# Tackling the Logiker 2023 Vintage Computing Christmas Challenge – part 3

I have to admit, this year’s Logiker challenge stumped me. I had a few “clever” ideas on how to reproduce the pattern, but the code was larger than just printing out the pattern from arrays of strings.

Meanwhile, Jason Pittman kept posting revisions of his concept in the comments. In a response to part 2, he shared this:

This is the last attempt I came up with. It uses the same “-3 to 2 … ABS()” from the other example I sent on part 1. I think one potential trick here is to treat it as if you are holding a rubber stamp that stamps out three asterisks that are six spaces apart. You want to make this pattern by stamping it twice on each line. You’re keeping track of a starting position on each row (1027 + 32 for each row) and an offset to add and subtract to the starting position for each row. On the first line, the offset is zero, so you stamp it twice on top of itself at the starting position. On the next line, the offset is 1, so you stamp it twice, but one time you add -1 and the other time you add +1 to the starting position. Does this make any sense?

1 CLS:P=1027:FORC=1TO3:FORX=-3TO2:FORS=0TO12STEP6:POKEP-3+ABS(X)+S,106:POKEP+3-ABS(X)+S,106:NEXT:P=P+32:NEXT:NEXT:GOTO1

– Jason Pittman

This was an optimization of his original approach, but it had one limitation that might prevent it from solving the challenge: The CoCo’s 32×16 screen is too small to display the entire image, and POKEing characters on the screen would be limited to just 16 lines of text. He was aware of this, and his program does POKE the full image, but it is POKEing past the end of the visible screen. Would this count? RUNning the program displays the pattern over and over again (which was done to avoid having a new line with an endless loop GOTO):

POKEing past the visible screen works here because I am emulating a Disk Extended BASIC CoCo, and the memory after the text screen is reserved for four pages of PMODE high resolution graphics. But, I suspect, if I ran this on a cassette based CoCo, it might be POKEing in to memory used for the BASIC program itself.

Perhaps the CoCo 3 could help, since it has 40×25 and 80×25 text modes? Jason tried that:

I may play around with LPOKE. This should get it on the 40 column screen. I bet there is a crafty way to (a) not do the last line manually outside of the loops (b) remove one of the FOR loops (c) Shoot, there’s probably some crafty wizard way to do it in one FOR loop with logical operators, but I wouldn’t ever find it.

1 WIDTH40:FORZ=0TO12STEP6:FORX=-3TO2:FORS=0TO12STEP6:LOCATEABS(X)+S,Z+X+4:PRINT”*”;:LOCATE6-ABS(X)+S,Z+X+4:PRINT”*”;:NEXT:NEXT:NEXT:FORX=3TO15STEP6:LOCATE X,19:PRINT”*”;:NEXT

– Jason Pittman

In this version, Jason uses LOCATE(x,y) to position the cursor. That is what the CoCo 3 used instead of PRINT@ for text positioning. And it works!

It also feels better to use built-in BASIC text commands versus POKEing in to memory.

But he wasn’t done! He added this version:

`10 WIDTH 4020 FOR S = -15 TO 33 STEP 630 FOR X = 0 TO 1840 IF S+X < 19 AND S+X >= 0 THEN LOCATE S+X,X:PRINT "*";41 IF S-X >= 0 AND S-X < 19 THEN LOCATE S-X,X:PRINT "*";50 NEXT X60 NEXT S70 GOTO 70`

This one draws the same pattern, but in a very different way. It draws diagonal lines going down from the points at the top. Try it! It’s cool!

And, then this odd one, which creates the pattern by drawing the asterisks RANDOMLY, eventually getting them all on the screen.

`0 WIDTH401 X=RND(3)*6:Y=RND(3)*6:RX=RND(7)-4:RY=ABS(RX)-3:D=RND(2)*2-3:LOCATE X+RX,Y+RY*D:PRINT"*";:GOTO 1`

Nice job, Jason!

But it is going to make my brain hurt to understand how this works…

Meanwhile, I received a message from Rick Adams on Facebook with an implementation he was working on for a (much more limited) PDP-8 BASIC.

To be continued…

# Tackling the Logiker 2023 Vintage Computing Christmas Challenge – part 2

As I write this, I have no idea how to make this work. Producing this pattern:

`   *     *     *  * *   * *   * * *   * *   * *   **     *     *     * *   * *   * *   *  * *   * *   * *   *     *     *  * *   * *   * * *   * *   * *   **     *     *     * *   * *   * *   *  * *   * *   * *   *     *     *  * *   * *   * * *   * *   * *   **     *     *     * *   * *   * *   *  * *   * *   * *   *     *     *`

…seems like it should be simple. Three asterisks, then six, then six, then four, then six, then six, then three… Spaces that go five, three, one, zero and then back up. Unhelpful.

But, it’s also just one pattern repeated across the screen three times…

`   *  * * *   **     *`

And then it’s reversed, so I think if we can do the above, we can do the whole pattern. We see spaces of three, two, one, zero on the left, and zero, one, three, and five in the inside.

Color BASIC does not have a SPC() option (my VIC-20 did, I think) for outputting spaces, but TAB() will go to a specific column. Maybe we can figure out which column the asterisks should be in:

`.........11111111111234567890123456789   *     *     *  * *   * *   * * *   * *   * *   **     *     *     *`

This gives us 4, 10 and 16. Then 3 and 5, 9 and 11, and 15 and 17. Then 2 and 6, 8 and 12, and 14 and 18. Finally, 1 and 7 and 13 and 19. I don’t know why, but I kind of like thinking about it as tab positions.

`10 FOR SP=3 TO 0 STEP-120 PRINT TAB(SP);"*";TAB(6-SP);30 IF SP<3 THEN PRINT "*";40 PRINT50 NEXT`

That would give us one of the pyramid shapes. To complete the bottom, we’d do another FOR/NEXT loop. At least, that’s what I would do. BUT, in a comment to part 1, Jason Pittman had a smarter idea:

Awesome! I’ve got an idea on this one but I’m not going to jump ahead this year and I’m just going to follow along.One thought here is that you could combine the two print loops on 100 and 110 by coming up with a series that goes “0 1 2 3 2 1”. I did it by replacing 100 and 110 with this: “100 FOR A=-3 TO 2:PRINT A\$(ABS(ABS(A)-3)):NEXT”

Or, you could shorten that a little if you reverse the direction of the array (so that it looks like “VVV”) and use “100 FOR A=-3 TO 2:PRINT A\$(ABS(A)):NEXT” – Jason Pittman

I could print one diamond like this:

`10 FOR A=-3 TO 3:SP=ABS(A)20 PRINT TAB(SP);"*";TAB(6-SP);30 IF SP<3 THEN PRINT "*";40 PRINT50 NEXT`

That prints almost the entire diamond, except for the final asterisk. Because, if I wanted to print three of them, I’d do this in a loop, then print the final asterisk row at the end.

Unfortunately, as I start going down this rabbit hole, I find the code of loops and such ends up looking larger than some much simpler approaches, like one shown to my by Rick Adams. His code was written for a PDP-8 BASIC, which lacks things like ELSE and MID\$. His technique was to have strings representing parts of the pyramid:

`"   *  ""  * * "" *   *""*     "`

…then to print each string three times across the screen. This produced:

`"   *     *     *  ""  * *   * *   * * "" *   * *   * *   *""*     *     *     "`

…and then do it backwards. There is a missing “*” on the right, that gets printed with an “IF”. In a chat, we bounced around some ideas to shrink the code, but looking at his approach, it seems everything I try to do gets larger:

• Try “run length encode” where DATA statements represent the spaces. Print that many spaces, then an asterisk, and repeat.
• Try DATA statements showing the positions of the asterisks. DATA is large than a string.
• Try simulating a “SET(x,y)” to draw it, but using PRINT@ on the CoCo. Alas, the CoCo 32×16 screen is too small to fit the whole pattern, so even if this was smaller, it would still require extra code at the end to scroll the screen and print the final few lines (as the top portion scrolls off). BUT, using a CoCo 3 40/80 column screen would work using LOCATE x,y instead. But still, larger.

Is there an elegant solution to this challenge that doesn’t involve just PRINTing strings?

We shall continue… Next time…

# Tackling the Logiker 2023 Vintage Computing Christmas Challenge – part 1

Special thanks to Jason Pittman for mentioning this year’s challenge in a comment…

Logiker is at it again, with a 2023 retro-programming Christmas challenge:

???? Vintage Computing Christmas Challenge (VC³) 2023 ???? – Logiker

This year, the pattern looks like this:

`   *     *     *  * *   * *   * * *   * *   * *   **     *     *     * *   * *   * *   *  * *   * *   * *   *     *     *  * *   * *   * * *   * *   * *   **     *     *     * *   * *   * *   *  * *   * *   * *   *     *     *  * *   * *   * * *   * *   * *   **     *     *     * *   * *   * *   *  * *   * *   * *   *     *     *`

This image is 19×19, so while it will fit on a Radio Shack Color Computer 1/2 screen width-wise, it’s a bit too tall to fit height-wise. The challenge allows for it to scroll off the screen, which is something we had to do for past challenges.

I can think of a number of ways to approach this.

The pattern is made up of only four unique lines, so you could print them A B C D B C A B C D and so on. There’s probably a simple way to do that with a FOR/NEXT loop and an array of those four lines.

`10 CLS50 A\$(0)="   *     *     *60 A\$(1)="  * *   * *   * *70 A\$(2)=" *   * *   * *   *80 A\$(3)="*     *     *     *90 FOR I=1 TO 3100 FOR A=0 TO 3:PRINT A\$(A):NEXT110 FOR A=2 TO 1 STEP-1:PRINT A\$(A):NEXT120 NEXT130 PRINT A\$(0)333 GOTO 333`

If we had a larger screen (like the 40 or 80 column screens on the Color Computer 3), we could use LOCATE x,y to plot the pattern using some line drawing type math.

We could try the RLE (run length encoding) compression from past years to see if we could compress it down to spaces and characters.

We could try using math to figure out a pattern.

These all seem fun.

I hope to find some time to experiment. I don’t plan to “enter,” since one of the asks for the challenge is to not share your work until after the challenge ends.

More to come…

# Color BASIC overflow bug – same as Commodore’s?

I just saw a tweet from Robin @ 8-Bit Show And Tell concerning a bug in Commodore BASIC that existed in the PET, C64 and VIC-20.

VAL() takes a string and converts it in to a floating point numerical variable. The value of “1E39” is a number in scientific notation, and this appears to cause a problem.

In Microsoft BASIC, the notation “1E39” represents the number 1 multiplied by 10 raised to the power of 39. This is also known as scientific notation, where the “E” indicates the exponent to which the base (10 in this case) is raised. So, “1E39” is equal to 1 * 10^39, which is an extremely large number:

1E39 = 1 * 10^39 = 1000000000000000000000000000000000000000

This number has 39 zeros after the 1, making it a very large value.

– ChatGPT

I was curious to see what the CoCo’s Color BASIC did, so I tried it…

It appears that port of BASIC to the 6809 also ported over this bug. Anyone want to take a look at the source and see what the issue is?

…to be continued, maybe…

# TIL: You can build C in Microsoft Visual Studio

I feel dumb for not knowing this, but I was under the impression that today’s Visual Studio only built things like C#, C++, etc. — since those are the things listed when you make a new project.

I happened to ask ChatGPT about building C in Studio, and it told me I could just make C++ project and save the file out as a .c and get a C project. I had no idea.

#TheMoreYouKnow

Though, it’s many more steps to do this than, say, popping out to https://www.onlinegdb.com/online_c_compiler, just to do a quick test, but it’s useful.

I have only tried this under Windows, but I plan to see if the Mac version of Visual Studio supports the same. I’ve tried to get C code building in VS Code, but it’s klunky (makefiles!).

# Tackling the Logiker 2022 Vintage Computing Christmas Challenge – part 7

See also: part 1, part 2, part 3, part 4, part 5, part 6 and part 7.

• 2022-12-30 – Update to Jason’s final version to make it two bytes smaller.

In this final (?) installment, I wanted to share some other approaches that were taken to by members of the CoCo community draw this:

…including one that immediately was smaller than the version I did.

Early on, a version was shared by legendary CoCo programmer Rick Adams. His version was not for the CoCo – he chose to do it “in a very primitive BASIC, BASIC8 on a simulated PDP8/I running the TSS8 OS”…

``````0 'RICK ADAMS
12 FOR B = 1 TO 4
14 GOSUB 2000
20 NEXT B
22 C = 0
24 D = 0
30 FOR I = 1 TO 9
34 GOSUB 1000
36 NEXT I
50 FOR B = 4 TO 1 STEP -1
52 GOSUB 2000
58 NEXT B
200 DATA 0, 17, 1, 15, 2, 13, 3, 11, 4, 9, 3, 11, 2, 13, 1, 15, 0, 17
300 STOP
1000 PRINT TAB(A);
1010 FOR J = 1 TO B
1020 PRINT "*";
1030 NEXT J
1040 PRINT TAB(A + B + C);
1050 FOR J = 1 TO D
1060 PRINT "*";
1070 NEXT J
1080 PRINT
1090 RETURN
2000 A = 4
2002 D = B
2010 C = 9 - 2 * B
2020 GOSUB 1000
2030 RETURN
2046 END``````

I am unfamiliar with the BASIC on this machine, but at least it doesn’t require using “LET“. This version can run on the CoCo as well, and correctly reproduces the pattern.

## Jim Gerrie – MC-10/CoCo

Next, take a look a this one by MC-10 BASIC-meister, Jim Gerrie:

His approach uses DATA statements and then draws the star in an interesting way.

## Jason Pittman

In the comments on an earlier installment, Jason shared his attempt. His approach was realizing that the shape was just “four overlapping right triangles.”

``1 FORX=64TO416STEP32:L=X/32:T\$=STRING\$(L,42):PRINT@X-28,T\$;:PRINT@(X-19-L),T\$;:PRINT@544-X+4,T\$;:PRINT@557-X-L,T\$;:NEXT:GOTO1``

This version is just 100 bytes! Due to the CoCo’s 32 column screen being too short, it doesn’t draw the top and end lines of the pattern, so it wouldn’t meet the challenge requirements. To fix that, he needed to add an IF:

``````1 FORX=32TO416STEP32:L=X/32:T\$=STRING\$(L,42):PRINT@X-28,T\$;:PRINT@(X-19-L),T\$;:IF X>32THEN PRINT@544-X+4,T\$;:PRINT@557-X-L,T\$;
2 NEXT
3 GOTO3``````

Since the CoC 3 also has a 40×24 and 80×24 screen, the entire pattern could fit on those screens. Version three looked like this:

``1 WIDTH40:FORX=1TO13:L\$=STRING\$(X,42):LOCATE14-X,X:PRINTL\$;:LOCATE14-X,18-X:PRINTL\$;:LOCATE5+L,X:PRINTL\$;:LOCATE5,18-X:PRINTL\$;:NEXT:GOTO1``

That one is a mere 88 bytes! And, the GOTO1 at the end is just to make it keep redrawing, else it stops near the top and would print the “OK” in the middle of the pattern.

I’d say the “WIDTH40:” is not required, since you could just say “run this from the 40 column screen.” And, to keep the loop, starting on LINE 0 allows just saying “GOTO” with no line number:

``0 FORX=1TO13:L\$=STRING\$(X,42):LOCATE14-X,X:PRINTL\$;:LOCATE14-X,18-X:PRINTL\$;:LOCATE5+L,X:PRINTL\$;:LOCATE5,18-X:PRINTL\$;:NEXT:GOTO``

By my count, that turns in to 83 bytes! Amazing.

UPDATE: L. Curtis Boyle pointed out there was an unnecessary “+L” left in the code, which can be removed to make this 81 bytes. More amazing!

``0 FORX=1TO13:L\$=STRING\$(X,42):LOCATE14-X,X:PRINTL\$;:LOCATE14-X,18-X:PRINTL\$;:LOCATE5,X:PRINTL\$;:LOCATE5,18-X:PRINTL\$;:NEXT:GOTO``

Here is what it looks like, though I paused it to capture the full image:

I’m really blown away by this.

Are we done? Is this as small as it gets?

Unless there are more ideas, I think that is the end.

Merry Christmas, everyone!

# Tackling the Logiker 2022 Vintage Computing Christmas Challenge – part 6

See also: part 1, part 2, part 3, part 4, part 5, part 6 and part 7.

Just when I thought I was out… they pull me back in.

Michael Corleon, Godfather III

Sometimes clever isn’t as good as brute force. In this installment, I’ll present a hybrid approach to the challenge of displaying the Logiker 2022 holiday image.

Instead of writing code to handle each section of the pattern, perhaps taking the simpler approach of just doing Run Length Encoding (sorta) might be smaller. I do not know where I first learned about RLE, but I implemented a simple version in Sub-Etha Software’s graphical “CoCoFEST Simulation” text adventure back in the early 1990s. The images in the graphical adventure were large, and I came up with a simple way to represent repeated data in them. (I do not know if someone gave me this idea, or if I worked it out on my own; in those pre-Internet days, knowledge was alot more organic.)

Basically, RLE looks for repeating sequences and replaces them with a count value and the sequence. The Wikipedia article gives this example:

Consider a screen containing plain black text on a solid white background. There will be many long runs of white pixels in the blank space, and many short runs of black pixels within the text. A hypothetical scan line, with B representing a black pixel and W representing white, might read as follows:

`WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW`

With a run-length encoding (RLE) data compression algorithm applied to the above hypothetical scan line, it can be rendered as follows:

`12W1B12W3B24W1B14W`

– https://en.wikipedia.org/wiki/Run-length_encoding

Above, if there are twelve “W” characters in a row, you would encode the value 12 followed by the “W”.

I wrote a simple program that will input a string and then scan it looking for repeating characters. When it finds a new character, it prints the count of the previous characters as well as what character that was.

``````0 ' RLE.BAS
10 LINE INPUT "STRING:";A\$
20 LC=ASC(LEFT\$(A\$,1)):L=1
30 FOR I=2 TO LEN(A\$)
40 C=ASC(MID\$(A\$,I,1))
50 IF C<>LC THEN PRINT L;LC:L=0:LC=C
60 L=L+1:NEXT
70 PRINT L;LC
``````

I pasted this code in to chat.openai.com and it correctly recognized it as an RLE routine! Here is how it described my program:

This looks like a BASIC program that encodes a string using run-length encoding (RLE). RLE is a simple form of data compression, where runs of repeated characters are replaced by a count and the character. For example, the string “AAABBCCC” would be encoded as “3A2B3C”.

The program starts by prompting the user to input a string, and then assigns the first character of the string to the variable `LC` (which stands for “last character”). The variable `L` is set to 1, and then a `FOR` loop is used to iterate through the remaining characters in the string.

For each character in the string, the program checks if it is different from the last character. If it is, it prints the length of the previous run and the character itself, and then resets the length and last character variables. If the character is the same as the last character, the length is incremented. This continues until the end of the string is reached, at which point the final run is printed.

For example, if the input string is “AAABBCCC”, the output would be:

3 65
2 66
3 67

This indicates that there were three instances of the character “A” (ASCII value 65), two instances of the character “B” (ASCII value 66), and three instances of the character “C” (ASCII value 67).

– chat.openai.com

Wow. Just wow.

Here is an example of the program running:

Since the Logiker Challenge image only contains spaces and asterisks, we don’t need a full RLE to encode it. I chose to use alternating numbers to indicate how many spaces, then how many asterisks.

Ultimately, I tried to save code space by breaking the image in to sections, and using two main routines to print that section within FOR/NEXT loops.

But … is that overkill? Instead of using multiple PRINT routines, what if I only needed one? By expanding the image data so each line covers the entire width of the CoCo’s 32-column screen, I could do away with the “end of line” markers in the data, and replace them with a larger series of spaces that goes from the end of the data on that line to the start of the data on the next line:

```[12345678901234567890123456789012]
"           *       *            "
"           **     **            "
"           ***   ***            "
"           **** ****            "
"       *****************        "
"        ***************         "
"         *************          "
"          ***********           "
"           *********            "
"          ***********           "
"         *************          "
"        ***************         "
"       *****************        "
"           **** ****            "
"           ***   ***            "
"           **     **            "
"           *       *            "
[12345678901234567890123456789012]```

Above, at the end of the first line’s asterisks, there are 12 spaces to the end of that line. For the next line, there are 11 spaces to get to the start of the next asterisks. That means after printing the last asterisks in line 1 we can just print 23 spaces and be at the start of the next line.

Assuming we start with a SPACE then an ASTERISK then a SPACE and do on, the data for the first two lines would look like this:

```11 - print11 spaces
1 - print 1 asterisk
7 - print 7 spaces
1 - print 1 asterisk
23 - print 23 spaces (to move to the start of data in the second line)
2 - print 2 asterisks
5 - print 5 spaces
2 - print 2 asterisks
...and so on...```

I was going to convert all the PRINT lines of the original version I started with to DATA statements and write a program to count this for me, but that sounded complicated. I just counted, and came up with the following numbers:

```11 1 7 1 23 2 5 2 3 3 23 4 4 4 23 5 16 15 16 17 18 16 5 2 3 3 23 4 4 4 23 5 18 2 5 2 1 1 7 1
```

I could store those in a DATA statement:

``DATA 11,1,7,1,23,2,5,2,3,3,23,4,4,4,23,5,16,15,16,17,18,16,5,2,3,3,23,4,4,4,23,5,18,2,5,2,1,1,7,1``

But, that takes up alot of room. There is a comma between each number, so for 50 numbers we’d be adding 49 commas, basically doubling the size of the data. Also, two digit numbers like 10 take up two bytes. I thought about using HEX numbers (0-15 turns in to 0-F) but the data has some values that are larger than 15 (the highest value that fits in a single character of a HEX value).

HEX is BASE-16 (0-F to represent 0-15) and what I really need is at least BASE-23 (0-23, the larger number I need). Since there are 26 letters in the alphabet, I could use all of them and get BASE-26 leaving me room to spare!

If A=1, B=2 and so on, the above series of numbers could be turned in to:

```K A G A W B E B W C C C W D A D S Q P O R M T K V I V K T M R O P Q S D A D W C C C W B E B W A G A
```

I could then turn those in to DATA:

``DATA K,A,G,A,W,B,E,B,W,C,C,C,W,D,A,D,S,Q,P,O,R,M,T,K,V,I,V,K,T,M,R,O,P,Q,S,D,A,D,W,C,C,C,W,B,E,B,W,A,G,A``

…and read them as a string (READ A\$) and then convert that string to a number by subtracting 63 (ASCII for A is 64, so if I read an A and get 64, subtracting 63 turns that in to 1):

``````READ A\$
V=ASC(A\$)-64``````

While this saves a byte for every number that was two digits, the extra code to convert from ASCII to a number may be larger than what we saved.

Since we have 49 commas, we could get rid of those and add code to parse a long string. As long as that code is smaller than 49 bytes, we come out ahead.

``DATA KAGAWBEBWCCCWDADSQPORMTKVIVKTMROPQSDADWCCCWBEBWAGA``

Now I could read that as a string and parse it in to numbers:

``````0 'STRTONUM.BAS
20 FOR I=1 TO LEN(A\$)
30 PRINT ASC(MID\$(A\$,I,1))-64;
40 NEXT
``````

And, if I want to use that series of numbers in a loop that prints alternating strings of spaces and asterisks, I don’t even need to bother with it being in a DATA statement. I could just embed it directly in the MID\$() command and hard code the lengthof the string, like this:

``````0 'STRTONUM2.BAS
20 FOR I=1 TO 50
40 NEXT
``````

And if I can do that, the only thing left is to figure out when to print a space and when to print an asterisks.

An easy way to do that is looking at the I variable in the FOR/NEXT loop. As it counts from 1 to 2 to 3 to 4, I can use AND to check bit 1. For odd numbers, that bit is set. For even numbers, it is not.

```0 = 0000000
1 = 0000001
2 = 0000010
3 = 0000011
4 = 0000100
5 = 0000101
...and so on...```

This means a simple check for “I AND 1” in an IF statement can help me decide which to print. Something like:

`IF (I AND 1) THEN PRINT space ELSE PRINT asterisk`

That gets me to something like this:

``````0 ' LOGIKER-ALPHA2.BAS
10 FORI=1TO50
30 IF I AND 1 THEN PRINT STRING\$(L,32); ELSE PRINT STRING\$(L,42);
40 NEXT
``````

Perhaps I can get rid of one of those PRINT STRING\$ commands… Since I know a space is ASCII 32 and an asterisk is ASCII 42, I could start with the 32 and add 10 if it’s the asterisk case. To do that, I need to see the result that comes back from AND:

```PRINT 1 AND 1
1

PRINT 2 AND 1
0```

So if the condition is TRUE (bit 1 is set, meaning the value is odd), I get a 1. If the condition is FALSE (bit 1 is clear, meaning the value is even), I get a 0.

Since I want to print spaces on the odd values, I need to use the 1 (odd) to mean 32, and the 0 (even) to mean 42. I’ll reverse my logic a bit and always start with 42 (asterisks) and multiply it by 10 times the result of (I AND 1). Something like this should work:

``````0 ' LOGIKER-ALPHA3.BAS
10 FOR I=1 TO 50
30 PRINT STRING\$(L,42-(I AND 1)*10);
40 NEXT
``````

And that gives me the pattern I want, with far less code. I can remove unneeded spaces and combine everything in to one line and see how big it is.

## Unneeded Spaces

A quick thing about unneeded spaces. There are spaces that BASIC itself doesn’t need, but the tokenizer that turns what you type in to the program DO need. For example:

`FOR I=100 TO 5000`

None of those spaces are needed, because BASIC knows where a keyword ends (FOR) and can tell the variable will be whatever is there before the “=”. The same is true for the numbers, since it can tell where a number ends and know to look for “TO”.

``FORI=100TO5000``

BUT, if you were using variables in that loop…

`FOR I=B TO E`

…and you took the spaces out:

`FORI=BTOE`

…how does BASIC know what your variable is? Is it “B”? Or “BT”? Or maybe “BTOE”? You will get an “?SN ERROR” if you try that because BASIC sees a non-number after the “=” and switches to parsing it as if it were a variable. To get around this, we have to put a space after it like this:

`FORI=B TOE`

That allows the tokenizer to work fine.

However If you were manually creating the BASIC program by packing bytes together in a file, you could omit that space and it will run just fine. Utilities such as Carl England’s CRUNCH do this trick to save a byte. BUT, if you were to CRUNCH the program then try to EDIT that line, you’d no longer have code that would run because updating the line requires it to be re-tokenized. #TheMoreYouKnow

## Why is that important?

I mention this because in my above program, I wanted to remove spaces from this line:

``30 PRINT STRING\$(L,42-(I AND 1)*10);``

I can remove all but one, since I need a space between “I” and “AND” for the same reason I just mentioned:

``30 PRINTSTRING\$(L,42-(I AND1)*10);``

But… instead of “I AND 1” I could change it to “1 AND I” and get the same result, but no longer need the space because BASIC can tell where a number stops:

``30 PRINTSTRING\$(L,42-(1ANDI)*10);``

And that, my friends, is how you save one more byte.

Would it be possible to also get rid of those parenthesis? Right now, I need to take my asterisk value (42) and subtract either 0 or 10. I need the results of “1 AND I” multiplied by 10, and if I removed the parens…

`42-1 AND I*10`

…BASIC would do the math first (42-1 and I*10) and if “I” was 3 at the time, I would get this:

```42-1 AND 3*10
41 AND 30```

…and that’s not at all what we want.

Can it be done? I moved things around but it really looks like the result of “1 AND I” has to be in parens. Can you figure a way to save those two bytes?

With that said, I present this version:

``````10 FOR I=1 TO 50
30 PRINT STRING\$(L,42-(1ANDI)*10);
40 NEXT
``````

…which can be packed in to this version:

``10 FORI=1TO50:L=ASC(MID\$("KAGAWBEBWCCCWDADSQPORMTKVIVKTMROPQSDADWCCCWBEBWAGA",I))-64:PRINTSTRING\$(L,42-(1ANDI)*10);:NEXT``

And that shows up as 114 bytes!

Oh, one thing I should also mention — during last year’s challenge, a comment was made about how ASC() works. If you give it a string, it returns the ASCII value of the first character. So ASC(“A”) returns 64, just like ASC(“ALLEN”) does. They said instead of using MID\$(A\$,I,1) to get one character, you can leave off that third parameterand MID\$ returns the rest of the string:

```A\$="HELLO"
PRINT MID\$(A\$,2,1)
C

PRINT MID\$(A\$,2)
ELLO```

If we were trying to print or use just one letter, we need that third parameter. But since I am passing it in to ASC, I could still give it the longer string and it would work fine:

```PRINT ASC("E")
69

PRINT ASC("ELLO")
69```

Thus, I can leave off that third parameter and save the two bytes that “,1” took up.

Neat!

Are we done? Can we save any more?

Until next time…

# Tackling the Logiker 2022 Vintage Computing Christmas Challenge – part 5

See also: part 1, part 2, part 3, part 4, part 5, part 6 and part 7.

The challenge continues. From humble beginnings of using PRINT, to fancier methods of encoding the image as a series of spaces and asterisks, we eventually ended up with an even fancier method that used only 1/4 of the image data to represent the entire symmetrical image.

That approach could work for any image that is symmetrical vertically and horizontally, and typically general purpose routines are not as small as custom routines that know what they will be doing.

## Knowing what we now know…

WIth that said, looking at this image, there is another shortcut that I missed:

The entire image is centered over one column… This means the amount of spaces on the left is unimportant — we just need to center the following lines:

```*       *
**     **
***   ***
**** ****
*****************
***************
*************
***********
*********
***********
*************
***************
*****************
**** ****
***   ***
**     **
*       *```

And, since we know it’s symmetrical both vertically and horizontally, we really only need to worry about this:

```*    | <- 1 asterisk, 4 spaces
**   | <- 2 asterisks, 3 spaces
***  | <- 3 asterisks, 2 space
**** | <- 4 asterisks, 1 space
*****************| 17
***************| 15
*************| 13
***********| 11
*********| 9```

I’m not sure what the pattern is as I type this, but I am expecting there is one. Here is a quick program that prints the rows of the shape using FOR/NEXT loops (uncentered):

``````0 ' LOGIKER13.BAS
10 FOR I=1 TO 4
20 PRINT STRING\$(I,"*");STRING\$(1+(4-I)*2," ");STRING\$(I,"*")
30 NEXT
40 FOR I=17 TO 9 STEP-2
50 PRINT STRING\$(I,"*")
60 NEXT
70 FOR I=11 TO 17 STEP 2
80 PRINT STRING\$(I,"*")
90 NEXT
100 FOR I=4 TO 1 STEP-1
110 PRINT STRING\$(I,"*");STRING\$(1+(4-I)*2," ");STRING\$(I,"*")
120 NEXT``````

If each of those lines were centered, we’d have our shape. Let’s try that by creating a string for the row, and then using the LEN() of that string to know how to center it using TAB().

``````0 ' LOGIKER14.BAS
10 FOR I=1 TO 4
20 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*")
25 PRINT TAB(16-LEN(A\$)/2);A\$
30 NEXT
40 FOR I=17 TO 9 STEP-2
50 A\$=STRING\$(I,"*")
55 PRINT TAB(16-LEN(A\$)/2);A\$
60 NEXT
70 FOR I=11 TO 17 STEP 2
80 A\$=STRING\$(I,"*")
85 PRINT TAB(16-LEN(A\$)/2);A\$
90 NEXT
100 FOR I=4 TO 1 STEP-1
110 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*")
115 PRINT TAB(16-LEN(A\$)/2);A\$
120 NEXT
130 GOTO 130``````

That produces our desired shape (though it does leave a blank line at the end, which our original version avoided by having a semi-colon on the PRINT and just breaking lines when we went to the next one).

The first thing I see it that the centering code on line 25, 55, 85 and 115 is the same. Subroutine!

``````0 ' LOGIKER15.BAS
10 FOR I=1 TO 4
20 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*")
25 GOSUB 150
30 NEXT
40 FOR I=17 TO 9 STEP-2
50 A\$=STRING\$(I,"*")
55 GOSUB 150
60 NEXT
70 FOR I=11 TO 17 STEP 2
80 A\$=STRING\$(I,"*")
85 GOSUB 150
90 NEXT
100 FOR I=4 TO 1 STEP-1
110 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*")
115 GOSUB 150
120 NEXT
130 GOTO 130
150 PRINT TAB(16-LEN(A\$)/2);A\$:RETURN``````

Next, we see that the string building code for the top and bottom are the same, so 20 and 110 are the same (it’s the value of I that changes how it prints), and then 50 and 80 are the same. Subroutines!

``````0 ' LOGIKER16.BAS
10 FOR I=1 TO 4
20 GOSUB 200
25 GOSUB 150
30 NEXT
40 FOR I=17 TO 9 STEP-2
50 GOSUB 250
55 GOSUB 150
60 NEXT
70 FOR I=11 TO 17 STEP 2
80 GOSUB 250
85 GOSUB 150
90 NEXT
100 FOR I=4 TO 1 STEP-1
110 GOSUB 200
115 GOSUB 150
120 NEXT
130 GOTO 130
150 PRINT TAB(16-LEN(A\$)/2);A\$:RETURN
200 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*"):RETURN
250 A\$=STRING\$(I,"*"):RETURN``````

Next, I notice the subroutines of 200 and 250 both have the centering PRINT called after them, so maybe we change it up a bit…

``````0 ' LOGIKER17.BAS
10 FOR I=1 TO 4
20 GOSUB 200
30 NEXT
40 FOR I=17 TO 9 STEP-2
50 GOSUB 250
60 NEXT
70 FOR I=11 TO 17 STEP 2
80 GOSUB 250
90 NEXT
100 FOR I=4 TO 1 STEP-1
110 GOSUB 200
120 NEXT
130 GOTO 130
200 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*"):GOTO 300
250 A\$=STRING\$(I,"*")
300 PRINT TAB(16-LEN(A\$)/2);A\$:RETURN``````

What else? The FOR/NEXT loops are basically all the same, except for the start and end value and the step value… Maybe we could come up with a way to have only one, and feed it those values using DATA statements?

```10 FOR I=1 TO 4
...
40 FOR I=17 TO 9 STEP-2
...
70 FOR I=11 TO 17 STEP 2
...
100 FOR I=4 TO 1 STEP-1

500 DATA 1,4,1
510 DATA 17,9,-2
520 DATA 11,17,2
530 DATA 4,1,-1```

If they all went to the same GOSUB routine this would be easy, but they don’t. The go 200, 250, 250, 200. We could add a fourth element in the DATA that tells it which routine to go to and “IF X=1 THEN GOSUB Y ELSE GOSUB Z” or something. That adds more code. Perhaps we don’t need the DATA since we know it alternates? Still, we’d have to track it ourselves with an IF or something. For now, let’s just try this:

``````0 ' LOGIKER18.BAS
10 FOR J=1 TO 4
30 FOR I=A TO B STEP C
40 IF D=0 THEN GOSUB 200 ELSE GOSUB 250
50 NEXT I
60 NEXT J
70 GOTO 70
200 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*"):GOTO 300
250 A\$=STRING\$(I,"*")
300 PRINT TAB(16-LEN(A\$)/2);A\$:RETURN
500 DATA 1,4,1,0
510 DATA 17,9,-2,1
520 DATA 11,17,2,1
530 DATA 4,1,-1,0``````

And that still produces your original shape. But is it any smaller?

In part 4, we had a version that (using my default XRoar emulator running DISK EXTENDED COLOR BASIC) showed 22499 bytes free after loading. This new version shows 22567 bytes free. So yes, it is smaller! And, we can pack those lines and make it even smaller than that. (And NEXT doesn’t near the variable — in fact, using “NEXT I” is slower than just saying “NEXT” so I’ll remove those here as well.)

``````0 ' LOGIKER19.BAS
10 FOR J=1 TO 4:READ A,B,C,D:FOR I=A TO B STEP C:IF D=0 THEN GOSUB 200 ELSE GOSUB 250
50 NEXT:NEXT
70 GOTO 70
200 A\$=STRING\$(I,"*")+STRING\$(1+(4-I)*2," ")+STRING\$(I,"*"):GOTO 300
250 A\$=STRING\$(I,"*")
300 PRINT TAB(16-LEN(A\$)/2);A\$:RETURN:DATA 1,4,1,0,17,9,-2,1,11,17,2,1,4,1,-1,0``````

That version shows me 22609 free, which is even smaller — and we could still make this a bit smaller by getting rid of unnecessary spaces in the code.

Side note: I am being lazy and just showing the BASIC “PRINT MEM” values rather than calculating the actual size of the program. On my configuration, 22823 is how much memory is there on startup. So, 22823-22609 shows that this program is 214 bytes. It uses more memory for the strings when running, but I don’t think that matters for this challenge.

What else can we do to save a few bytes? Well, STRING\$() takes two parameters. The first is the count of how many times to repeat the second parameter. The second parameter can be a quoted character like “*”, or a number like 42 (the ASCII value of the asterisk). 42 is one by smaller than “*” so we can do that as well as use 32 (the ASCII value for space) instead of ” “:

``````200 A\$=STRING\$(I,42)+STRING\$(1+(4-I)*2,32)+STRING\$(I,42):GOTO 300
250 A\$=STRING\$(I,42)``````

Another thing we know is that in the shape there are always the same number of spaces before the top and bottom sections, so we really don’t need to center it. We could just hard code a PRINT TAB for that instead of building a string and calling a center subroutine:

``````200 PRINT TAB(11);STRING\$(I,42);STRING\$(1+(4-I)*2,32);STRING\$(I,42)
210 RETURN``````

The middle section is similar. Since we know the length, we could calculate how many spaces to tab using that number:

``````250 PRINT TAB(16-I/2);STRING\$(I,42)
260 RETURN``````

And that removes a subroutine, leaving us with this (not line packed yet):

``````0 ' LOGIKER20.BAS
10 FOR J=1 TO 4
30 FOR I=A TO B STEP C
40 IF D=0 THEN GOSUB 200 ELSE GOSUB 250
50 NEXT
60 NEXT
70 GOTO 70
200 PRINT TAB(11);STRING\$(I,42);STRING\$(1+(4-I)*2,32);STRING\$(I,42)
210 RETURN
250 PRINT TAB(16-I/2);STRING\$(I,42)
260 RETURN
500 DATA 1,4,1,0
510 DATA 17,9,-2,1
520 DATA 11,17,2,1
530 DATA 4,1,-1,0``````

Two FOR/NEXT loops, a READ, an IF, and two PRINT subroutines.

Maybe we don’t need those subroutines, now that we have an “IF” in line 40 that decides what to do?

``````0 ' LOGIKER21.BAS
10 FOR J=1 TO 4
30 FOR I=A TO B STEP C
40 IF D=0 THEN PRINT TAB(11);STRING\$(I,42);STRING\$(1+(4-I)*2,32);STRING\$(I,42) ELSE PRINT TAB(16-I/2);STRING\$(I,42)
50 NEXT
60 NEXT
70 GOTO 70
500 DATA 1,4,1,0
510 DATA 17,9,-2,1
520 DATA 11,17,2,1
530 DATA 4,1,-1,0``````

That’s an ugly line 40, but it got rid of two GOSUBs and two RETURNS. Plus every line takes up 5 bytes, so by removing two of those (each subroutine must start on its own line) we saved even more.

Not only is this smaller, it’s faster, too, since it doesn’t have to spend time seeking through the program looking for a subroutine line number for each line it prints.

If we pack this version, removing unnecessary spaces and semicolons and such, and remove the “GOTO” loop (the challenge does not require the screen to be cleared, and says it can return to a prompt after running), we get:

``````0 ' LOGIKER22.BAS (don't include this line)

2 NEXT:NEXT:DATA1,4,1,0,17,9,-2,1,11,17,2,1,4,1,-1,0``````

151 bytes of BASIC code!

Is there more that can be done? Comment with your ideas! I think I’m out of them, here.

Until next time…

# Tackling the Logiker 2022 Vintage Computing Christmas Challenge – part 4

See also: part 1, part 2, part 3, part 4, part 5, part 6 and part 7.

The story so far…

In the beginning Logiker offered up a programming challenge to produce this simple image:

From its humble beginnings as a bunch of PRINT statements, to a DATA driven output routine, we have now ended up with code that can print the top half of the symmetrical image by only having data to describe one quarter of the image.

When we last left off, I had code that would read this data, print the left half, then print the right half, mirroring the same data. It worked well, but only printed the top half of the image.

I showed that code to OpenAI and asked how it would accomplish the task of reversing the data to print the bottom portion:

To make the program print the rows in reverse order after it has finished printing all of the rows, you will need to make some changes to the code. One way to do this would be to add another loop after the main loop that reads the data from the DATA statements and prints the rows in reverse order. You can use the “REVERSE” statement in BASIC to reverse the order of the rows in the DATA statements, and then use the same code as in the main loop to read the data and print the rows. This will cause the program to print the rows in the original order, and then again in reverse order.

– chat.openai.com

I’ve never heard of the REVERSE keyword, and the CoCo’s BASIC doesn’t have it. But if I did have it, this task would be pretty easy. I informed the A.I. that I had no such command, and it said:

If your BASIC does not have a REVERSE command, you can still print the rows in reverse order by using a different approach. One way to do this would be to store the data from the DATA statements in an array, and then loop through the array in reverse order to print the rows. In BASIC, you can use the “DIM” statement to create an array, and the “READ” statement to read data from the DATA statements into the array. You can then use a loop to iterate through the array in reverse order, using the same code as in the main loop to print the rows. This will cause the program to print the rows in the original order, and then again in reverse order.

– chat.openai.com

That’s the approach I was thinking of taking, so… okay!

Side Note: I did ask the A.I. to modify my program to do this, and it gave me something that would have been a good starting point for the rest of this article. I could have then discussed things with the A.I. a bit more and encouraged it to evolve the program until it met the requirements. (One issue with BASIC is that the A.I. inevitably tries to do things from other dialects of BASIC that Color BASIC does not have, such as the REVERSE command.) I hope to demonstrate one of these co-programming sessions in a future article series, provided access to ChatGPT remains available to me.

I changed my routine so it loaded all the DATA values in to a dimension. Since I knew how many entries there would be (27), I could eliminate the need for having a -1 to mark the end of data.

Then, in the PRINT routine, instead READing A, B and C and looking at them, I’d use the D() dimension variable as D(x), D(x+1) and D(x+2) to get to the same three values.

At the end of each line (three numbers, or two numbers followed by a zero), I would increment my index by three so it starts at the next set of numbers in the array.

I used a new variable, M, to indicate the mode (forward or reverse). When it reaches the end of the data (D goes past the 27 entries), it sets that variable to 1, and other code now runs decrementing the index by 3 each time until it reaches the top (0).

Here is my new code:

``````0 ' LOGIKER11.BAS
10 CLS:DIM D(26)
11 FOR I=0 TO 26:READ D(I):NEXT:D=0:M=0
15 PRINTTAB(6);
20 PRINT STRING\$(D(D),32);
25 IF D(D+2)=0 THEN PRINT STRING\$(D(D+1)*2-1,42);STRING\$(D(D),32);:GOTO 40
30 PRINT STRING\$(D(D+1),42);STRING\$(D(D+2)*2-1,32);STRING\$(D(D+1),42);
40 IF M=0 THEN D=D+3:IF D<27 THEN PRINT:GOTO 15 ELSE M=1
41 IF D>0 THEN PRINT:D=D-3:GOTO 15
50 DATA 5,1,4
60 DATA 5,2,3
70 DATA 5,3,2
80 DATA 5,4,1
90 DATA 1,9,0
100 DATA 2,8,0
110 DATA 3,7,0
120 DATA 4,6,0
130 DATA 5,5,0
220 GOTO 220``````

OpenAI describes it as follows:

This program is similar to the previous ones, but it uses a slightly different approach to read the data from the DATA statements and print the rows. The program starts by defining an array `D` to store the data from the DATA statements, and then reads the data into the array. The program then enters a loop that prints the rows of the pattern. In each iteration of the loop, the program prints a row of the pattern by printing a string of spaces, followed by a string of asterisks, and then another string of spaces. The number of spaces and asterisks in each string is determined by the data in the array.

After the program has printed all of the rows in the original order, it enters another loop that prints the rows in reverse order. In each iteration of the loop, the program prints the same row as in the previous loop, but in reverse order. This causes the program to print the rows of the pattern in the original order, and then again in reverse order. However, without more information about the data in the DATA statements, it is not possible to say exactly what the output of the program would look like.

– chat.openai.com

Nicely done, A.I.

When I run this program, I get the desired output!

It is, of course, slower than the original brute-force PRINT version. But, we are going for size. And speaking of size, lets try to pack this down as much as possible and see where we stand:

``````0 ' LOGIKER12.BAS
10 CLS:DIM D(26):FOR I=0 TO 26:READ D(I):NEXT:D=0:M=0
15 PRINTTAB(6);STRING\$(D(D),32);:IF D(D+2)=0 THEN PRINT STRING\$(D(D+1)*2-1,42);STRING\$(D(D),32);:GOTO 40
30 PRINT STRING\$(D(D+1),42);STRING\$(D(D+2)*2-1,32);STRING\$(D(D+1),42);
40 IF M=0 THEN D=D+3:IF D<27 THEN PRINT:GOTO 15 ELSE M=1
41 IF D>0 THEN PRINT:D=D-3:GOTO 15
50 GOTO 50:DATA 5,1,4,5,2,3,5,3,2,5,4,1,1,9,0,2,8,0,3,7,0,4,6,0,5,5,0``````

Loading the first version gives me 22431 bytes free. Loading the second gives me 22499 – about 68 bytes smaller! Compare that to the original brute-force PRINT version (which was 22309 free), we have saved 190 bytes so far.

And, we could remove spaces, get rid of the REM comment at the start, and save even more.

But is that enough for the challenge? One thing the challenge says it you are not required to clear the screen, and you can return to an OK/Ready prompt. That means I could remove the CLS and the GOTO loop at the end, saving even more.

But saving even more is not enough. I think there’s a few more things we can do, especially now that I understand what it takes to draw this shape.

Until next time, take a look at what I have done in LOGIKER11.BAS and see what suggestions you can come up with.

# Tackling the Logiker 2022 Vintage Computing Christmas Challenge – part 3

See also: part 1, part 2, part 3, part 4, part 5, part 6 and part 7.

So far, we’ve taken a brute force PRINT program and turned it in to a less-brute force program that did the same thing using DATA statements:

``````0 ' LOGIKER8.BAS
10 CLS
15 CH=32:PRINTTAB(6);
20 READ A:IF A=-1 THEN 220
25 IF A=0 THEN PRINT:GOTO 15
30 PRINT STRING\$(A,CH);
35 IF CH=32 THEN CH=42 ELSE CH=32
40 GOTO 20
50 DATA 5,1,7,1,0
60 DATA 5,2,5,2,0
70 DATA 5,3,3,3,0
80 DATA 5,4,1,4,0
90 DATA 1,17,0
100 DATA 2,15,0
110 DATA 3,13,0
120 DATA 4,11,0
130 DATA 5,9,0
140 DATA 4,11,0
150 DATA 3,13,0
160 DATA 2,15,0
170 DATA 1,17,0
180 DATA 5,4,1,4,0
190 DATA 5,3,3,3,0
200 DATA 5,2,5,2,0
210 DATA 5,1,7,1,0
215 DATA -1
220 GOTO 220``````

All of this in an effort to try to print out this image:

While there are still many BASIC optimizations we could do (removing spaces, combining lines even further, renumbering by 1, etc.), those would apply to any version of the code we create. Instead of doing that, let’s look at some other ways we can represent this data.

## Simpsons Atari 2600 did it first.

Let no meme go to waste, I always say.

When the Atari VCS came out in 1977 (you younguns may only know it as the 2600, but it didn’t get that name until 1982 — five years after its release later), it required clever tricks to make games run in only 1K or 2K of ROM and with just 128 bytes of RAM.

The game Adventure was quite the challenge, since it had multiple screens representing different mazes, castles and areas.

Each screen was represented by only 21 bytes of ROM! If you follow that link, you can read more about my efforts to understand how this worked. Here is an example of how the castle room was represented:

```;Castle Definition
CastleDef:
.byte \$F0,\$FE,\$15 ;XXXXXXXXXXX X X X      R R R RRRRRRRRRRR
.byte \$30,\$03,\$1F ;XX        XXXXXXX      RRRRRRR        RR
.byte \$30,\$03,\$FF ;XX        XXXXXXXXXXRRRRRRRRRR        RR
.byte \$30,\$00,\$FF ;XX          XXXXXXXXRRRRRRRR          RR
.byte \$30,\$00,\$3F ;XX          XXXXXX    RRRRRR          RR
.byte \$30,\$00,\$00 ;XX                                    RR
.byte \$F0,\$FF,\$0F ;XXXXXXXXXXXXXX            RRRRRRRRRRRRRR   ```

There are three bytes to represent each line. Three bytes would only be able to represent 24 pixels (8 bits per byte), and the ASCII art shows the screen width is actually 40. Those three bytes cannot represent the entire row of pixels.

In fact, 4-bits of that isn’t used. Each set of three bytes represents halfa row (20 bits out of the 24 the three bytes represent). Look at the first entry:

`  .byte \$F0,\$FE,\$15 ;XXXXXXXXXXX X X X      R R R RRRRRRRRRRR   `

If you turn those bytes into binary, you get this pattern:

``` byte 1  byte 2  byte 3|
--------========--------
111100001111111000010101```

The Atari drew the first 8-bits from least significant bit to most. the second 8-bits from most significant to least, then the third from least significant to most. That makes it look like this, matching the ASCII art (skipping the unused 4-bits):

```000011111111111010101000
XXXXXXXXXXX X X X   ```

To represent a full screen, the Atari had a trick that would mirror or duplicate the other half of the screen. In the case of the castle, the right side was a mirror image. In the case of certain mazes, the data was duplicated.

Looking at our image here, since it is symmetrical, we could certainly use the same trick and only store half of the image.

```+----------------+
|           *    |
|           **   |
|           ***  |
|           **** |
|       *********|
|        ********|
|         *******|
|          ******|
|           *****|
|          ******|
|         *******|
|        ********|
|       *********|
|           **** |
|           ***  |
|           **   |
|           *    |
+----------------+```

Also, since the top and bottom are also mirror images, we could mirror those, too, and get away with only storing 1/4 of the image:

```+----------------+
|           *    |
|           **   |
|           ***  |
|           **** |
|       *********|
|        ********|
|         *******|
|          ******|
|           *****|
+----------------+```

Since the image is 17×17 (an odd number, so there is a halfway row and column), we’d actually need to just draw to that halfway row/column, then reverse back through the data.

We should be able to take our existing data and crop it down from this, which represents the full image:

```50 DATA 5,1,7,1,0
60 DATA 5,2,5,2,0
70 DATA 5,3,3,3,0
80 DATA 5,4,1,4,0
90 DATA 1,17,0
100 DATA 2,15,0
110 DATA 3,13,0
120 DATA 4,11,0
130 DATA 5,9,0
140 DATA 4,11,0
150 DATA 3,13,0
160 DATA 2,15,0
170 DATA 1,17,0
180 DATA 5,4,1,4,0
190 DATA 5,3,3,3,0
200 DATA 5,2,5,2,0
210 DATA 5,1,7,1,0
```

…to this, which represents the top left quarter-ish of the image:

```50 DATA 5,1,4,0   '     X    '
60 DATA 5,2,3,0   '     XX   '
70 DATA 5,3,2,0   '     XXX  '
80 DATA 5,4,1,0   '     XXXX '
90 DATA 1,9,0     ' XXXXXXXXX'
100 DATA 2,8,0    '  XXXXXXXX'
110 DATA 3,7,0    '   XXXXXXX'
120 DATA 4,6,0    '    XXXXXX'
130 DATA 5,5,0    '     XXXXX'```

That represents all the data up to the center row/column, and that seems to be a considerable savings in code space (removing eight lines).

But how do we draw that forward, then in reverse? There is no way to back up when using the READ command, so we’d have to remember what we just did. For a general purpose “compress 1-bit image” routine it would be more complex, but since we know the image we are going to produce, we can make an assumption:

1. The image never has more than three transitions (space, asterisk, space) in a line.
2. No line entry has more than 4 numbers total.

Knowing that, we could simply save up to three numbers in variables, so we would print them out A B C and then C B A. We won’t even need the zeros now, since we can read A,B,C and act on them (stopping if C is 0).

Neat!

A quick bit of trial and error gave me this code that will print the top half of the image:

``````0 ' LOGIKER10.BAS
10 CLS
15 CH=32:PRINTTAB(6);
20 READ A:IF A=-1 THEN 220
21 PRINT STRING\$(A,32);
25 IF C=0 THEN PRINT STRING\$(B*2-1,42);STRING\$(A,32):GOTO 15
30 PRINT STRING\$(B,42);STRING\$(C*2-1,32);STRING\$(B,42)
40 GOTO 15
50 DATA 5,1,4
60 DATA 5,2,3
70 DATA 5,3,2
80 DATA 5,4,1
90 DATA 1,9,0
100 DATA 2,8,0
110 DATA 3,7,0
120 DATA 4,6,0
130 DATA 5,5,0
215 DATA -1
220 GOTO 220``````

It creates this:

I can now say “we are halfway there!”

But now I have another issue to solve. How do I back up? There is no way to READ data in reverse. It looks like I’m going to need to load all those numbers in to memory so I can reverse back through them.

To be continued…