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

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
 

Exploring the Goldbach Conjecture using a Commodore 64

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
 

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

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$(147)
 200 DEF FN F(A) = A*A*A - 2*A - 5 
 210 DEF FN F1(A) = 3*A*A - 2 
 220 DEF FN F2(A) = 6*A
 230 E$ = "XXX - 2X - 5" 
 235 PRINT "--------------------------------------" 
 240 GOSUB 10000 
 245 PRINT "--------------------------------------" 
 250 GOSUB 20000 
 300 DEF FN F(A) = EXP(A) - A*A
 310 DEF FN F1(A) = EXP(A) - 2*A 
 320 DEF FN F2(A) = EXP(A) - 2  
 330 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 Commodore 64 and the Kolakoski Sequence

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:

An AND gate from discrete components

I decided to build an AND gate from discrete components. I used this (https://www.electronics-tutorials.ws/logic/logic_2.html) extremely helpful page as a guide.

Here is the schematic:

Basically, here’s how it works. If Input A is low (the switch is open so the input is pulled down to ground by the pull down resistor R4), Q1 is off since no current is flowing into its base. Therefore current does not flow to the output, which is pulled low by R3.

The same with Input B.

But if both buttons are pressed, both inputs are high, and both transistors are on, so current flows from +5 to the LED.

Here is the breadboard design:



Here is the device, with no buttons pushed:

Here is the device with button one pushed:

Here is the device with button two pushed:

Here is the device with both buttons pushed, i.e. the AND gate activated:

And that’s all …

A keyboard sonata …

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:

C64 Keyboard Matrix

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.

my-keyboard-matrix

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.

keyboard-codes

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.:

self-list

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.

ender's_game_cover_isbn_0312932081

 

 

 

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

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.

primetime

 

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

A Sieve of Eratosthenes in Assembler

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