How to generate checksum using micropython?

[AmirJ]

How to generate checksum using micropython?

[jimmo]

Which checksum algorithm do you need to use? CRC32?

[AmirJ]

With thanks for your attention

I do not want CRC Check.
I don't know the check algorithm. but i have very example:

"String" => Checksum

"ABC"=>85
"CBA"=>85

"A"=>84
"BBB"=>87

"CCC"=>86
"C"=>86

"123"=>37
"321"=>37
"213"=>37
"132"=>37

[pythoncoder]

That is a very poor algorithm. I can't deduce how it works from those examples but maybe someone else can.

[Roberthh]

Yes, that looks like some variant of an LRC. Do you also have an example for two identical characters, like "AA"?

[jimmo]

It's

Code: Select all

def checksum(msg):
  v = 21
  for c in msg:
    v ^= ord(c)
  return v

[Roberthh]

good!

[pythoncoder]

So it is! Nice one

[AmirJ]

It's

Code: Select all

def checksum(msg):
  v = 21
  for c in msg:
    v ^= ord(c)
  return v

That's great, thank you
how did you know?
How did you find
Thank you again.

[jimmo]

How did you find

There are three clues:
- Ordering doesn't matter. (e.g. ABC == CBA) so it's likely that the checksum is processed byte at a time.
- Making it longer (especially repeating, and even more so, triple repeats) doesn't change the result dramatically. Not likely to be adding the bytes together, like many other simple checksums. This leaves XOR as the likely candidate. Triple AAA giving the same result as single A is a good confirmation of XOR.
- A, B, C checksums clustered around a different value to 1, 2, 3 based one. Means that it's possibly based on the ascii value (i.e. ord). Also A and C going to values two apart (84 and 86) suggests there's no non-linear change to the byte value before whatever accumulation happens.

XOR-based ones often start with a non-zero value, this is so that checksumming a message of NUL bytes (i.e. all zeros) doesn't also give zero as the result.

At that point it's just algebra to solve for the starting value -- ord('A') ^ 84 --> 21. Check with 21^ord('C') --> 86, and 21^ord('1')^ord('2')^ord('3') --> 37.