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Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex
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Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex
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\section{Quantum Gates}
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\definition{}
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The quantum analog of a logic gate is a \textit{quantum gate,} \par
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which we'll define as a linear map from $\mathbb{B}^n$ to $\mathbb{B}^n$. \par
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\note[Note]{This definition is temporary, we'll extend it soon.}
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\problem{}
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Consider the CNOT (controlled not) gate. \par
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When applied to a two-bit state $\ket{ab}$, CNOT inverts $b$ iff $a$ is $\ket{1}$. \par
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Find the matrix that represents the CNOT gate. \par
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\hint{what are the dimensions of this matrix?}
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\begin{solution}
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\begin{equation*}
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\text{CNOT} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 0 \\
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\end{bmatrix}
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\end{equation*}
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\vspace{4mm}
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If $\ket{a}$ is $\ket{0}$, $\ket{a} \otimes \ket{b}$ is
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$
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\begin{bmatrix}
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\begin{bmatrix}
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b_1 \\ b_2
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\end{bmatrix} \\ 0 \\ 0
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\end{bmatrix}
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$, and the \say{not} portion of the matrix is ignored.
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\vspace{4mm}
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If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is
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$
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\begin{bmatrix}
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0 \\ 0 \\
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\begin{bmatrix}
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b_1 \\ b_2
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\end{bmatrix}
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\end{bmatrix}
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$, and the \say{identity} portion of the matrix is ignored.
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The state of $\ket{a}$ is always preserved, since it's determined by the position of
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$\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ in the tensor product.
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If $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on top, $\ket{a}$ is $\ket{0}$,
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and if $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on the bottom, $\ket{a}$ is $\ket{1}$.
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\end{solution}
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\vfill
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\problem{}
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Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
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\begin{solution}
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\begin{equation*}
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\text{CNOT}_{\text{mod}} = \begin{bmatrix}
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0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 0 & 0 & 1
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\end{bmatrix}
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\end{equation*}
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\end{solution}
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\vfill
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\pagebreak
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\iffalse
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\problem{}
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Finally, modify the original CNOT gate so that the roles of its bits are reversed: \par
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$\text{CNOT}_{\text{flip}} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$.
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\begin{solution}
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\begin{equation*}
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\text{CNOT}_{\text{flip}} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 0 \\
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0 & 1 & 0 & 0 \\
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\end{bmatrix}
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\end{equation*}
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\end{solution}
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\vfill
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\fi
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\problem{}
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The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par
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Find its matrix.
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\begin{solution}
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\begin{equation*}
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\text{SWAP} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par
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In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par
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Find the $T$ gate's matrix. \par
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\note{
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This gate is particularly interesting because it's a \textit{universal quantum gate}: \\
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like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates.
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}
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\begin{solution}
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\begin{equation*}
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\text{T} = \begin{bmatrix}
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1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
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0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
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\end{bmatrix}
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\end{equation*}
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\end{solution}
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\vfill
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\pagebreak
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The last thing we need is a way to draw complex sequences of gates. \par
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We already know how to do this with classical gates, \'a la logic circuit:
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\begin{center}
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\begin{tikzpicture}[circuit logic US, scale=1.5]
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\node[and gate] (and) at (0,-0.8) {\tiny\texttt{and}};
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\draw ([shift={(-0.5, 0)}] and.input 1) node[left] {\texttt{1}} -- (and.input 1);
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\draw ([shift={(-0.5, 0)}] and.input 2) node[left] {\texttt{0}} -- (and.input 2);
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\draw (and.output) -- ([shift={(0.5, 0)}] and.output) node[right] {\texttt{0}};
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\end{tikzpicture}
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\end{center}
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We draw quantum circuits in a very similar way. \par
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For example, here a simple three-bit circuit consisting of a CNOT gate on the first bit, \par
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controlled by the third. The first bit is inverted iff the third bit is $\ket{1}$:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\node[qubit] (c) at (0, -2) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$};
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\draw[wire] (c) -- ([shift={(4, 0)}] c.center) node[qubit] {$\ket{1}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{3}{CNOT}
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\end{tikzpicture}
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\end{center}
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\problem{}
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Draw the CNOT gate as a classical logic circuit. \par
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\hint{This can be done with one gate.}
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[circuit logic US, scale=2]
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\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
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\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
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\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
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\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$a$};
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\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$b$};
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\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
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\draw (dot) -- (dot -| 1,0) node[right] {$b_\text{out}$};
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\draw (xor.output) -- (xor.output -| 1,0) node[right] {$a_\text{out}$};
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\vfill
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Gate controls may be marked with a filled circle or an empty circle. \par
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Empty circles denote \textit{inverse controls,} which (of course) have an inverse effect. \par
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For example, the two circuits below are identical:
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\null\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[
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qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
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qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
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;
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\draw[wireijoin]
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($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{3}{CNOT}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[
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qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
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qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1.5,0)}] a)!0.5!([shift={(2.5,0)}] a)$) --
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($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{3}{CNOT}
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\qubox{b}{0.5}{b}{1.5}{X}
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\qubox{b}{2.5}{b}{3.5}{X}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill\null
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\problem{}
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What are $\ket{a}$ and $\ket{b}$ in the diagrams above?
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[
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qubit] {$\ket{1}$};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
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qubit] {$\ket{0}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
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;
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\draw[wireijoin]
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($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{3}{CNOT}
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\vfill
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\pagebreak
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As we noted before, quantum gates don't \textit{consume} bits, they \textit{transform} them. \par
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Thus, quantum circuits are drawn with a fixed set of bits, whose states change with time: \par
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\note[Note]{
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In this diagram, CNOT and SWAP are drawn as $\oplus$ and \rotatebox[origin=c]{90}{$\leftrightarrows$} to save space.
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}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\node[qubit] (a) at (0, 0) {$\ket{a}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\node[qubit] (c) at (0, -2) {$\ket{c}$};
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\node[qubit] (d) at (0, -3) {$\ket{d}$};
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\node[qubit] (e) at (0, -4) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$};
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\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{b}$};
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\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$};
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\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$};
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\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{e}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\draw[wireijoin]
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($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{T}
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\qubox{a}{4}{a}{5}{X}
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\draw[wire]
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($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
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($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
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;
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\draw[wireijoin]
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($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{e}{2}{e}{3}{$\oplus$}
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\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}}
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\draw[wire]
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($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
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($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
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;
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\draw[wirejoin]
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($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
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circle[radius=0.1] coordinate(dot)
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;
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\draw[wirejoin]
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($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{b}{5}{b}{6}{T}
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\end{tikzpicture}
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\end{center}
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In a quantum circuit, the ONLY way two bits can interact is through a gate. \par
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We cannot add \say{branches} in quantum circuits like we do in classical circuits:
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\begin{center}
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\begin{tikzpicture}[circuit logic US, scale=1.2]
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\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
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\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
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\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
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\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$x$};
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\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$y$};
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\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
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\draw (dot) -- (dot -| 1,0) node[right] {$y_\text{out}$};
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\draw (xor.output) -- (xor.output -| 1,0) node[right] {$x_\text{out}$};
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\draw[->, line width = 1, ogrape]
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([shift={(0.3,-0.5)}] dot) node[right] {This is a branch}
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-| ([shift={(0,-0.2)}] dot)
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;
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\end{tikzpicture}
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\end{center}
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\problem{}
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Find the values of $\ket{a}$ through $\ket{e}$ in the above circuit.
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\node[qubit] (a) at (0, 0) {$\ket{1}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{1}$};
|
||||
\node[qubit] (c) at (0, -2) {$\ket{1}$};
|
||||
\node[qubit] (d) at (0, -3) {$\ket{0}$};
|
||||
\node[qubit] (e) at (0, -4) {$\ket{1}$};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[
|
||||
qubit] {$\ket{0}$};
|
||||
\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[
|
||||
qubit] {$\ket{1}$};
|
||||
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[
|
||||
qubit] {$\ket{1}$};
|
||||
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[
|
||||
qubit] {$\ket{0}$};
|
||||
\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[
|
||||
qubit] {$\ket{0}$};
|
||||
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
|
||||
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\draw[wireijoin]
|
||||
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{a}{1}{a}{2}{T}
|
||||
|
||||
\qubox{a}{4}{a}{5}{X}
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
|
||||
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
|
||||
;
|
||||
\draw[wireijoin]
|
||||
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{e}{2}{e}{3}{$\oplus$}
|
||||
|
||||
\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}}
|
||||
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
|
||||
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{b}{5}{b}{6}{T}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
Loading…
x
Reference in New Issue
Block a user