String Figures and Knot Theory
- mathematics of the unknot under tension

by Martin Probert

PROOFS

LEMMA

F is a string figure implies that the mirror image of F is a string figure.

LEMMA - Proof

If a string figure F is constructed in view of a mirror, a mirror image is constructed by the person in the mirror. The mirror and the person in the mirror may be replaced by a second person exchanging movements of left and right hand throughout the construction. Thus the mirror image of F is constructable.

THEOREM 1 - Proof

Suppose a two-dimensional subset S1 of motifs is completely unravelled (leaving no string contacts) to leave a subset S2. S2 is a string figure, so the mirror image s2 of S2 (which exists by the lemma above) can be formed. The two-dimensional set of unravelled string segments resulting from the unravelling of S1, apart from the orientation of the remainder of the figure in relation to it, is identical in S2 or s2. The subset S1 is now reconstructed on s2. The resultant string figure is SÈ s2.

The mirror image of SÈ  s2, that is, sÈ  S2, exists by the above lemma. Each motif of s1 is in the same position in relation to S2 as was the corresponding motif of S1 but is of opposite parity. Hence s1 is the reflection of S1.

Suppose that the action described above takes place in front of a mirror: as S1 is reconstructed on s2 to give SÈ  s2 an image in the mirror is forming s1 (the mirror image of S1) on S2 to give sÈ  S2. Similarly, given S2 in a loop of string, s1 may be added to it. That is, given any string figure, if a two-dimensional subset S1 may be unravelled to leave S2, then s1 may be formed on S2 to give sÈ  S2. Thus any such two-dimensional subset of two-dimensional motifs that can be unravelled can be reformed in the reflected state.

THEOREM 2 - Proof

The unravelling and reforming in the same or the reflected state of a subset of motifs is a procedure that affects only the string contacts that constitute the motifs of the subset, not the segments of string left without string contacts by earlier unravellings. Hence the successive unravelled segments of string left by a sequence of unravellings may each in turn, by theorem 1, be reformed in the original or the reflected state (or, in the case of three-dimensional subsets of string contacts consisting of entire motifs together with part or all of the 3D framework of the string figure, in the original state).

Since there are two choices for how each of the U two-dimensional subsets may be reformed, the procedure can lead to any one of 2U look-alikes.

THEOREM 3 - Proof

Every look-alike, through some unravelling process, generates a discoverable subset of look-alikes. (Even the trivial string figure generates such a subset, a subset containing 20 = 1 look-alike, namely the trivial string figure itself.) Each discoverable subset so generated includes the generating look-alike: hence the union of the discoverable subsets generated by all look-alikes of a string figure is equal to the set of all look-alikes. In practice a union of a smaller number of discoverable subsets than those generated by all look-alikes will be sufficient.

THEOREM 4 - Proof

We assume there is at least one non-empty proper subset which is a shard. Let this shard be called S with catalogue {S1, S2, ... Sm}. Let the remainder of the string figure be a member of the set {T1, T2, ... Tn} where each Tj is a distinct valid state. We shall show that the remainder of the figure is composed of shards.

S is a shard, and so (by the definition of shard), for each valid state assumed by the remainder of the figure, S can independently assume any of its valid states. Hence for each Tj (that is, for each valid state of the remainder of the figure) we can construct look-alikes {S1, Tj}, {S2, Tj}, ... {Sm, Tj}. By listing the look-alikes for each Tj we obtain {S1, T1}, {S2, T1}, ... {Sm, T1}; {S1, T2}, {S2, T2}, ... {Sm, T2}; ... ; {S1, Tn}, {S2, Tn}, ... {Sm, Tn}. Rearranging we can list the look-alikes as {S1, T1}, {S1, T2}, ... {S1, Tn}; {S2, T1}, {S2, T2}, ... {S2, Tn}; ... ; {Sm, T1}, {Sm, T2}, ... {Sm, Tn}. Thus, for each Si, Tj can assume all valid states. Hence either the remainder of the figure is a shard, or contains a shard. Continuing the argument in this fashion the entire string figure is shown to be composed of shards.