I have created a very simple Excel sheet to calculate sprite values for monochrome sprites.

Here’s what the sheet looks like:

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Commodore 64

## An Excel sheet to generate hi-res Sprite data for the Commodore 64

## Plotting a function in hi-res graphics mode on a Commodore 64

## Exploring the Goldbach Conjecture using a Commodore 64

## Commodore 64: Using the Newton/Raphson method and Halley’s method to find the zeros of a function

## The Commodore 64 and the Kolakoski Sequence

## A keyboard sonata …

## A comparison of using a Sieve of Eratosthenes vs. Trial Division to find primes on a C64 in BASIC

## A Sieve of Eratosthenes in Assembler

I have created a very simple Excel sheet to calculate sprite values for monochrome sprites.

Here’s what the sheet looks like:

I wrote a program to plot a function in hi-res graphics mode on the C64. It is entirely in BASIC, so it is mind-numbingly slow, but it works, and can be used to plot some fairly interesting functions.

This mode only allows one background color and one foreground color for each 8×8 block of pixels. It would, I suppose, be possible to plot multiple functions in multiple colors, but the intersections would be strange, as each 8×8 block can only have two colors, including the background. An obvious next step is to use the multi-color bitmap mode, but which this would allow four colors in any given 8×8 block, it would reduce the resolution from 320×200 down to 160×200. I will try this soon.

Another obvious improvement will be to move much of the setup stuff (like erasing the graphics ram ($2000-$3fff)) down into machine language, which would speed things up significantly.

*The Graphics Book for the Commodore 64* was indispensable.

Here is the plot of f(x) = sin(x) + sin(1.5*x + pi/2)

And here is the code:

10 rem *** funcplot 20 rem 30 pi = 3.141592653 40 yu = 2.2 : yd = -2.2 50 xu = 6pi : xd = -6pi 60 xr = xu - xd : yr = yu - yd 70 xs = xr/320 : ys = yr/200 80 def fn f(x) = sin(x) + sin(1.5*x + pi/2) 100 rem set vic address 110 v = 53248 120 rem set graphic ram address 130 ga = 8192 140 rem set video ram address 150 vr = 1024 160 rem set border to black 170 poke v+32,0 500 gosub 20000 : rem turn on hires 510 gosub 21000 : rem graphic ram area 520 gosub 22000 : rem set color ram 530 gosub 23000 : rem clr graphic ram 1000 rem draw x axis 1005 y=0 : yc=int(200-(y-yd)*(200/yr)) 1010 for xc = 0 to 319 1020 gosub 30000 : rem set (xc,yc) 1030 next xc 1100 rem draw y axis 1110 x = 0 1120 for yc = 1 to 200 1130 xc = int((x-xd)*(320/xr)) 1140 gosub 30000 : rem set (xc,yc) 1150 next yc 1300 rem plot function 1310 for x = xd to xu step xs 1320 y = fn f(x) 1330 xc = int((x-xd)*(320/xr)) 1340 yc = int(200-(y-yd)*(200/yr)) 1350 gosub 30000 : rem set (xc,yc) 1360 next x 1400 poke 198,0 : wait 198,1 1450 gosub 24000 1500 end 20000 rem turn on hi res graphics 20010 rem 1. set bits 5/6 of v+17 20020 rem 2. clr bit 4 of v+22 20030 poke v+17,peek(v+17) or (11*16) 20040 poke v+22,peek(v+22) and (255-16) 20050 return 21000 rem set graphic ram area 21010 rem 1. set bit 3 of v+24 21020 poke v+24, peek(v+24) or 8 21030 return 22000 rem set color ram 22010 rem 1. color ram is 1024-2023 22020 rem 2. set background 1 - white 22030 rem 3. set foreground 0 - black 22040 co = 0*16 + 1 22050 for i = vr to vr+1000 22060 poke i,co 22070 next i 22080 return 23000 rem clear graphic ram 23010 rem 1. graphic ram is ga to 23020 rem ga + 8000 23030 for i = ga to ga +8000 23040 poke i,0 23050 next i 23060 return 24000 rem turn graphics off 24010 rem 1. clr bits 5/6 of v+17 24020 rem 2. clr bit 4 of v+22 24030 rem 3. clr bit 3 of v+24 24040 poke v+17,peek(v+17) and (255-96) 24050 poke v+22,peek(v+22) and (255-16) 24060 poke v+24,peek(v+24) and (255-8) 24070 return 30000 rem set pixel 30010 ra = 320*int(yc/8)+(yc and 7) 30020 ba = 8*int(xc/8) 30030 ma = 2^(7-(xc and 7)) 30040 ad = ga + ra + ba 30050 poke ad,peek(ad) or ma 30060 return 31000 rem clr pixel 31010 ra = 320*int(yc/8)+(yc and 7) 31020 ba = 8*int(xc/8) 31030 ma = 255-2^(7-(xc and 7)) 31040 ad = ga + ra + ba 31050 poke ad,peek(ad) or ma 31060 return

The Goldbach Conjecture states that every even integer greater than two can be expressed as a sum of two primes.

I programmed my C64 in Basic to ask the question “How many ways can a particular even integer be expressed as the sum of two primes?” So, for instance, the number 10 can be expressed as 3+7 or as 5+5, so two ways. The number 12 can be expressed as 5+7, so only one way. The number 22 can be expressed as 11+11 or as 19+3 or as 17+5, so three ways.

I left the program running for twelve hours, to calculate the Goldbach partitions of all even integers from six to one thousand. I’m very sure that there are much more efficient routines, which would run faster, but it is at least clear that it can be done.

Note that the Goldbach conjecture is not proven. The Wikipedia article here states that the conjecture has been tested by computers for numbers up to 4 x 10^18. But I only went as high as 1000.

Here is a plot of my results (not on the C64, but rather using gnuplot on Windows.

And here is the code:

50 qm = 1000 100 rem get primes up to 1000 200 dim p%(400) 220 p%(1) = 2 230 pc = 1 240 n = 3 300 for i = 1 to pc 310 w = n/p%(i) 320 if w = int(w) then 400 330 next i 340 pc = pc + 1 350 p%(pc) = n 355 print n 400 for i = 1 to 1: next i: n = n+1 410 if n < qm then goto 300 1000 rem get golbach numbers 1010 dim gb%(501) 1020 rem gb%(1) = 0 and gb%(2) = 0 1030 rem because 2 and 4 have no 1040 rem goldbach partitions 1050 gb%(1) = 0 : gb%(2) = 0 1060 q2 = int(qm/2) 1100 for m = 3 to q2 1110 n = m*2 1120 ct = 0 1130 for i = 1 to pc 1140 if p%(i) > n then goto 1200 1150 pr = n - p%(i) 1160 pf = 0 1162 for j = 1 to pc 1164 if p%(j) = pr then pf = 1:goto 1195 1166 if p%(j) > pr then goto 1200 1168 next j 1195 if pf = 1 then ct = ct + 1 1200 for j = 1 to 1:nextj:next i 1210 if (ct/2) <> int(ct/2) then ct = ct + 1 1220 ct = int(ct/2) 1250 print n,ct 1260 gb%(m) = ct 1300 next m 2000 open 2,8,2,"@:goldlist,s,w" 2010 for q = 1 to int(qm/2) 2020 print#2,q*2,gb%(q) 2030 next q 2040 close 2

Click for good descriptions of Newton’s method and Halley’s method.

I decided to write my C64 Basic code for two different functions:

*f(x) = x*x*x – 2*x – 5**f(x) = exp(x) – x*x*

Here are the plots for these functions, with the roots indicated by a blue dot:

These methods require choosing a first guess for the root. In both cases I chose *x = 2* as a first guess.

Here is the code:

10 REM =============================== 20 REM 30 REM FIND THE ZEROS OF A FUNCTION 35 REM USING NEWTON/RAPHSON METHOD 40 REM AND HALLEY'S METHOD 60 REM 70 REM =============================== 100 PRINT CHR$(147) 200 DEF FN F(A) = A*A*A - 2*A - 5210 DEF FN F1(A) = 3*A*A - 2220 DEF FN F2(A) = 6*A 230 E$ = "XXX - 2X - 5"235 PRINT "--------------------------------------"240 GOSUB 10000245 PRINT "--------------------------------------"250 GOSUB 20000300 DEF FN F(A) = EXP(A) -A*A 310 DEF FN F1(A) = EXP(A) - 2*A320 DEF FN F2(A) = EXP(A) - 2330 E$ = "EXP(X) - X*X" 335 PRINT "--------------------------------------" 340 GOSUB 10000 345 PRINT "--------------------------------------" 350 GOSUB 20000 1000 END 10000 REM NEWTON/RAPHSON 10002 X = 2 10005 S = 0 : DI = 10 : TL = 0.000001 10010 IF DI < TL THEN GOTO 10100 10020 G = FN F(X) : G1 = FN F1(X) 10030 X1 = X - G/G1 10040 DI = ABS(X - X1) 10050 X = X1 10060 S = S + 1 10070 GOTO 10010 10100 PRINT "NEWTON'S METHOD FOR: ";E$ 10110 PRINT "ROOT = ";X 10120 PRINT "F(ROOT) = "; FN F(X) 10130 PRINT "FOUND IN ";S;" STEPS" 10200 RETURN 20000 REM HALLEY 20002 X = 2 20005 S = 0 : DI = 10 : TL = 0.000001 20010 IF DI < TL THEN GOTO 20100 20020 G = FN F(X) : G1 = FN F1(X) 20025 G2 = FN F2(X) 20030 NU = 2 * G * G1 20040 DE = 2 * G1 * G1 - G * G2 20050 X1 = X - NU/DE 20060 DI = ABS(X - X1) 20070 X = X1 20080 S = S + 1 20090 GOTO 20010 20100 PRINT "HALLEY'S METHOD FOR ";E$ 20110 PRINT "ROOT = ";X 20120 PRINT "F(ROOT) = "; FN F(X) 20130 PRINT "FOUND IN ";S;" STEPS" 20200 RETURN

Here is the screen shot:

The Kolakoski sequence is a strange but kind of fun sequence of integers.

All the members of the sequence are either 1 or 2. The sequence is made up of successive “runs” of 1’s and 2’s. Sometimes a run is one element long, and sometimes it is two elements long. If a particular run is made up of 1’s, the next run must be made up of 2’s. And vice versa.

The nth element of the sequence tells you how long the nth run of the sequence is.

The first element of the sequence is a 1.

So the first run has only one 1 in it, which means that the next element has to be a two. So now the sequence is 1, 2

But the second element of the sequence being a 2 tells us that the second run has to be two elements long, so the sequence has to be 1, 2, 2.

So we have first three elements (1, 2, 2), and the first two runs (1, and then 2,2).

Now let’s work on the third run. The third element of the sequence is a 2, so the third run must be two elements long, and since the second run was made up of 2’s, the third run must be made up of 1’s (two of them). So now the sequence is 1, 2, 2, 1, 1

Now let’s go on to the fourth run. It has to be made up of 2’s, since the third run was made up of 1’s, and the fourth element in the sequence, which tells us how long the fourth run is, is a 1. So now the sequence is 1, 2, 2, 1, 1, 2.

One more time. Let’s look at the fifth run. It has to be made up of 1’s, since the fourth run was made up of 2’s. How many? Well, the fifth element in the sequence is a 1, so the fifth run is one element long, and that element has to be a 1. So now the sequence is 1, 2, 2, 1, 1, 2, 1.

And so on.

The sequence is described in the Online Encyclopedia of Integer Sequences, in which it is sequence A00002, and in this Wikipedia article.

It has not yet been proved whether the sequence, as it gets longer and longer, converges to having closer and closer to 1/2 of its elements be 1’s, but it has been proved that it is very close, and in computation it seems to get closer the larger the sequence gets. See the Wikipedia article for more on that.

So …

I programmed my C64 to calculate the sequence out to a limit of 1000 elements, and to print the first 100 elements, and then to report on the number of 1’s and 2’s in the first 1000 elements.

First, here is the program:

5 rem ================================ 6 rem 10 rem exploring the kolakoski sequence 15 rem 20 rem ================================ 50 m = 1000 100 dim a%(m+1) 120 a%(1) = 1 130 c = 2 140 cc = 1 150 if cc >= m then goto 5000 160 d = a%(cc) 170 e = 2 180 if d = 2 then e = 1 185 cc = cc + 1 190 a%(cc) = e 210 if a%(c) = 2 then cc = cc + 1:a%(cc)= e 220 c = c + 1 230 goto 150 220 c = c + 1 230 goto 150 5000 ct1 = 0 : ct2 = 0 5020 print chr$(147) 5025 print " kolakoski sequence": print 5030 print "total: ";m 5035 print 5040 mm = m : if mm > 100 then mm = 100 5045 print "first ";mm; " elements" 5050 for i = 1 to m 5100 if i <= mm then print a%(i);" "; 5150 if a%(i) = 1 then t1 = t1 + 1 5170 if a%(i) = 2 then t2 = t2 + 1 5200 next i 5250 print:print 5300 print "1 count: ";t1 5310 print "2 count: ";t2

Here is the screen shot:

This post is a consideration of the C64 keyboard. It is not exhaustive, and it is not systematic. I just want to consider a couple of topics: the keyboard matrix, keyboard codes, the conversion tables, and the keyboard buffer.

When a key is pressed on the keyboard on the Commodore 64, the kernal tries to read the keyboard matrix.

First, here is what the matrix looks like – the image is from the Commodore 64 Service Manual:

We can watch the keyboard matrix work ourselves, using a BASIC program.

First, we need to turn off the kernal IRQ handler. This link tells us how to do that. Basically, we need to put a zero in the first bit of byte 56334 ($DC0E). Doing this will disable the keyboard, so we want to turn the IRQ handler back on at the end of the program, by putting a one in the first bit of byte 56334 ($DC0E). I turn the handler off in line 10 and back on in line 1000.

Now we are ready to probe the keyboard matrix. Here’s how that works. First, we poke a byte into $DC00 (56320) that indicates which set of eight possible characters we want to test for. In that byte, seven of the eight bits will be high, and one of the bits will be low – the low byte indicates which set of characters we are looking for (which row in the matrix we are probing). So the possible values are:

- 254 (b11111110) for row 0
- 253 (b11111101) for row 1
- 251 (b11111011) for row 2
- 247 (b11110111) for row 3
- 239 (b11101111) for row 4
- 223 (b11011111) for row 5
- 191 (b10111111) for row 6
- 127 (b01111111) for row 7

In the program below, I set the row byte in line 100. In the example, we are reading row 0.

Then, from 120 to 150 a FOR/NEXT loop waits for a key to be pressed. We can read the column number of the key pressed in byte $DC01 (56321). If no key is being pressed, that position will hold an $FF (255). If a key is being pressed, it will hold the column number of the pressed key, using the same format as for the rows above (254 -> 0, 253 -> 1, 251 -> 2, etc.) Thus, the number we poked into $DC00 and the number we read peeked from $DC01 gives us 8 x 8 or 64 possible characters as we see in the image from the service manual above.

Here is the code:

10 POKE 56334, PEEK(56334) AND 254 100 POKE 56320, 254 120 FOR I = 1 TO 5000 130 X = PEEK(56321) 140 IF X <> 255 THEN PRINT X : GOTO 1000 150 NEXT I 1000 POKE 56334, PEEK(56334) OR 1

Here is the matrix that I worked out, by putting all eight row values into $DC00 in line 100, then pushing keys until one of them gives a value that is not $FF in $DC01. I used my real C64C to be sure I was seeing what the real keyboard matrix looked like, rather than an artifact of WinVICE.

It is the same as the one from the manual, except that I reorganized the rows and columns to go in numeric order of the byte they contain, rather than trying to preserve any of the geography of the actual keyboard.

When the matrix has been read, the keyboard code is placed in the zero page memory position 203 ($CB). Then, before the keyboard is scanned again, the value in 203 is moved to 197 ($C5). When no key is being pressed, the value 64 ($40) is placed in 203, then is moved to 197 before the next scan. So, the situation that indicates “a new key has been pressed” is that 197 contains 64 and 203 contains a value less than 64.

This little program tests this. Because the keyboard matrix scan is fast and BASIC is slow, I have to press a key several times in order to have BASIC catch the transition between not pressed and pressed – awkward, but it demonstrates the concept.

100 X = PEEK(197) : Y = PEEK(203) 120 IF X = 64 AND Y <> 64 THEN PRINT Y 140 GOTO 100

Here is the table I developed using my real C64C. It is pleasingly alike to the table found on the c64-wiki here.

The table on the wiki indicates that code 63 ($3F) is RUN/STOP but I didn’t include that as the program in its current state can’t test that.

Having acquired the keyboard code for the pressed key, we now need to determine whether or not a SHIFT or CTRL or C= key was simultaneously pressed. To do that we can read memory position 653. It will equal:

- 1 if SHIFT is pressed
- 2 if C= is pressed
- 4 if CTRL is pressed
- 0 if none of those

So, after reading the matrix and assigning a keyboard code, and determining if a SHIFT or CTRL or C= is simultaneously pressed, the next step is to determine what character should be entered into the keyboard buffer as a result of the key press.

There are four lookup tables:

- 60289-60353 ($EB81-$EBC1) – for no simultaneous key press
- 60354-60418 ($EBC2-$EC02) – for SHIFT pressed
- 60419-60483 ($EC03-$EC43) – for C= pressed
- 60536-60600 ($EC78-$ECB8) – for CTRL pressed

Let’s look at one or two elements of these lookup tables. I know that if I press A, the keyboard code is 10. PEEK(60289 + 10) gives me 65, which is in fact the PETSCII code for A.

PEEK(60354 + 10) gives me 193, which is the PETSCII code for a playing card spades symbol in a box, which is what I would get on the Commodore if I typed in SHIFT-A.

You can find an excellent PETSCII table here.

A statement like IF PEEK(653) = 0 THEN A$ = CHR$(PEEK(60289 + PEEK(197))) shows how the lookup tables work.

Let’s take a look now at the keyboard buffer, which is ten bytes from 631-640 ($0277-$0280) that hold as yet unprocessed key presses. Zero page byte 198 ($C6) holds the number of entries in the buffer – from 0 to 10. The little program below lets you watch the keyboard buffer being filled in, including tracking the number of entries. When an input statement is encountered, the buffer is read in, and the number entries reset to zero. So run the program, and press keys while the loops are running. Be sure that your last key is a RETURN.

100 FOR K = 1 TO 20

105 PRINT CHR$(147)

110 FOR J = 1 TO 5

120 FOR I = 631 TO 640

130 PRINT PEEK(I),

140 NEXT I

150 PRINT PEEK(198) : PRINT 160 NEXT J

170 NEXT K

Here is a kind of fun program that lists itself when run. It puts the characters “L”,”I”,”S”,”T” and RETURN into the buffer, and then the number of entries (5) into 198. When the program is finished, the buffer, including the RETURN, is read and processed by BASIC, and the program lists itself.:

That’s all for now. For a cultural dive into the 8-bit era, I recommend dusting off your copy of Orson Scott Card’s *Ender’s Game* – it’s a lot of fun.

I wrote a BASIC program for the Commodore 64 to compare run times for finding primes using both a Sieve of Eratosthenes and Trial Division.

Here is the screen shot, with the sieve (clearly the winner) at the top, and the trial division run at the bottom.

Here is a disk image: sievevstrial

Here is the code:

5 print chr$(147) 10 u=10 20 dim a%(2^u + 1) 25 m = int((2^u)/log(2^u)) 30 dim b%(int(1.5 * m)) 50 print "sieve" 60 print "=====" 70 print "limit","primes"," time" 80 print "-----","------"," ----" 100 for n = 8 to u 105 t = time 110 lm = 2^n : sm = int(sqr(lm)) 200 for i = 1 to lm : a%(i)=1 : next i 210 a%(1) = 0 500 for p = 2 to sm 520 if a%(p) = 0 then 620 560 for ct = 2*p to lm step p 580 a%(ct) = 0 600 next ct 620 next p 630 pc = 0 640 for i = 1 to lm 660 if a%(i) = 0 then 700 680 pc = pc + 1 700 next i 705 s = (time-t)/60 706 s = int(100*s)/100 500 for p = 2 to sm 520 if a%(p) = 0 then 620 560 for ct = 2*p to lm step p 580 a%(ct) = 0 600 next ct 620 next p 630 pc = 0 640 for i = 1 to lm 660 if a%(i) = 0 then 700 680 pc = pc + 1 700 next i 705 s = (time-t)/60 706 s = int(100*s)/100 710 print "2^";n,pc,s 720 next n 1000 print : print "Trial" 1010 print "=====" 1020 print "limit","primes"," time" 1030 print "-----","------"," ----" 1040 for n = 8 to u 1042 lm = 2^n 1045 b%(1) = 2 1050 t = time 1060 pc = 1 1070 for x = 3 to lm 1080 for y = 1 to pc 1090 w = x/b%(y) 1095 if w = int(w) then 1200 1100 if (b%(y)*b%(y)) > x then goto 1120 1110 next y 1120 pc = pc + 1 1130 b%(pc) = x 1200 next x 1270 s = (time-t)/60 1275 s = int(100*s)/100 1300 print "2^";n,pc,s 1400 next n

I wrote a short program in assembler, using Turbo Macro Pro, to implement a Sieve of Eratosthenes to find the 8 bit prime numbers.

Right click here to download the disk image: tmpsieve

Here is the screenshot:

The ML program places a basic program in memory that is just:

10 SYS 4096

So, once SIEVE is loaded, (LOAD “SIEVE”,8,1), all one has to do is type RUN to run the sieve.

chrout = $ffd2 sieve = $0900 ; 8 bit sieve *= $0800 ;the 00 byte that starts basic .byte $00 ; pointer to the next basic line ; which is at $080c .byte $0c,$08 ; line 10 .byte $0a,$00 ; the basic token for sys .byte $9e ; a space .byte $20 ; "4096" .byte $34,$30,$39,$36 ; 00 for the end of the basic line .byte $00 ; 00 00 for the end of the basic ; program .byte $00,$00 w1 = $0880; working variable w2 = $0881; working variable w3 = $0882; working variable w4 = $0883; working variable w5 = $0884; working variable hx = $0885; holder for x hy = $0886; holder for y hdig = $0887; hundreds digit tdig = $0888; tens digit odig = $0889; ones digit head = $0890 ; header ; this is the header ; it is 68 bytes *= $0890 ;[clr][sp][sp][sp] .byte $93,$20,$20,$20 ;[sp][sp][sp][sp] .byte $20,$20,$20,$20 ;[sp]eig .byte $20,$45,$49,$47 ;ht[sp]b .byte $48,$54,$20,$42 ;it[sp]p .byte $49,$54,$20,$50 ;rime .byte $52,$49,$4d,$45 ;[sp]num .byte $20,$4e,$55,$4d ;bers .byte $42,$45,$52,$53 ;[cr][sp][sp][sp] .byte $0d,$20,$20,$20 ;[sp][sp][sp][sp] .byte $20,$20,$20,$20 ;by[sp]a .byte $42,$59,$20,$41 ;[sp]sie .byte $20,$53,$49,$45 ;ve[sp]o .byte $56,$45,$20,$4f ;f[sp]er .byte $46,$20,$45,$52 ;atos .byte $41,$54,$4f,$53 ;then .byte $54,$48,$45,$4e ;es[cr][cr] .byte $45,$53,$0d,$0d *= $1000 jmp main mult10 ; multiply by 10 subroutine asl a sta w1 asl a asl a adc w1 rts ; now, in order to output base-10 ; values to the screen we need ; to get a char which represents ; each digit in the base-10 ; representation of the 8-bit ; number to be output ; we will store the hundreds place ; digit in hdig, the tens plae ; digit in tdig, and the ones ; place digit in odig ; get hundreds digit subroutine gethundig cmp #$c8 ; is a gt 200? bcc hless200 ldx #$02 ; if here, hdig is stx hdig ; 2 so store it sec sbc #$c8 ; subtract 200 from rts ; a and return hless200 cmp #$64 ; is a gt 100? bcc hless100 ldx #$01 ; if here, hdig is stx hdig ; 1 so store it sec sbc #$64 ; subtract 100 from rts ; a and return hless100 ldx #$00 ; if here, hdig is stx hdig ; 0 so store it rts gettendig ; when this sub is run, a is 0-99 ; start by putting a 9 in x, ; multiplying that by 10, then ; seeing if 90 is higher than the ; test number - if not, tdig is 9, ; and so on sta w2; preserve a ldx #$09 tryaten txa jsr mult10 sta w3; w3 = 90,80,70,etc lda w2 cmp w3 bcc tennotfound txa ; we hav found our ten sta tdig lda w2 sec sbc w3 rts ; we've found the digit, ; and left the ones place ; in a - so return tennotfound dex ; try the next lower bne tryaten ldy #$00; if here, tdig=0 sty tdig lda w2 rts getonedig sta odig; by this time, the ; ones digit is all that ; is left in a rts getdigs jsr gethundig jsr gettendig jsr getonedig rts printdigs jsr getdigs lda hdig ora #$30; convert to ascii jsr chrout lda tdig ora #$30 jsr chrout lda odig ora #$30 jsr chrout lda #$20; two spaces jsr chrout jsr chrout rts dosieve ldx #$00; use x to index loop1 lda #$01 loop1 sta sieve,x; fill the sieve wit inx bne loop1 lda #$00 ldx #$00 sta sieve,x; 0 is not prime inx sta sieve,x; 1 is not prime ; w1 is counter as we step ; w2 is step size lda #$01 sta w1 sta w2 nextpass inc w2; if we are on first pass ; through sieve, we just ; increased step from 1 to 2 ; which is what we want lda w2; the step size is also ; the first element in ; any pass sta w1 cmp #$10; have we reached 16? ; if so, we ar done bne not16 rts not16 ldx w2 lda sieve,x; does x point to a ; zero in the sieve? beq nextpass ; ifso x composite ; so move on nextstep lda w1 txa beq nextpass; we've stepped all ; the way through ; sieve so increase ; step lda w1 adc w2 ; add the step bcs nextpass; if the carry flag ; is set we're past ; the end of the ; sieve so nextpass sta w1 tax ; update x pointer lda #$00 sta sieve,x; x not prime so ; mark with a zero jmp nextstep printheader ldx #$00 phloop lda head,x jsr chrout inx cpx #$44 bne phloop rts main jsr dosieve jsr printheader ;sieve now marks primes ;with a 1 - step and ;print primes ldy #$00 sty w5; w5 is primes on ; this current line probesieve lda sieve,y beq movetonext tya stx hx sty hy jsr printdigs ldy hy ldx hx inc w5 lda w5 cmp #$08; end of line? bne movetonext lda #$0d jsr chrout lda #$00 sta w5; reset line index movetonext iny bne probesieve lda #$0d jsr chrout jsr chrout rts