The Order $N$ Improper Flexagons

In places where a right pat was slit instead of a left pat, it is impossible to add more (for a new cycle) sides without suppressing others, for the hinge joining two pats resulting from a $2-$flex joins the topmost leaves in each pat and since the added leaf must be at the bottom of the pat there would have to be a $1-$hinge connecting the top and bottom leaves of the pat. This arrangement would suppress all in between sides. All of the places in the cycle of an improper flexagon at which there are no degree 1 pats represent shortcuts, made by leaving at least two leaves in the left pat. Thus, none of the diagonals of the map, which are shortcuts or just plain ``cuts'' ^9.1 can be added to in the conventional manner. That is, after $1-$flexing along an out-side edge or $0-$cut there will be a single leaf in either the left or right pat. This leaf may be slit and a new side revealed. There is a choice in slitting for all the slits after the first until $(G-2)$ slits have been made. Thus, to any improper cycle it is possible to add at a $1-$flex $0-$cut another cycle of any nature, so long as it too has a $1-$flex $0-$cut in common with the first cycle (see figure 9.5).

Because of the nature of a shortcut, any $0-$cut tubulation also has a single leaf in either the left or right pat. We have seen that we can force the tubulation, rotate, and open up to the next side along the outside edge or $0-$cut cycle ^9.2. What if we decide not to rotate but to slit the single leaf instead? This opens up a new side which can only be reached through the tubulation. Take, for example the single cycle improper pentaflexagon shown in figure 9.6a. If we flex from 6 to 5, we may tubulate to 3 by removing all but one leaf from the right hand pat (after rotation). We can now force this tubulation, then, instead of rotating and opening up to side 2, by a $-1$ flex or side 6 with a +1 flex, we may slit the single right leaf and open up to a new side, 4 (see figure 9.6b). From 4, it happens that we can rotate and $1-$flex to 5. From 5, tubulate to 3, and if we again force the tubulation and rotate, we can continue to the three-cycle $5-3-4$. If instead of rotating after forcing the tubulation, the same flexing axis is maintained, we can flex to side 6.

There is, however, a much easier way to turn up the new side. If we flex from 2 to 3, we can tubulate to 5, but if we force this tubulation we cannot open up to 4 with a $1-$flex (although this may be accomplished with a backwards flex). We can, however, open up to 4 without forcing the tubulation but instead after tubulating, rotate the tubulation, (by exchan-ging right for left pats) and open up with a forwards flex to 4. Similarly, after we flex from 4 to 5, we may tubulate to 3, but without forcing, we can rotate the tubulation and open up to side 1. This method is much easier than the method involving forcing the tubulation. The operation just described is called ``tubulating through''. In constructing this type of flexagon in the first place, it is much easier to tubulate say from 3 to 5, rotate the tubulation, and then slit the single leaf in the right hand pat to open up to 4.

Returning to figure 9.6a, if we flex to 5 from 6, we could tubulate to 3. If instead of forcing the tubulation, we were to rotate it and slit the single left leaf, we would open up to a side, 4, which would $2-$flex instead of $1-$flex to 3 (see figure 9.6c). After rotation, we may open up this tubulation by slitting one of two leaves depending whether we wish a $0-$cut flex or a shortcut flex. Then, from this side, we may $1-$flex to 3 (see figure 9.6 d & e). Now let us reconsider the flexagon shown in figure 9.6b. If we flex from side 4 to side 5, we find that after rotation there is a single leaf in the left hand pat. Also, this single leaf is connected to the lowest leaf in the right pat, which means that either leaf may be slit without suppressing another side (call this choice ``A''). Whichever one we slit, we may flex to a new side which then tubulates to side 3. This tubulation may be opened in three ways, (choice ``B'') in the manner we have used for all tubulations. Whichever way we do it, we can flex back to side 3.

This completes a separate improper cycle joined to the original by a tubulation. Choice ``B'' decides which of three improper cycles will be added, and choice ``A'' determines how it will be oriented with respect to the original cycle. If we agree to slit the left most leaf in choice ``B'', slitting the left leaf in choice ``A'' will give the map in figure 9.7a, while slitting the lowest right leaf will produce the map in Figure 9.7b.

This method of adding more cycles attached by tubulations may be extended to include the $G-$flexagon. We may tubulate through from any tubulation which is on the $0-$cut cycle by tubulating, rotating, and slitting the single leaf. After we have completed a series of $1-$flexes (and if nece-ssary, slittings) which leads back to the side across the tubulation, we have a choice of two possible leaves to slit to start the series of flexes which will complete the cycle. The orientation of the new cycle will be determined by this slitting (in symmetrical cycles, such as in tetraflexagons, this di-fference in orientation will appear to be a reversal of vectors - see figure 9.8). When all cycles are completed, the flexagon must be renumbered by folding the unit together and numbering the leaves consecutively down the unit $1-2,\; 2-3,\; \ldots,\; (N-1)-n,\; N-1$.