Is it possible to find an AES key given part of the key, the ciphertext, initialization vector and mode of encryption?

Question

The following details are given:

• Partial Key: XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX11000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011
• Initialization Vector: E898EF8E91F8C9B201E6E29DF87EE152
• Ciphertext Block 1: 14B8D1412766A8520BACE4598F8AFAEE
• Ciphertext Block 2: 7E687A49015FA6F1B914635325A6361B
• Ciphertext Block 3: 8AD191394EF79CEC4B5A256313632CD4
• Ciphertext Block 4: 8BB4D49F3FA7A917CDF02ECCAA8C4765
• CBC Mode No padding
• Character set: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz1234567890⌴‛~@#$%^&*(){}[]+=-_<>,.?/!:\;’|"
• Character set info: range [32,127), ASCII encoded, each character is 8 bits

My thinking is that somehow you will need to decrypt each ciphertext block with each of the known 32 bits of the key and work down. The problem is the first ciphertext block has an unmatched key (XXX ...)
So as a result you will get a plain text of all known characters except the first 32 bits.

Answer 1

Can one achieve to search 32-bit keyspace?

OpenSSL with AES-NI can operate 61510120 iterations for 64-byte blocks AES-128 in CBC mode per 3 seconds on my machine.

run openssl speed -evp AES128 to see in your machine.

This makes $2^{26}$ keyspace for three seconds. One needs $2^{6} =64*3$ seconds to find the key candidates with good coding.

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For your cause create 3 functions;

• $P = \operatorname{AES-Dec-CBC}(k, IV, C)$ where $C$ is the ciphertext blocks and $P$ is the decrypted plaintext under the current key $k$ of $C$.
• $k =\operatorname{GetNextCandidateKey(current)}$ This simple increments the current and creates a key $k = current\mathbin\|1100\cdots011$ in the binary form. You need to convert this into binary to fit the standard encryption librarires
• $b = \operatorname{CheckTheMessage}(P)$. This function gets a plaintext and checks that the bytes are in the range. If not returns 0 else returns 1

Now with these 3 functions;

current = -1

while current < 2^32 do:
    k = GetNextCandidateKey(current)
    P = AES-Dec-CBC(k, IV, C)
    b = CheckTheMessage(P)

    if b == 1
        print(current)
    current++

Answer 2

Is it possible to find an AES key given part of the key, the ciphertext, initialization vector and mode of encryption?

Sometimes it is possible. Sometimes it is not possible. It depends on how much of the key you already know.

In your case you seem to know all but 33 bits of the key. So, you should be able to brute force it to figure out the rest of the key.