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You've constructed a (somewhat artificial) special case where what you call the "plaintexts" also meet the requisites that one-time pad keys are supposed to meet:
Secret
Chosen uniformly at random
Never reused
So we can analyze this by flipping the labels, so that we regard what you label the "key" and the "plaintexts" the other way around, and the ...
4
The usual padding for block ciphers ("PKCS#7 padding") is not a sequence of zeroes, but a sequence of P = N - (X % N) bytes each with value P.
If the message is a multiple of the block size, then a full block of padding is added (where each byte value is the block size).
For example, is the message is 15-byte long and the block size is 16 bytes, than one ...
4
Alice and Bob's keys are used to compute a shared key.
Both parties compute the same key, which is used both encryption and decryption.
This is a traditional, hybrid encryption system.
Using only asymmetric cryptography for encryption would be painfully slow.
As a mitigation, the sender can create an ephemeral key pair, deleted right after the message ...
2
All encryption methods deal with bit or numbers not strings. You can convert your strings into bits and numbers to process. As I understood from your question you are looking for detection of mail provider on the FHE encrypted data.
One way to this is by encoding your data. You can separate the e-mail as the user_name and domain then encrypt and use integer ...
1
I wrote several papers on such a topic. I called it end-to-display encryption. But it is more commonly called "pixel-domain encryption" There are several easy ways to do this. My way included adding a header with RSA encrypted AES Keys.
In essence, you take the picture, extract the pixels, XOR with the bitstream created through AES in Countermode, increase ...
1
If you're like me, you're probably wondering what to do in this case
... 02 02
... 02 01
and you've probably already heard the solution that tells you to just modify the last two bytes instead of just the last one. You're probably wondering why that actually works (which it does).
You will find two valid paddings when trying all 256 values for the last ...
1
Would a pictographic, phoneme-based, hand cipher be difficult to crack with computer crypt-analysis?
Yes, of course it would deter computers...
... if it were done as substitution and every instantiation of a symbol had a new meaning. This would avoid frequency analysis.
But you are going to be drawing a lot of pictures, and your correspondent is going to need a thick book of hand-drawn symbols if you have a lot of traffic (will each of your hand-drawn ...
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