= Solution
Use the ordered <computational basis> $(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$, with qubit 1 written first. \b[The half-time gate requires a branch convention.] Even stationary <Hamiltonian> evolution with $W(2\tau)=\mathrm{SWAP}$ does not uniquely specify $W(\tau)$: different accumulated <eigenstate> phases can agree at $2\tau$ and differ at $\tau$. In particular, both $S$ and $S^\dagger$ below square to <SWAP gate> and leave $|00\rangle,|11\rangle$ unchanged. Adopt the conventional continuous isotropic <exchange interaction> branch, with a relative singlet phase increasing from $0$ to $\pi$ during the full pulse. This is the <exchange square-root branch convention>.
The symmetric and antisymmetric <eigenstates>
$$
|\psi_+\rangle=\frac{|01\rangle+|10\rangle}{\sqrt2},\qquad
|\psi_-\rangle=\frac{|01\rangle-|10\rangle}{\sqrt2}
$$
have <SWAP gate> <eigenvalues> $+1,-1$. Let $P_-=|\psi_-\rangle\langle\psi_-|$ and $P_+=I_4-P_-$ be their antisymmetric and symmetric <orthogonal projectors>. The adopted exchange evolution is
$$
W(t)=P_++e^{i\pi t/(2\tau)}P_-,\qquad
H_{\mathrm{ex}}=-\frac{\pi\hbar}{2\tau}P_-.
$$
For effective spin-one-half <qubits>, $H_{\mathrm{ex}}=\pi\hbar(X_1X_2+Y_1Y_2+Z_1Z_2-I_4)/(8\tau)$, an isotropic <exchange interaction> plus an energy offset that fixes the common phase. At half time choose $+1$ on the symmetric sector and $+i$ on the antisymmetric sector, the conventional principal square-root value of $-1$. Hence the <square root of the SWAP gate> is
$$
\boxed{S=\sqrt{\mathrm{SWAP}}=P_++iP_-
=\begin{pmatrix}
1&0&0&0\\
0&(1+i)/2&(1-i)/2&0\\
0&(1-i)/2&(1+i)/2&0\\
0&0&0&1
\end{pmatrix}.}
$$
Thus $S|00\rangle=|00\rangle$, $S|11\rangle=|11\rangle$, $S|01\rangle=a|01\rangle+b|10\rangle$ and $S|10\rangle=b|01\rangle+a|10\rangle$, where $a=(1+i)/2$ and $b=(1-i)/2$. The identities $a^2+b^2=0$ and $2ab=1$ verify $S^2=\mathrm{SWAP}$, and the projector form proves <unitarity>. The endpoint truth table alone would allow other <matrix square roots>; the stated exchange branch supplies the additional physical assumption.
For the network, fix the <Pauli Z gate>, <Hadamard gate> and <S gate> conventions
$$
Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad
H=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad
P=\sqrt Z=\begin{pmatrix}1&0\\0&i\end{pmatrix}.
$$
The letter $P$ distinguishes this local <S gate> from the two-<qubit> exchange gate $S$. In particular $P^2=Z$; replacing $P$ by $P^\dagger$ changes the network. Subscripts specify the acted-on <qubit>: $Z_1=Z\otimes I_2$, $P_2=I_2\otimes P$ and $H_2=I_2\otimes H$. Operators act \b[rightmost first], so the printed <quantum circuit> is
$$
U=H_2SZ_1SP_2H_2.
$$
First multiply the central two exchange pulses. Since $a^2-b^2=i$, their off-diagonal terms cancel:
$$
K=SZ_1S=\operatorname{diag}(1,i,-i,-1),\qquad
D=KP_2=\operatorname{diag}(1,-1,-i,-i).
$$
This reduces the entire network to $U=H_2DH_2$. Its action on all four <computational basis> states, retaining every relative <quantum phase>, is
$$
\begin{array}{c|c|c|c}
\text{input}&\text{after rightmost }H_2&\text{after }D&\text{after leftmost }H_2\\\hline
|00\rangle&(|00\rangle+|01\rangle)/\sqrt2&(|00\rangle-|01\rangle)/\sqrt2&|01\rangle\\
|01\rangle&(|00\rangle-|01\rangle)/\sqrt2&(|00\rangle+|01\rangle)/\sqrt2&|00\rangle\\
|10\rangle&(|10\rangle+|11\rangle)/\sqrt2&-i(|10\rangle+|11\rangle)/\sqrt2&-i|10\rangle\\
|11\rangle&(|10\rangle-|11\rangle)/\sqrt2&-i(|10\rangle-|11\rangle)/\sqrt2&-i|11\rangle
\end{array}
$$
Equivalently,
$$
\boxed{U=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&-i&0\\0&0&0&-i\end{pmatrix}
=|0\rangle\langle0|\otimes X-i|1\rangle\langle1|\otimes I_2.}
$$
It flips the second <qubit> when the first is zero and contributes $-i$ when the first is one. That conditional phase cannot be discarded as a <global phase>.
To obtain the standard <CNOT gate> controlled by qubit 1, append an $X$ gate to qubit 2 and $P$ to qubit 1. They commute, and
$$
\boxed{\mathrm{CNOT}_{1\to2}=P_1X_2U
=P_1X_2H_2SZ_1SP_2H_2
=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}.}
$$
For control zero, $X_2$ undoes the original flip. For control one, it supplies the desired flip and $P_1$ cancels the phase because $i(-i)=1$. This <phase-corrected exchange network for a CNOT gate> realizes the standard gate exactly with the declared conventions.
Finally, $S|01\rangle=(e^{i\pi/4}|01\rangle+e^{-i\pi/4}|10\rangle)/\sqrt2$ has equal <Schmidt coefficients> and is maximally entangled: a half exchange creates <quantum entanglement> although a full <SWAP gate> does not. More constructively, the network above produces a <CNOT gate> from two half exchanges and local gates. Since <CNOT gates> and arbitrary single-<qubit> <unitary operators> form a <universal quantum gate set>, \b[the <square root of the SWAP gate> and arbitrary single-<qubit> gates are universal]. The alternative inverse exchange branch remains entangling and universal, but its phases and local correction must be recomputed consistently.
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