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To solve this puzzle you must wind the string around the grey pegs in such a way that removing any single peg will allow the string to fall. Imagine the start and finish nodes pulling on the ends of the string: the string must come free when any one peg is removed, but must not come free if no pegs are removed.

Solution

author: Diggory Blake

This puzzle is similar to stringy, but is a more complicated version. Now we must either grok the arXiv paper or use group theory.

To use group theory, consider the two-peg stringy problem, and label the left peg A and the right B. Now we can denote the top side of peg A A and the bottom a, and similarly for B. A and a are inverses. We then find a string such that removing all the as and As results in Bs and bs which cancel, and the same for A, e.g. ABab, not including the start and finish pegs.

To extend to multiple pegs, add further elements, C and c for a third peg. Now append a C, and invert the preceding elements such that the removal of the C causes the entire string to collapse, and finally terminate with a c: ABabCBAbac.

Repeating the procedure, we get the full solution for four pegs: ABabCBAbacDCABabcBAbad.

All that remains is to actually construct the string interactively, like so:

You can choose any peg-label mapping (I used top-left for A, bottom-left for B, bottom-right for C, top-right for D). It may help to edit the javascript source code in addPeg like so:

//addHoverEffect(e1, {"fill": "#888"}, {"fill": "#ff0"})
addHoverEffect(e1, {"fill": "#aaa"}, {"fill": "#ff0"})

so that you can keep track of which side of each peg is which, although it should be soluble without this.