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+The first quantum algorithm we will look at is the Bernstein-Vazirani algorithm.
+
+Given a secret s $\in \{0,1\}^n$ and the function f : $\{0,1\}^n \rightarrow \{0,1\}$ defined as f(x) = x * s
+\* denotes the inner product of two bitstrings here
+For Bitstrings x and y of length n the inner product x * y is $x_1y_1 + ... + x_ny_n$ mod 2
+The goal is to find the secret s using as few queries of f as possible.
+So as few evaluations of f as possible.
+
+We will look at a quantum algorithm that will find s with only one evaluation of f.
+
+![[Pasted image 20260730144713.png]]
+
+The top wire consists of n qubits in state 0.   $\ket{0}^n = \ket{0} \otimes ... \otimes \ket{0}$ 
+The bottom wire is in state 1.
+Both wires together are in state $\ket{0^n1} = \ket{0}^n \otimes \ket{1}$ 
+
+First, we apply the Hadamard gate on all qubits.
+The resulting state is calculated as follows:
+
+![[Pasted image 20260730145119.png]]
+
+We are now in superposition between all classical possibilities on the top wire
+and in $\ket{-}$ on the bottom wire.
+
+
+Next we apply the unitary $U_f$ on both wires.
+The Unitary is defined as:
+$U_f \ket{x,y} = \ket{x,y \otimes f(x)}$ 
+This unitary applies the function f to the bottom wire y.
+It can be calculated as follows:
+
+![[Pasted image 20260730153305.png]]
+
+And we can rewrite $U_f(\ket{x} \otimes \ket{-})$ as
+
+![[Pasted image 20260730154444.png]]
+
+The bottom wire has not changed.
+The top wire now has f(x) encoded in its quantum state.
+The phenomenon that the output of f is encoded as a -1 is called phase kickback.
+We need one final step before measuring.
+
+
+Now, we perform one final hadamard gate on the top wire.
+We want the result to be $\ket{s} \otimes \ket{-}$
+To check, we calculate $(H^{\otimes n})^\dagger \ket{s} \otimes \ket{-}$ and check if it is equal to $\psi_2$   :
+
+![[Pasted image 20260730155153.png]]
+
+Because $HH^\dagger = I$ we now can measure s on the top wire.