[Cryptography] End-to-End, One-to-Many, Encryption Question

[Kent Borg]

I have a question, I think it is about concatenated encryption and 
convolved keys, but I am not sure. It is the sort of question that lots 
of people should be asking these days, so forgive me it lots of people 
have, I have been behind in my reading.


Alice lives on the far end of a single DSL line, and produces data on a 
regular basis, she encrypts it with a key only she knows, and she sends 
it to Bob.

Bob lives in the cloud (and so has lots of bandwidth), but Bob is in the 
cloud, and therefore is only partially trusted, so he is given no 
ability to directly decrypt the data. There is also lot of data 
accumulated, he doesn't can't store unique copies for each client.

Charley is a client, one of many (Charley-1, Charley-2, Charley-3, etc., 
clients can come and go), he lives in a smart phone, say. He asks Bob 
for a specific piece of data, Bob encrypts it with a Charley-1-specific 
key and sends it off.

Charley-1 decrypts the data with a key that Bob does not know.

If Alice discovers Charley-1 is compromised, she can instruct Bob to 
delete Charley-1-specific data, destroying his ability to read data from 
Bob. Alice probably knows everyone's keys, but Bob and Charley do not 
know each other's keys, and again only Alice knows her key.

If Charley-1 and Bob collude, the system is, unfortunately, broken, but 
that seems unavoidable.

An attempt to restate the question:

  Is there a way to encrypt once with key A, super-encrypt with key B1 
(not knowing any other keys), and finally decrypt with key C1 (not 
knowing any other keys)?  Or, super-encrypt with key B2, then decrypt 
with key C2?

In some respect this is a satellite TV problem subscription problem, 
with an on-demand component.

Is there a canonical answer here? Is it a stupid question?


Thanks,

-kb, the Kent who Googled some on this but the closest PDF seemed to 
want to know all the Charlies in advance, and was too encrypted a paper 
for him to really understand anyway.