Recall the security definition of encryption $($IND$-$CPA variant$).$ The game is as follows: Let $n$ be the plaintext length and $$ a security parameter $($key length$).$ The challenger chooses a key $ki{n}_R\{0,1{\}}^{}$ and a bit $bi{n}_R\{0,1\}.$ The adversary chooses two messages, $m_0,m_{1}in\{0,1{\}}^n$ and gives them to the challenger. The challenger encrypts $m_b$ and returns $E_k(m_b)$ to the adversary. This $($Adv$,$ in $i-$th iteration, sending $m_0^i,m_1^i$ and receiving $E_k(m_b^i)$ for the same $b)$ is repeated a polynomial $($in $)$ number of times. Then, the adversary produces a guess $b^{\text{'}}.$ The adversary wins if $b=b^{\text{'}}.$ We say that an encryption scheme is secure if adversary success $($i$.$e$.$ probability of winning$)$ is close to $\frac{1}{2}($Recall$,\frac{1}{2}$ corresponds to the success probability of a random guess, and can always be achieved by a na$$ve adversary simply choosing $b^{\text{'}}$ at random$).$

Prove that a deterministic encryption scheme $($which uses no randomness$)$ cannot be secure according to this definition.

$($Hint: Construct an adversary that always wins the game no matter what deterministic encryption scheme the challenger chooses.$)$