= Solution
Label the four crossings $A,B,C,D$ in top, left-middle, right-middle, bottom order. Label the bounded regions $d$ (left), $e$ (right), $a$ (upper middle), $b$ (central bigon), $c$ (lower middle), and the unbounded region $o$. The drawing preserves the crossings and the orientation of the supplied <figure-eight knot> diagram. Its overpassing strands are the outer-left/upper-right strand at $A$, the upper-left/lower-central strand at $B$, the upper-central/lower-right strand at $C$, and the lower-left/outer-right strand at $D$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-141-dehn.png]
{title=The figure-eight graph, regions and Heegaard state}
Here is the <Dehn presentation Heegaard diagram> specified by this drawing. Replace the underlying four-valent planar graph $G$ by a small three-dimensional regular neighborhood $N(G)$ and put $\Sigma=\partial N(G)$. Since $G$ has four vertices and eight edges, $\Sigma$ has <genus> $8-4+1=5$. On $\Sigma$, take the five $\alpha$ curves bordering the bounded faces, pushed slightly onto the boundary of $N(G)$. They bound a complete disk system in the complementary <handlebody> $S^3\setminus\operatorname{int}N(G)$; their labels are $\alpha_d,\alpha_e,\alpha_a,\alpha_b,\alpha_c$.
The four $\beta$ curves bound the disks dual to the four short crossing tunnels in $N(G)\setminus\operatorname{int}\nu K_8$. Lift the crossing circles shown in red onto the upper and lower sheets of $\Sigma$, routing them along the corresponding face boundaries; the crossing's overpassing strand determines the lift. Orient the $\alpha$ disk normals and the $\beta$ curves so that their signed successive intersections with the $\alpha$ curves are
$$
\begin{array}{c|l}
\beta_A&a\ e^{-1}\ d^{-1}\\
\beta_B&a\ d^{-1}\ c\ b^{-1}\\
\beta_C&b\ a^{-1}\ e\ c^{-1}\\
\beta_D&c\ d^{-1}\ e^{-1}.
\end{array}
$$
At $A,D$ the missing fourth corner is the unbounded region $o$, whose generator is omitted. This sheet-lifting prescription, together with the projected graph, specifies the curves on the <Heegaard surface>; the red circles alone are only their local projections.
To recover the <knot exterior> from $\Sigma\times I$, attach five <two-handles> along the $\alpha$ curves on one side and cap the resulting <sphere> with a <three-handle>. On the other side attach four <two-handles> along the $\beta$ curves. Compressing that side leaves the boundary <torus> of the <knot exterior>, which is retained. In particular, the counts are five $\alpha$ compressions and four $\beta$ compressions, rather than two complete genus-five disk systems.
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