#set page(
  paper: "a4",
  header: align(center)[
    QCS - DM2 - Augustin LUCAS
  ],
)

== Q1

We want to *simplify* the following circuit:

#{
  import "@preview/quill:0.4.0": *

  quantum-circuit(
    lstick($|v_0〉$), $H$, ctrl(1), $H$, [\ ],
    lstick($|v_1〉$), $H$, targ(), $H$
  )
}

The gates 

#{
  import "@preview/quill:0.4.0": *

  quantum-circuit(
    lstick($$), $H$, rstick($$), [\ ],
    lstick($$), $H$, rstick($$)
  )
}

correspond to the following matrice: $1/2 mat(H, H; H, -H;)$

And CNOT = $mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0;)$

The circuit then corresponds to the following operation:

$A&=1/2 mat(1, 1, 1, 1; 1, -1, 1, -1; 1, 1, -1, -1; 1, -1, -1, 1)
        mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0;)
    1/2 mat(1, 1, 1, 1; 1, -1, 1, -1; 1, 1, -1, -1; 1, -1, -1, 1) \

 &=1/4 mat(4, 0, 0, 0; 0, 0, 0, 4; 0, 0, 4, 0; 0, 4, 0, 0) \
 &=mat(1, 0, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0; 0, 1, 0, 0)$

Which gives $cases(
  |00〉 => |00〉,
  |01〉 => |11〉,
  |10〉 => |10〉,
  |11〉 => |01〉
)$ that corresponds to an *inversed CNOT* that we can denote:

#{
  import "@preview/quill:0.4.0": *

  quantum-circuit(
    lstick($|v_0〉$), targ(), rstick($$), [\ ],
    lstick($|v_1〉$), ctrl(-1), rstick($$)
  )
}


== Q2

In the *classical case*, we need $2^(n-1)+1$ queries to determine if $f$ is constant.
Using only $2^n$ queries, all queries could have the same value with $f$ balanced.

In the *quantum version*, we may use the following circuit:
#{
  import "@preview/quill:0.4.0": *

  quantum-circuit(
    lstick($|0〉$), $H$, mqgate($U_f$, n:4), $H$, meter(), setwire(2), rstick($$), [\ ],
    lstick($|0〉$), $H$, targ(), $H$, meter(), setwire(2), rstick($$), [\ ],
    lstick($...$), [\ ],
    lstick($|1〉$), $H$, targ()
  )
}