does CFB support random read access? According to Wikipedia block cipher modes of operation, simplified CFB supports random read access. but what about the real CFB where we have a shift register? below are pictures extracted from the book &quot;Cryptography and Network Security by William Stallings&quot; for CFB.

And

Does it support random read access? if yes, how?
Yes, with some more ciphertext.
The CFB mode encryption equations;
\begin{align}
I_0 &amp;= \text{IV}\\
I_i &amp;= \big((I_{i-1} \ll s) + C_i\big) \bmod 2^b,\\
C_i &amp;= \operatorname{MSB}_s\big(E_K(I_{i-1})\big) \oplus P_i\\
\end{align}
and decryption equations;
\begin{align}
I_0 &amp;= \text{IV}\\
I_i &amp;= \big((I_{i-1} \ll s) + C_i\big) \bmod 2^b,\\
P_i &amp;= \operatorname{MSB}_s\big(E_K(I_{i-1})\big) \oplus C_i\\
\end{align}
As you can see, to read one value one needs more than one ciphertext and that is defined by the CFB mode is used. NIST has CFB-1, CFB-8, and CFB-128. The $s$ in the CFB-$s$ represents the encryption size per block cipher encryption/decryption.

CFB-1 needs to read 129 ciphertexts block, one can be IV, to decrypt one block
CFB-8 needs 9 to read 129 ciphertexts block, one can be IV, to decrypt one block
CFB-128 needs 2 to read 129 ciphertexts block, one can be IV, to decrypt one block

The good is that if you read sequentially then the need is not linear, for example, 2 sequential ciphertext blocks will need only 3 on CFB-128, 10 in CFB-8, and 120 in CFB-1.
Actually, even in the reduced sizes like CFB-1, this is not really a problem, since the 128 ciphertexts are bits and that is only two reads if the data can be access 128-bit per reading. The memories can handle this and harddisk has wider read access. This, however, has a drawback on the decryption time. AES-CFB-1 needs 128 decryptions, AES-CFB-8 needs 16 decryptions.
Note that we don't use CFB or CTR on-disk encryption, we rather prefer XTS or XEX modes.