Hi all,
I have not programmed in about 6+ years and need to start getting back into it.
Does anyone have any code for encrypting a plaintext file into DES and then back again to plain text. I have attached a doc with what i want to try to do.
I have had a look on the web but i am so out of touch with programming i have not got a clue.
Any help would be great.
Thanks
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\pnlcltr\pnstart1\pnindent720\pnhang {\pntxta )}}{\*\pnseclvl5\pndec\pnstart1\pnindent720\pnhang {\pntxtb (}{\pntxta )}}{\*\pnseclvl6\pnlcltr\pnstart1\pnindent720\pnhang {\pntxtb (}{\pntxta )}}{\*\pnseclvl7\pnlcrm\pnstart1\pnindent720\pnhang {\pntxtb (}
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\fs24\cf1\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\b\cf0\insrsid9454416 Network Securit}{\cf0\insrsid9454416 y \line
\par }\pard\plain \s17\ql \li0\ri0\sa548\nowidctlpar\faauto\rin0\lin0\itap0 \fs24\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\b\fs28\insrsid9454416 Implementation of Data Encryption Standard (DES). }{\fs28\insrsid9454416
\par }\pard\plain \s18\ql \li0\ri0\sa270\sl276\slmult0\nowidctlpar\faauto\rin0\lin0\itap0 \fs24\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\insrsid9454416
Most encryption schemes used today are based on the Data Encryption Standard (DES). Recommended back in 1977 by the National Bureau of Standards. The algorithm is based on encrypting 64-bit blocks using a 5
6-bit key to produce a 64-bit encrypted output. The encryption process goes through a number of steps. To decrypt the same output back into the same plane text, the same steps are performed using the same key.
\par }\pard \s18\ql \li0\ri260\sa270\sl276\slmult0\nowidctlpar\faauto\rin260\lin0\itap0 {\insrsid9454416 The encryption process has two main inputs: t
he plain text to be encrypted and the key. The plain text must be a 64-bit block and the key must be a 56-bit in length.
\par \sect }\sectd \sbknone\linex0\sectdefaultcl\sftnbj \pard\plain \s15\ql \li0\ri0\nowidctlpar\faauto\rin0\lin0\itap0 \fs24\cf1\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\cf0\insrsid9454416
\par }\pard\plain \s18\ql \li0\ri0\sa270\nowidctlpar\faauto\rin0\lin0\itap0 \fs24\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\b\fs28\insrsid9454416 The Encryption process: }{\fs28\insrsid9454416
\par }\pard \s18\ql \li0\ri0\sa270\sl276\slmult0\nowidctlpar\faauto\rin0\lin0\itap0 {\insrsid9454416 There are three major stages the encryption process performs on the plain text:
\par }\pard \s18\ql \fi-360\li720\ri0\sa270\sl276\slmult0\nowidctlpar\faauto\rin0\lin720\itap0 {\insrsid9454416 1)\tab Initial P
ermutation (IP): The order of bit in the input plain text is rearranged using a special permutation table (IT-1) producing a 64-bit message block. The 64-bit message is split into two halves, a Left and Right 32-bit each.
\par }\pard \s18\ql \fi-360\li720\ri125\sa270\sl276\slmult0\nowidctlpar\faauto\rin125\lin720\itap0 {\insrsid9454416 2)\tab A Permutation and a substituti
on function are repeated 16 times. A number of 16 sub-keys are created from the main key. Each sub-key will be used in each of the 16 repeated steps. Every step consists of the following:
\par }\pard\plain \s15\ql \fi-285\li993\ri155\sl276\slmult0\nowidctlpar\faauto\rin155\lin993\itap0 \fs24\cf1\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\cf0\insrsid9454416 a) Every step starts with the previous (32-bit) left and right halv
es or parts of the message.
\par }\pard\plain \s18\ql \fi-285\li993\ri0\sa270\sl268\slmult0\nowidctlpar\faauto\rin0\lin993\itap0 \fs24\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\insrsid9454416 b) The following two function are applied to generate the next two halves: }{
\i\fs23\insrsid9454416 L}{\i\fs15\insrsid9454416 n }{\fs23\insrsid9454416 = }{\i\fs23\insrsid9454416 R}{\i\fs15\insrsid9454416 n-1 }{\i\fs23\insrsid9454416 R}{\i\fs15\insrsid9454416 n }{\fs23\insrsid9454416 = }{\i\fs23\insrsid9454416 L}{
\i\fs15\insrsid9454416 n-1 }{\fs23\insrsid9454416 + }{\i\fs23\insrsid9454416 f}{\fs23\insrsid9454416 (}{\i\fs23\insrsid9454416 R}{\i\fs15\insrsid9454416 n-1}{\fs23\insrsid9454416 ,}{\i\fs23\insrsid9454416 K}{\i\fs15\insrsid9454416 n}{\fs23\insrsid9454416
) where 1< n <16
\par }\pard\plain \s19\ql \li993\ri0\nowidctlpar\faauto\rin0\lin993\itap0 \fs24\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\fs23\insrsid9454416 The new left half (L}{\i\fs23\insrsid9454416 n}{\fs23\insrsid9454416
) is easy; The previous right half becomes the new left half.
\par To work out the right half, first consider the function }{\i\fs23\insrsid9454416 f()}{\fs23\insrsid9454416 in the above formulae.
\par }\pard\plain \s18\ql \li993\ri0\sa270\nowidctlpar\faauto\rin0\lin993\itap0 \fs24\lang2057\langfe2057\cgrid\langnp2057\langfenp2057 {\fs23\insrsid9454416 The function }{\i\fs23\insrsid9454416 f() }{\fs23\insrsid9454416
is applied on to the right half only and it involves the following:
\
I have an old zip archive with many encryption related code for vb6, but I can not send it to you until tomorrow. Try this one:
http://www.codearchive.com/list.php?go=0206
and find "Author: Saravanan Krishnaraj" approximately in the middle of page.
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