE ╘AYLOR --- AA601@CFN.CS.DAL.CA ╫HEW! ╫HAT A BUSY TIME IT'S BEEN -- RESEARCH TO GET DONE, CONFERENCES, CLASSES... BETWEEN GETTING THINGS DONE AND BLOWING OTHER THINGS OFF, ╔ ONE DAY REFLECTED FOR A MOMENT AND REALIZED THAT ╔ HAD THREE DAYS LEFT TO GET THE NEXT ARTICLE TOGETHER FOR ├=╚ACKING! ╙O EVERYTHING HAS BEEN SLAPPED TOGETHER AT THE LAST MINUTE, AND ╔ HOPE YOU'LL FORGIVE ANY BUGS OR UNCLEAR CONCEPTS. >>> ┴╬┼├─╧╘┼ ┴╠┼╥╘ <<< ┴ND THAT REMINDS ME: ╔ JUST GOT ╩IFFY─╧╙ AND AN ╞─-2000 DRIVE -- WHAT A WONDERFUL DEVICE. ╔ HAVE A 1.6 MEGABYTE DISK FORMATTED INTO THREE PARTITIONS. ╘HE FIRST CONTAINS MY ═ERLIN 128 ASSEMBLER, THE SECOND IS SOME 4000 BLOCKS LARGE AND ╔ USE IT FOR ALL MY VARIOUS VERSIONS OF CODE WHILE DEBUGGING, AND THE THIRD IS MAYBE 1000 BLOCKS, AND CONTAINS ONLY FINISHED CODE -- NO MORE SWAPPING DISKS, NO MORE DELETING OLD VERSIONS THAT ╔ HOPE ╔ DON'T NEED TO MAKE ROOM ON THE DISK. ┴LSO, WHEN ╔ INSTALLED ╩IFFY─╧╙ ╔ FOUND A SERIOUS BUG IN MY 128─ -- A CRICKET, DEAD AMONG THE ╔├'S. ╘HIS TIME WE WILL COVER A LOT OF GROUND WHICH ISN'T SO MUCH CUTTING-EDGE AS IT IS VERY USEFUL. ╠ET'S FACE IT: CUBES ARE GETTING MORE THAN A LITTLE DULL. ┴ WORTHY END GOAL IS TO HAVE A COMPLETELY GENERAL ROUTINE FOR PLOTTING A SERIES OF POLYGONS -- THAT IS, YOU SUPPLY A LIST OF (X,Y,Z) COORDINATES FROM WHICH THE PROGRAM CAN FORM A LIST OF POLYGONS. ╘HESE POLYGONS MAY THEN BE DISPLAYED IN 2─, ROTATED, MAGNIFIED, FILLED, ETC. ┴ND, MUCH TO MY THREE-DAY ASTONISHMENT, THAT IS EXACTLY WHAT WE ARE GOING TO DO. ┬UT FIRST, A LITTLE EXCURSION. ╧NE THING WE ARE OF COURSE ALWAYS THINKING ABOUT IS OPTIMIZATION POSSIBILITIES: IN THE SHOWER, WHILE SLEEPING/DREAMING, OUT ON DATES, ETC. ╙O, WHERE TO BEGIN? ╘HE BIGGEST CYCLE HOGS IN THE PROGRAM ARE LINE DRAWING AND FACE FILLING -- WELL, FILLING FACES IS PRETTY STRAIGHTFORWARD. ╫HAT ABOUT LINE DRAWING? ╫ELL, ONE DOWNER OF THE ROUTINE IS THAT EVERY SINGLE PIXEL IS PLOTTED. ┬UT AS WE KNOW, ON A COMPUTER ANY GIVEN LINE IS MADE UP OF SEVERAL SMALLER VERTICAL AND HORIZONTAL LINES -- WOULDN'T IT BE NEAT IF WE COULD THINK OF A WAY TO PLOT THESE LINE CHUNKS ALL AT ONCE, INSTEAD OF A PIXEL AT A TIME? ╚ECK YES IT WOULD! ╙O HERE WE GO: ╬EAT-O ┼NHANCED ├HUNKY ╠INE ─RAWING ╥OUTINE ------------------------------------------- ╞IRST WE NEED TO BE IN THE RIGHT MINDFRAME. ╠ET'S SAY YOU'RE DRAWING A LINE WHERE YOU MOVE THREE PIXELS IN X BEFORE IT'S TIME TO TAKE A STEP IN Y. ╔NSTEAD OF PLOTTING ALL THREE PIXELS IT WOULD OF COURSE BE MUCH MORE EFFICIENT TO JUST STICK A NUMBER LIKE %00011100 IN THE DRAWING BUFFER. ┬UT SOMEHOW WE NEED TO KEEP TRACK OF A) HOW LARGE THE CHUNK NEEDS TO BE, AND B) WHERE EXACTLY THE CHUNK IS. ╔N THE ABOVE EXAMPLE, WE STARTED AT A PARTICULAR X-VALUE: %00010000 AND WE WANT TO KEEP ADDING ONES TO THE RIGHT OF THE STARTING POINT; THREE, TO BE EXACT. ╚MMM... WE NEED TO SOMEHOW ROTATE THE STARTING BIT IN A WAY THAT LEAVES A TRAIL OF ONES BEHIND IT. ═AYBE ROTATE AND ╧╥┴ WITH THE ORIGINAL BIT? ┬UT WHAT HAPPENS WHEN YOU TAKE A STEP IN ┘? ╬O, WE NEED SOMETHING FAR SNEAKIER. ╠ET'S SAY THAT INSTEAD OF %00010000 WE START WITH X = %00011111 ╬OW, WITH EACH STEP IN THE X DIRECTION, WE DO AN ARITHMETIC SHIFT ON X. ╙O AFTER ONE STEP WE HAVE X = %00001111 AND AFTER TWO STEPS X = %00000111 AND AT THE THIRD STEP OF COURSE X = %00000011 ╬OW IT IS TIME TO TAKE A STEP IN ┘. ┬UT NOW LOOK: IF WE ┼╧╥ X WITH ITS ORIGINAL VALUE XOLD = %00011111, WE GET X ┼╧╥ XOLD = %00011100 WHICH IS EXACTLY THE CHUNK WE WANTED. ═OREOVER, X STILL REMEMBERS WHERE IT IS, SO WE DON'T HAVE TO DO ANYTHING SPECIAL EACH TIME A STEP IS TAKEN IN THE Y-DIRECTION. ╙O HERE IS THE ALGORITHM FOR DRAWING A LINE IN THE X-DIRECTION: INITIALIZE X, DX, ETC. XOLD = X TAKE A STEP IN X: ╠╙╥ ╪ HAVE WE HIT THE END OF A COLUMN? ╔F SO, THEN PLOT AND CHECK ON Y IS IT TIME TO TAKE A STEP IN Y? IF NOT, TAKE ANOTHER STEP IN X IF IT IS, THEN LET A=X ┼╧╥ XOLD PLOT A INTO THE BUFFER LET XOLD=X KEEP ON GOING UNTIL WE'RE FINISHED ╘HIS SIMPLE MODIFICATION GIVES US A SUBSTANTIAL SPEED INCREASE -- ON THE OLD FILLED HIRES CUBE3D PROGRAM, ╔ MEASURED A GAIN OF ONE FRAME PER SECOND. ╬OT EARTH-SHATTERING, BUT NOT BAD EITHER! ╫HEN FACES ARE NOT FILLED, THE DIFFERENCE IS OF COURSE MUCH MORE NOTICABLE. ╘HERE ARE A FEW THINGS TO BE CAREFUL OF. ╘HERE WAS A BUG IN THE OLD ROUTINE WHEN THE LINE WAS A SINGLE POINT. ╔N THAT CASE DX=DY=0, AND THE PROGRAM WOULD DRAW A VERTICAL LINE ON THE SCREEN. ╘HERE ARE PROBABLY SOME OTHER THINGS TO BE CAREFUL OF, BUT SINCE ╔ WROTE THIS PART OF THE CODE THREE MONTHS AGO ╔ REALLY DON'T REMEMBER ANY OF THEM! ╘HIS TAKES CARE OF HORIZONTAL LINE CHUNKS -- WHAT ABOUT VERTICAL CHUNKS? ╫ELL, BECAUSE OF THE WAY POINTS ARE PLOTTED THERE IS NOTHING WE CAN DO ABOUT THEM. ┬UT, AS WE SHALL SOON SEE, IF WE USE AN ┼╧╥-BUFFER TO FILL FACES WE WILL BE FORCED TO TAKE CARE OF THE VERTICAL CHUNKS! ╟ENERAL ╨OLYGON ╥OUTINE ----------------------- ╬OW WE CAN BEGIN THINKING ABOUT A GENERAL POLYGON ROUTINE. ╞IRST WE NEED A LIST OF SETS OF POINTS, WHERE EACH SET CORRESPONDS TO A POLYGON. ╘HE FIRST NUMBER IN A SET COULD BE THE NUMBER OF (X,Y,Z) POINTS IN THAT SET, AND THE POINTS COULD THEN FOLLOW. ╙O A TRIANGLE COULD BE GIVEN BY THE DATA SET: 3 -1 0 0 0 1 0 1 0 0 ╘HIS WOULD BE A TRIANGLE WITH VERTICES AT (-1,0,0), (0,1,0), AND (1,0,0). ╫E CAN MASH A BUNCH OF THESE SETS TOGETHER, BUT SOMEHOW WE HAVE TO KNOW WHEN WE'VE HIT THE END -- FOR THIS WE CAN USE A ZERO, SINCE WE DON'T WANT TO PLOT POLYGONS WITH ZERO POINTS IN THEM. ╞OR THAT MATTER, HOW MANY POINTS SHOULD THERE BE IN A POLYGON? ╘HERE MUST BE AT LEAST THREE, OTHERWISE IT MAKES NO SENSE. ╙INCE WE WANT OUR POLYGONS TO BE CLOSED, THE COMPUTER SHOULD BE SMART ENOUGH TO CONNECT THE LAST POINT TO THE FIRST POINT -- IN OUR TRIANGLE ABOVE, THE COMPUTER WOULD JOIN (-1,0,0) TO (0,1,0), (0,1,0) TO (1,0,0), AND (1,0,0) TO (-1,0,0). ╬OW THAT WE HAVE A POLYGON, WE WANT TO ROTATE IT. ┘OU WILL RECALL THAT WE HAVE CALCULATED A ROTATION MATRIX ═, WHICH ACTS ON POINTS. ╙O WE NEED APPLY OUR ROTATION TRANSFORM TO EACH OF THE POINTS IN THE POLYGON, I.E. MULTIPLY ═ TIMES EACH POINT OF THE POLYGON. ╞URTHERMORE, WE NEED TO PROJECT EACH OF THESE POINTS. ╒H-OH: MATRIX MULTIPLICATION. ╔N THE PAST WE HAVE AVOIDED THIS ISSUE BY PUTTING THE VERTICES OF OUR CUBE AT 1 OR -1. ╙O WE NEED TO USE OUR MULTIPLICATION ROUTINE FROM LAST TIME. ┬UT WAIT! ┴S YOU RECALL, THE LAST PROGRAM USED A SPECIALLY MODIFIED MULTIPLICATION TABLE. ╘O GET A WIDER RANGE OF NUMBERS TO MULTIPLY WE WILL NEED ANOTHER SET OF MULTIPLICATION TABLES -- NO BIG WHOOP. ╬OW, IF YOU REVIEW THE MULTIPLICATION ROUTINE FROM LAST TIME, IT ADDS TWO NUMBERS AND SUBTRACTS TWO NUMBERS. ╫HAT KINDS OF NUMBERS WILL WE BE DEALING WITH? ╘HE MATRIX ELEMENTS VARY BETWEEN -64..64. ╘HIS THEN FIXES OUR RANGE OF POLYGON COORDINATES FROM -64..64. ╫HY? ╔F THE MATRIX ELEMENT IS 64, AND WE MULTIPLY IT BY 64, THE MULTIPLICATION ROUTINE WILL ADD 64 AND 64 AND GET 128, WHICH IS RIGHT ON THE EDGE OF OUR MULTIPLICATION TABLE. ├AN WE IMPROVE THIS ROTATION PROCESS IN ANY WAY? ╔N FACT, WE CAN CUT DOWN ON THE NUMBER OF MULTIPLICATIONS (I.E. DO EIGHT OR EVEN SEVEN INSTEAD OF NINE MULTIPLICATIONS). ╚OWEVER, THERE IS A FAIR AMOUNT OF OVERHEAD INVOLVED IN DOING SO, AND OUR MULTIPLY ROUTINE IS FAST ENOUGH THAT THE EXTRA OVERHEAD AND COMPLEXITY REALLY GAIN US VERY LITTLE IN ALL BUT THE MOST COMPLICATED OF POLYGONS. ╔N OTHER WORDS, ╔ DIDN'T BOTHER. ╫HAT ABOUT HIDDEN FACES? ┴GAIN, FROM LAST TIME YOU MAY RECALL THAT A METHOD WAS DESCRIBED WHICH USED THE CROSS-PRODUCT OF THE PROJECTED VECTORS. ╚OW DO WE IMPLEMENT THIS IN THE PROGRAM? ╫ELL, IF WE TAKE THE FIRST THREE POINTS OF THE POLYGON, WE HAVE TWO VECTORS. ╠ET'S SAY THESE POINTS ARE ╨1 ╨2 AND ╨3. ╘HEN ╓1=╨1-╨2 AND ╓2=╨3-╨2 ARE TWO VECTORS IN THE PLANE OF THE POLYGON WHICH ARE CONNECTED AT THE POINT ╨2 (THIS ANALYSIS WILL OF COURSE ONLY WORK IF THE POLYGON LIES IN SOME PLANE). ─EPENDING ON HOW WE TAKE THE CROSS PRODUCT, THE SIGN WILL BE POSITIVE OR NEGATIVE, AND THIS WILL TELL US IF THE POLYGON IS VISIBLE. ─EPENDING ON HOW WE TAKE THE CROSS PRODUCT? ┴BSOLUTELY. V1 X V2 = -V2 X V1. ╫HAT IT REALLY BOILS DOWN TO IS HOW YOU DEFINE THE POINTS IN YOUR POLYGON. ╙PECIFICALLY, WHAT ORDER THEY ARE IN. ╨OINTS THAT ARE SPECIFIED IN A CLOCKWISE MANNER WILL GIVE A FACE POINTING IN THE OPPOSITE DIRECTION OF A POLYGON WITH THE SAME POINTS SPECIFIED IN A COUNTER-CLOCKWISE ORDER. ╔N MY PROGRAM, THE POLYGONS MUST BE ENTERED IN COUNTER-CLOCKWISE ORDER (WITH YOU FACING THE POLYGON) FOR HIDDEN FACES TO WORK THE WAY YOU WANT THEM TO ;-). ╧NE OTHER NEAT THING TO HAVE IS THE ABILITY TO ZOOM IN AND OUT. ╫E KNOW FROM THE VERY FIRST ARTICLE THAT ZOOMING CORRESPONDS TO MULTIPLYING THE PROJECTED POINTS BY A NUMBER, SO THAT'S WHAT WE'LL DO. ╘HE MULTIPLICATION ROUTINE RETURNS ┴=┴*┘/64, SO A ZOOM FACTOR OF 64 WOULD BE LIKE MULTIPLYING THE POINT BY ONE. ┴LL THE PROGRAM DOES IS MULTIPLY THE PROJECTED POINTS BY A NUMBER ZOOM, UNLESS ZOOM=64, IN WHICH CASE THE PROGRAM SKIPS THE ZOOM MULTIPLY. ┬E WARNED! ╬O CHECKS OF ANY SORT ARE MADE IN THE PROGRAM, SO YOU CAN ZOOM AT YOUR OWN RISK! ╘HE IMPORTANT THINGS TO REMEMBER ARE: WHEN ENTERING POLYGONS, MAKE SURE THE NUMBERS RANGE FROM -64 TO 64, AND THAT YOU ENTER POINTS IN COUNTERCLOCKWISE. ╧UR TRIANGLE EXAMPLE ABOVE REALLY SHOULD HAVE BEEN ENTERED AS, SAY, 3 -64 0 0 64 0 0 0 64 0 ╞ILLED ╞ACES -- ╒SING AN ┼╧╥ BUFFER ----------------------------------- ╫ELL WE STILL HAVE ONE THING LEFT, WHICH WAS ALLUDED TO IN THE PREVIOUS ARTICLE: USING ┼╧╥ TO MAKE A FILLED FACE. ╙OME POSSIBLE DIFFICULTIES WERE RAISED, BUT WHEN YOU PLOT A SINGLE POLYGON AT A TIME, THE PROBLEM BECOMES VASTLY SIMPLIFIED. ╞IRST ╔ SHOULD PERHAPS REMIND YOU WHAT EXCLUSIVE-OR IS: EITHER ┴ OR ┬, BUT NOT BOTH. ╙O 1 ┼╧╥ 0 = 1, AS DOES 0 ┼╧╥ 1, BUT 0 ┼╧╥ 0 = 0 AND 1 ┼╧╥ 1 = 0. ┴S A SIMPLE INTRODUCTION TO USING THIS FOR FILLING FACES, CONSIDER THE FOLLOWING PIECE OF THE DRAWING BUFFER: 00001011 ═1 00000000 ═2 00000001 ═3 00001010 ═4 ╠ETS SAY WE MOVE DOWN MEMORY, ┼╧╥ING AS WE GO. ╠ET ═2 = ═1 ┼╧╥ ═2. ╘HEN LET ═3 = ═2 ┼╧╥ ═3. ╘HEN LET ═4 = ═3 ┼╧╥ ═4. ╧UR LITTLE PIECE OF MEMORY IS NOW: 00001011 ═1 00001011 ═2 00001010 ═3 00000000 ═4 ╫HAT JUST HAPPENED? ╔MAGINE THAT THE ORIGINAL MEMORY WAS A SERIES OF PIECES OF LINE SEGMENTS. ╫E HAVE JUST FILLED IN THE AREA BETWEEN THE TWO LINE SEGMENTS, LIKE MAGIC! ╔F YOU STILL AREN'T GETTING IT, DRAW A LARGE SECTION OF MEMORY, AND THEN DRAW AN OBJECT IN IT, LIKE A TRIANGLE, OR A TRAPAZOID, AND ┼╧╥ THE MEMORY BY HAND, STARTING FROM THE TOP AND MOVING DOWNWARDS. ┼╧╥ FLIPS BITS. ╔F YOU START WITH A ZERO, IT STAYS ZERO UNTIL IT HITS A ONE. ╔T WILL THEN STAY ONE UNTIL IT HITS ANOTHER ONE. ╙O YOU CAN SEE THAT IF YOU HAVE AN OBJECT BOUNDED BY ONES, ┼╧╥ING SUCCESSIVE MEMORY LOCATIONS WILL AUTOMAGICALLY FILL THE OBJECT. ╥IGHT? ╫ELL, WE HAVE TO BE CAREFUL. ╧NE MAJOR PROBLEM IS A VERTICAL LINE: 1 1 1 GOES TO 0 1 1 1 0 ╬OT ONLY IS THE RESULTANT LINE DASHED, BUT IF THERE ARE AN ODD NUMBER OF POINTS IN THE LINE SEGMENT, THE LAST ONE WILL HAPPILY MOVE DOWNWARDS IN MEMORY, AND GIVE YOU A MUCH LONGER VERTICAL LINE THAN YOU EXPECTED! ╙INCE ANY LINE WITH SLOPE GREATER THAN ONE IS MADE UP OF A SERIES OF LINE SEGMENTS, THIS IS A MAJOR CONSIDERATION. ┴NOTHER PROBLEM ARISES WITH SINGLE POINTS: A ONE JUST SITTING ALL BY ITSELF WILL ALSO GENERATE A NICE STREAK DOWN YOUR DRAWING AREA. ╔F YOU THINK ABOUT IT, WHAT WE IDEALLY WANT TO HAVE IS AN OBJECT THAT AT ANY GIVEN VALUE OF X THERE ARE EXACTLY TWO POINTS, ONE DEFINING THE TOP OF THE OBJECT, AND THE OTHER DEFINING THE BOTTOM. ╘HIS GIVES US THE INSIGHT TO SOLVE THE ABOVE TWO PROBLEMS. ╞IRST LET'S THINK ABOUT VERTICAL LINES. ╔N PRINCIPLE WE COULD PLOT THE FIRST AND LAST ENDPOINTS OF EACH VERTICAL LINE CHUNK, BUT THAT IS EXACTLY WHAT WE DON'T WANT! ╥EMEMBER THAT THESE ARE CLOSED POLYGONS, WHICH MEANS THAT THERE ARE _TWO_ LINES WE NEED TO THINK ABOUT. ╔F ╔ PLOT JUST A SINGLE POINT IN EACH VERTICAL LINE SEGMENT, THERE MUST BE ANOTHER POINT SOMEHWERE, EITHER ABOVE OR BELOW IT, FROM ANOTHER LINE SEGMENT, WHICH WILL CLOSE THE POINT TO ┼╧╥-FILLING. ╥EMEMBER, WE WANT EXACTLY TWO POINTS AT EACH VALUE OF X: ONE WILL COME FROM THE LINE, AND THE OTHER WILL COME FROM THE OTHER LINE WHICH MUST LIE ABOVE OR BELOW THE CURRENT ONE. ╞URTHERMORE, WITH ANY CONVEX POLYGON THERE ARE EXACTLY TWO LINES WHICH COME TOGETHER AT EACH VERTEX OF THE POLYGON. ╘HIS MEANS THAT THERE ARE ONLY CERTAIN CASES WHICH WE NEED TO WORRY ABOUT. ╞OR INSTANCE, TWO LINES MIGHT JOIN IN ANY OF THE FOLLOWING WAYS: \ / \ / \ / \ / \_____ _____/ \/ ETC. ╔F YOU DRAW OUT THE DIFFERENT CASES INVOLVING VERTICAL LINES, YOU CAN SEE THAT YOU HAVE TO BE CAREFUL ABOUT PLOTTING THE LINES. ╧NE TRICKY ONE IS WHERE TWO VERTICAL LINES WITH DIFFERENT SLOPES OVERLAP AT THE POINT OF INTERSECTION. ╙O AFTER STARING AT THESE PICTURES FOR A WHILE, YOU CAN FIND A CONSISTENT METHOD WHICH SOLVES THESE DIFFICULTIES. ┴S LONG AS YOU FOLLOW THE FOLLOWING RULES, THE PROBLEMS ALL DISAPPEAR; THE LINE ROUTINE NEEDS TO BE MODIFIED SLIGHTLY: 1) ╫HEN PLOTTING A VERTICAL LINE (I.E. BIG STEPS IN ┘ DIRECTION), DON'T PLOT THE ENDPOINTS (I.E. X1,Y1 AND X2,Y2). 2) ╫HEN PLOTTING A VERTICAL LINE, CONSISTENTLY PLOT EITHER THE FIRST PART OF EACH CHUNK OR THE LAST PART OF EACH CHUNK (EXCLUDING THE ENDPOINTS OF COURSE). ╔N OTHER WORDS, ONLY PLOT A POINT WHEN YOU TAKE A STEP IN X, AND THEN PLOT ONE AND ONLY ONE POINT. ╬OW ╔ DEDUCED THESE BY STARING AT PICTURES FOR A FEW HOURS AND TRYING DIFFERENT THINGS LIKE TOP/BOTTOM OF CHUNK, LEFT/RIGHT, FIRST/LAST, ETC. ┘OU CAN SEE THAT IN SOME CASES THIS ENSURES THAT ONLY ONE POINT APPEARS ON A GIVEN LINE SEGMENT. ┬UT TO ME THE ONLY WAY TO CONVINCE YOURSELF THAT THIS REALLY DOES WORK IS TO DRAW A BUNCH OF PICTURES, AND TRY IT OUT! ┘OU HAVE CASES WHERE TWO VERTICAL LINES INTERSECT, AND WHERE A VERTICAL LINE INTERSECTS A HORIZONTAL LINE. ┬UT THERE IS STILL ONE THING WHICH WE HAVE FORGOTTEN -- THE CASE OF A SINGLE POINT. ╘HIS CAN HAPPEN IN, FOR INSTANCE, A POINTY TRIANGLE, POINTING IN THE X-DIRECTION. ╚OW DO WE FIX THIS? ┬Y SIMPLY AVOIDING THE POINT: IN THE LINE DRAWING ROUTINE, USE ┼╧╥ TO PLOT THE POINTS INSTEAD OF ╧╥┴. ╙INCE VERTICAL LINES SKIP THE ENDPOINTS, VERTICAL-HORIZONTAL INTERSECTIONS ARE ╧╦. ╚ORIZONTAL- HORIZONTAL INTERSECTIONS WILL FORCE THE POINT OF INTERSECTION TO BE ZERO. ╒H-OH, WHAT ABOUT INTERSECTIONS LIKE -----*------. ╤UITE FRANKLY ╔ JUST THOUGHT OF IT, AND ╔ THINK MY PROGRAM WILL FAIL ON INTERSECTIONS LIKE THESE. ─RAT. ╫ELL, THAT JUST GIVES US SOMETHING FOR NEXT TIME! ╧NE OTHER THING NEEDS TO BE MENTIONED: FOR ┼╧╥-FILLING TO BE USEFUL YOU NEED TO DRAW THE POLYGON IN A SPECIAL BUFFER, AND THEN ┼╧╥ THIS BUFFER INTO THE MAIN DISPLAY BUFFER. ╔F YOU TRY TO ┼╧╥ THE DISPLAY BUFFER DIRECTLY YOU ARE GOING TO HAVE A WHOLE HEAP OF TROUBLE, SUCH AS THE CONCERNS RAISED LAST TIME. ╞INALLY, THIS GIVES A SIMPLE WAY OF FILLING WITH PATTERNS INSTEAD OF BORING MONOCOLOR. ╔NSTEAD OF ┼╧╥ (┼╧╥┬╒╞),┘ : ╧╥┴ (─╥┴╫┬╒╞),┘ YOU CAN USE ┼╧╥ (┼╧╥┬╒╞),┘ : ┴╬─ ╨┴╘╘┼╥╬,┘ : ╧╥┴ (─╥┴╫┬╒╞),┘ (AS LONG AS YOU PRESERVE THE ORIGINAL ┼╧╥ (┼╧╥┬╒╞),┘). ╫ELL ╔ AM EXTREMELY TIRED AND ╔ HOPE ├RAIG HASN'T SENT OUT ├=╚ACKING WITHOUT ME! ╔ HOPE YOU HAVE FUN PLAYING WITH THE PROGRAM, AND ╔ WOULD BE VERY INTERESTED IN SEEING ANY NEAT GEOMETRIC SHAPES YOU MIGHT DESIGN! ╨ROGRAM NOTES: -------------- - ╚IDDEN FACES DEFAULTS TO "ON". ╔F YOU ENTER A SHAPE AND A BLACK SCREEN COMES UP, HIT 'H' TO TURN OFF HIDDEN FACES (YOU PROBABLY ENTERED THE POLYGON CLOCKWISE). - ╘HERE IS NO PATTERN FILLING -- JUST SIMPLE ┼╧╥ WITH A TWIST: THE ┼╧╥ BUFFER IS ┼╧╥'D INTO THE DRAWING BUFFER. - ┘OU MIGHT START HOSING MEMORY IF YOU ZOOM TOO LARGE. ╙╠╩ 6/15/95 ┴DDENDUM -------- ╙TEPHEN ╩UDD SJUDD@NWU.EDU ╠AST TIME WE PUT A CIRCLE INTO THE 2─ GRAPHICS TOOLBOX. ├HRIS ═C┬RIDE HAS POINTED SOMETHING OUT TO ME ABOUT THE ALGORITHM, WHICH MAKES IT COMPLETE. ┴S YOU MAY RECALL, THE ALGORITHM GAVE A VERY SQUARISH CIRCLE FOR SMALL RADII. ├HRIS TOLD ME THAT SETTING THE INITIAL COUNTER VALUE TO ╥/2, INSTEAD OF ╥, GAVE A PERFECT CIRCLE. ╫HAT IS GOING ON? ╔F YOU RECALL THE ALGORITHM, WE ARE COMPUTING A FRACTIONAL QUANTITY, AND WHEN THAT QUANTITY BECOMES LARGER THAN ONE, WE DECREASE ╪. ╫OULDN'T IT BE A WHOLE LOT SMARTER TO ROUND OFF THAT FRACTION INSTEAD OF TRUNCATE IT? ╧F COURSE IT WOULD, AND THAT IS WHAT STARTING THE COUNTER AT ╥/2 DOES. ╙O, TO UPDATE THE PREVIOUS ALGORITHM, ┴ SHOULD BE INITIALIZED TO ╥/2 INSTEAD OF ╥, WHICH MEANS THAT WE CHANGE ╠─┴ ╥ TO ╠─┴ ╥ ╠╙╥ FOR A PERFECT CIRCLE EVERY TIME. 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HIDDEN SURFACES, FILLED * * FACES, AND EXTRA TOP SECRET * * TEXT MESSAGES! * * * * V3.0 + ╞AST CHUNKY LINE * * ROUTINE. * * * * V3.1 + ╟ENERAL POLYGON PLOT * * WITH HIDDEN FACES (╪-PRODUCT)* * AND ZOOM FEATURE. * * * * V3.2 + ┼╧╥-BUFFER FILLING * * * * ╘HIS PROGRAM IS INTENDED TO * * ACCOMPANY THE ARTICLE IN * * ├=╚ACKING, ╩UN. 95 ISSUE. * * ╞OR DETAILS ON THIS PROGRAM, * * READ THE ARTICLE! * * * * ╫RITE TO US! * * * * ═YSELF WHEN YOUNG DID * * EAGERLY FREQUENT * * ─OCTOR AND ╙AINT, AND HEARD * * GREAT ┴RGUMENT * * ┴BOUT IT AND ABOUT: BUT * * EVERMORE * * ├AME OUT BY THE SAME ─OOR * * AS IN ╔ WENT. * * - ╥UBAIYAT * * * * ╘HOUGH ╔ SPEAK WITH THE * * TONGUES OF MEN AND OF ANGLES * * AND HAVE NOT LOVE, ╔ AM * * BECOME AS SOUNDING BRASS, OR * * A TINKLING CYMBAL. * * - 1 ├ORINTHIANS 13 * * * * ╨.╙. ╘HIS WAS WRITTEN USING * * ═ERLIN 128. * ******************************** ╧╥╟ $8000 * ├ONSTANTS ┬╒╞╞1 ┼╤╒ $3000 ;╞IRST CHARACTER SET ┬╒╞╞2 ┼╤╒ $3800 ;╙ECOND CHARACTER SET ┼╧╥┬╒╞ ┼╤╒ $4000 ;┼╧╥-BUFFER ┬╒╞╞┼╥ ┼╤╒ $┴3 ;╨RESUMABLY THE TAPE WON'T BE RUNNING ╪1 ┼╤╒ $╞┬ ;╨OINTS FOR DRAWING A LINE ┘1 ┼╤╒ $╞├ ;╘HESE ZERO PAGE ADDRESSES ╪2 ┼╤╒ $╞─ ;DON'T CONFLICT WITH ┬┴╙╔├ ┘2 ┼╤╒ $╞┼ ╧╠─╪ ┼╤╒ $╞─ ├╚╒╬╦ ┼╤╒ $╞┼ ─╪ ┼╤╒ $67 ;╘HIS IS SHARED WITH ╘1 BELOW ─┘ ┼╤╒ $68 ╘┼═╨1 ┼╤╒ $╞┬ ;╧F COURSE, COULD CONFLICT WITH X1 ╘┼═╨2 ┼╤╒ $╞├ ;╘EMPORARY VARIABLES ┌╘┼═╨ ┼╤╒ $02 ;╒SED FOR BUFFER SWAP. ─ON'T TOUCH. ┌1 ┼╤╒ $22 ;╒SED BY MATH ROUTINE ┌2 ┼╤╒ $24 ;─ON'T TOUCH THESE EITHER! ┌3 ┼╤╒ $26 ┌4 ┼╤╒ $28 ╦ ┼╤╒ $┬6 ;├ONSTANT USED FOR HIDDEN ;SURFACE DETECTION - DON'T TOUCH ╚╔─┼ ┼╤╒ $┬5 ;┴RE SURFACES HIDDEN? ╞╔╠╠ ┼╤╒ $50 ;┴RE WE USING ┼╧╥-FILL? ┴╬╟═┴╪ ┼╤╒ 120 ;╘HERE ARE 2*PI/ANGMAX ANGLES * ╓╔├ ╓═├╙┬ ┼╤╒ $─018 ┬╦╟╬─ ┼╤╒ $─020 ┬╧╥─┼╥ ┼╤╒ $─021 ╙╙╘┴╥╘ ┼╤╒ 1344 ;ROW 9 IN SCREEN MEMORY AT 1024 * ╦ERNAL ├╚╥╧╒╘ ┼╤╒ $╞╞─2 ╟┼╘╔╬ ┼╤╒ $╞╞┼4 * ╙OME VARIABLES ╟╠╧┬╪═╔╬ = $3╞ ;╘HESE ARE USED IN CLEARING THE ╟╠╧┬╪═┴╪ = $40 ;DRAWING (GLOBAL) BUFFER ╟╠╧┬┘═╔╬ = $41 ╟╠╧┬┘═┴╪ = $42 ╠╧├╪═╔╬ = $57 ;╘HESE ARE USED IN CLEARING THE ╠╧├╪═┴╪ = $58 ;┼╧╥ (LOCAL) BUFFER ╠╧├┘═╔╬ = $59 ╠╧├┘═┴╪ = $60 ╨1╪ = $92 ;╘HESE ARE TEMPORARY STORAGE ╨1┘ = $93 ;╒SED IN PLOTTING THE PROJECTION ╨1┌ = $94 ╨2╪ = $95 ;╘HEY ARE HERE SO THAT WE ╨2┘ = $96 ;DON'T HAVE TO RECALCULATE THEM. ╨2┌ = $┴┼ ╨3╪ = $┴╞ ;╘HEY MAKE LIFE EASY. ╨3┘ = $┬0 ╨3┌ = $┬1 ;╫HY ARE YOU LOOKING AT ME LIKE THAT? ╨1╘ = $┬2 ;─ON'T YOU TRUST ME? ╨2╘ = $┬3 ╨3╘ = $┬4 ;╚AVING ANOTHER CHILD WASN'T MY IDEA. ╔╬─┼╪ = $51 ├╧╒╬╘╨╘╙ = $52 ┌╧╧═ = $71 ;┌OOM FACTOR ─╙╪ = $61 ;─╙╪ IS THE INCREMENT FOR ;ROTATING AROUND X ─╙┘ = $62 ;╙IMILAR FOR ─╙┘, ─╙┌ ─╙┌ = $63 ╙╪ = $64 ;╘HESE ARE THE ACTUAL ANGLES IN X Y AND Z ╙┘ = $65 ╙┌ = $66 ╘1 = $67 ;╘HESE ARE USED IN THE ROTATION ╘2 = $68 ╘3 = $69 ;╙EE THE ARTICLE FOR MORE DETAILS ╘4 = $6┴ ╘5 = $6┬ ╘6 = $6├ ╘7 = $6─ ╘8 = $6┼ ╘9 = $6╞ ╘10 = $70 ┴11 = $┴5 ;╘HESE ARE THE ELEMENTS OF THE ROTATION MATRIX ┬12 = $┴6 ;╪┘┌ ├13 = $┴7 ─21 = $┴8 ;╘HE NUMBER DENOTES (ROW,COLUMN) ┼22 = $┴9 ╞23 = $┴┴ ╟31 = $┴┬ ╚32 = $┴├ ╔33 = $┴─ *** ═ACROS ═╧╓┼ ═┴├ ╠─┴ ]1 ╙╘┴ ]2 <<< ╟┼╘╦┼┘ ═┴├ ;╫AIT FOR A KEYPRESS ╫┴╔╘ ╩╙╥ ╟┼╘╔╬ ├═╨ #00 ┬┼╤ ╫┴╔╘ <<< ─┼┬╒╟ ═┴├ ;╨RINT A CHARACTER ─╧ 0 ;─ON'T ASSEMBLE ╠─┴ #]1 ╩╙╥ ├╚╥╧╒╘ ├╠╔ >>> ╟┼╘╦┼┘ ;┴ND WAIT TO CONTINUE ├═╨ #'S' ;═Y SECRECT SWITCH KEY ┬╬┼ ╠1 ╩╙╥ ├╠┼┴╬╒╨ ╩═╨ ─╧╬┼ ╠1 ├═╨ #'X' ;═Y SECRET ABORT KEY ┬╬┼ ─╧╬┼ ╩═╨ ├╠┼┴╬╒╨ ╞╔╬ ─╧╬┼ <<< ─┼┬╒╟┴ ═┴├ ─╧ 0 ╠─┴ ]1 ╙╘┴ 1024 ╞╔╬ ─╧╬┼┴ <<< *------------------------------- ╠─┴ #$00 ╙╘┴ ┬╦╟╬─ ╙╘┴ ┬╧╥─┼╥ ╠─┴ ╓═├╙┬ ┴╬─ #%00001111 ;╙CREEN MEMORY TO 1024 ╧╥┴ #%00010000 ╙╘┴ ╓═├╙┬ ╠─┘ #00 ╠─┴ #<╘╘┼╪╘ ╙╘┴ ╘┼═╨1 ╠─┴ #>╘╘┼╪╘ ╙╘┴ ╘┼═╨2 ╩═╨ ╘╔╘╠┼ ╘╘┼╪╘ ╚┼╪ 9305111111 ;CLEAR SCREEN, WHITE, CRSR DN ╘╪╘ ' CUBE3D V3.2',0D,0D ╘╪╘ ' BY',0D ╚┼╪ 9╞ ;CYAN ╘╪╘ ' STEPHEN JUDD' ╚┼╪ 99 ╘╪╘ ' GEORGE TAYLOR',0D,0D ╚┼╪ 9┬ ╘╪╘ ' CHECK OUT THE JAN. 95 ISSUE OF',0D ╚┼╪ 96 ╘╪╘ ' C=HACKING' ╚┼╪ 9┬ ╘╪╘ ' FOR MORE DETAILS!',0D ╚┼╪ 0─1─1─9┼12 ╘╪╘ 'F1/F2',92 ╘╪╘ ' - INC/DEC X-ROTATION',0D ╚┼╪ 1─1─12 ╘╪╘ 'F3/F4',92 ╘╪╘ ' - INC/DEC Y-ROTATION',0D ╚┼╪ 1─1─12 ╘╪╘ 'F5/F6',92 ╘╪╘ ' - INC/DEC Z-ROTATION',0D ╚┼╪ 1─1─12 ╘╪╘ ' F7 ',92 ╘╪╘ ' - RESET',0D ╚┼╪ 1─1─12 ╘╪╘ ' +/- ',92 ╘╪╘ ' - ZOOM IN/OUT',0D ╚┼╪ 1─1─12 ╘╪╘ ' H ',92 ╘╪╘ ' - TOGGLE HIDDEN SURFACES',0D ╚┼╪ 1─1─12 ╘╪╘ 'SPACE',92 ╘╪╘ ' - TOGGLE SURFACE FILLING',0D,0D ╘╪╘ ' PRESS Q TO QUIT',0D ╚┼╪ 0─05 ╘╪╘ ' PRESS ANY KEY TO BEGIN',0D ╚┼╪ 00 ╘╔╘╠┼ ╠─┴ (╘┼═╨1),┘ ┬┼╤ :├╧╬╘ ╩╙╥ ├╚╥╧╒╘ ╔╬┘ ┬╬┼ ╘╔╘╠┼ ╔╬├ ╘┼═╨2 ╩═╨ ╘╔╘╠┼ :├╧╬╘ >>> ╟┼╘╦┼┘ **** ╙ET UP TABLES(?) * ╘ABLES ARE CURRENTLY SET UP IN ┬┴╙╔├ * AND BY THE ASSEMBLER. ╘┴┬╠┼╙ ╠─┴ #>╘═┴╘╚1 ╙╘┴ ┌1+1 ╙╘┴ ┌2+1 ╠─┴ #>╘═┴╘╚2 ╙╘┴ ┌3+1 ╙╘┴ ┌4+1 **** ├LEAR SCREEN AND SET UP "BITMAP" ╙┼╘╒╨ ╠─┴ #$01 ;╫HITE ╙╘┴ $─021 ;╘HIS IS DONE SO THAT OLDER ╠─┴ #147 ;MACHINES WILL SET UP ╩╙╥ ├╚╥╧╒╘ ╠─┴ #$00 ;CORRECTLY ╙╘┴ $─021 ╠─┴ #<╙╙╘┴╥╘ ┴─├ #12 ;╘HE GOAL IS TO CENTER THE GRAPHICS ╙╘┴ ╘┼═╨1 ;├OLUMN 12 ╠─┴ #>╙╙╘┴╥╘ ;╥OW 9 ╙╘┴ ╘┼═╨1+1 ;╙╙╘┴╥╘ POINTS TO ROW 9 ╠─┴ #00 ╠─┘ #00 ╠─╪ #00 ;X WILL COUNT 16 ROWS FOR US ├╠├ :╠╧╧╨ ╙╘┴ (╘┼═╨1),┘ ╔╬┘ ┴─├ #16 ┬├├ :╠╧╧╨ ├╠├ ╠─┴ ╘┼═╨1 ┴─├ #40 ;╬EED TO ADD 40 TO THE BASE POINTER ╙╘┴ ╘┼═╨1 ;╘O JUMP TO THE NEXT ROW ╠─┴ ╘┼═╨1+1 ┴─├ #00 ;╘AKE CARE OF CARRIES ╙╘┴ ╘┼═╨1+1 ╠─┘ #00 ╔╬╪ ╘╪┴ ;╪ IS ALSO AN INDEX INTO THE CHARACTER NUMBER ├╨╪ #16 ┬╬┼ :╠╧╧╨ ;╬EED TO DO IT 16 TIMES **** ├LEAR BUFFERS ╠─┴ #<┬╒╞╞1 ╙╘┴ ┬╒╞╞┼╥ ╠─┴ #>┬╒╞╞1 ╙╘┴ ┬╒╞╞┼╥+1 ╠─┘ #$00 ╠─╪ #24 ;┴SSUMING ALL THREE BUFFERS ARE ╠─┴ #$00 ;BACK-TO-BACK :┬╠╧╧╨ ╙╘┴ (┬╒╞╞┼╥),┘ ╔╬┘ ┬╬┼ :┬╠╧╧╨ ╔╬├ ┬╒╞╞┼╥+1 ─┼╪ ┬╬┼ :┬╠╧╧╨ **** ╙ET UP BUFFERS ╠─┴ #<┬╒╞╞1 ╙╘┴ ┬╒╞╞┼╥ ╠─┴ #>┬╒╞╞1 ╙╘┴ ┬╒╞╞┼╥+1 ╙╘┴ ┌╘┼═╨ ;ZTEMP WILL MAKE LIFE SIMPLE FOR US ╠─┴ ╓═├╙┬ ┴╬─ #%11110001 ;╙TART HERE SO THAT SWAP BUFFERS WILL WORK RIGHT ╧╥┴ #%00001110 ╙╘┴ ╓═├╙┬ **** ╙ET UP INITIAL VALUES ╔╬╔╘ ╠─┴ #00 ╙╘┴ ╠╧├╪═╔╬ ╙╘┴ ╠╧├╪═┴╪ ╙╘┴ ╠╧├┘═╔╬ ╙╘┴ ╠╧├┘═┴╪ ╙╘┴ ╟╠╧┬╪═╔╬ ╙╘┴ ╟╠╧┬┘═╔╬ ╙╘┴ ╟╠╧┬╪═┴╪ ╙╘┴ ╟╠╧┬┘═┴╪ ╙╘┴ ─╙╪ ╙╘┴ ─╙┘ ╙╘┴ ─╙┌ ╙╘┴ ╙╪ ╙╘┴ ╙┘ ╙╘┴ ╙┌ ╙╘┴ ╞╔╠╠ ╠─┴ #01 ╙╘┴ ╚╔─┼ ╠─┴ #64 ╙╘┴ ┌╧╧═ *------------------------------- * ═AIN LOOP **** ╟ET KEYPRESS ═┴╔╬ ├╠╔ ╦╨╥┼╙╙ ╩╙╥ ╟┼╘╔╬ ├═╨ #133 ;╞1? ┬╬┼ :╞2 ╠─┴ ─╙╪ ├═╨ #┴╬╟═┴╪/2 ;╬O MORE THAN PI ┬┼╤ :├╧╬╘1 ╔╬├ ─╙╪ ;OTHERWISE INCREASE X-ROTATION ╩═╨ :├╧╬╘ :╞2 ├═╨ #137 ;╞2? ┬╬┼ :╞3 ╠─┴ ─╙╪ ┬┼╤ :├╧╬╘1 ─┼├ ─╙╪ ╩═╨ :├╧╬╘ :╞3 ├═╨ #134 ┬╬┼ :╞4 ╠─┴ ─╙┘ ├═╨ #┴╬╟═┴╪/2 ┬┼╤ :├╧╬╘1 ╔╬├ ─╙┘ ;╔NCREASE Y-ROTATION ╩═╨ :├╧╬╘ :╞4 ├═╨ #138 ┬╬┼ :╞5 ╠─┴ ─╙┘ ┬┼╤ :├╧╬╘1 ─┼├ ─╙┘ ╩═╨ :├╧╬╘ :╞5 ├═╨ #135 ┬╬┼ :╞6 ╠─┴ ─╙┌ ├═╨ #┴╬╟═┴╪/2 ┬┼╤ :├╧╬╘1 ╔╬├ ─╙┌ ;Z-ROTATION ╩═╨ :├╧╬╘ :╞6 ├═╨ #139 ┬╬┼ :╞7 ╠─┴ ─╙┌ ┬┼╤ :├╧╬╘1 ─┼├ ─╙┌ ╩═╨ :├╧╬╘ :╞7 ├═╨ #136 ┬╬┼ :╨╠╒╙ ╩═╨ ╔╬╔╘ :├╧╬╘1 ╩═╨ :├╧╬╘ :╨╠╒╙ ├═╨ #'+' ┬╬┼ :═╔╬╒╙ ╔╬├ ┌╧╧═ ;┬AH, WHO NEEDS ERROR CHECKING? ╔╬├ ┌╧╧═ ╩═╨ :├╧╬╘ :═╔╬╒╙ ├═╨ #'-' ┬╬┼ :╚ ─┼├ ┌╧╧═ ─┼├ ┌╧╧═ ┬╨╠ :├╧╬╘ ╔╬├ ┌╧╧═ ╔╬├ ┌╧╧═ ╩═╨ :├╧╬╘ :╚ ├═╨ #'H' ┬╬┼ :╙╨┴├┼ ╠─┴ ╚╔─┼ ┼╧╥ #$01 ╙╘┴ ╚╔─┼ ╩═╨ :├╧╬╘ :╙╨┴├┼ ├═╨ #' ' ┬╬┼ :╤ ╠─┴ ╞╔╠╠ ┼╧╥ #$01 ╙╘┴ ╞╔╠╠ ╩═╨ :├╧╬╘ :╤ ├═╨ #'Q' ;Q QUITS ┬╬┼ :├╧╬╘ ╩═╨ ├╠┼┴╬╒╨ :├╧╬╘ ╙┼╔ ;╙PEED THINGS UP A BIT **** ╒PDATE ANGLES ╒╨─┴╘┼ ├╠├ ╠─┴ ╙╪ ┴─├ ─╙╪ ├═╨ #┴╬╟═┴╪ ;┴RE WE >= MAXIMUM ANGLE? ┬├├ :├╧╬╘1 ╙┬├ #┴╬╟═┴╪ :╔F SO, RESET :├╧╬╘1 ╙╘┴ ╙╪ ├╠├ ╠─┴ ╙┘ ┴─├ ─╙┘ ├═╨ #┴╬╟═┴╪ ┬├├ :├╧╬╘2 ╙┬├ #┴╬╟═┴╪ ;╙AME DEAL :├╧╬╘2 ╙╘┴ ╙┘ ├╠├ ╠─┴ ╙┌ ┴─├ ─╙┌ ├═╨ #┴╬╟═┴╪ ┬├├ :├╧╬╘3 ╙┬├ #┴╬╟═┴╪ :├╧╬╘3 ╙╘┴ ╙┌ **** ╥OTATE COORDINATES ╥╧╘┴╘┼ *** ╞IRST, CALCULATE T1,T2,...,T10 ** ╘WO MACROS TO SIMPLIFY OUR LIFE ┴──┴ ═┴├ ;┴DD TWO ANGLES TOGETHER ├╠├ ╠─┴ ]1 ┴─├ ]2 ├═╨ #┴╬╟═┴╪ ;╔S THE SUM > 2*PI? ┬├├ ─╧╬┼ ╙┬├ #┴╬╟═┴╪ ;╔F SO, SUBTRACT 2*PI ─╧╬┼ <<< ╙╒┬┴ ═┴├ ;╙UBTRACT TWO ANGLES ╙┼├ ╠─┴ ]1 ╙┬├ ]2 ┬├╙ ─╧╬┼ ┴─├ #┴╬╟═┴╪ ;╧OPS, WE NEED TO ADD 2*PI ─╧╬┼ <<< ** ╬OW CALCULATE T1,T2,ETC. >>> ╙╒┬┴,╙┘;╙┌ ╙╘┴ ╘1 ;T1=SY-SZ >>> ┴──┴,╙┘;╙┌ ╙╘┴ ╘2 ;T2=SY+SZ >>> ┴──┴,╙╪;╙┌ ╙╘┴ ╘3 ;T3=SX+SZ >>> ╙╒┬┴,╙╪;╙┌ ╙╘┴ ╘4 ;T4=SX-SZ >>> ┴──┴,╙╪;╘2 ╙╘┴ ╘5 ;T5=SX+T2 >>> ╙╒┬┴,╙╪;╘1 ╙╘┴ ╘6 ;T6=SX-T1 >>> ┴──┴,╙╪;╘1 ╙╘┴ ╘7 ;T7=SX+T1 >>> ╙╒┬┴,╘2;╙╪ ╙╘┴ ╘8 ;T8=T2-SX >>> ╙╒┬┴,╙┘;╙╪ ╙╘┴ ╘9 ;T9=SY-SX >>> ┴──┴,╙╪;╙┘ ╙╘┴ ╘10 ;T10=SX+SY * ┼T VOILA! *** ╬EXT, CALCULATE ┴,┬,├,...,╔ ** ┴NOTHER USEFUL LITTLE MACRO ─╔╓2 ═┴├ ;─IVIDE A SIGNED NUMBER BY 2 ;╔T IS ASSUMED THAT THE NUMBER ┬╨╠ ╨╧╙ ;IS IN THE ACCUMULATOR ├╠├ ┼╧╥ #$╞╞ ;╫E NEED TO UN-NEGATIVE THE NUMBER ┴─├ #01 ;BY TAKING IT'S COMPLEMENT ╠╙╥ ;DIVIDE BY TWO ├╠├ ┼╧╥ #$╞╞ ┴─├ #01 ;═AKE IT NEGATIVE AGAIN ╩═╨ ─╧╬┼─╔╓ ╨╧╙ ╠╙╥ ;╬UMBER IS POSITIVE ─╧╬┼─╔╓ <<< ═╒╠2 ═┴├ ;═ULTIPLY A SIGNED NUMBER BY 2 ┬╨╠ ╨╧╙═ ├╠├ ┼╧╥ #$╞╞ ┴─├ #$01 ┴╙╠ ├╠├ ┼╧╥ #$╞╞ ┴─├ #$01 ╩═╨ ─╧╬┼═╒╠ ╨╧╙═ ┴╙╠ ─╧╬┼═╒╠ <<< ** ╬OTE THAT WE ARE CURRENTLY MAKING A MINOR LEAP ** OF FAITH THAT NO OVERFLOWS WILL OCCUR. :├┴╠├┴ ├╠├ ╠─╪ ╘1 ╠─┴ ├╧╙,╪ ╠─╪ ╘2 ┴─├ ├╧╙,╪ ╙╘┴ ┴11 ;┴=(COS(T1)+COS(T2))/2 :├┴╠├┬ ╠─╪ ╘1 ╠─┴ ╙╔╬,╪ ╙┼├ ╠─╪ ╘2 ╙┬├ ╙╔╬,╪ ╙╘┴ ┬12 ;┬=(SIN(T1)-SIN(T2))/2 :├┴╠├├ ╠─╪ ╙┘ ╠─┴ ╙╔╬,╪ >>> ═╒╠2 ╙╘┴ ├13 ;├=SIN(SY) :├┴╠├─ ╙┼├ ╠─╪ ╘8 ╠─┴ ├╧╙,╪ ╠─╪ ╘7 ╙┬├ ├╧╙,╪ ╙┼├ ╠─╪ ╘5 ╙┬├ ├╧╙,╪ ├╠├ ╠─╪ ╘6 ┴─├ ├╧╙,╪ ;─I=(COS(T8)-COS(T7)+COS(T6)-COS(T5))/2 >>> ─╔╓2 ├╠├ ╠─╪ ╘3 ┴─├ ╙╔╬,╪ ╙┼├ ╠─╪ ╘4 ╙┬├ ╙╔╬,╪ ╙╘┴ ─21 ;─=(SIN(T3)-SIN(T4)+─I)/2 :├┴╠├┼ ╙┼├ ╠─╪ ╘5 ╠─┴ ╙╔╬,╪ ╠─╪ ╘6 ╙┬├ ╙╔╬,╪ ╙┼├ ╠─╪ ╘7 ╙┬├ ╙╔╬,╪ ╙┼├ ╠─╪ ╘8 ╙┬├ ╙╔╬,╪ ;┼I=(SIN(T5)-SIN(T6)-SIN(T7)-SIN(T8))/2 >>> ─╔╓2 ├╠├ ╠─╪ ╘3 ┴─├ ├╧╙,╪ ├╠├ ╠─╪ ╘4 ┴─├ ├╧╙,╪ ╙╘┴ ┼22 ;┼=(COS(T3)+COS(T4)+┼I)/2 :├┴╠├╞ ╠─╪ ╘9 ╠─┴ ╙╔╬,╪ ╙┼├ ╠─╪ ╘10 ╙┬├ ╙╔╬,╪ ╙╘┴ ╞23 ;╞=(SIN(T9)-SIN(T10))/2 :├┴╠├╟ ╠─╪ ╘6 ╠─┴ ╙╔╬,╪ ╙┼├ ╠─╪ ╘8 ╙┬├ ╙╔╬,╪ ╙┼├ ╠─╪ ╘7 ╙┬├ ╙╔╬,╪ ╙┼├ ╠─╪ ╘5 ╙┬├ ╙╔╬,╪ ;╟I=(SIN(T6)-SIN(T8)-SIN(T7)-SIN(T5))/2 >>> ─╔╓2 ├╠├ ╠─╪ ╘4 ┴─├ ├╧╙,╪ ╙┼├ ╠─╪ ╘3 ╙┬├ ├╧╙,╪ ╙╘┴ ╟31 ;╟=(COS(T4)-COS(T3)+╟I)/2 :├┴╠├╚ ├╠├ ╠─╪ ╘6 ╠─┴ ├╧╙,╪ ╠─╪ ╘7 ┴─├ ├╧╙,╪ ╙┼├ ╠─╪ ╘5 ╙┬├ ├╧╙,╪ ╙┼├ ╠─╪ ╘8 ╙┬├ ├╧╙,╪ ;╚I=(COS(T6)+COS(T7)-COS(T5)-COS(T8))/2 >>> ─╔╓2 ├╠├ ╠─╪ ╘3 ┴─├ ╙╔╬,╪ ├╠├ ╠─╪ ╘4 ┴─├ ╙╔╬,╪ ╙╘┴ ╚32 ;╚=(SIN(T3)+SIN(T4)+╚I)/2 :╫╚┼╫ ├╠├ ╠─╪ ╘9 ╠─┴ ├╧╙,╪ ╠─╪ ╘10 ┴─├ ├╧╙,╪ ╙╘┴ ╔33 ;╔=(COS(T9)+COS(T10))/2 ** ╔T'S ALL DOWNHILL FROM HERE. ─╧╫╬╚╔╠╠ **** ├LEAR BUFFER * ┴ LITTLE MACRO ╙┼╘┬╒╞ ═┴├ ;╨UT BUFFERS WHERE THEY CAN BE HURT ╠─┴ #00 ╙╘┴ ┬╒╞╞┼╥ ╠─┴ ┌╘┼═╨ ;╚IGH BYTE ╙╘┴┬╒╞ ╙╘┴ ┬╒╞╞┼╥+1 <<< >>> ╙┼╘┬╒╞ ├╠╥─╥┴╫ ╠─╪ #08 ╠─┴ #00 :╞╧╧╠ ╠─┘ #00 :─╧╨┼ ╙╘┴ (┬╒╞╞┼╥),┘ ╔╬┘ ┬╬┼ :─╧╨┼ ╔╬├ ┬╒╞╞┼╥+1 ─┼╪ ┬╬┼ :╞╧╧╠ **** ═Y GOODNESS BUT ╔'M A DOPE *├╠╥─╥┴╫ ╠─┴ ╟╠╧┬╪═╔╬ * ╠╙╥ ;╬EED TO GET INTO THE RIGHT COLUMN * ┬├├ :┼╓┼╬ ;┼XPLAINED IN MORE DETAIL BELOW * ╠─┘ #$80 * ╙╘┘ ┬╒╞╞┼╥ ;╨RESUMABLY THIS WILL BE A LITTLE * ├╠├ ;MORE EFFICIENT. *:┼╓┼╬ ┴─├ ┬╒╞╞┼╥+1 * ╙╘┴ ┬╒╞╞┼╥+1 * ╠─┴ ╟╠╧┬╪═┴╪ * ╙┼├ * ╙┬├ ╟╠╧┬╪═╔╬ * ╘┴╪ * ╔╬╪ * ╠─┘ ╟╠╧┬┘═┴╪ * ┬┼╤ :╥┼╙┼╘ *:┘┴┘ ╠─┴ #$00 * ╠─┘ ╟╠╧┬┘═┴╪ *:┬╠┴╚ ╙╘┴ (┬╒╞╞┼╥),┘ * ─┼┘ * ├╨┘ ╟╠╧┬┘═╔╬ * ┬├╙ :┬╠┴╚ * ╠─┴ ┬╒╞╞┼╥ * ┼╧╥ #$80 * ╙╘┴ ┬╒╞╞┼╥ * ┬╬┼ :╫╚╧╨┼┼ * ╔╬├ ┬╒╞╞┼╥+1 *:╫╚╧╨┼┼ ─┼╪ * ┬╬┼ :┘┴┘ *:╥┼╙┼╘ ╠─┴ #0 ;╬EED TO RESET THESE GUYS * ╙╘┴ ╟╠╧┬╪═┴╪ * ╙╘┴ ╟╠╧┬┘═┴╪ * ╠─┴ #$╞╞ * ╙╘┴ ╟╠╧┬╪═╔╬ * ╙╘┴ ╟╠╧┬┘═╔╬ **** ╬EXT, READ AND DRAW POLYGONS ╥┼┴──╥┴╫ ╠─┘ #00 ╙╘┘ ╔╬─┼╪ ╧┬╩╠╧╧╨ ╠─┘ ╔╬─┼╪ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ;╞IRST, THE NUMBER OF POINTS ┬╬┼ :├╧╬╘ ;┬UT IF NUMPOINTS IS ZERO THEN ╩═╨ ╧┬╩─╧╬┼ ;WE ARE AT THE END OF THE LIST :├╧╬╘ ╙╘┴ ├╧╒╬╘╨╘╙ ╔╬├ ╔╬─┼╪ * ╥OTATE PROJECT AND DRAW THE POLYGON * ═AKE SURE BUFFER BEING DRAWN TO IS CLEAR! :─╧╔╘ ╩╙╥ ╥╧╘╨╥╧╩ * ├ONVERT XMIN AND XMAX TO COLUMNS ╠─┴ ╠╧├╪═╔╬ ╠╙╥ ╠╙╥ ╠╙╥ ;X MOD 8 ╙╘┴ ╠╧├╪═╔╬ ├═╨ ╟╠╧┬╪═╔╬ ┬├╙ :╬┴╚ ╙╘┴ ╟╠╧┬╪═╔╬ :╬┴╚ ╠─┴ ╠╧├┘═╔╬ ├═╨ ╟╠╧┬┘═╔╬ ┬├╙ :╒╚╒╚ ╙╘┴ ╟╠╧┬┘═╔╬ :╒╚╒╚ ╠─┴ ╠╧├╪═┴╪ ╠╙╥ ╠╙╥ ╠╙╥ ╙╘┴ ╠╧├╪═┴╪ ├═╨ ╟╠╧┬╪═┴╪ ┬├├ :╬╧╫┴┘ ╙╘┴ ╟╠╧┬╪═┴╪ :╬╧╫┴┘ ╠─┴ ╠╧├┘═┴╪ ├═╨ ╟╠╧┬┘═┴╪ ┬├├ ┼╧╥╞╔╠╠ ╙╘┴ ╟╠╧┬┘═┴╪ * ╔F USING THE ┼╧╥-BUFFER, COPY INTO DRAWING BUFFER * ┴ND THEN CLEAR THE ┼╧╥-BUFFER ┼╧╥╞╔╠╠ ╠─┴ ╞╔╠╠ ┬┼╤ ╧┬╩╠╧╧╨ >>> ╙┼╘┬╒╞ ╠─┴ #<┼╧╥┬╒╞ ╙╘┴ ╘┼═╨1 ╠─┴ #>┼╧╥┬╒╞ ╙╘┴ ╘┼═╨1+1 ╠─┴ ╠╧├╪═╔╬ ;╠╧├╪═╔╬ NOW CONTAINS COLUMN ╠╙╥ ;┼ACH COLUMN IS 128 BYTES ┬├├ :┼╓┼╬ ;╙O THERE MIGHT BE A CARRY ╠─┘ #$80 ╙╘┘ ┬╒╞╞┼╥ ╙╘┘ ╘┼═╨1 ├╠├ :┼╓┼╬ ╙╘┴ ╘2 ┴─├ ┬╒╞╞┼╥+1 ╙╘┴ ┬╒╞╞┼╥+1 ;┼ACH COLUMN IS 128 BYTES ╠─┴ ╘2 ┴─├ ╘┼═╨1+1 ;╬OW WE WILL START AT THE ╙╘┴ ╘┼═╨1+1 ;COLUMN ╠─┴ ╠╧├╪═┴╪ ╙┼├ ╙┬├ ╠╧├╪═╔╬ ╘┴╪ ;╘OTAL NUMBER OF COLUMNS TO DO ╔╬╪ ;E.G. FILL COLUMNS 1..3 ╠─┘ ╠╧├┘═┴╪ ┬╬┼ :╞╧╧╨ ╔╬├ ╠╧├┘═┴╪ :╞╧╧╨ ╠─┘ ╠╧├┘═┴╪ ╠─┴ #00 :╟╧╧╨ ┼╧╥ (╘┼═╨1),┘ ;┼╧╥-BUFFER ╨╚┴ * ═AYBE PUT AN ┼╧╥ BELOW? ┼╧╥ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╠─┴ #00 ;═IGHT AS WELL CLEAR IT NOW ╙╘┴ (╘┼═╨1),┘ ╨╠┴ ─┼┘ ├╨┘ ╠╧├┘═╔╬ ┬├╙ :╟╧╧╨ ╠─┴ ┬╒╞╞┼╥ ┼╧╥ #$80 ╙╘┴ ┬╒╞╞┼╥ ╙╘┴ ╘┼═╨1 ┬╬┼ :┬╧╧╨ ╔╬├ ┬╒╞╞┼╥+1 ╔╬├ ╘┼═╨1+1 :┬╧╧╨ ─┼╪ ┬╬┼ :╞╧╧╨ ╩═╨ ╧┬╩╠╧╧╨ ╧┬╩─╧╬┼ **** ╙WAP BUFFERS ╙╫┴╨┬╒╞ ╠─┴ ╓═├╙┬ ┼╧╥ #$02 ;╨RETTY TRICKY, EH? ╙╘┴ ╓═├╙┬ ╠─┴ #$08 ┼╧╥ ┌╘┼═╨ ;ZTEMP=HIGH BYTE JUST FLIPS ╙╘┴ ┌╘┼═╨ ;BETWEEN $30 AND $38 ╩═╨ ═┴╔╬ ;┴ROUND AND AROUND WE GO... ╘╪╘ '╟EE ┬RAIN, WHAT DO YOU WANT TO DO ' ╘╪╘ 'TONIGHT?' ** ╥OTATE, PROJECT, AND STORE THE POINTS * * ╘HIS PART IS A SIGNIFICANT CHANGE SINCE * V2.0. ╬OW IT IS A COMPLETELY GENERAL POLYGON PLOTTER. * ┴ SET OF POINTS IS READ IN, ROTATED AND PROJECTED, AND * PLOTTED INTO THE DRAWING BUFFER (┼╧╥ OR NORMAL). ╥╧╘╨╥╧╩ * ┴ NEAT MACRO ╬┼╟ ═┴├ ;├HANGE THE SIGN OF A TWO'S COMPLEMENT ├╠├ ╠─┴ ]1 ;NUMBER. ┼╧╥ #$╞╞ ┴─├ #$01 <<< *------------------------------- * ╘HESE MACROS REPLACE THE PREVIOUS PROJECTION * SUBROUTINE. ╙═╒╠╘ ═┴├ ;═ULTIPLY TWO SIGNED 8-BIT ;NUMBERS: ┴*┘/64 -> ┴ ╙╘┴ ┌3 ├╠├ ;╘HIS MULTIPLY IS FOR NORMAL ┼╧╥ #$╞╞ ;NUMBERS, I.E. X=-64..64 ┴─├ #$01 ╙╘┴ ┌4 ╠─┴ (┌3),┘ ╙┼├ ╙┬├ (┌4),┘ <<< ;┴LL DONE :) ╙═╒╠╘┌ ═┴├ ;═ULTIPLY TWO SIGNED 8-BIT ;NUMBERS: ┴*┘/64 -> ┴ ╙╘┴ ┌1 ├╠├ ;┴ND THIS MULTIPLY IS SPECIFICALLY ┼╧╥ #$╞╞ ;FOR THE PROJECTION PART, WHERE ┴─├ #$01 ;NUMBERS ARE -110..110 AND 0..40 ╙╘┴ ┌2 ╠─┴ (┌1),┘ ╙┼├ ╙┬├ (┌2),┘ <<< ;┴LL DONE :) ╨╥╧╩┼├╘ ═┴├ ;╘HE ACTUAL PROJECTION ROUTINE ;╘HE ROUTINE TAKES THE POINT ;]1 ]2 ]3, ROTATES AND ;PROJECTS IT, AND STORES THE ;RESULT IN ]1 ]2 ]3. ╠─┘ ]1 ;═ULTIPLY FIRST ROTATION COLUMN ╠─┴ ┴11 >>> ╙═╒╠╘ ╙╘┴ ╨1╘ ╠─┴ ─21 >>> ╙═╒╠╘ ╙╘┴ ╨2╘ ╠─┴ ╟31 >>> ╙═╒╠╘ ╙╘┴ ╨3╘ ╠─┘ ]2 ;╙ECOND COLUMN ╠─┴ ┬12 >>> ╙═╒╠╘ ├╠├ ┴─├ ╨1╘ ╙╘┴ ╨1╘ ╠─┴ ┼22 >>> ╙═╒╠╘ ├╠├ ┴─├ ╨2╘ ╙╘┴ ╨2╘ ╠─┴ ╚32 >>> ╙═╒╠╘ ├╠├ ┴─├ ╨3╘ ╙╘┴ ╨3╘ ╠─┘ ]3 ;╘HIRD COLUMN ╠─┴ ├13 >>> ╙═╒╠╘ ├╠├ ┴─├ ╨1╘ ╙╘┴ ╨1╘ ╠─┴ ╞23 >>> ╙═╒╠╘ ├╠├ ┴─├ ╨2╘ ╙╘┴ ╨2╘ ╠─┴ ╔33 >>> ╙═╒╠╘ ├╠├ ┴─├ ╨3╘ ╙╘┴ ]3 ;╥OTATED ┌ ╘┴╪ ╠─┘ ┌─╔╓,╪ ;╘ABLE OF D/(Z+Z0) ;╬OW ┘ CONTAINS PROJECTION CONST ╠─┴ ╨1╘ >>> ╙═╒╠╘┌ ╠─╪ ┌╧╧═ ├╨╪ #64 ┬┼╤ ├╧╬╘╪ ╙╘┘ ╘┼═╨1 ╠─┘ ┌╧╧═ >>> ╙═╒╠╘ ╠─┘ ╘┼═╨1 ├╧╬╘╪ ├╠├ ┴─├ #64 ;╧FFSET THE COORDINATE ╙╘┴ ]1 ;╥OTATED AND PROJECTED ├═╨ ╠╧├╪═╔╬ ;╙EE IF IT IS A LOCAL MINIMUM ┬├╙ ╬╧╘╪═╔╬ ╙╘┴ ╠╧├╪═╔╬ ╬╧╘╪═╔╬ ├═╨ ╠╧├╪═┴╪ ┬├├ ╬╧╘╪═┴╪ ╙╘┴ ╠╧├╪═┴╪ ╬╧╘╪═┴╪ ╠─┴ ╨2╘ >>> ╙═╒╠╘┌ ├╨╪ #64 ┬┼╤ ├╧╬╘┘ ╠─┘ ┌╧╧═ >>> ╙═╒╠╘ ├╧╬╘┘ ├╠├ ┴─├ #64 ╙╘┴ ]2 ;╥OTATED AND PROJECTED ┘ ├═╨ ╠╧├┘═╔╬ ┬├╙ ╬╧╘┘═╔╬ ╙╘┴ ╠╧├┘═╔╬ ╬╧╘┘═╔╬ ├═╨ ╠╧├┘═┴╪ ┬├├ ╬╧╘┘═┴╪ ╙╘┴ ╠╧├┘═┴╪ ╬╧╘┘═┴╪ <<< ;┴LL DONE * ╠─┴ #<┼╧╥┬╒╞ ;╞IRST WE NEED TO CLEAR THE * ╙╘┴ ┬╒╞╞┼╥ ;┼╧╥ BUFFER * ╠─┴ #>┼╧╥┬╒╞ * ╙╘┴ ┬╒╞╞┼╥+1 ╠─┴ #0 ;╥ESET ┘MIN AND ┘MAX ╙╘┴ ╠╧├┘═┴╪ ╙╘┴ ╠╧├╪═┴╪ ╠─┴ #$╞╞ ╙╘┴ ╠╧├┘═╔╬ ╙╘┴ ╠╧├╪═╔╬ ╥┼┴─╨╘╙ ╠─┘ ╔╬─┼╪ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨1╪ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨1┘ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨1┌ ╔╬┘ ─┼├ ├╧╒╬╘╨╘╙ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨2╪ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨2┘ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨2┌ ╔╬┘ ─┼├ ├╧╒╬╘╨╘╙ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨3╪ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨3┘ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨3┌ ╔╬┘ ╙╘┘ ╔╬─┼╪ >>> ╨╥╧╩┼├╘,╨1╪;╨1┘;╨1┌ >>> ╨╥╧╩┼├╘,╨2╪;╨2┘;╨2┌ >>> ╨╥╧╩┼├╘,╨3╪;╨3┘;╨3┌ ╠─┴ ╚╔─┼ ┬┼╤ :─╧╔╘ ╠─┴ ╨2╪ ;╚IDDEN FACE CHECK ╙┼├ ╙┬├ ╨1╪ ╘┴┘ ;┘=(X2-X1) ╠─┴ ╨3┘ ╙┼├ ╙┬├ ╨2┘ ;┴=(Y3-Y2) >>> ╙═╒╠╘ ╙╘┴ ╘┼═╨1 ╠─┴ ╨3╪ ╙┼├ ╙┬├ ╨2╪ ╘┴┘ ╠─┴ ╨2┘ ╙┼├ ╙┬├ ╨1┘ >>> ╙═╒╠╘ ├═╨ ╘┼═╨1 ;╔F X1*Y2-Y1*X2 > 0 THEN FACE ┬═╔ :─╧╔╘ ;IS VISIBLE ─┼├ ├╧╒╬╘╨╘╙ ;╧THERWISE READ IN REMAINING ┬┼╤ :┴┬╧╥╘ ;POINTS AND RETURN :╨╧╧╨ ╔╬├ ╔╬─┼╪ ╔╬├ ╔╬─┼╪ ╔╬├ ╔╬─┼╪ ─┼├ ├╧╒╬╘╨╘╙ ┬╬┼ :╨╧╧╨ :┴┬╧╥╘ ╥╘╙ :─╧╔╘ ╠─┴ ╨1╪ ╙╘┴ ╪1 ╠─┴ ╨1┘ ╙╘┴ ┘1 ╠─┴ ╨2╪ ╙╘┴ ╪2 ╠─┴ ╨2┘ ╙╘┴ ┘2 ╩╙╥ ─╥┴╫ ╠─┴ ╨2╪ ╙╘┴ ╪1 ╠─┴ ╨2┘ ╙╘┴ ┘1 ╠─┴ ╨3╪ ╙╘┴ ╪2 ╠─┴ ╨3┘ ╙╘┴ ┘2 ╩╙╥ ─╥┴╫ ─┼├ ├╧╒╬╘╨╘╙ ┬╬┼ ╨╧╠┘╠╧╧╨ ;╔S IT JUST A TRIANGLE? ╩═╨ ╨╧╠┘─╧╬┼ ╨╧╠┘╠╧╧╨ ╠─┘ ╔╬─┼╪ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨2╪ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨2┘ ╔╬┘ ╠─┴ ╨╧╠┘╠╔╙╘,┘ ╙╘┴ ╨2┌ ╔╬┘ ╙╘┘ ╔╬─┼╪ >>> ╨╥╧╩┼├╘,╨2╪;╨2┘;╨2┌ ╠─┴ ╨2╪ ╙╘┴ ╪1 ╠─┴ ╨2┘ ╙╘┴ ┘1 ╠─┴ ╨3╪ ╙╘┴ ╪2 ╠─┴ ╨3┘ ╙╘┴ ┘2 ╩╙╥ ─╥┴╫ ╠─┴ ╨2╪ ╙╘┴ ╨3╪ ╠─┴ ╨2┘ ╙╘┴ ╨3┘ ─┼├ ├╧╒╬╘╨╘╙ ┬┼╤ ╨╧╠┘─╧╬┼ ╩═╨ ╨╧╠┘╠╧╧╨ ╨╧╠┘─╧╬┼ ╠─┴ ╨1╪ ;├LOSE THE POLYGON ╙╘┴ ╪2 ╠─┴ ╨1┘ ╙╘┴ ┘2 ╠─┴ ╨3╪ ╙╘┴ ╪1 ╠─┴ ╨3┘ ╙╘┴ ┘1 ╩╙╥ ─╥┴╫ ╥╘╙ ╘╪╘ '╙AME THING WE DO EVERY NIGHT, ╨INKY: ' ╘╪╘ 'TRY TO TAKE OVER THE WORLD!' *------------------------------- * ╟ENERAL QUESTIONABLE-VALUE ERROR PROCEDURE *├╚╧╦┼ ╠─╪ #00 *:╠╧╧╨ ╠─┴ :├╘┼╪╘,╪ * ┬┼╤ :─╧╬┼ * ╩╙╥ ├╚╥╧╒╘ * ╔╬╪ * ╩═╨ :╠╧╧╨ *:─╧╬┼ ╥╘╙ *:├╘┼╪╘ ╚┼╪ 0─ ;├╥ * ╘╪╘ 'SOMETHING CHOKED :(' * ╚┼╪ 0─00 * ╘╪╘ '╬ARF!' *------------------------------- * ─RAWIN' A LINE. ┴ FAHN LAHN. *** ╙OME USEFUL MACROS ├╔╬╔╘ ═┴├ ;═ACRO TO INITIALIZE THE COUNTER ╠─┴ ]1 ;DX OR DY ╠╙╥ <<< ;╘HE DX/2 MAKES A NICER LOOKING LINE ***** ═ACRO TO TAKE A STEP IN ╪ ╪╙╘┼╨ ═┴├ ╠─╪ ─╪ ;╬UMBER OF LOOP ITERATIONS >>> ├╔╬╔╘,─╪ ╪╠╧╧╨ ╠╙╥ ├╚╒╬╦ ┬┼╤ ╞╔╪├ ;╒PDATE COLUMN ╙┬├ ─┘ ┬├├ ╞╔╪┘ ;╘IME TO STEP IN ┘ ─┼╪ ┬╬┼ ╪╠╧╧╨ ─╧╬┼ ╠─┴ ╧╠─╪ ;╨LOT THE LAST CHUNK ┼╧╥ ├╚╒╬╦ ╧╥┴ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╥╘╙ ╞╔╪├ ╨╚┴ ╠─┴ ╧╠─╪ ╧╥┴ (┬╒╞╞┼╥),┘ ;╨LOT ╙╘┴ (┬╒╞╞┼╥),┘ ╠─┴ #$╞╞ ;╒PDATE CHUNK ╙╘┴ ╧╠─╪ ╙╘┴ ├╚╒╬╦ ╠─┴ #$80 ;╔NCREASE THE COLUMN ┼╧╥ ┬╒╞╞┼╥ ╙╘┴ ┬╒╞╞┼╥ ┬╬┼ ├2 ╔╬├ ┬╒╞╞┼╥+1 ├2 ╨╠┴ ╙┬├ ─┘ ┬├╙ ├╧╬╘ ┴─├ ─╪ ╔╞ ╔,]1 ;─O WE USE ╔╬┘ OR ─┼┘? ╔╬┘ ┼╠╙┼ ─┼┘ ╞╔╬ ├╧╬╘ ─┼╪ ┬╬┼ ╪╠╧╧╨ ╩═╨ ─╧╬┼ ╞╔╪┘ ┴─├ ─╪ ╨╚┴ ╠─┴ ╧╠─╪ ┼╧╥ ├╚╒╬╦ ╧╥┴ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╠─┴ ├╚╒╬╦ ╙╘┴ ╧╠─╪ ╨╠┴ ╔╞ ╔,]1 ;╒PDATE ┘ ╔╬┘ ┼╠╙┼ ─┼┘ ╞╔╬ ─┼╪ ┬╬┼ ╪╠╧╧╨ ╥╘╙ <<< ;┼ND OF ═ACRO XSTEP ***** ╘AKE A STEP IN ┘ ┘╙╘┼╨ ═┴├ ╠─╪ ─┘ ;╬UMBER OF LOOP ITERATIONS ┬┼╤ ─╧╬┼ ;╔F DY=0 IT'S JUST A POINT >>> ├╔╬╔╘,─┘ ╙┼├ ┘╠╧╧╨ ╨╚┴ ╠─┴ ╧╠─╪ ╧╥┴ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╨╠┴ ╔╞ ╔,]1 ╔╬┘ ┼╠╙┼ ─┼┘ ╞╔╬ ╙┬├ ─╪ ┬├├ ╞╔╪╪ ─┼╪ ┬╬┼ ┘╠╧╧╨ ─╧╬┼ ╠─┴ ╧╠─╪ ╧╥┴ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╥╘╙ ╞╔╪╪ ┴─├ ─┘ ╠╙╥ ╧╠─╪ ╙┼├ ;╔MPORTANT! ┬┼╤ ╞╔╪├ ─┼╪ ┬╬┼ ┘╠╧╧╨ ╩═╨ ─╧╬┼ ╞╔╪├ ╨╚┴ ╠─┴ #$80 ╙╘┴ ╧╠─╪ ┼╧╥ ┬╒╞╞┼╥ ╙╘┴ ┬╒╞╞┼╥ ┬╬┼ ├2 ╔╬├ ┬╒╞╞┼╥+1 ├2 ╨╠┴ ─┼╪ ┬╬┼ ┘╠╧╧╨ ╩═╨ ─╧╬┼ <<< ;┼ND OF ═ACRO YSTEP * ╘AKE AN X STEP IN THE ┼╧╥ BUFFER * ╘HE SOLE CHANGE IS TO USE ┼╧╥ INSTEAD OF ╧╥┴ ┼╧╥╪╙╘┼╨ ═┴├ ╠─╪ ─╪ ;╬UMBER OF LOOP ITERATIONS >>> ├╔╬╔╘,─╪ ╪╠╧╧╨ ╠╙╥ ├╚╒╬╦ ┬┼╤ ╞╔╪├ ;╒PDATE COLUMN ╙┬├ ─┘ ┬├├ ╞╔╪┘ ;╘IME TO STEP IN ┘ ─┼╪ ┬╬┼ ╪╠╧╧╨ ─╧╬┼ ╠─┴ ╧╠─╪ ;╨LOT THE LAST CHUNK ┼╧╥ ├╚╒╬╦ ┼╧╥ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╥╘╙ ╞╔╪├ ╨╚┴ ╠─┴ ╧╠─╪ ┼╧╥ (┬╒╞╞┼╥),┘ ;╨LOT ╙╘┴ (┬╒╞╞┼╥),┘ ╠─┴ #$╞╞ ;╒PDATE CHUNK ╙╘┴ ╧╠─╪ ╙╘┴ ├╚╒╬╦ ╠─┴ #$80 ;╔NCREASE THE COLUMN ┼╧╥ ┬╒╞╞┼╥ ╙╘┴ ┬╒╞╞┼╥ ┬╬┼ ├2 ╔╬├ ┬╒╞╞┼╥+1 ├2 ╨╠┴ ╙┬├ ─┘ ┬├╙ ├╧╬╘ ┴─├ ─╪ ╔╞ ╔,]1 ;─O WE USE ╔╬┘ OR ─┼┘? ╔╬┘ ┼╠╙┼ ─┼┘ ╞╔╬ ├╧╬╘ ─┼╪ ┬╬┼ ╪╠╧╧╨ ╩═╨ ─╧╬┼ ╞╔╪┘ ┴─├ ─╪ ╨╚┴ ╠─┴ ╧╠─╪ ┼╧╥ ├╚╒╬╦ ┼╧╥ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╠─┴ ├╚╒╬╦ ╙╘┴ ╧╠─╪ ╨╠┴ ╔╞ ╔,]1 ;╒PDATE ┘ ╔╬┘ ┼╠╙┼ ─┼┘ ╞╔╬ ─┼╪ ┬╬┼ ╪╠╧╧╨ ╥╘╙ <<< ;┼ND OF ═ACRO XSTEP * ╘AKE A Y-STEP IN THE ┼╧╥-BUFFER * ├HANGES FROM ABOVE ARE: ONLY PLOT LAST PART OF EACH * VERTICAL CHUNK, DON'T PLOT LAST POINT, PLOT WITH ┼╧╥ ┼╧╥┘╙╘┼╨ ═┴├ ╠─╪ ─┘ ;╬UMBER OF LOOP ITERATIONS ┬┼╤ ─╧╬┼ ;╔F DY=0 IT'S JUST A POINT >>> ├╔╬╔╘,─┘ ╙┼├ *┘╠╧╧╨ ╨╚┴ * ╠─┴ ╧╠─╪ * ╧╥┴ (┬╒╞╞┼╥),┘ * ╙╘┴ (┬╒╞╞┼╥),┘ * ╨╠┴ ┘╠╧╧╨ ╔╞ ╔,]1 ╔╬┘ ┼╠╙┼ ─┼┘ ╞╔╬ ╙┬├ ─╪ ┬├├ ╞╔╪╪ ─┼╪ ┬╬┼ ┘╠╧╧╨ *─╧╬┼ ╠─┴ ╧╠─╪ * ╧╥┴ (┬╒╞╞┼╥),┘ * ╙╘┴ (┬╒╞╞┼╥),┘ ─╧╬┼ ╥╘╙ ╞╔╪╪ ┴─├ ─┘ ╨╚┴ ;╫E ONLY PLOT THE LAST PART OF EACH CHUNK ╠─┴ ╧╠─╪ ┼╧╥ (┬╒╞╞┼╥),┘ ╙╘┴ (┬╒╞╞┼╥),┘ ╨╠┴ ╠╙╥ ╧╠─╪ ╙┼├ ;╔MPORTANT! ┬┼╤ ╞╔╪├ ─┼╪ ┬╬┼ ┘╠╧╧╨ ╩═╨ ─╧╬┼ ╞╔╪├ ╨╚┴ ╠─┴ #$80 ╙╘┴ ╧╠─╪ ┼╧╥ ┬╒╞╞┼╥ ╙╘┴ ┬╒╞╞┼╥ ┬╬┼ ├2 ╔╬├ ┬╒╞╞┼╥+1 ├2 ╨╠┴ ─┼╪ ┬╬┼ ┘╠╧╧╨ ╩═╨ ─╧╬┼ <<< ;┼ND OF ═ACRO YSTEP **** ╔NITIAL LINE SETUP ** ╘HE COMMENTED LINES BELOW ARE NOW TAKEN CARE OF BY THE ** CALLING ROUTINE. *─╥┴╫ >>> ═╧╓┼,╘╪1;╪1 ;═OVE STUFF INTO ZERO PAGE * >>> ═╧╓┼,╘╪2;╪2 ;╫HERE IT CAN BE MODIFIED * >>> ═╧╓┼,╘┘1;┘1 * >>> ═╧╓┼,╘┘2;┘2 ─╥┴╫ ╠─┴ ╞╔╠╠ ┬╬┼ :╙┼╘┼╧╥ >>> ╙┼╘┬╒╞ ╩═╨ :╙┼╘╒╨ :╙┼╘┼╧╥ ╠─┴ #<┼╧╥┬╒╞ ;╒SE ┼╧╥ BUFFER INSTEAD OF ╙╘┴ ┬╒╞╞┼╥ ;DISPLAY BUFFER FOR DRAWING ╠─┴ #>┼╧╥┬╒╞ ╙╘┴ ┬╒╞╞┼╥+1 :╙┼╘╒╨ ╙┼├ ;═AKE SURE X1<X2 ╠─┴ ╪2 ╙┬├ ╪1 ┬├╙ :├╧╬╘ ╠─┴ ┘2 ;╔F NOT, SWAP ╨1 AND ╨2 ╠─┘ ┘1 ╙╘┴ ┘1 ╙╘┘ ┘2 ╠─┴ ╪1 ╠─┘ ╪2 ╙╘┘ ╪1 ╙╘┴ ╪2 ╙┼├ ╙┬├ ╪1 ;╬OW ┴=DX :├╧╬╘ ╙╘┴ ─╪ ╠─╪ ╪1 ;╨UT X1 INTO ╪, NOW WE CAN TRASH ╪1 ├╧╠╒═╬ ╘╪┴ ;╞IND THE FIRST COLUMN FOR ╪ ╠╙╥ ╠╙╥ ;╘HERE ARE X1/8 128 BYTE BLOCKS ╠╙╥ ;╫HICH MEANS X1/16 256 BYTE BLOCKS ╠╙╥ ┬├├ :┼╓┼╬ ;╫ITH A POSSIBLE EXTRA 128 BYTE BLOCK ╠─┘ #$80 ;IF SO, SET THE HIGH BIT ╙╘┘ ┬╒╞╞┼╥ ├╠├ :┼╓┼╬ ┴─├ ┬╒╞╞┼╥+1 ;┴DD IN THE NUMBER OF 256 BYTE BLOCKS ╙╘┴ ┬╒╞╞┼╥+1 ╙┼├ ╠─┴ ┘2 ;├ALCULATE DY ╙┬├ ┘1 ┬├╙ :├╧╬╘2 ;╔S Y2>Y1? ┼╧╥ #$╞╞ ;╧THERWISE DY=Y1-Y2 ┴─├ #$01 :├╧╬╘2 ╙╘┴ ─┘ ├═╨ ─╪ ;╫HO'S BIGGER: DY OR DX? ┬├├ ╙╘┼╨╔╬╪ ;╔F DX, THEN... ╩═╨ ╙╘┼╨╔╬┘ ╙╘┼╨╔╬╪ ╠─┘ ┘1 ├╨┘ ┘2 ╠─┴ ┬╔╘╨,╪ ;╪ CURRENTLY CONTAINS X1 ╙╘┴ ╧╠─╪ ╙╘┴ ├╚╒╬╦ ┬├├ ╪╔╬├┘ ;─O WE STEP FORWARDS OR BACKWARDS IN ┘? ╩═╨ ╪─┼├┘ ╪╔╬├┘ ╠─┴ ╞╔╠╠ ┬┼╤ ╬╧╥═╪╔╬├ >>> ┼╧╥╪╙╘┼╨,╔╬┘ ╬╧╥═╪╔╬├ >>> ╪╙╘┼╨,╔╬┘ ╪─┼├┘ ╠─┴ ╞╔╠╠ ┬┼╤ ╬╧╥═╪─┼├ >>> ┼╧╥╪╙╘┼╨,─┼┘ ╬╧╥═╪─┼├ >>> ╪╙╘┼╨,─┼┘ ╙╘┼╨╔╬┘ ╠─┘ ┘1 ╠─┴ ┬╔╘╨,╪ ;╪=X1 ╙╘┴ ╧╠─╪ ╠╙╥ ;┘ DOESN'T USE CHUNKS ┼╧╥ ╧╠─╪ ;╙O WE JUST WANT THE BIT ╙╘┴ ╧╠─╪ ├╨┘ ┘2 ┬├╙ ┘─┼├┘ ┘╔╬├┘ ╠─┴ ╞╔╠╠ ┬┼╤ ╬╧╥═╔╬├ >>> ┼╧╥┘╙╘┼╨,╔╬┘ ╬╧╥═╔╬├ >>> ┘╙╘┼╨,╔╬┘ ┘─┼├┘ ╠─┴ ╞╔╠╠ ┬┼╤ ╬╧╥═─┼├ >>> ┼╧╥┘╙╘┼╨,─┼┘ ╬╧╥═─┼├ >>> ┘╙╘┼╨,─┼┘ *------------------------------- * ├LEAN UP ├╠┼┴╬╒╨ ╠─┴ ╓═├╙┬ ;╙WITCH CHAR ROM BACK IN ┴╬─ #%11110101 ;DEFAULT ╙╘┴ ╓═├╙┬ ╥╘╙ ;BYE! ╘╪╘ 'SPINAL CRACKER ' ╘╪╘ 'SLJ 6/95' *------------------------------- * ╙ET UP BIT TABLE ─╙ ^ ;├LEAR TO END OF PAGE ;╙O THAT TABLES START ON A PAGE BOUNDARY ┬╔╘╨ ╠╒╨ 16 ;128 ┼NTRIES FOR ╪ ─╞┬ %11111111 ─╞┬ %01111111 ─╞┬ %00111111 ─╞┬ %00011111 ─╞┬ %00001111 ─╞┬ %00000111 ─╞┬ %00000011 ─╞┬ %00000001 --^ ╙╔╬ ;╘ABLE OF SINES, 120 BYTES ├╧╙ ┼╤╒ ╙╔╬+128 ;╘ABLE OF COSINES ;┬OTH OF THESE TRIG TABLES ARE ;CURRENTLY SET UP FROM ┬┴╙╔├ ┌─╔╓ ┼╤╒ ├╧╙+128 ;─IVISION TABLE ╘═┴╘╚1 ┼╤╒ ┌─╔╓+384 ;═ATH TABLE OF F(X)=X*X/256 ╘═┴╘╚2 ┼╤╒ ╘═┴╘╚1+512 ;╙ECOND MATH TABLE ╨╧╠┘╠╔╙╘ ┼╤╒ ╘═┴╘╚2+512 ;╠IST OF POLYGONS ======================================================================== ╙ECOND ╙╔─ ├HIP ╔NSTALLATION ├OPYRIGHT 1988 ═ARK ┴. ─ICKENSON ╘HIS INFORMATION AND SOFTWARE IS ├╧╨┘╥╔╟╚╘┼─ AND MADE AVAILABLE ON A ╙╚┴╥┼╫┴╥┼ BASIS. ╘HIS FILE CAN BE FREELY COPIED AND DISTRIBUTED AS LONG AS IT IS NOT ╙╧╠─. ╘HIS INFORMATION CANNOT BE USED TO CONSTRUCT AND SELL A HARDWARE DEVICE WITHOUT RECEIVING PRIOR PERMISSION FROM THE AUTHOR. ╘HERE IS NOT A SET FEE FOR THE USE OF THIS INFORMATION. ╩UST SEND IN WHATEVER YOU FEEL THE INFORMATION IS WORTH. ╔F YOU HAVE ANY GRIPES, COMPLAINTS, SUGGESTIONS, ├╧═╨╠╔═┼╬╘╙ OR ─╧╬┴╘╔╧╬╙ OF ANY SORT PLEASE SEND THEM TO: ═ARK ─ICKENSON 600 ╙OUTH ╫EST ╙TREET ╬EVADA, ═ISSOURI 64772 ┴DDING AN EXTRA ╙╔─ 6581/6582 CHIP ╘HIS IS NOT A PROJECT TO BE TACKLED BY THE SQEAMISH OR PEOPLE WHO ARE DEATHLY AFRAID OF OPENING THEIR COMPUTER JUST TO TAKE A PEEK INSIDE. ╬OW LET'S GET RID OF THE NASTY STUFF FIRST. ╬O LIABILITY IS ASSUMED WITH RESPECT TO THE USE OF THE FOLLOWING INFORMATION. ╔N OTHER WORDS IF YOU SCREW-UP TRYING TO INSTALL THIS MODIFICATION, THEN IT'S YOUR RESPONSABILITY. ┘╧╒ ─╧ ╘╚╔╙ ┴╘ ┘╧╒╥ ╧╫╬ ╥╔╙╦!!!! ╔F YOU DO NOT FEEL UP TO IT ╨╠┼┴╙┼ TAKE IT TO A ├OMMODORE REPAIR CENTER OR A REPAIR SERVICE THAT CAN WORK ON COMPUTERS AND LET THEM DO THE INSTALLATION. ╔ WILL WARN YOU THAT MOST ├OMMODORE ╥EPAIR ├ENTERS WILL NOT OR DO NOT LIKE TO DO THIS MODIFICATION. ╫HEN THEY DO, IT CAN BE EXPENSIVE. ╔F YOU BELONG TO A ╒SERS ╟ROUP, TELL THEM ABOUT THE PROJECT AND ASK IF THERE IS ANYONE THERE THAT COULD PERFORM THE OPERATION. ╘HIS MODIFICATION WILL ╬╧╘ HURT THE COMPUTER IN ANY WAY, UNLESS IT IS INSTALLED ╫╥╧╬╟. ┘OU CAN MAKE YOUR OWN PIGGY BACK BOARD OR YOU CAN DO WHAT ╔ AM GOING TO DESCRIBE (SINCE IT IS A LITTLE HARD TO PUT A SCHEMATIC IN A TEXT FILE). ┘OU SHOULD GROUND YOURSELF WITH A STATIC GUARD WRISTBAND (SUCH AS WHAT ╥ADIO ╙HACK SELLS). ┼VEN THOUGH THE CHIP IS QUITE DURABLE, JUST THE RIGHT STATIC DISCHARGE CAN RUIN ALL OR PART OF THE ╙╔─ CHIP. ╞OR THOSE OF YOU THAT ARE NOT FAMILIER WITH THE WAY PINS ARE NUMBERED ON AN ╔├ CHIP HERE IS A SHORT EXPLANATION. ╧N ONE END OF THE ╔├ YOU SHOULD FIND A LITTLE NOTCH, LOOKING AT THE CHIP WITH THE NOTCH AT THE TOP THE NUMBERING GOES THIS WAY. ╘HE UPPER LEFT CORNER OF THE CHIP IS PIN 1 AND THEY ARE NUMBERED CONSECUTIVELY, COUNTER-CLOCKWISE AROUND THE CHIP. ╙OME CHIPS DO NOT HAVE A NOTCH IN ONE END, BUT INSTEAD DOT IS PLACED IN ONE OF THE CHIP CORNERS TO DESIGNATE THAT PIN 1 STARTS IN THAT LOCATION. NOTCH ----,,---- 1-!. !-8 2-! DOT !-7 3-! !-6 4-! !-5 ---------- ╔ HAVE INCLUDED THE INFORMATION THAT IS NEEDED TO INSTALL THIS MODIFICATION ON THE ├OMMODORE 64, 64├ AND 128. ╔ HAVEN'T BEEN ABLE TO LOOK INSIDE THE 128─, SO ╔ CANNOT PROVIDE THE INFORMATION WITH ANY ACCURACY. ╘HERE ARE ╘╫╧ DIFFERENT 64├ CIRCUIT BOARDS AND BOTH USE ─╔╞╞┼╥┼╬╘ ╙╔─ CHIPS. ┘OU CAN TELL THE DIFFERENCE BY OPENING THE 64├. ╔F YOU SEE A 64-PIN CHIP ON THE BOARD AND THE BOARD IS ONLY 5.5-6 INCHES WIDE THEN YOU HAVE THE NARROW BOARD 64├ AND MUST USE THE 9 VOLT 6582 ╙╔─ CHIP. ╘HE NUMBER OF THE CHIP IN THE 64├ NARROW IS AN 8520 AND IS THE SAME AS THE 6582. ---------------------------------- ╨ARTS ├OMMODORE 64, 64├ (WIDE) & 128 1 - 6581 ╙╔─ CHIP FROM ╩AMCO OR ╦ASSARA ═ICROSYSTEMS 1 - 2╬2222 TRANSISTOR ╥ADIO ╙HACK 276-1617 2 - 220PF CAPACITORS ╥ADIO ╙HACK 272-124 ----------------------------------- ╨ARTS ├OMMODORE 64├ ╬ARROW ┬OARD 1 - 6582 ╙╔─ ├HIP ╞ROM ╩AMCO OR ╦ASSARA ═ICROSYSTEMS 1 - 2222┴ TRANSISTOR ╥ADIO ╙HACK 276-2009 2 - .022UF CAPACITORS ╥ADIO ╙HACK 272-1066 2 - 1K OHM 1/4 WATT RESISTORS ╥ADIO ╙HACK 271-1321 ----------------------------------- ╨ARTS 64, 64├ (ALL) & 128 2 - 1K OHM 1/4 WATT RESISTORS ╥ADIO ╙HACK 271-1321 1 - 1000 PF CAPACITOR ╥ADIO ╙HACK 272-126 LISTED AS .001 MF THIS IS THE SAME AS 1000PF 1 - 10K OHM 1/4 WATT RESISTOR ╥ADIO ╙HACK 271-1335 1 - 10 UF ELECTROLITIC CAPACITOR ╥ADIO ╙HACK 272-1025 1 - 5 INCH LENGTH OF WIRE 1 - 5 INCH LENGTH OF SHIELDED CABLE 1 - SURFACE MOUNT FEMALE ╥├┴ PLUG (THIS IS WHAT YOU NORMALLY FIND ON THE BACK OF YOUR STEREO. ╧N THE ├-64 AND 64├ (WIDE) THE ╙╔─ IS ╔├ ╒18 (THE ╔├ NUMBER WILL BE MARKED IN WHITE ON THE CIRCUIT BOARD). ╔T IS USUALLY LOCATED IN THE MIDDLE OF THE CIRCUIT BOARD, NEXT TO THE METAL VIDEO CHIP CASE OR UP BETWEEN AND JUST BELOW THE SERIAL AND MONITOR JACKS. ╧N THE ├-64├ (NARROW BOARD) THE ╙╔─ CHIP IS ╔├ ╒9. ╔T IS LOCATED IN THE MIDDLE OF THE BOARD, JUST A LITTLE TO THE RIGHT OF CENTER) AND CALLED 520. ╧N THE ├-128 THE ╙╔─ IS ╔├ ╒5. ╔T IS LOCATED AT THE BACK OF THE CIRCUIT BOARD JUST TO THE RIGHT OF THE METAL HOUSING FOR THE 40 AND 80 COLUMN VIDEO CHIPS. ╞IRST BEND OUT PINS 23, 24 AND 26 AND CUT THEM OFF OF THE 6581/6582 ╙╔─ CHIP. ╘HESE ARE FOR THE TWO ANALOG AND ONE AUDIO INPUT LINES. ╘HEY WILL CAUSE PROBLEMS IF CONNECTED AND SINCE THEY WILL NOT BE USED IT IS BEST TO REMOVE THEM. ╬OW BEND OUT PINS 1, 2, 3, 4, 8, AND 27. ╙OLDER ONE OF THE 220PF CAPACITORS (64├ NARROW USES .022 UF) TO PINS 1 AND 2 THEN SOLDER THE OTHER 220PF (64├ NARROW - .022UF) CAPACITOR TO PINS 3 AND 4. ╘HE CAPACITORS CONTROL THE UPPER AND LOWER FREQUENCY RANGE AND FILTERS OF THE ╙╔─ CHIP. ╘HE REASON ╔ AM USING 220PF CAPACITORS IS BECAUSE OF PROBLEMS WITH THE FILTERS IN THE ╙╔─ CHIP. ╘HE ├-64 FIRST CAME OUT WITH 2200PF CAPACITORS, BUT THEY WERE CHANGED TO 470PF. ╘HE REASON FOR THIS WAS BECAUSE THE FILTERS OF THE ╙╔─ VARY FROM CHIP TO CHIP AND USING 2200PF CAUSED A LOT OF THEM TO SOUND MUFFELED WHEN THE FILTERS WERE ON. ╔ HAVE FOUND THAT BY LOWERING THE CAPACITOR VALUE TO 220 PF HELPS EVEN MORE. ╔F YOU WISH, YOU CAN USE 470S IF YOU FEEL IT WOULD BE BETTER, BUT ─╧ ╬╧╘ USE 2200PF. ╘HE 6582 ╙╔─ CHIP FOR THE 64├ NARROW MUST USE THE .022UF CAPACITORS, AS THE FILTER RANGE IS MUCH DIFFERENT. ╙OLDER ONE END OF YOUR WIRE TO PIN 8 OF THE ╙╔─ CHIP. ╘HIS IS FOR THE CHIP SELECT LINE. ╫E WILL CONNECT THIS TO THE CARTRIDGE PORT. ╘HIS TELLS THE COMPUTER WHERE IN MEMORY THE CHIP RESIDES (DESCRIBED LATER). ╬OW SOLDER THE REMAINING PINS (EXCLUDING THE ONES WE HAVE BENT OUT AND/OR REMOVED 1, 2, 3, 4, 8, 23, 24, 26 AND 27) TO THE SID CHIP CURRENTLY IN YOUR COMPUTER. ┘OU MAY HAVE TO BEND THOSE PINS INWARD JUST A LITTLE FOR THEM TO GET A GOOD GRIP ON THE ╙╔─ CHIP. ┬E VERY CAREFUL NOT LEAVE THE SOLDERING IRON ON THE CHIP ╘╧╧ LONG AS YOU COULD RUIN ┬╧╘╚ ╙╔─ CHIPS. ╔ WOULD PUT SOME HEAT SINK (SILICON GREASE) BETWEEN THE TWO CHIPS BEFORE SOLDERING THEM TOGETHER. ╘HIS WILL PROVIDE BETTER HEAT DISPERSAL ON THE BOTTOM CHIP. ╬OW THAT YOU HAVE THE CHIPS SOLDERED TOGETHER (PLACE THE ╙╔─ CHIPS BACK IN THE SOCKET IF YOU REMOVED THEM), SOLDER THE WIRE FROM PIN 8 (ON THE ╙╔─ CHIP) TO PIN 7 OF THE CARTRIDGE PORT ON THE BACK OF THE COMPUTER. ╙ET THE COMPUTER INFRONT OF YOU LIKE TO ARE GETTING READY TO TYPE, WITH THE BACK OF THE COMPUTER AWAY FROM YOU. ╠OOK AT THE CARTRIDGE PORT (LOCATED IN THE UPPER RIGHT CORNER OF THE CIRCUIT BOARD). ┘OU WILL SEE TWO ROWS OF PINS CONNECTING THE CARTRIDGE PORT TO THE CIRCUIT BOARD. ┘OU WANT THE ROW OF PINS CLOSEST TO THE FRONT OF THE COMPUTER. ╬OW, COUNT THE PINS STARTING AT THE ╠┼╞╘ SIDE AND COUNTING TO THE RIGHT. ┘OU WANT TO SOLDER THE WIRE FROM PIN 8 OF THE EXTRA ╙╔─ CHIP TO PIN NUMBER 7 OF THE CARTRIDGE PORT. ╘HIS IS THE SAME PLACE ON ALL OF THE MODELS ├-64, 64├ AND 128. ╘HIS WILL TELL THE COMPUTER THAT THE EXTRA ╙╔─ CHIP IS AT ADDRESS $─┼00 HEX OR 56832 DECIMAL. ┘OU WILL ACCESS IT JUST LIKE YOU WOULD THE REGULAR SID CHIP BUT STARTING AT THIS ADDRESS. ╔ AM NO LONGER DESCRIBING HOW TO CONNECT FOR ADDRESS $─╞00. ╘HIS ADDRESS CAUSES PROBLEMS WITH THE ╥┴═ ┼XPANSION ╒NITS AND NUMEROUS OTHER CARTRIDGES. ╞ROM NOW ON ADDRESS $─┼00 IS THE ╧╬╠┘ ADDRESS FOR THE ╙╔─ CHIP. ╬OW PARTIALLY REASSEMBLE YOUR COMPUTER (BE CAREFUL THAT NOTHING SHORTS OUT THE PINS STILL STICKING OUT). ╘URN THE COMPUTER ON AND LOAD THE PLAYER PROGRAM PROVIDED AND TELL IT TO LOAD IN '╘┼╙╘'. ╔F YOU GET SOUND THEN SO FAR SO GOOD. ╘URN OFF THE COMPUTER AND DISASSEMBLE THE CASE. ─RILL A HOLE IN THE BACK END OF THE COMPUTER JUST LARGE ENOUGH TO ANCHOR THE ╥├┴ PLUG. ╘HEN SOLDER THE CENTER WIRE OF THE SHIELDED CABLE TO THE CENTER POST OF THE ╥├┴ PLUG. ╔NSERT THE WIRE THROUGH THE HOLE YOU HAVE JUST DRILLED AND ANCHOR THE PLUG TO THE CASE. ╬OW SOLDER THE GROUND WIRE TO THE GROUND TAB ON THE ╥├┴ PLUG. ╚ERE COMES THE DIFFICULT PART TO EXPLAIN. ╘HIS IS THE COUPLING CIRCUIT FOR THE AUDIO OUTPUT. ╚ERE IS A ROUGH SCHEMATIC. ╨IN 27 ON 12VOLTS DC, 9VOLTS 64├ (NARROW) ╙╔─ CHIP RESISTOR !--. 10K OHM !COLLECTOR 27!----.--/!/!/--.-----╧ 2N2222 OR 2222┴ --' ! ! !EMITTER ! ! ! <RESISTOR ! ! >1K ! ! + =========================================================================== ╙╧╠╓╔╬╟ ╠┴╥╟┼ ╙┘╙╘┼═╙ ╧╞ ╠╔╬┼┴╥ ┼╤╒┴╘╔╧╬╙ ╧╬ ┴ ├64 ╫╔╘╚╧╒╘ ═┼═╧╥┘ BY ┴LAN ╩ONES (ALAN.JONES@QCS.ORG) ╧╦, NOW THAT ╔ HAVE YOUR ATTENTION, ╔ LIED. ┘OU CAN'T SOLVE DENSE LINEAR SYSTEMS OF EQUATIONS BY DIRECT METHODS WITHOUT USING MEMORY TO STORE THE PROBLEM DATA. ╚OWEVER, ╔'LL COME BACK TO THIS MEMORY FREE ASSERTION LATER. ╘HE MAIN PURPOSE OF THIS ARTICLE IS TO RESCUE A USEFULL NUMERICAL ALGORITHM, "╤UARTERSOLVE", AND ALSO TO PROVIDE A BRIEF LOOK AT THE ├╧═┴╠ PROGRAMMING LANGUAGE AND ┬╠┴╙ ROUTINES. ╠INEAR SYSTEMS OF EQUATIONS, ┴(,)*X()=B(), WHERE ┴ IS A SQUARE MATRIX AND X AND B ARE VECTORS (OR ARRAYS), MUST OFTEN BE SOLVED FOR X IN THE SOLUTION OF A VARIETY OF PROBLEMS. ╘HE SIZE OR DIMENSION OF THE PROBLEM IS N AND JUST STORING ┴ REQUIRES 5*N*N BYTES OF MEMORY, ASSUMING ├64/128 5 BYTE REAL VARIABLES. ╘HE PREFERED SOLUTION METHOD IS A FORM OF ╟AUSSIAN ┼LIMINATION WHICH REQUIRES 1/3 N*N*N MULTIPLICATIONS. ╔'LL IGNORE THE ADDITIONAL N*N AND N MULTIPLIES. ╞OR LARGE PROBLEMS OUR ├64 HAS TWO SERIOUS LIMITATIONS, SMALL MEMORY SIZE AND SLOW FLOATING POINT ARITHMETIC. ╨ROBLEMS WITH N=10 CAN BE COMPUTED EASILY. ╨ROBLEMS WITH N=100 WILL REQUIRE 100 TIMES MORE MEMORY AND 1000 TIMES MORE COMPUTING TIME. ╘HE COMPUTING TIME IS NOT A REAL PROBLEM. ╔ DON'T MIND LETTING MY COMPUTER RUN WHILE ╔ WATCH A MOVIE, SLEEP, OR GO ON A TRIP. ├ALCULATING OR SETTING UP THE PROBLEM MAY TAKE MUCH LONGER THAN ITS SOLUTION ANYWAY. ┴VAILABLE MEMORY IS THE PRACTICAL LIMITING FACTOR. ┴FTER WE USE UP AVAILABLE ╥┴═ WE HAVE TO RESORT TO OTHER ALGORITHMS THAT WILL USE THE DISK DRIVE TO MOVE DATA IN AND OUT OF THE COMPUTER. ╘HE 1541 DRIVE IS PARTICULARLY SLOW AND ╔ WOULD NOT WANT TO SUBJECT IT TO UNDUE WEAR AND TEAR. ╚OW BIG A PROBLEM DO WE NEED TO BE ABLE TO SOLVE? ╔N MANY CASES THE PROBLEM ITSELF WILL FIX N AND THERE IS NO WAY TO REDUCE IT. ╔N OTHER CASES YOU MIGHT BE MODELING A REAL CONTINUOUS PROBLEM WITH A DISCRETE NUMBER OF ELEMENTS. ╬ SHOULD BE INFINITY BUT FOR PROBLEM SOLUTION N=50 MIGHT BE BIG ENOUGH. ├ONSIDER CALCULATING THE AERODYNAMIC POTENTIAL FLOWFIELD AROUND A BODY OF REVOLUTION. ┘OU COULD FIX POINTS ON THE SURFACE OF THE BODY (A MERIDIAN) AND HAVE A SERIES OF SORT LINE SEGMENTS MAKE UP ELEMENTS TO APPROXIMATE THE SHAPE. ╘HE LAGER N IS THE CLOSER THE SMOOTH SHAPE IS APROXIMATED AND THE MORE ACCURATE THE COMPUTED SOLUTION BECOMES. N=100 MIGHT BE A GOOD CHOICE FOR A SIMPLE SHAPE. ╫E COULD ALSO USE A "HIGHER ORDER" MENTHOD. ╔N THIS CASE WE CAN SUBSTITUTE A CURVED LINE ELEMENT FOR THE STRAIGHT LINE SEGMENT. ├ALCULATING THE MATRIX ELEMENTS WILL BE MORE DIFFICULT BUT N=40 CURVED ELEMENTS MIGHT GIVE A MORE ACCURATE SOLUTION THAN 100 FLAT ELEMENTS. ┴NOTHER CONSIDERATION IS THE RESOLUTION OF THE SOLUTION. ┘OU MIGHT WANT TO PLOT THE SOLUTION ON THE 200 X 320 PIXEL HI-RES ├64 SCREEN. 40 POINTS MIGHT BE TOO COARSE AND 320 MIGHT BE OVERKILL. ╫E MIGHT ALSO NEED TO CALCULATE THE SLOPE OR DERIVATIVES FROM THE CALCULATED SOLUTION WHICH WILL REQUIRE MORE CLOSELY SPACED SOLUTION POINTS. ╘HERE ARE OFTEN CHOICES THAT YOU CAN MAKE IN MODELING A SYSTEM AND SELECTING A SOLUTION ALGORITHM SO THAT A PROBLEM CAN BE SOLVED WITHIN THE LIMITS OF A ├64. ╘HERE ARE OFTEN INTERESTING TRADEOFFS IN MEMORY REQUIREMENTS AND EXECUTION SPEED. ╚OW BIG A PROBLEM CAN WE SOLVE WITH A ├64? ╒SING ╤UARTERSOLVE WITH ASSEMBLY LANGUAGE WE CAN PROBABLY DO N=200 OR MORE. ╔F WE ARE GOING TO STORE THE PROBLEM DATA ON A SINGLE 1541 DISKETTE AND READ IT IN A ROW AT TIME WE CAN ONLY DO N=182 OR SO. ┴CTUALLY ╔ THINK N SHOULD BE WELL UNDER 100. ─IFFERENT OPERATING SYSTEMS AND LANGUAGES LIMIT THE AMOUNT OF USEABLE ╥┴═; ┬┴╙╔├ 40╦, ├╧═┴╠ 2.0 30╦, ╟┼╧╙ 23╦, THE INITIAL DISK LOADED ├╧═┴╠ 0.14 10╦... ╙OLVING A LINEAR SYSTEM MAY ONLY BE A SMALL SUBPROBLEM INSIDE A LARGE APPLICATION PROGRAM. ╘HE IDEA IS TO BE ABLE TO SOLVE REASONABLE SIZED PROBLEMS USING YOUR PREFERED COMPUTING ENVIRONMENT WITHOUT HAVING TO DO A LOT OF CHAINING OR LOADING OF SEPARATE PROGRAMS. ╤UARTERSOLVE CAN FREE UP A LOT OF MEMORY FOR OTHER ROUTINES OR ALLOW N TO BE DOUBLED. ╙╨┼┼─ ╘HERE ARE A FEW THINGS THAT WE CAN DO TO SPEED UP THE CALCULATIONS. ╞IRST WE CAN SELECT A FAST PROGRAMMING LANGUAGE. ╔ PREFERE ├╧═┴╠ 2.0 WHICH IS A FAST THREE PASS INTERPRETER. ╒SING AN ASSEMBLER COULD BE THE FASTEST AND PROVIDE THE MOST USEABLE MEMORY. ┴ TRUE COMPILER SUCH AS ├ OR ╨ASCAL COULD ALSO BE A GOOD CHOICE. ┬┴╙╔├ IS A POOR CHOICE EXCEPT THAT IT IS BUILT IN AND FREE. ╔N MOST CASES EXECUTION CAN BE SPED UP WITH SOME MACHINE LANGUAGE ROUTINES LIKE THE ┬╠┴╙ (┬ASIC ╠INEAR ┴LGEBRA ╙UBROUTINES). ├ALCULATION SPEED IS MEASURED IN ╞╠╧╨╙/SEC (╞LOATING ╨OINT ╧╨ERATION╙) WHERE, C(I#,J#):=C(I#,J#) + A(I#,K#)*B(K#,J#) IS THE OPERATION. ╔T IS ONE ╞╨ MULTIPLY, ONE ╞╨ ADD, AND SOME INDEXING OVERHEAD. ╫ITH SOME INTERPRETERS THE INDEXING AND INTERPRETING OVERHEAD CAN FAR EXCEED THE ACTUAL ╞╨ MULTIPLY TIME. ╫ITH ASSEMBLED CODE THE ╞╨ MULTIPLY TIME SHOULD DOMINATE. ╔ USE A ═╠ LEVEL 1 ┬╠┴╙ PACKAGE WITH ├╧═┴╠ 2.0. ╞OR EXAMPLE: C(I#,╩#):+SDOT(N#,A(I#,1),1,B(1,J#),SDB#) ╞╧╥ K#:=1 TO N# DO C(I#,J#):+A(I#,K#)*B(K#,J#) BOTH CALCULATE THE SAME THING, A DOT PRODUCT WITH N# ╞╠╧╨╙. ╞OR LARGE N# ON A ├64 THE ┬╠┴╙ APPROACH ABOUT 320 ╞╠╧╨╙/SEC., ╘HE OVERHEAD OF CALLING THE PROCEDURE FROM THE INTERPRETER IS ABOUT THE EQUIVALENT OF 4 ╞╠╧╨╙. ╧F COURSE MODERN COMPUTER PERFORMANCE IS MEASURED IN ═EGA╞╠╧╨╙/SEC. WITH 8 BYTE REALS (SUPER COMPUTERS RUN HUNDREDS OR THOUSANDS OF ═╞╠╧╨╙/SEC.). ╘HEY ALSO INFLATE THE PERFORMANCE BY COUNTING THE MULTIPLY AND ADD AS TWO ╞╠╧╨╙. ╔N HIS ARTICLE ╔ USE THE "OLD FLOPS" OR NUMBER OF MULTIPLIES. ╔T MAY ALSO BE POSSIBLE TO CODE 6502 ╞╨ ARITHMETIC ROUTINES USING LOOKUP TABLES THAT MAY PERFORM FASTER THAN THE BUILT IN ROUTINES. ╫E COULD ALSO USE THE ├╨╒ IN THE DISK DRIVES TO DO DISTRIBUTED PROCESSING. ┬UT THIS IS BEYOND THE SCOPE OF THIS ARTICLE. ╙╧╠╒╘╔╧╬ ═┼╘╚╧─╙ ├ONSIDER THE FOLLOWING CHOICES FOR NUMERICAL SOLUTION ALGORITHMS: ═┼╘╚╧─ ═┼═╧╥┘ ╞╠╧╨╙ ╟AUSSIAN ┼LIMINATION N*N 1/3 N*N*N ├HOLESKY ─ECOMPOSITION 1/2 N*N 1/6 N*N*N ╤╥ DECOMPOSITION N*N 2/3 N*N*N ╤╥ UPDATING 1/2 N*N 2 N*N*N ╟AUSS-╩ORDAN N*N 1/2 N*N*N ╤UARTERSOLVE 1/4 N*N 1/2 N*N*N ╟AUSSIAN ┼LIMINATION IS THE PREFERED METHOD WHEN ENOUGH MEMORY IS AVAILABLE. ╔N MODERN TERMINOLOGY THIS IS ╠╒ DECOMPOSITION WHERE ┴ IS DECOMPOSED OR FACTORED INTO A LOWER TRIANGULAR MATRIX AND AN UPPER TRIANGULAR MATRIX. ╨ARTIAL PIVOTING OF ROWS OR COLUMNS IS AN ADDITIONAL COMPLICATION OFTEN REQUIRED FOR A STABLE SOLUTION. ┴FTER THE ╠╒ DECOMPOSTION YOU CAN READILY SOLVE FOR ANY NUMBER OF RIGHT HAND SIDE VECTORS IN N*N FLOPS EACH. ╔N ADDITION YOU CAN CALCULATE MATRIX CONDITION NUMBER ESTIMATES AND USE ITERATIVE IMPROVEMENT TECHNIQUES. ╘HE ╠╒ DECOMPOSITION IS DONE IN PLACE OVERWRITING THE PROBLEM MATRIX ┴. ├HOLESKY ─ECOMPOSITION IS A SPECIALIZED VERSION OF ╟AUSSIAN ┼LIMINATION FOR SYMETRIC POSITIVE DEFINITE MATRICES ONLY. ╙INCE ┴ IS SYMETRIC WE ONLY NEED N*(N+1)/2 MEMORY STORAGE LOCATIONS. ╘HE ╠ AND ╒ TRIANGULAR MATRICES ARE SIMPLY TRANSPOSES OF THE OTHER SO ONLY ONE NEEDS TO BE STORED AND IS COMPUTED IN PLACE OVERWRITING THE ORIGINAL STORAGE USED FOR ┴. ╬O PIVOTING IS REQUIRED. ╘HIS ALGORITHM CANNOT SOLVE GENERAL NONSYMETRIC PROBLEMS AND IS INCLUDED ONLY FOR COMPARISON. ╤╥ DECOMPOSITION FACTORS ┴ INTO AN ORTHOGONAL MATRIX ╤ AND A TRIANGULAR MATRIX ╥. ╤╥ DECOMPOSITION IS VERY STABLE AND CAN BE PERFORMED WITHOUT PIVOTING. ╙INCE ╤ IS ORTHOGONAL ITS INVERSE IS JUST ╤ TRANSPOSE. ╘O SOLVE THE LINEAR SYSTEM WE MULTIPLY THE RIGHT HAND SIDE VECTOR BY ╤ TRANSPOSE THEN SOLVE THE TRIANGULAR SYSTEM ╥. ╤ IS COMPUTED IN A SPECIAL COMPACT FORM AND STORED IN THE SPACE BELOW ╥. ╘HE DECOMPOSITION IS DONE IN PLACE IN THE STORAGE USED FOR ┴, PLUS AN ADDITIONAL N STORAGE LOCATIONS. ╤╥ DECOMPOSITION REQUIRES ABOUT TWICE AS MANY FLOPS AS ╟AUSSIAN ┼LIMINATION. ╘HERE IS A VARIATION OF THE ╤╥ SOLUTION KNOWN AS ╤╥ UPDATING. ╘HE PROBLEM IS SOLVED A ROW AT A TIME. ┴ ╥OW OF ┴ CAN BE READ IN FROM DISK STORAGE OR CALCULATED AS NEEDED. ╧NLY ╥ NEEDS TO BE STORED IN MAIN MEMORY, N*(N+1)/2 MEMORY LOCATIONS. ╥ IS INITIALY THE IDENTITY MATRIX AND IS UPDATED AS EACH ROW OF ┴ AND ITS RIGHT HAND SIDE ELEMENT ARE PROCESSED. ╤ IS NOT STORED, BUT THE RIGHT HAND SIDE VECTOR IS SEQUENTIALY MULTIPLIED BY ╤ TRANSPOSE. ┴FTER ALL N ROWS HAVE BEEN PROCESSED THE SOLUTION IS FOUND BY SIMPLY SOLVING THE TRIANGULAR SYSTEM ╥. ╙INCE THIS METHOD ONLY NEEDS HALF AS MUCH MEMORY STORAGE AS ╠╒ DECOMPOSITION, WE CAN SOLVE PROBLEMS 40% LARGER IN A LIMITED MEMORY SPACE. ╚OWEVER, THE COST IN FLOPS IS HIGH. ┴CTUALLY ╤╥ UPDATING IS BEST USED FOR SOLVING LARGE OVERDETERMINED LEAST SQUARES PROBLEMS. ╟AUSS-╩ORDAN IS A VARIATION OF ╟AUSSIAN ┼LIMINATION THAT REDUCES ┴ TO THE ╔DENTITY MATRIX INSTEAD OF TO ╠╒ FACTORS. ┬Y APPLYING THE SAME TRANSFORMATIONS TO TO THE RIGHT HAND SIDE THAT REDUCE ┴ TO THE IDENTITY MATRIX, THE RIGHT HAND SIDE BECOMES THE SOLUTION AT COMPLETION. ╨IVOTING IS REQUIERED. ╟AUSS-╩ORDAN REQUIRES ABOUT 50% MORE FLOPS THAN ╟AUSSIAN ┼LIMINATION AND MOST CODES USE N*N MEMORY STORAGE. ╙INCE THE ╠╒ FACTORS ARE NOT COMPUTED WE CAN'T SOLVE ADDITIONAL RIGHT HAND SIDE VECTORS LATER, OR ESTIMATE THE MATRIX CONDITION NUMBER, OR USE ITERATIVE IMPROVEMENT TECHNIQUES. ╔T WILL SOLVE MULTIPLE RIGHT HAND SIDES THAT ARE AVAILABLE FROM THE START. ╤UARTERSOLVE IS A CLEVER IMPLEMENTATION OF ╟AUSS-╩ORDAN(?) THAT SOLVES THE PROBLEM A ROW AT A TIME LIKE ╤╥ UPDATING BUT ONLY REQUIRES 1/4 N*N MEMORY STORAGE. ╫ITH FIXED AVAILABLE MEMORY ╤UARTERSOLVE CAN SOLVE A PROBLEM TWICE AS LARGE AS ╟AUSSIAN ┼LIMINATION BUT WITH A MODEST PERFORMANCE PENALTY. ╙OLVING A 2N PROBLEM WITH ╤UARTERSOLVE WOULD TAKE 12 TIMES LONGER (INSTEAD OF 8) THAN ╟AUSSIAN ┼LIMINATION ON A SIZE N PROBLEM. ═Y RECOMMENDATION IS TO USE ╟AUSSIAN ELIMINATION FOR SOLVING DENSE GENERAL SYSTEMS OF LINEAR EQUATIONS WHEN ENOUGH MAIN MEMORY IS AVAILABLE AND SWITCH TO ╤UARTERSOLVE FOR LARGER PROBLEMS. ╞OR SOLVING HUGE PROBLEMS REQUIERING EXTERNAL STORAGE A BLOCKED VERSION OF ╤╥ DECOMPOSITION MIGHT WORK BEST. ├HOLESKY DECOMPOSITION SHOULD BE USED FOR SYMETRIC POSITIVE DEFINITE PROBLEMS. ╠ARGE PROBLEMS ARE OFTEN SPARSE, CONTAINING LOTS OF ZEROS THAT NEED NOT BE STORED. ╙PECIALIZED CODE EXISTS FOR SOLVING MANY SPARCE PROBLEMS, PARTICULARLY BANDED MATRICES, AND MANY OF THESE METHODS CAN BE USED ON A ├64. ├ODES FOR SOLVING UNSTRUCTURED SPARCE PROBLEMS ARE NOT VERY SUITABLE FOR THE ├64 SINCE THEY ARE COMPLEX AND REDUCE THE AMOUNT OF MEMORY AVAILABLE FOR SOLVING THE PROBLEM. ╚OWEVER, LARGE SPARCE PROBLEMS CAN ALSO BE SOLVED ON THE ├64 BY ITERATIVE METHODS SUCH AS ╟AUSS-╙IEDEL AND ├ONJUGATE ╟RADIENT ALGORITHMS. ╤╒┴╥╘┼╥╙╧╠╓┼ ╤UARTERSOLVE IS A USEFUL METHOD FOR SOLVING GENERAL DENSE SYSTEMS OF LINEAR EQUATIONS THAT ╔ DISCOVERED ALMOST BY ACCIDENT WHILE DOING RANDOM RESEARCH IN THE LIBRARY. ╔ HAVE NOT SEEN ANY RECENT TEXTS OR PAPERS MENTIONING THIS ALGORITHM. ╔ HAVE NOT SEEN ANY REFERENCE TO IT IN THE ├64 LITERATURE EITHER. ┴T LEAST ONE OLDER TEXT MENTIONED IT IN PASSING SAYING THAT THE CODE WAS TOO LONG OR COMPLEX. ╘HIS IS A VALID POINT SINCE USUALY THE CODE SIZE DIRECTLY SUBTRACTS FROM THE PROBLEM STORAGE. ╘HE CODE IS LONGER THAN THE ╟AUSSIAN ┼LIMINATION CODE BUT IN MY IMPLEMENTATION IT ONLY TAKES ABOUT 2╦ OF MAIN MEMORY STORAGE AND IT IS A REAL ADVANTAGE ON THE ├64. ╫ITH A ├64 WE CAN ALSO PUT THE ENTIRE CODE IN AN ┼╨╥╧═ ON A CARTRIDGE SO THE CODE SIZE IS OF LITTLE CONCERN. ╔ FOUND ╤UARTERSOLVE IN ╥EF. 1 (╥. ┴. ┌AMBERDINO, 1974), WHICH CREDITED THE ALGORITHM TO ╥EF. 2 (┴. ╧RDEN, 1960). ╔ AM A LITTLE UNEASY DESCRIBING THE ALGORITHM SINCE ╔ HAVE NOT SEEN ╥EF. 2 OR ANALYZED THE ALGORITHM. ╔ HAVE CODED THE ALGORITHM, TESTED IT, AND USED IT TO SOLVE SOME LARGE PROBLEMS ON A ├64, UP TO N=90. ┌AMBARDINO MAKES TWO INTERESTING STATEMENTS IN ╥EF 1. "╘HE NUMBER OF ARITHMETIC OPERATIONS IS THE SAME AS FOR THE ╟AUSSIAN ┼LIMINATION METHOD." ╔ AM REASONABLY SURE FROM THE DESCRIPTION THAT HE MEANT ╟AUSS-╩ORDAN WHICH REQUIRES ABOUT 50% MORE ARITHMETIC THAN ╟AUSSIAN ┼LIMINATION. ┴FTER PROCESSING THE ITH ROW ONLY I(N-I) STORAGE LOCATIONS ARE REQUIRED TO STORE THE REDUCED MATRIX. ═AX[I(N-I)] = N*N/4. ╘HIS MAXIMUM MEMORY REQUIREMENT OCCURS AT I = N/2. ┴S I INCREASES FURTHER MEMORY REQUIRED IS REDUCED. ┴LTHOUGH N*N/4 MEMORY LOCATIONS MUST BE ALLOCATED AND DIMENSIONED IN AN ARRAY AT THE START, ╤UARTERSOLVE ALWAYS USES THE FIRST PORTION OF THE ARRAY CONTINUOUSLY AND DOES NOT FREE UP MEMORY IN HOLES SCATTERED THROUGHOUT THE ARRAY. ╘HE ├ LANGUAGE COULD POSSIBLY USE "HEAP STORAGE" AND RELEASE THE MEMORY FOR ANY OTHER USE AS THE PROCEDURE ADVANCES. ╬OW BACK TO MY INITIAL MEMORY FREE CLAIM. ╘HE LARGE PROBLEM THAT ╔ ACTUALLY WANTED TO SOLVE WAS: ┴*X=B, ┬*X=R, FOR R GIVEN B AND THE SQUARE MATRICES ┴ AND ┬. ┼LEMENTS OF ┴ AND ┬ ARE MOST EFFICIENTLY CALCULATED AT THE SAME TIME. ╔ COULD WRITE ┬ TO THE DRIVE AND READ IT BACK IN AFTER X IS COMPUTED TO CALCULATE R, BUT ╔ ACTUALLY WANTED TO SOLVE THIS REPEATEDLY INSIDE ANOTHER LOOP AND ╔ DID NOT WANT TO READ AND WRITE TO A LOUSY 1541 THAT MUCH. ╒SING ╟AUSSIAN ELIMINATION WOULD REQUIRE 2N*N STORAGE. ╒SING ╤UARTERSOLVE COULD REQUIRE 1.25N*N STORAGE. ╚OWEVER, ONLY N*N STORAGE IS NEEDED, THAT FOR ┬. ┴T THE ITH STEP THE ITH ROW OF ┴ AND ┬ ARE CALCULATED. ╘HE ROW OF ┴ IS PROCESSED INTO THE THE N*N DIMENSIONED ARRAY ┬ FILLING IT FROM THE FRONT. ╘HE CORRESPONDING ROW OF ┬ IS STORED IN THE SAME ARRAY ┬ FILLING FROM FROM THE END OF ARRAY ┬. ┴S THE PROCESS CONTINUES ╤UARTERSOLVE "DISSUSES" ARRAY ┬ SO THAT ROWS OF ┬ NEVER OVERWRITE STORAGE NEEDED BY ╤UARTERSOLVE. ┴T THE END WE HAVE COMPUTED X AND ALL OF ┬ IS STORED IN THE ARRAY ┬. ╙IMPLE MULTIPLICATION PRODUCES R. ╙O ╔ CAN SAY WITH PRIDE, AT THE EXPENSE OF HONESTY, THAT ╔ HAVE SOLVED ┴*X=B WITHOUT ANY ADDITIONAL MEMORY STORAGE FOR ┴. ╨╥╧├ SLV(N#,NR#,I#,╥┼╞ A(),╥┼╞ C(),╥┼╞ B(,),SDB#,╥┼╞ SW#(),╥┼╞ FAIL#) ├╠╧╙┼─ // ╘HIS ROUTINE SOLVES A SYSTEM OF EQUATIONS USING THE QUARTERSOLVE // ALGORITHM WITH PARTIAL PIVOTING. // ╔T IS CALLED A "LINE AT A TIME" AND USES ONLY // 0.25*NN MEMORY LOCATIONS WHICH ENABLES LARGER PROBLEMS TO BE SOLVED. // ╘HE ╠╒ FACTORIZATION IS NOT AVAILABLE, NOR A CONDITION ESTIMATE. // N# IS THE DIMENSION OF THE PROBLEM // NR# IS THE NUMBER OF RIGHT HAND VECTORS TO BE SOLVED FOR. // B(,) IS THE RIGHT HAND SIDE COLUMNS // SDB# IS THE SECOND DIMENSION OF THE ARRAY B(,) ╒╙┼ BLAS ╒╙┼ STRINGS Q#:=I#-1; M#:=N#-Q#; MM1#:=M#-1; FAIL#:=╘╥╒┼; IP1#:=I#+1 ╔╞ I#=1 ╘╚┼╬ //INITIALIZE PIVOT ARRAY ╞╧╥ J#:=1 ╘╧ N# ─╧ SW#(J#):=J# ┼╬─╔╞ ╞╧╥ J#:=1 ╘╧ Q# ─╧ //ADJUST FOR PREVIOUS PIVOTING K#:=SW#(J#) ╫╚╔╠┼ K#<J# ─╧ K#:=SW#(K#) ╔╞ K#>J# ╘╚┼╬ SWAP'REAL(C(J#),C(K#)) ┼╬─╞╧╥ J# ╞╧╥ J#:=I# ╘╧ N# ─╧ C(J#):-SDOT(Q#,C(1),1,A(J#-Q#),M#) P#:=Q#+ISAMAX#(M#,C(I#),1) R:=┴┬╙(C(P#)) ╔╞ R=0 ╘╚┼╬ ╥┼╘╒╥╬ FAIL#:=╞┴╠╙┼ ╔╞ P#<>I# ╘╚┼╬ SWAP'REAL(C(I#),C(P#)) SWAP'INTEGER(SW#(I#),SW#(P#)) SSWAP(Q#,A(1),M#,A(P#-Q#),M#) ┼╬─╔╞ R:=1/C(I#) ╔╞ MM1#<>0 ╘╚┼╬ SSCAL(MM1#,R,C(IP1#),1) ╞╧╥ J#:=1 ╘╧ NR# ─╧ B(I#,J#):=R*(B(I#,J#)-SDOT(Q#,C(1),1,B(1,J#),SDB#)) ╞╧╥ K#:=1 ╘╧ NR# ─╧ SAXPY(Q#,-B(I#,K#),A(1),M#,B(1,K#),SDB#) ╔╞ MM1#>0 ╘╚┼╬ T#:=1 ╞╧╥ J#:=1 ╘╧ Q# ─╧ R:=A(T#); T#:+1 SCOPY(MM1#,A(T#),1,A(T#-J#),1) SAXPY(MM1#,-R,C(IP1#),1,A(T#-J#),1) T#:+MM1# ┼╬─╞╧╥ J# SCOPY(MM1#,C(IP1#),1,A(MM1#*Q#+1),1) ┼╠╙┼ //UNSCRAMBLE SOLUTION FROM PIVOTING ╞╧╥ J#:=1 ╘╧ NR# ─╧ ╞╧╥ K#:=1 ╘╧ N# ─╧ C(SW#(K#)):=B(K#,J#) SCOPY(N#,C(1),1,B(1,J#),SDB#) ┼╬─╞╧╥ J# ┼╬─╔╞ ┼╬─╨╥╧├ SLV // N#:=8; SDRH#:=1; NRH#:=1; NR#:=1 // A IS OF DIMENSION N*N/4 ─╔═ A(16), B(N#), RHS(N#,NRH#), SW#(N#) ╞╧╥ I#:=1 ╘╧ N# ─╧ ╞╧╥ J#:=1 ╘╧ N# ─╧ B(J#):=2297295/(I#+J#-1) S:=0 ╞╧╥ J#:=N# ╘╧ 1 ╙╘┼╨ -1 ─╧ S:+B(J#) RHS(I#,1):=S SLV(N#,NR#,I#,A(),B(),RHS(,),SDRH#,SW#(),FAIL#) ╔╞ FAIL# ╘╚┼╬ ╨╥╔╬╘ "SINGULARITY DETECTED AT I=";I# ╙╘╧╨ ┼╬─╔╞ ┼╬─╞╧╥ I# ╞╧╥ J#:=1 ╘╧ N# ─╧ ╨╥╔╬╘ RHS(J#,1); ┼╬─ ╘HE ╤UARTERSOLVE ALGORITHM IS PRESENTED HERE AS A ├╧═┴╠ 2.0 PROCEDURE "SLV". ├╧═┴╠ IS PRETTY MUCH A DEAD LANGUAGE AND ╔ DON'T EXPECT ANYONE TO RUN THIS CODE. ╚OWEVER, ├╧═┴╠ IS A STRUCTURED ALGORITHMIC LANGUAGE THAT IS EASY TO READ. ┘OU CAN READILY TRANSLATE IT INTO THE PROGRAMMING LANGUAGE OF YOUR CHOICE. ╙LV IS CODED AS A ├╠╧╙┼─ PROCEDURE AS A PERSONAL MATTER OF STYLE. ┴N OPEN PROCEDURE WOULD EXECUTE FASTER. ╘HE ARRAYS ARE PASSED BY ╥┼╞┼╥┼╬├┼ AND DO NOT ALLOCATE ADDITIONAL LOCAL STORAGE. ╘HE MAIN PROGRAM IS JUST AN EXAMPLE FOR TESTING. ╔T CALLS SLV N TIMES TO SOLVE THE LINEAR SYSTEM. ╘HE TEST PROBLEM SOLVES A SCALED ╚ILBERT MATRIX WHICH IS ILL CONDITIONED. ╔N THE ABSENCE OF ROUNDOFF ERROR THE SOLUTION SHOULD BE A VECTOR OF ONES. ╔ USUALLY DIMENSION A() TO N#*(N#+1)/4. ╙LV IS PRESENTED IN ITS FULL GENERALITY, BUT YOU MAY WANT TO MAKE SOME SIMPLIFICATIONS. ╙LV CAN HANDLE MULTIPLE RIGHT HAND SIDE VECTORS WITH THE TWO DIMENSIONAL ARRAY B(,) IN MOST APPLICATIONS YOU WILL ONLY USE A SINGLE VECTOR, NR#=1, AND YOU CAN MAKE SOME SIMPLIFICATIONS BY JUST USING A ONE DIMENSIONAL ARRAY. ╨IVOTING ALSO COMPLICATES THE CODE. ╨ROBLEMS WHICH ARE POSITIVE DEFINITE, OR DIAGONALLY DOMINANT, OR SOMETIMES JUST WELL CONDITIONED CAN BE SAFELY SOLVED WITHOUT PIVOTING. ╙TRIPPING OUT THE PIVOTING CODE IS STRAIGHT FORWARD AND WILL SHORTEN THE CODE AND SPEED EXECUTION. ┴NYTHING FOLLOWING // IS A COMMENT AND CAN BE DELETED FROM YOUR RUNNING CODE. ╔N ├╧═┴╠ 2.0 YOU CAN ALSO "PROTECT" THE CODE WHICH WILL STRIP OUT COMMENTS AND OTHER INFORMATION TO MAKE A SHORTER RUNNING VERSION. ╘HE REMAINING DISCUSSION WILL CONCERN ├╧═┴╠ 2.0 AND THE ┬╠┴╙. ├╧═┴╠ 2.0 ├╧═┴╠ 2.0 IS AN EXCELLENT PROGRAMMING LANGUAGE FOR THE ├64/128 AND ╔ CAN'T DESCRIBE ALL OF ITS VIRTUES HERE. ╔T HAS ONE SERIOUS LIMITATION. ╔T DOES NOT USE "CONTINUATION" LINES, SO LINE LENGTH IS LIMITED. ╘HIS IS MOST RESTRICTIVE IN FUNCTION AND PROCEDURE LINES WHERE IT LIMITS THE NUMBER OF PARAMETERS THAT CAN BE PASSED. ╠INE LENGTH IS LIMITED TO 80 CHARACTERS. ╚OWEVER, IF YOU USE A SEPARATE TEXT EDITOR OR WORD PROCESSOR YOU CAN ENTER 120 CHARACTER LINES. ├OMAL WILL ACTUALLY EXECUTE TOKENIZED LINES UP TO 256 CHARACTERS SO THE LIMITATION IS REALLY IN THE EDITOR RATHER THAN ├╧═┴╠. ╨ROCEDURE AND VARIABLE NAMES CAN BE QUITE LONG IN ├OMAL, BUT ARE KEPT SHORT BECAUSE OF THE LINE LENGTH LIMITATION. "╤UARTERSOLVE" WAS SHORTENED TO "SLV" FOR THIS REASON. A:+T IS A SHORTER FASTER VERSION A:=A+T, AND A:-T IS A SHORTER FASTER VERSION OF A:=A-T. ╘HIS IS MOST USEFULL WHEN "A" IS AN ARRAY ELEMENT OR AN INTEGER. ├OMAL 2.0 SUPPORTS ═╠ PACKAGES. ┴ PACKAGE IS A COLLECTION OF FUNCTIONS OR PROCEDURES THAT CAN BE CALLED AND EXECUTED. ┴ PACKAGED CAN BE ╥╧══┼─ AND STORED IN ┼╨╥╧═ ON THE ├OMAL 2.0 CARTRIDGE. ┴ PACKAGE CAN ALSO BE LOADED FROM DISK AND WILL NORMALLY BE STORED IN A ╥┴═ LOCATION THAT ├╧═┴╠ DOES NOT USE FOR NORMAL PROGRAMS. ╠╔╬╦ "FILENAME" WILL LOAD AND LINK THE ═╠ PACKAGE TO A ├OMAL PROGRAM. ╔T WILL STAY ATTACHED TO THE PROGRAM WHEN THE PROGRAM IS SAVED AND LOADED, UNLESS IT IS MARKED AS ╥╧══┼─. ╘HE ENTIRE SLV PROCEDURE COULD BE CODED IN ASSEMBLY LANGUAGE AND BE PLACED IN A PACKAGE. ╘HE SLV PROCEDURE USES TWO PACKAGES, STRINGS AND BLAS. ╘HE COMMAND ╒╙┼ PACKAGENAME MAKES ALL OF THE FUNCTIONS AND PROCEDURES OF THE PACKAGE KNOWN. ┴LTERNATIVELY, YOU COULD PLACE THE ╒╙┼ PACKAGENAME COMMAND IN THE MAIN PROGRAM AND PUT ╔═╨╧╥╘ PROCEDURENAME INSIDE ALL OF THE CLOSED PROCEDURES THAT CALL PROCEDURENAME. ╙LV CALLS THE SWAP'REAL AND SWAP'INTEGER PROOCEDURES FROM THE STRINGS PACKAGE. ╘HE STRINGS PACKAGE IS A ╥╧══┼─ PACKAGE ON THE ╙UPER ├HIP ╥╧═. ╔T DOES EXACTLY WHAT IT SAYS, E.G. SWAP'REAL(A,B) IS THE SAME AS: T:=A; A:=B; B:=T. ╙LV CALLS THE SDOT, ISAMAX#, SSWAP, SSCAL, SAXPY, AND SCOPY ROUTINES FROM THE BLAS PACKAGE. ╘HE BLAS PACKAGE IS ╠╔╬╦ED TO THE PROGRAM, BUT IT COULD, AND SHOULD, BE PLACED ON ┼╨╥╧═. ┬ASIC ╠INEAR ┴LGEBRA ╙UBROUTINES, ┬╠┴╙ ╘HE ┬╠┴╙ WERE ORIGINALLY WRITTEN FOR THE ╞ORTRAN LANGUAGE TO SPEED EXECUTION AND STREAMLINE CODE USED FOR SOLVING LINEAR ALGEBRA AND OTHER MATRIX PROBLEMS. ╘HE ╠╔╬╨┴├╦ ROUTINES, ╥EF. 3, USE THE ┬╠┴╙ AND ARE PERHAPS THE BEST KNOWN. ╘HE IDEA IS THAT THE ┬╠┴╙ ROUTINES WILL BE HIGHLY OPTIMIZED FOR A PARTICULAR COMPUTER, CODED IN ═╠ OR A ╚IGH ╧RDER ╠ANGUAGE. ╙OME OPERATING SYSTEMS EVEN INCLUDE ┬╠┴╙ LIKE ROUTINES. ╫RITING FAST EFFICIENT PROGRAMS IS THEN A SIMPLE MATTER OF SELECTING THE BEST SOLUTION ALGORITHM AND CODING IT IN A MANNER THAT MAKES BEST USE OF THE BLAS ROUTINES. ╘HERE ARE BLAS ROUTINES FOR SINGLE PRECISION, DOUBLE PRECISION, AND COMPLEX NUMBERS. ╘HE LEVEL 1 ┬╠┴╙ PERFORM OPERATIONS ON ROWS OR COLUMNS OF AN ARRAY AND TYPICALY DO N SCALAR OPERATIONS REPLACING THE INNER MOST LOOP OF CODE. ╘HERE ARE ALSO LEVEL 2 ┬╠┴╙ THAT PERFORM N*N OPERATIONS AND ╠EVEL 3 ┬╠┴╙ THAT PERFORM N*N*N OPERATIONS. ╬ICHOLAS ╚IGHAM HAS CODED MOST OF THE SINGLE PRECISION LEVEL 1 BLAS ROUTINES AND PUT THEM IN A ├OMAL 2.0 PACKAGE. ╘HE ├OMAL BLAS PACKAGE IS INCLUDED ON THE ╨ACKAGES ╠IBRARY ╓OLUME 2 DISK. ╔ AM NOT AWARE OF ═╠ BLAS ROUTINES CODED FOR ANY OTHER ├64/128 LANGUAGES ALTHOUGH THIS IS CERTAINLY POSSIBLE AND RECOMMENDED. ╘HE ├OMAL BLAS ROUTINES BEHAVE EXACTLY THE SAME WAY THAT THE ╞ORTRAN BLAS ROUTINES DO EXCEPT THAT ╞ORTRAN CAN PASS THE STARTING ADDRESS OF AN ARRAY WITH JUST "A", WHILE ├OMAL REQUIRES "A(1)". ╘HE ├OMAL BLAS WILL ALLOW YOU PASS AN ARRAY, BY REFERENCE, OF SINGLE OR MULTIPLE DIMENSIONS AND START FROM ANY POSITION IN THE ARRAY. ╔F YOU CODE THE BLAS ROUTINES AS ORDINARY ├OMAL ROUTINES YOU HAVE TO PASS ADDITIONAL PARAMETERS AND HAVE SEPARATE ROUTINES FOR SINGLE DIMENSIONED ARRAYS AND TWO DIMENSIONAL ARRAYS. ╬OTE ALSO THAT ╞ORTRAN STORES TWO DIMENSIONAL ARRAYS BY COLUMNS, AND ├OMAL (LIKE MANY OTHER LANGUAGES) STORES TWO DIMENSIONAL ARRAYS BY ROWS. ╔F YOU TRANSLATE CODE BETWEEN ╞ORTRAN AND ├OMAL USING BLAS ROUTINES YOU WILL HAVE TO CHANGE THE INCREMENT VARIABLES. ╞ORTRAN ├OMAL DIMENSION C(N), A(ILDA,ISDA) ─╔═ C(N#), A(LDA#,SDA#) SCOPY(N,C,1,A(I,1),ILDA) SCOPY(N#,C(1),1,A(I#,1),1) SCOPY(N,C,1,A(1,J),1) SCOPY(N#,C(1),1,A(1,J#),SDA#) ╘HE FIRST SCOPY COPIES ARRAY C INTO THE ITH ROW OF ARRAY A. ╘HE SECOND SCOPY COPIES ARRAY C INTO THE JTH COLUMN OF ARRAY A. ╘HIS IS WHAT SCOPY DOES IN ╞ORTRAN: SUBROUTINE SCOPY(N,SX,INCX,SY,INCY) REAL SX(1),SY(1) IX=1 IY=1 DO 10 I = 1,N SY(IY) = SX(IX) IX = IX + INCX IY = IY + INCY 10 CONTINUE RETURN END ╘HE ├OMAL ┬╠┴╙ DOES EXACTLY THE SAME THING. ╔F CODED ENTIRELY IN ├╧═┴╠ RATHER THAN AS A PACKAGE IT WOULD HAVE TO BE DIFFERENT. ╘HE CALL WOULD CHANGE. SCOPY(N#,C(1),1,A(1,J#),SDA#) WOULD HAVE TO BECOME, SCOPY(N#,C(),1,1,A(,),1,J#,SDA#,SDA#) AND THE ├OMAL PROCEDURE MIGHT BE: ╨╥╧├ SCOPY(N#, ╥┼╞ X(), IX#, INCX#, ╥┼╞ Y(,), IY#, JY#, SDY#, INCY#) ├╠╧╙┼─ IYINC#:=INCY# ─╔╓ SDY# //ASSUMING Y IS DIMENSIONED Y(?,SDY#) JYINC#:=INCY# ═╧─ SDY# ╞╧╥ I#=1 ╘╧ N# ─╧ Y(IY#,JY#):=X(IX#) IX#:+INCX#; IY#:+IYINC#; JY#:+JYINC# ┼╬─╞╧╥ ┼╬─╨╥╧├ SCOPY ╬OTE THAT MORE INFORMATION HAS TO BE PASSED TO THE PROCEDURE AND USED THAT THE ═╠ BLAS PICKS UP AUTOMATICALLY. ┴LSO WE WOULD NEED SEPARATE PROCEDURES TO HANDLE EVERY COMBINATION OF SINGLE AND MULTI DIMENSIONAL ARRAYS. ╘HE ├OMAL ═╠ BLAS ARE INDEED WONDERFUL. ╞OR SPEED CONSIDERATIONS THIS SHOULD ALSO BE LEFT AS AN OPEN PROCEDURE OR BETTER YET JUST USE IN LINE CODE. ╚ERE IS A VERY SIMPLIFIED DESCRIPTION OF WHAT EACH OF THE ROUTINES IN THE ├OMAL ┬╠┴╙ PACKAGE DOES. SUM:=SASUM(N#,X(1),1) ╥ETURNS SUM OF ABSOLUTE VALUES IN X(). SUM:=0 ╞╧╥ I#:=1 ╘╧ N# ─╧ SUM:+┴┬╙(X(I#)) SAXPY(N#,SA,X(1),1,Y(1),1) ┴DD A MULTIPLE OF X() TO Y(). ╞╧╥ I#:=1 ╘╧ N# ─╧ Y(I#):+SA*X(I#) PROD:=SDOT(N#,X(1),1,Y(1),1) ╥ETURNS DOT PRODUCT OF X() AND Y(). PROD:=0 ╞╧╥ I#:=1 ╘╧ N# ─╧ PROD:+X(I#)*Y(I#) SSWAP(N#,X(1),1,Y(1),1) ╙WAPS X() AND Y(). ╞╧╥ I#:=1 ╘╧ N# ─╧ T:=X(I#); X(I#):=Y(I#); Y(I#):=T SCOPY(N#,X(1),1,Y(1),1) ├OPY X() TO Y(). ╞OR I#:=1 ╘╧ N# ─╧ Y(I#):=X(I#) MAX#:=ISAMAX#(N,X(1),1) ╥ETURNS INDEX OF THE ELEMENT OF X() WITH THE LARGEST ABSOLUTE VALUE. T:=0; MAX#:=1 ╞╧╥ I#:=1 ╘╧ N# ╔╞ ┴┬╙(X(I#))>T ╘╚┼╬ T:=┴┬╙(X(I#)); MAX#:=I# ┼╬─╞╧╥ I# SSCAL(N#,SA,X(1),1) ╙CALE X() BY A CONSTANT SA. ╞╧╥ I#:=1 ╘╧ N# ─╧ X(I#):=SA*X(I#) SNRM2(N#,X(1),1) ╥ETURNS THE 2 NORM OF X(). NORM2:=0 ╞╧╥ I#:=1 ╘╧ N# ─╧ NORM2:+X(I#)*X(I#) NORM2:=╙╤╥(NORM2) SROT(N#,X(1),1,Y(1),1,C,S) ┴PPLY ╟IVENS ROTATION. ╞╧╥ I#:=1 ╘╧ N# ─╧ T:=C*X(I#) + S*Y(I#) Y(I#):=S*X(I#) + C*Y(I#) X(I#):=T ┼╬─╞╧╥ I# ┬EAR IN MIND THAT EACH OF THESE SIMPLE EXAMPLES CAN BE MORE COMPLEX AS WAS GIVEN FOR SCOPY. ┘OU NOW HAVE ENOUGH INFORMATION TO WRITE YOUR OWN ┬╠┴╙ ROUTINES IN ═╠ OR THE PROGRAMMING LANGUAGE OF YOUR CHOICE, OR TO EXPAND THE ┬╠┴╙ ROUTINE CALLS IN SLV TO ORDINARY IN LINE CODE. ┘OU CAN ALSO APPLY THE ┬╠┴╙ ROUTINES IN CREATIVE WAYS BESIDES JUST OPERATING ON ROWS OR COLUMNS. ╞OR EXAMPLE YOU COULD CREATE THE IDENTITY MATRIX WITH: ─╔═ A(N#,N#) A(1,1):=1; A(1,2):=0 SCOPY(N#*N#-2,A(1,2),0,A(1,3),1) // ZERO THE REST OF THE MATRIX SCOPY(N#-1,A(1,1),0,A(2,2),N#+1) // COPY ONES TO THE DIAGONAL. ╥EFERENCES 1. ┌AMBARDINO, ╥. ┴., "╙OLUTIONS OF ╙YSTEMS OF ╠INEAR ┼QUATIONS WITH ╨ARTIAL ╨IVOTING AND ╥EDUCED ╙TORAGE ╥EQUIREMENTS", ╘HE ├OMPUTER ╩OURNAL ╓OL. 17, ╬O. 4, 1974, PP. 377-378. 2. ╧RDEN ┴., "═ATRIX ╔NVERSION AND ╥ELATED ╘OPICS BY ─IRECT ═ETHODS", IN ═ATHEMATICAL ═ETHODS FOR ─IGITAL ├OMPUTERS, ╓OL. 1, ┼DITED BY ┴. ╥ALSTON AND ╚. ╫ILF, ╩OHN ╫ILEY AND ╙ONS ╔NC., 1960. 3. ─ONGARRA, ╩. ╩., ═OELER, ├. ┬., ┬UNCH, ╩. ╥., ╙TEWART, ╟. ╫., ╠INPACK ╒SERS' ╟UIDE, ╙╔┴═ ╨RESS, ╨HILADELPHIA, 1979.