Game Text Encoding Problem

[DanBoris]

I need some help figuring out something, and thought some of the people here might spot whatever I am missing...

 

I'm playing around with decoding the format of the MSDOS Adventure Construction Set game files. Some of it has been pretty easy to figure out but when I came to the object names I found out that they were compressing this text, but I can't quite figure out the logic. Here is what I know:

 

- Each character can be one of 40 characters (26 letters, 10 digits, and 4 symbols)

- Every three characters is encoded into 2 bytes

- Here are some of the encodings I have seen, showing the character, hex values and binary values:

 

	1 = C0 A8	        11000000 10101000
A = 40 06		01000000 00000110
B = 80 0C               10000000 00001100
C = C0 12               11000000 00010010
D = 00 19		00000000 00011001
E = 40 1F		01000000 00011111
F = 80 25               10000000 00100101

       A =   40 06		01000000 00000110
AA =  38 13             00111000 00010011
AAA = 69 06	        01101001 00000110

ABC = 93 06             10010011 00000110

 

Anyone have any ideas on this?

[GroovyBee]

Its some form of radix 40

 

e.g.

 

(X-'A')*1

+ (Y-'A')*40

+ (Z-'A')*1600

 

That'll fit into an unsigned 16 bit word.

[SeaGtGruff]

Its some form of radix 40

 

e.g.

 

(X-'A')*1

+ (Y-'A')*40

+ (Z-'A')*1600

 

That'll fit into an unsigned 16 bit word.

Yes, I said the same thing, but I lost my internet connection while I was typing my reply, so it got lost when I tried to post it.

 

Notice the pattern with some of the values shown:

 

b = 00 00 (I'm guessing that 00 00 is a space) ?

A = 40 06 (add 40 06)

B = 80 0C (add 40 06)

C = C0 12 (add 40 06)

D = 00 19 (add 40 06, then add the carry flag)

E = 40 1F (add 40 06)

F = 80 25 (add 40 06)

G = C0 2B (add 40 06) ?

H = 00 32 (add 40 06, then add the carry flag) ?

etc.

 

Michael

 

Edit: Also, notice that it takes the same number of bytes to code 1, 2, or 3 characters, further pointing to a base-40 system.

 

The only other way I know to code 3 characters in 2 bytes is to split the bits, 5 bits per character, with 1 bit left over, but that gives only 32 characters.

Edited July 17, 2010 by SeaGtGruff

[SeaGtGruff]

Anyone have any ideas on this?

Adding to my previous comments, I suggest looking at the encoding systematically (b = SPACE):

 

bbb = ?? ??

bbA = ?? ??

bbB = ?? ??

bbC = ?? ??

etc.

 

That should give you the values for the 1s place (0 to 39).

 

The rest should be a matter of just multiplying by decimal 40 for the 10s place, or by decimal 1600 for the 100s place, but you could verify that systematically:

 

bAb = ?? ?? (should be the same as bbA times decimal 40)

bBb = ?? ?? (should be the same as bbB times decimal 40)

bCb = ?? ?? (should be the same as bbC times decimal 40)

etc.

 

Abb = 40 06 (should be the same as bAb divided by decimal 40, or bbA divided by decimal 1600)

Bbb = 80 0C

etc.

 

But you'd have to take the carry into consideration, since it appears that the carry might be getting added back to the lo byte?

 

Michael

[SeaGtGruff]

Another thought:

 

I think the values shown are lo byte first:

 

Abb = hex 40 06 = $0640 = decimal 1600 = 1*1600

Bbb = hex 80 0C = $0C80 = decimal 3200 = 2*1600

Cbb = hex C0 12 = $12C0 = decimal 4800 = 3*1600

Dbb = hex 00 19 = $1900 = decimal 6400 = 4*1600

etc.

 

Michael

[GroovyBee]

I think the values shown are lo byte first:

 

Makes sense since the files come from an x86 based machine which is little endian.

[SeaGtGruff]

This seems to work for some, but not all, of the examples you posted:

 

bbb = $0000 = 0

bbA = $0001 = 1

bbB = $0002 = 2

bbC = $0003 = 3

 

bAb = $0028 = 1*40

bBb = $0050 = 2*40

bCb = $0078 = 3*40

 

Abb = $0640 = 1*1600

Bbb = $0C80 = 2*1600

Cbb = $12C0 = 3*1600

 

AAb = $0640+$0028=$0668 -- you gave $1338, or 38 13

AAA = $0640+$0028+$0001=$0669

ABC = $0640+$0050+$0003=$0693

 

By my figuring, $1338 should be CCb, not AAb.

 

Michael

Edited July 17, 2010 by SeaGtGruff

[SeaGtGruff]

Since 1 (presumably 1bb) is encoded as C0 A8, or $A8C0, which is decimal 43200, which is 27*1600, I'm guessing the characters have the following values:

 

b = 0 (space)

A = 1

B = 2

C = 3

D = 4

E = 5

F = 6

G = 7

H = 8

I = 9

J = 10

K = 11

L = 12

M = 13

N = 14

O = 15

P = 16

Q = 17

R = 18

S = 19

T = 20

U = 21

V = 22

W = 23

X = 24

Y = 25

Z = 26

1 = 27

2 = 28

3 = 29

4 = 30

5 = 31

6 = 32

7 = 33

8 = 34

9 = 35

0 = 36

? = 37 (unknown symbol)

? = 38 (unknown symbol)

? = 39 (unknown symbol)

 

These are multipled by 40^0=1, 40^1=40, or 40^2=1600, depending on their position. In example ABC, A is in the 100s place, B is in the 10s place, and C is in the 1s place.

 

Michael

[DanBoris]

You guys rock! Thanks!

 

Yes, "AAb = $1338" was a mistake, $0668 is the correct value.

 

Dan

Edited July 17, 2010 by DanBoris