$7 \cdot 20$ Given that all vertices of a broken line are located on the surface of a cube with an edge length of 2, each edge of the broken line is 3 units long, and the two endpoints are precisely the two farthest vertices of the cube. How many edges does this broken line have at minimum?

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[Solution] Let $P, Q, R$ be the midpoints of the three edges extending from point $B_{3}$. Clearly, $A_{1} P=A_{1} Q=A_{1} R=3$. Thus, the set of points on the surface of the cube that are 3 units away from $A_{1}$ consists of three arcs passing through $P, Q$, and $R$ on the three sides with $B_{3}$ as the vertex (see the right figure). By symmetry, we only need to consider the case of arc $P R$.

It is easy to see that, except for the endpoints $P, Q$, all other points on arc $P R$ are 3 units away from $A_{1}$ and are less than 3 units away from the other points on the surface of the cube. Therefore, if $A_{1}$ is the starting point of the polyline and the other endpoint of the first edge is on $P R$, it can only be one of the points $P, R$ (otherwise, it is impossible to draw a second edge of length 3 with endpoints on the surface of the cube). Without loss of generality, assume the first edge is $A_{1} P$. Then the second edge can only be $P A_{4}$. This indicates that after the polyline has passed through two segments, it has moved from one endpoint of an edge to the other. Since it takes three edges to go from $A_{1}$ to $B_{3}$, the polyline must have at least 6 edges.

By connecting the points $A_{1}, P, A_{4}, S, B_{4}, T, B_{3}$ in sequence in the previous figure, we obtain a polyline with 6 edges that meets the requirements of the problem. Therefore, the polyline that satisfies the requirements has at least 6 segments.