# With DES, is ECB encryption mode equivalent to CBC with null IV for short data blocks?

I got to this by experimentation, but can someone maybe provide an explanation for this?

• The CBC mode processed as $$c_i = E_k(p_i \oplus c_{i-1})$$ with $$c_0 = IV$$

• The ECB mode processed as $$c_i = E_k(p_i)$$

If you set the first block with $$c_o = IV = 0$$ in the CBC mode, than it is calculated as $$c_1 = E_k(p_1).$$ This is exactly as ECB mode.

The next blocks, however, will not be equal;

• in ECB mode $$c_i = E_k(p_i)$$
• whereas in CBC mode $$c_i = E_k(p_i \oplus c_{i-1}) \neq E_k(p_i) \text{ for } 1 < i \leq m$$ where $$m$$ is the number of blocks.

Therefore the equality is only valid for the first block.

• Exactly as I suspected, thank you for a more formal explanation <3 – bbozo Oct 15 '18 at 10:51