The 2012 result says a hanging exists for every k and every n [3]. It says nothing about how much string the hanging costs, and the wrappings people actually write down can be very elaborate [10]. That gap is where the work went.
Take the 2-out-of-4 case as the ledger. The camp session brought the shortest known solution from 80 wraps to 58 [5]. Verhoeff got it to 18 on his own, then to 16 with a program that checked every shorter hanging [6][7]. Subtracting, 80 minus 16 leaves 64 wraps gone, so the final hanging is a fifth the length of the one the workshop inherited [1]; measured against the 58 the schoolchildren reached, 16 is about 72 percent shorter [2].
Only the last step is a different kind of claim. The 58 and the 18 are records, meaning somebody exhibited a hanging that short. The 16 is a statement about the case, because the search ruled out everything below it [7]. The arXiv posting lists the shortest known explicit solutions for large families of these problems [8], and "known" is doing real work in that phrase.
The tooling story is the one practitioners will recognise. Verhoeff first showed Heuseveldt a program that solved the problem in about two hours; Heuseveldt says he replied that his own program took two seconds, and that Verhoeff's is now the faster of the two [9]. Two hours is 7,200 seconds, so that first jump was a factor of roughly 3,600 [3]. It is the difference between checking a case you already suspect and sweeping a family you do not.
What keeps the family from being a novelty is its classification. You cannot demand that pulling nail A alone drops the painting while pulling A and B together leaves it hanging, but any reasonable rule set is realisable, and the reasonable ones are exactly the monotone Boolean functions, which cryptography and voting theory both lean on [12][13]. In the 1-out-of-n case a solution can be drawn as a loop along the edges of an n-dimensional cube passing through every corner [11], which pulls graph theory, group theory and knot theory together in the toolkit here [14].
The 16 covers only this one case; it says nothing about 3-out-of-5 or about a general formula for minimal length. It is one certified number for one case, obtained by enumeration rather than by argument, and enumeration has to be paid for again for each new case. Verhoeff's answer to the usefulness question is that it is the wrong question: we do not know where our spaceship is going or what we will need to survive, and play is one of the ways we learn [15].
