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Does anyone know specifically how the checksum for gba pokemon file is calculated? Im trying to create a pokemon I can trade from Emerald into xD: Gale of Darkness for sweeping Mt. Battle. My problem is that I have to hex edit the 3gpkm file. Where my problem comes in is that I'm using a move several indexes past Shadow Rush which itself is 356. The move I'm trying to teach is index 362, and no editor has support for moves past 354, Psycho Boost, the move with the highest legally obtainable index, and the upper limit for all editors I've seen.

EDIT: According to http://www.ppnstudio.com/maker/PokemonMakerHelp.txt it's calculated by summing all the unencrypted words, but the problem lies in the fact that the definition of a word changes between system dependent on the processor.

Edited by Delta Blast Burn
Got some new information.
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What I know is that the cksm itself is 16-bit but the data it references is four blocks of 12 bytes each, encrypted 32-bits at a time, via the PID Xor'ed with the OTID AND the SID. But my problem is not the cryptography, it's the math. It seems that whenever change more than one byte ...

You know what i'll just attach my work so far. If i had any programming knowledge, I would probably write a command line tool.

May post more info later.

Here are my notes from various sources.

3gpkm File structure.txt

3gpkm File structure.txt

Edited by Delta Blast Burn
clarified my thoughts
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  • 7 months later...

Is it a problem with endianness?

I do the calculation on lines 453-470 here https://github.com/Zazcallabah/PokeSave/blob/master/PokeSave/MonsterEntry.cs

Maybe no help if you cant read code, so basically this happens:

* Group the entire array of data into 12 sections with 4 bytes in each section. (Not 4 sections with 12 each.)

* The key you want to xor each section with is, as you say, the PID xor:ed with the original trainer id.

* In every section, after xor:ing with the key, take the high 16 bits and the low 16 bits and add them together. (You now have a number that is the sum of 24 other numbers)

* The low 16 bits of that number is what you want

I can give you an example, if you want, I just need to look through my notes for a good sample

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Example:

lets use the pokemon provided in section 11 in this document http://www.ppnstudio.com/maker/PokemonMakerHelp.txt

9de847ff

e1dd6e3b

bdbbcdbd

c9c9c8ff

80430202

c5d9e2ff

ffffff00

a4f10000

7c3529c4

7c3529c4

7c3529c4

593429c4

013529c4

7c7329c4

7c0eace4

5875f8c9

7c3529c4

163529c4

7c3529c4

623529c4

Remember the endianness. Its little-endian, so when you do maths, the pid isn't 9de847ff. You have to reverse it one byte at a time. The pid is ff47e89d.

Same with the next line, the original trainer id. (The SID and TID) is 3b6edde1.

Xor these two to get the key: C429357C

Now we take the subsection data:

7c3529c4

7c3529c4

7c3529c4

593429c4

013529c4

7c7329c4

7c0eace4

5875f8c9

7c3529c4

163529c4

7c3529c4

623529c4

Each line is one part. Remember it is little-endian, so each line has to be reversed before we do maths:

c429357c

c429357c

c429357c

c4293459

c4293501

c429737c

e4ac0e7c

c9f87558

c429357c

c4293516

c429357c

c4293562

Now xor each line separately with the key c429357c

00000000

00000000

00000000

00000125

0000007D

00004600

20853B00

0DD14024

00000000

0000006A

00000000

0000001E

Until now the endianness actually didnt really matter since we did it both to the data and the key. But now we start adding, so now it gets confusing if you didnt do it at the start.

Split each line in two.

0000 0000

0000 0000

0000 0000

0000 0125

0000 007D

0000 4600

2085 3B00

0DD1 4024

0000 0000

0000 006A

0000 0000

0000 001E

And add them together and take the lowest 16 bits, the result in this case is F1A4

Remember that you have done the little-endian conversion, and you have to convert this number back before you write it to the file.

9de847ff

e1dd6e3b

bdbbcdbd

c9c9c8ff

80430202

c5d9e2ff

ffffff00

a4f10000 <-- Here it is

7c3529c4

7c3529c4

7c3529c4

593429c4

013529c4

7c7329c4

7c0eace4

5875f8c9

7c3529c4

163529c4

7c3529c4

623529c4

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  • 5 years later...

So, to confirm I'm reading this right:

for each line that in the original data was ordered:
wwxxyyzz,

I reverse it to:

zzyyxxww

decrypt it to :

ZZYYXXWW

split it into:

ZZYY XXWW

Take the sum:

ZZYY + XXWW = S

Repeat that each of the three lines in each of the 4 blocks, then sum the 12 S's to:

ABCDEFGH 

Then take the last four hex characters:

EFGH as the checksum?

Edited by ABZB
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