Some of the party games that young teenagers play have surprisingly rich mathematical appease "Entanglement" is one such game in which couplings are linked with strings tied tightly to their wrists.

Some of the party games that young teenagers play have surprisingly rich mathematical appease "Entanglement" is one such game in which couplings are linked with strings tied tightly to their wrists, as shown in Figure 1a, and then challenged to disentangle themselves, ofttimes leading to Figure 1b. In fact there is a quick solution (needles to say missing the point of the game): simply push a bit of the girl's string in subordination to the boy's at his wrist, pass the resulting bight over his hand, and then hap it free on the other side, as shown in Figure 2

One way to formulate this mathematically is to ask for an isotopy to displace the meridian loop m from a rigid wire w embedded in the 3-sphere, as shown in Figure 3a. (The wire shows the boy with his attached string, where the upper noose of the wire is his right hand, while the meridian portrays the girl with her string.) Of course this is easy: just slide m along w to a point just below the upper turn of w, and then tighten it out and pull it not upon the top. Or put differently, detect that w is isotopic to the trivial wire *____* and in the same manner the meridian slips right opposite

A similar puzzle attributed to Stewart Coffin [2] was considered at Inta Bertuccioni in a latter issue of this MONTHLY [1] Here the wire w is configured slightly differently (as in Figure 3b) and there is in fact no solution, that is, m cannot be isotoped along w. Bertuccioni proves this through an explicit calculation showing that m portray by actions a nontrivial element in the fundamental arrange of the complement of if.

The drift of this note is to give a more conceptual trial of the impossibility of solving Coffin's embarrass We then generalize the proof-sheet using an elementary but nontrivial consequence in knot theory to point out to that all but one of the vast "menagerie" of possible poses suggested by entanglement and Coffin's labyrinth are also unsolvable.

First consider Coffin's perplex The wire w consists of brace unknots ?« joined by an arc ?± as illustrated in Figure 4a. From this united obtains a knot k on banding the components of ? together along a, shown in Figure 4b Actually there are infinitely many knots that can be formed in this way, for there may be twists in the band. The common pictured is readily seen to be the square knot.

Now say that if m could be isotoped opposite w, then it certainly could be isotoped distant from k, since the complement of k contains the consummation of w (with a thickened a bit). however then the lift k of k to the infinite cyclic guard of the complement of m would consist of an infinite number of copies of k (Note that the tale of m is a solid torus, and the shroud is an infinite solid cylinder.) However, it is easily seen that the composings of k are unknotted, contradicting the fact that k is a nontrivial knot (the square knot). Indeed, viewing k as the closure of a tangle with axis m the link k is obtained through composing infinitely many copies of this tangle, as illustrated in Figure 5 (see for example, [4]) Thus each constituent of k looks like a prolonged worm that can be contracted by pushing from its "free" extremity

For a general embarrass in the menagerie we allow any embedding of the wire for which ? is an unlink, while the arc ?± can be arbitrary. single in kind such puzzle is shown in Figure 6a. The foregoing argument demonstrates that there is no trust for a solution unless the associated knot k is trivial, because each ingredient of the lift k is unknotted (as seen in Figure 6b; in general slide m bring to a period to an endpoint of ?± before taking the cover) However, by the agency of a theorem of Marty Scharlemann [5] k is trivial if and and nothing else if the arc ?± is isotopic (fixing ?) to a trivial arc. Thus "most" of the labyrinths in the menagerie, indeed all if it be not that those equivalent to entanglement, have no solution.

From a practical point of view the same might ask how to recognize when a is trivial. This is actually a difficult question. However there is a simple proof for the triviality of the homotopy class of ?±: The fundamental form into groups of the complement of ? is released of rank two, with generators x and y corresponding to the sum of two units components of ?. The arc ?± oriented from the x-loop to the y-loop of ? determines a word in x and y also denoted ?± Since the arc ?± can be made to spiral around ? at its endpoints, its homotopy class corresponds to an equivalence class of words: ?± ~ ?? if and barely if there exist integers m and n for which ?± = x^sup m^??^sup n^ Thus the arc a is homotopically trivial if and no other than if the word ?± ~ 1 For example, the entanglement mystify has ?± = xy ~ 1 as look forward toed whereas Coffin's puzzle has ?± - yx * 1 and the stagger in Figure 6a has a = yxyx * 1

Of course there are many homotopically trivial arcs ?± that are isotopically nontrivial, or equivalently, whose associated knots are nontrivial. Consider, for example, the perplex in Figure 7, which has a = xy ~ 1 The associated knot has Jone polynomial t^sup -5^ - t^sup -4^ - t^sup -1^ + 2 - t + t^sup 2^ + t^sup 5^ - t^sup 6^ (see [3]; its Alexander polynomial is trivial), in like manner this puzzle has no solution.