# Is the one-time pad still perfectly secret if all-zero keys are excluded?

I'm trying to solve this question related to one-time pads and perfect secrecy:

My solution is: I assumed that the current message space is $M = \left\{ 0,1 \right\}^l$ and the new keyspace after removing $0^l$ from the set is $K = \left\{ 0,1 \right\}^l - 0^l$. Since clearly $|K| < |M|$, the new scheme is not perfectly secret.

Is my solution correct?

• Do you expect anything but a simple "yes" as the answer? – SEJPM Mar 31 '18 at 9:12
• Hi Mitch and welcome. Just a small remark: try to make your titles as precise as possible; there are many ways in which the OTP's perfect secrecy can be broken: the title should clearly indicate what is asked. – Maarten - reinstate Monica Mar 31 '18 at 13:46

## 1 Answer

Yes your solution is correct; the flawed OTP scheme is no longer perfectly secret if there are fewer keys than messages.

Once you remove the all zero key the property $$\mathbb{P}[M=m|C=c]=\mathbb{P}[M=m],\qquad \forall m,\forall c,\quad(1)$$ required for perfect secrecy will no longer necessarily

• I've removed the word "also" from the second part of your answer, could you check if this still good for your answer? – Maarten - reinstate Monica Mar 31 '18 at 13:44