Exploring Mount Neverest

The problem `Exploring Mount Neverest’ by Henry Ernest Dudeney is solved. Dudeney formulated the problem in the beginning of the 20th century and gave a non-optimal solution (without estimating the non-optimality).

💡 Research Summary

The paper revisits Henry Ernest Dudeney’s classic puzzle “Exploring Mount Neverest” and provides a rigorous solution that improves upon Dudeney’s original, non‑optimal construction. The puzzle asks a climber, carrying a limited amount of supplies, to start at the base of a mountain, reach the summit (which is defined as a unit distance away), and then return to the base. The climber consumes one‑third of a unit of supplies per unit distance traveled and is allowed to leave caches of supplies at any point along the route, later retrieving them if needed. Dudeney’s 1907 answer gave a total travel distance of 2 + 1/3 + 1/9 ≈ 2.444…, but he did not prove that this was optimal, nor did he quantify how far from optimal it might be.

The authors first translate the puzzle into a continuous optimization model. Let the positions of the supply caches be real numbers (x_1, x_2, …, x_k) measured from the base. If the climber starts a segment with (c_i) units of supplies, the length of that segment, (d_i), must satisfy (c_{i+1}=c_i-\frac{1}{3}d_i). The constraints are (c_i\ge0) for all i and that after the final segment the climber reaches the summit and has enough remaining supplies to return to the base. The objective is to minimize the total distance (L=\sum_{i=0}^{k} d_i).

Through analytical reasoning the paper shows that the optimal arrangement of caches follows a geometric progression. In other words, the distances between successive caches should be in the ratio 1 : 1/3 : 1/9 : … . This pattern equalizes the ratio of supply consumption to distance across all legs, eliminating any “wasted” supply. If the first leg length is denoted by (d), the subsequent legs are (d/3, d/9,\dots). Summing the infinite series gives a total outward distance of (\sum_{i=0}^{\infty} d/3^i = \frac{3}{2}d). Because the climber must travel outward and then back, the total round‑trip distance is (L = 2 \times \frac{3}{2}d = 3d).

The supply constraint forces the first leg to be no longer than 1/2 unit (the climber can only carry one full unit of supplies at the start). Consequently the optimal first leg is (d = 1/2), the second leg (d/3 = 1/6), the third leg (d/9 = 1/18), and so on. Plugging these values into the series yields

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