MD5 for text [Resolved]

[Kwai chang caine]

Hello at all

It's surely again a "Dyingoflaughter" question for you, but i dare when even

Exist it ...a style of MD5 for a text, that is... a way for crypting a text in another text, but the most shorter possible and also unique a style of "KEY" what, for be sure this two text are the same
Like this i can compare two texts of (for exampler) 20 characters with a (for exampler) key of 4 to 6 characters or less if possible
I hope i have good explained, what cow pat idea i have again, in the bidet that serves as my brain

Have a good day

[infratec]

In former days a password was stored as MD5 hash, not the real password.
Today a Sha256 or 512 hash is stored.
So your idea is not new

[Kwai chang caine]

Hello INFRATEC glad to read you

It's not an idea (A good idea from KCC... that would be too good to be true ) ...it's a lack of knowledge, like usually

What idiot i am
Since the begining..i use only FileFingerprint() for testing same files, so i believe it's the only one function
And when i read your answer... i guess "Perhaps there are another function that "FileFingerprint()" in the cipher library ?"
I open the help... and see "StringFingerprint()"

Thanks a lot for your precious help, i test that immediately
I not need a strong encryption, but the shorter KEY possible, it's for create a "reference"
Have a very good day

[Kwai chang caine]

Unfortunately the MD5 is really longer than the original text
That not works for what i need to do

For the moment..i have create a cow pat code..
But it's not perfect (KCC trademark ), i have the same REF for two different titles

DIBAEL4
DIBAEL4

Code: Select all

Procedure.s CreateUniqueShortKey(NomElement.s)

 NormElement$ = RemoveString(NomElement, "'")
 Interprete$ = Trim(StringField(NormElement$, 1, "("))
 
 If Interprete$ <> ""
  
  Interprete$ = ReplaceString(Interprete$, "  ", " ")
  MaxEspaces = CountString(Interprete$, " ") + 1
  Ref$ = ""
  
  For i = 1 To MaxEspaces
   Ref$ + UCase(Left(StringField(Interprete$, i, " "), 2))
  Next
  
 EndIf
 
 Titre$ = Trim(StringField(StringField(NormElement$, 2, "("), 1, ")"))
 Titre$ = Trim(StringField(Titre$, 1, "'"))
 
 If Titre$ <> ""
  
  Titre$ = ReplaceString(Titre$, "  ", " ")
  MaxEspaces = CountString(Titre$, " ") + 1
       
  For i = 1 To MaxEspaces
   Ref$ + UCase(Left(StringField(Titre$, i, " "), 2))
  Next
  
  Ref$ + Trim(Str(Len(Titre$)))
  
 EndIf
 
 ProcedureReturn Ref$

EndProcedure

Title1$ = "Didier Barbelivien (Elle) Slow.mp3"
Title2$ = "Didier Barbelivien (Elsa) Slow.mp3"

Debug CreateUniqueShortKey(Title1$)
Debug CreateUniqueShortKey(Title2$)

[Bitblazer]

Maybe CRC32Fingerprint is good enough for you.

[normeus]

The help example for PB 6.21 (x64) is different from the one on the website, the one on the web is for older versions of PB:

Code: Select all

UseCRC32Fingerprint()
Debug StringFingerprint("Didier Barbelivien (Elsa) Slow.mp3", #PB_Cipher_CRC32) ; KCC's text
Debug StringFingerprint("Didier Barbelivien (Elle) Slow.mp3", #PB_Cipher_CRC32) ; KCC's text

Norm.

[BarryG]

Unfortunately the MD5 is really longer than the original text

Maybe use Left() on it to crop it? Should still be a unique result if not used for private or critical situations.

[Kwai chang caine]

You have right the CRC32 is really shorter than MD5, that's incredible, i can use it, because it divided the len of text by more than 4
I would have thought that a newer and more powerful encryption would necessarily take longer, but not at all

On the other hand, it's funny a CRC32 of CRC32 have the same lenght, i don't understand why

Code: Select all

UseMD5Fingerprint()
UseCRC32Fingerprint()

Text1$ = "Didier Barbelivien (Elsa) Slow.mp3"
Text2$ = "Didier Barbelivien (Elle) Slow.mp3"

Debug "Len text 1 = " + Len(Text1$)
Debug "Len text 2 = " + Len(Text2$)
Debug ""

MD5_1$ = StringFingerprint(Text1$,  #PB_Cipher_MD5) ; KCC's text
MD5_2$ = StringFingerprint(Text2$,  #PB_Cipher_MD5) ; KCC's text

Debug "Len MD5 1 = " + Len(MD5_1$)
Debug "Len MD5 2 = " + Len(MD5_2$)
Debug ""

CRC32_1$ = StringFingerprint(Text1$,  #PB_Cipher_CRC32) ; KCC's text
CRC32_2$ = StringFingerprint(Text2$,  #PB_Cipher_CRC32) ; KCC's text

Debug "Len CRC32 1 = " + Len(CRC32_1$)
Debug "Len CRC32 2 = " + Len(CRC32_2$)
Debug ""

CRC32_1x2$ = StringFingerprint(CRC32_1$,  #PB_Cipher_CRC32) ; KCC's text
CRC32_2x2$ = StringFingerprint(CRC32_2$,  #PB_Cipher_CRC32) ; KCC's text

Debug "Len CRC32_1 x 2 = " + Len(CRC32_1x2$)
Debug "Len CRC32_2 x 2 = " + Len(CRC32_2x2$)

Thanks a lot at you three and have a very good day

[infratec]

CRC32 means that the result is always 32 bit wide.
If you let run it above 1 byte or 1 tera byte the result is always 4 byte long.

There is also CRC16, CRC8, CRC7 and CRC1 available.

https://en.wikipedia.org/wiki/Cyclic_redundancy_check

Code example:

https://de.wikipedia.org/wiki/Zyklische ... zw._Delphi

ModBus CRC in PB:
viewtopic.php?p=625650#p625650

[Kwai chang caine]

Thanks a lot for your explanation

But i believe you forgot you talk to KCC
KCC understand each word, but it's all the words together who do the sentence that KCC not really understood

There is also CRC16, CRC8, CRC7 and CRC1 available

Apparently Fred have not coded the CRC under 32 probably because it's pointless to have weaker encryption than the one everyone else uses.
Believe you a CR1 (for example) can return a shorter result than the CRC32 ?

[kenmo]

Hello, CRC (and MD5) are not "encryption", meaning they cannot be turned back into the original text. I hope you understand that! Look up the differences between checksum/hash/digest vs. encryption/decryption.

[Kwai chang caine]

Yes, you're right, I hadn't really made the distinction in my head.