The conjecture is true.
Indeed, you can always obtain an optimal strategy by iteratively sending the two slowest people across, using a ratchet (1,2),(1),(p,q),(2) or a pair of quick escorts (1,p),(1),(1,q),(1), whichever is faster. When only three people remain, you finish the puzzle with escorts: (1,3),(1),(1,2).
A graph-based solution
Solution adapted from Crossing the Bridge at Night, Günter Rote
Defining the graph. In reasoning about the proof, a key trick is to represent each candidate solution as a graph rather than a sequence of moves. To represent a solution involving $n$ people, use a graph with vertices $1\ldots n$, and a distinct edge $i-j$ for each time $i$ and $j$ travel together. This unordered representation abstracts away from messy problems of interleaving.
The total trip cost can be read off the graph
You can prove that, in an optimal solution, the boat always carries two people going forward and one person going back. This means that in the graph of an optimal solution, the edges correspond exactly to the forward trips.
This gives us a way to calculate the total trip cost from the graph. Let's suppose we have the graph of an optimal solution. Then:
The edges of the graph correspond to the forward trips.
Hence the number of forward trips that $k$ makes is equal to the degree of vertex $k$.
In any complete solution, each person travels forward exactly one more time than they travel back. Hence the number of $k$'s return trips is one less than the degree of vertex $k$.
Hence, suppose we have a problem involving $n$ people with travel times $t_1,\ldots,t_n$. Given a graph representation of an optimal solution, the total cost $C$ of the solution can be computed from the vertices $V$ and edges $E$ as:
$$C \equiv \sum_{\langle i, j\rangle \in E} \max{(t_i,t_j)} + \sum_{k \in V} (\deg{(k})-1) t_k$$
The first term represents forward trips and the second term represents backward trips. (Note that if there are multiple edges between $i$ and $j$, they count multiple times in this sum.)
Of course, instead of counting the degree separately, we could fold it into our sum over edges. Hence another way to write the cost $C$ of the solution is:
$$C \equiv \sum_{\langle i, j\rangle \in E} \left[\max{(t_i,t_j) } + t_i + t_j \right] - \sum_{k\in V} t_k$$
Note that this second term doesn't depend on the choice of solution; it's just the sum of everyone's individual travel times. Hence we'll minimize cost $C$ if we minimize this first term.
Cost-cutting surgery on optimal solutions
In this section, we will start with a graph representing a candidate solution. I will introduce several graph transformations that will convert the graph into one with a ~lower cost, while preserving the number of edges. Although we will have no prior guarantee that the resulting transformed graph will correspond to a realizable, actual sequence of moves across the river, I will prove in the end that it does. In fact, the transformed graph can always be realized as a sequence of ratchets and escorts; this will complete the proof.
Let's start with the graph of a candidate solution. We'll prove that this solution is either suboptimal, or that it can be transformed without raising its cost.
Fixed number of edges. We know that in an optimal solution, the boat takes two people across each time and one person back. Hence an optimal solution for $n\geq 2$ people must end in exactly $2n-3$ steps—$(n-1)$ going forward, and $(n-2)$ going back.
We can therefore assume that this graph has exactly $n-1$ edges.
Let the fastest people escort. Let $e$ be any edge that shares a vertex $v$ with another edge. If the shared edge $v=1$ or if both edges are equal to $\langle 1,2\rangle$, we'll leave them alone. Otherwise, we can always reroute $e$ so that instead of going to vertex $v$, it goes to vertex 1 or 2 instead, lowering the total cost of this solution.
When you apply this process repeatedly, to exhaustion, you convert the graph into one with the following properties: (1) If two edges have exactly one vertex $v$ in common, that vertex is $v=1$. (2) If two edges are parallel—that is, they have two vertices in common— then those edges are $\langle 1,2\rangle$. Hence in this transformed graph, (3) Each vertex has degree one, except perhaps vertices 1 and 2.
In terms of boat trips, these properties translate into a solution where (1) Each person makes exactly one trip across the river, except perhaps the two fastest people "1" and "2". (2) Although the second-fastest person "2" can make multiple trips across the river, that person only travels with the fastest person. (3) Hence the only person who escorts multiple different people across the river is the fastest person "1".
Hence
Each person $k>2$ travels across the river exactly once, either with the fastest person (1,k) or with some other $\ell>2$, (k,l).
Reassign pairs Pick any two edges with no vertices in common. Without loss of generality, say the vertices are $i < j < k < \ell$. Observe that out of all possible ways the edges might be arranged, you'll get the lowest cost if the edges are $i-j$ and $k-\ell$; therefore, arrange the edges to be $i-j$ and $k-\ell$.
In terms of boat trips, this means that whenever you have four distinct people in two boat trips, the two fastest and two slowest people are paired up.
Here's an interesting consequence: Each person $k>2$ travels across the river exactly once, either with the fastest person (1,k) or with some other $\ell>2$, (k,l).
Now suppose $a$ is a person who is escorted (1,a), and $b$ is a person who travels with someone else, (b,c). As a result of our pair-swapping transformation above, we must have that $a<b$ and $a<c$.
This means that in this graph, all the escorted people are faster than the paired-up people.
We have transformed the graph, preserving the number of edges and preserving (or lowering) the total cost. But is the result even realizable anymore as a sequence of moves?
The streamlined solution can be realized using ratchets and escorts
We have to show that this transformed graph can be realized as an ordered list of trips across the river— more specifically, we want to show that it can be realized as a sequence of ratchets and escorts. This will establish that for every candidate solution, there is a cheaper (no-more-expensive) solution in terms of ratchets and escorts, which was to be shown.
Indeed, let's tally up the edges. There are $n-1$ in total. Consider the people k>2. Each of them travels across the river exactly once. Some of them are paired up the fastest person, (1,k). Call these escort-type edges. Others are paired up with someone else l>2, (k,l). Call these ratchet-type edges.
- There are $n$ people and $n-1$ edges in total.
- Let $m$ denote the number of ratchet-type edges.
- Then there are $n-2m-2$ escort-type edges. This is because there are $n-2$ people besides 1 and 2, and $2m$ of them have a trip already accounted for by a ratchet edge.
- This leaves $m + 1$ edges: we started with $n-1$ edges and have accounted for $m$ ratchet-type edges and $n-2m-2$ escort-type edges. All the remaining edges must be of type (1,2).
You can see how any such trip can be realized as a sequence of moves using $m$ ratchets (1,2),(1),(p,q),(2) and $n-2m-2$ escorts (1,p) (their order doesn't matter), finishing with a final (1,2). Q.E.D.
Qualitatively, it turns out that a ratchet is better for slower people, and an escort is better for faster people. In fact, the cutoff time, $\Delta \equiv 2t_2 - t_1$ determines numerically which is better: Suppose you're sending (p,q) across the river. If neither individual could cross the river in time $\Delta$, you're better off with a ratchet. Otherwise, you're better off with two quick escorts. Hence the strategy above, which considers the slowest people first, will generally start by sending people using ratchets, then eventually transition to sending people using escorts.
As for the proof that in an optimal solution, two people travel forward and one person travels back:
Use the notation $+\{a,b\}$ to mean $a$ and $b$ travel forward, and $-\{a,b\}$ to mean $a$ and $b$ travel back. Given a sequence of moves, consider the first place where it deviates from the pattern of "two forward, one back".
Suppose the first deviation is that one person travels forward $+\{a\}$. This cannot be the first move in an optimal solution, because the return trip would make a loop. Therefore, there must have been a return move, just before this: $-\{b\}$. And note that $a\neq b$ to avoid a loop.
Before $b$ returned, one of the following must have happened most recently: $a$ returning, or $b$ setting out with someone. In either case, we can simplify the path, lowering its cost. If the path looked like this, we transform it like that:
$$\ldots +\{b,c\}, \ldots, +\{a\}, -\{b\} \Longrightarrow \ldots +\{a,c\}, \ldots, \emptyset, \emptyset$$
$$\ldots -\{a\}, \ldots, +\{a\}, -\{b\} \Longrightarrow \ldots -\{b\}, \ldots, \emptyset, \emptyset$$
Both of these adjustments strictly lower the cost.
Suppose the first deviation is that two people travel back. Then find which one of them traveled forward most recently and you can safely delete them from that entire round trip:
$$+\{a,c\}, \ldots, -\{a,b\} \Longrightarrow +\{c\}, \ldots, -\{b\}$$
The main point is convincing yourself that these adjustments preserve the legality of all the moves, e.g. don't end up trying to move someone who isn't on the proper side.