I need a one-way encryption algorithm!
in [Cobol]
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From: ellie.odonnell.iei8 on 1 Jul 2005 12:13 Hi, I need to write a very simple one-way encryption alogorithm in COBOL. It must not be able to be decrypted. I thought I could simply use the UNSTRING function to separate all the characters in the field, convert them to something else, and then STRING all the characters back together into the field, but that is not working. I also thought there was a way to count all the characters in a string, but it seems like the COUNT verb is to count fields in a record, not individual characters. I have an encryption algorithm to use for numeric values, but am having a lot of trouble figuring out what to do about the alpha character strings. If only there were a RANDOMIZE function for character strings!! Please help!! Thanks! Ellie
From: docdwarf on 1 Jul 2005 12:38 In article <1120234427.622603.211450(a)g43g2000cwa.googlegroups.com>, <ellie.odonnell.iei8(a)statefarm.com> wrote: >Hi, > >I need to write a very simple one-way encryption alogorithm in COBOL. >It must not be able to be decrypted. So you just need to put garbage in a particular field? DD
From: Arnold Trembley on 1 Jul 2005 12:57 ellie.odonnell.iei8(a)statefarm.com wrote: > Hi, > > I need to write a very simple one-way encryption alogorithm in COBOL. > It must not be able to be decrypted. I thought I could simply use the > UNSTRING function to separate all the characters in the field, convert > them to something else, and then STRING all the characters back > together into the field, but that is not working. I also thought there > was a way to count all the characters in a string, but it seems like > the COUNT verb is to count fields in a record, not individual > characters. > > I have an encryption algorithm to use for numeric values, but am having > a lot of trouble figuring out what to do about the alpha character > strings. > > If only there were a RANDOMIZE function for character strings!! > > Please help!! > > Thanks! > Ellie > A "hashing algorithm" might do what you want. From Joel C. Ewing in bit.listserv.ibm-mail on March 16, 2001 (you can find it with google advanced newsgroup search): "A trivial hashing function that works adequately in many cases is to take your data entry "key" field 4-bytes at a time, XOR these together, take absolute value, and then mod N' (i.e., remainder of division by N' - Constraining N' to a prime tends to improve the dispersion) and then use this..." Of course, XOR'ing binary fields in COBOL might be a little bit difficult to do without a library routine or assembler subprogram. If your function is truly trivial, you might just break up your "key" field into two-byte subfields, coerce each one to be a PIC S9(4) COMP field (I'm assuming IBM mainframe COBOL here), and simply add each number together, truncating the results. The final two-byte binary number cannot possibly be decrypted back to the original value, and will give you approximately 32,767 semi-random results. Perhaps someone else can suggest a simpler or more secure hashing algorithm. I hope that helps. -- http://arnold.trembley.home.att.net/
From: HeyBub on 1 Jul 2005 16:24 ellie.odonnell.iei8(a)statefarm.com wrote: > Hi, > > I need to write a very simple one-way encryption alogorithm in COBOL. > It must not be able to be decrypted. I thought I could simply use the > UNSTRING function to separate all the characters in the field, convert > them to something else, and then STRING all the characters back > together into the field, but that is not working. I also thought there > was a way to count all the characters in a string, but it seems like > the COUNT verb is to count fields in a record, not individual > characters. > > I have an encryption algorithm to use for numeric values, but am > having a lot of trouble figuring out what to do about the alpha > character strings. > > If only there were a RANDOMIZE function for character strings!! > > Please help!! > > Thanks! > Ellie Check your mail
From: ellie.odonnell.iei8 on 1 Jul 2005 16:36
Yes, I need to put garbage in most of the fields on the file. But the resulting fields need to be the same data type and length as the original fields. CHAR (15) needs to be CHAR(15) when I'm done with it, etc. And the file itself needs to remain the same length. |