The Flexing Operation and Tubulating Proper Flexagons

The flexing operation, we have seen, requires a thumbhole with a $1-$hinge. In the case of the proper flexagon, this is the number one thumbhole. This thumbhole splits the left hand pat into two groups, one composed of a single subpat and the other composed of $G-2$ subpats. The complete flexing operation inverts the top-most subpat of the left pat, leaving it on the left side, while it deposits the remaining subpats on the inverted right pat (see figure 8.1). The number one thumbhole has been opened out to display the next side. The two new pats are now joined by what was originally the $1-$hinge. When the flexagon is rotated, the hinges must be renumbered. The original $1-$hinge will become the new $0-$hinge, and each of the other hinges will have values one lower than before the flex and rotation. In the normal flexing operation of a proper flexagon, which has a consecutive subpat hinge sequence and a consecutive subpat structure, the order of turning up sides must also be consecutive (as has been shown empirically). This is because each successive thumbhole is associated with a correspondingly numbered hinge, (i.e. the $1-$hinge with the number one thumbhole, etc.). Each flexing operation subtracts one from the value of each hinge, thus bringing the thumbhole with which any particular hinge is associated closer to the position for being opened up next. A subpat hinge which in in position $R$ will be opened after $R$ flexes and rotations.

If we flex along a given cycle of a flexagon, we notice that we always progress in the same direction, either clockwise or counterclockwise, along the path of the map. If we draw vectors along the edges of the map (indicating in which direction we are progressing), the vectors for a given cycle will always point consistently clockwise or counterclockwise. Whichever way they do point, their direction may be reversed by turning the flexagon over. If a second cycle is added and vectors are drawn on the map, this new cycle will have one vector in common with the first cycle, but the direction of all the vectors around the center of the map polygon representing this cycle will be just opposite from that of the vectors in the first cycle. In fact, the vectors of any map cycle which has one edge in common with any other given cycle will point in the opposite direction with respect to that cycle. This means that all of the polygons in the map of a given flexagon will be oriented; the sense of the orientation may be changed by turning the flexagon over, since in so doing all vectors are reversed.

The reason for the reversed pat structure of a subpat with respect to its large pat can now be explained. Consider the history of a proper subpat of a proper flexagon. The pat structure of this subpat will remain unchanged throughout the flexing operation (assuming the flexagon flexes left) provided it is in the left pat and provided it is below thumbhole 1, which has a hinge in the ``1'' position. This is because a $1-$flex moves all of the subpats which are below thumbhole 1 from the left pat to the right pat, unchanged in any way (i.e. uninverted). However, if this subpat is on the top of the left pat, and the flexagon is flexed, the subpat will remain in the left pat but will be inverted. A rotation and flex will reinvert it and place $G-2$ subpats on top of it, leaving it undisturbed for $G-2$ flexes thereafter. However, if we decide not to rotate, but to flex along a new cycle, this subpat alone will remain in the left hand pat. As the flexagon must always flex left (we built it that way), the first flexing operation will open up the first thumbhole to a side, $(a-l)$, the last side up being (a) if the flexing was proceeding in an ascending order around the map (see figure 8.2a). For instance, in an order 6 tetraflexagon as shown in figure 8.2b, if we flex from 1 to 2 to 3 to 4 to 1, and then decide to change over to a new cycle, we must flex next to side 6. Although the numbering of the map is still counterclockwise, the flexing vectors have changed direction, and the flexing must proceed against the numbering. In order for the side following side (a) to be side $(a-1)$, the $(a-1)$ thumbhole must be the lowest thumbhole in the subpat containing the new cycle, for that subpat will be inverted, making the $(a-1)$ thumbhole the top most one when the subpat is left along in the left pat. Similarly, in flexing about the second cycle (assuming there are no other cycles attached to the second one) $(a-2)$ will follow $(a-1)$ and will in turn be followed by $(a-3)$, and so on until $(a-G-1)$ is reached. Since side $(a-1)$ used the first thumbhole in the subpat, sides $(a-2)$ through $(a-G-1)$ must use the others in order, and when the subpat constant order is inverted by the subpat's incorporation into the large pat, $(a-G-1)$ will be the side nearest the top. Since each thumbhole in a subpat can be associated with a single leaf, the thumbhole order becomes reversed also. Since the basic number sequence must increase consecutively when read down the pat, the pat structure for the subpat must be inverted with respect to the large pat.

Since we are on the subject of flexing operation, let us consider flexing operations other than the $1-$flex; i.e., the tubulations. For all proper flexagons, the $2-, 3-, 4- \ldots, (G-1)-$flexes remove all but $2, 3, 4, \ldots (G-1)$ leaves respectively from the left pat and deposit them from the right pat; As has been stated before, a tubulation acts like a flex. For instance, in a tetraflexagon of order 3 (see figure 8.3a) which has a tubulation from 1 to 3, we can 1 flex from face $2-1$ to face $3-2$ and when we tubulate, we can cut the hinge and lay the flexagon out in the form of a straight strip of squares with one on the top and three on the bottom. This, then, is face $1-3$ (see figure 8.3b). When we want to flex from face $1-3$ back to $2-1$, we fold the three's so that they face each other and tape the cut hinge back together. The flexagon will now open up to side 2. We should notice, however, that this process of turning the tubulation in side out has also exchanged the position of the two pats of a unit with respect to each other. This is equivalent to a rotation so in this operation, we have both flexed and rotated.