Bill Trost <trost at cloud.rain.com> wrote:
> Open questions:
>
> 1. Where does the "real" MIMEType of the message go? Does the
> entire message itself constitute a message with its own headers,
> or can all the headers of the multipart message be included as
> part of part 1's headers?
>
> 2. Are the parts encrypted individually, or is the whole message
> encrypted and then split up? Does it matter?
I think the entire "real" message (data plus "real" private metadata)
should be first constructed as usual and encrypted as a whole, as if it
were about to be inserted in the normal way. The resulting block should
then be treated as a blob with no internal structure and cut into parts.
Each part has private metadata added to it indicating its part number, and
gets inserted as if it were a normal piece of data. This means each part
gets encrypted again by its own part-key, just as all inserts are. The
public headers indicate nothing.
So it looks like this:
1. When we retrieve the key intended by the author, we get an encrypted
block of data. At this point it does not look like a multipart document.
Upon decrypting, we see this:
MIMEType=multipart/linear
Multipart.PartNumber=1
Multipart.PartCount=3
Mulitpart.Part2.Key=D013F88300744B8926D89661DDFD2E93
Mulitpart.Part3.Key=9C858CDC8C3129073785AEB5682079BE
Multipart.TotalLength=38362
encrypted-block-1
2. We retrieve key D013F88300744B8926D89661DDFD2E93. Again it does not
look like a multipart document. We decrypt it to reveal:
MIMEType=multipart
Multipart.PartNumber=2
Mulitpart.Part1.Key=E1B8115F246A85D93AE2731F0C51691D
encrypted-block-2
3. Repeat. Once we have retrieved all the parts, we concatenate encrypted
blocks 1-n. This now looks like we did a normal request and got the
unsplit document. We decrypt the whole thing, using the author-supplied
decryption key, to reveal:
MIMEType=application/real
Author=Brandon
plaintext
Maybe this is one encryption too many, though. Is it ok if concatenating
blocks 1-n gives this directly?
theo
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