Compound Faces

Just as we had met heterocyclic flexagons before they were discussed as such, so among the heterocyclic flexagons will find flexagons with compound faces.

A compound face is one in which the two pats are of different face degrees. Now, the face degree of a pat is the sum of the signs for the leaves in that pat, except that alternate pats will have the sign of the face degree reversed. The reason for this latter fact is that the sum of the entire sign sequence is the angle between the perpendiculars on the hinges entering and leaving the folded-together unit (see fig. 13.1). When one pat is unfolded from the unit, as shown, it is turned upside down, so that its face degree becomes negative in respect to that of the other pat. If the face degrees of the two pats $i$ and $j$ are equal, we obtain the followings: ^13.1 $\qquad \qquad \qquad \qquad$ $f_i\equiv D_i\;(\mbox{mod } 360^{\mbox{o}} )=f_j\equiv -D_j\;(\mbox{mod } 360^{\mbox{o}} )\;;\; f_i-f_j = D \equiv 0^{\mbox{o}} \;(\mbox{mod }360^{\mbox{o}} )$

Here, then, is merely a restatement of the rule that the sum of the sign sequence must be congruent to zero, $\mbox{mod } 360^{\mbox{o}}$; either this, or its faces will be compound. Thus the magnitude of $D$ is a measure of how compound a given face is; $D$ is the difference in face degree between the two pats making up any given face of the flexagon.

In our discussion of heterocyclic flexagons, it was said in effect that compound faces would be allowed only so long as $D \equiv 0^{\mbox{o}} (\mbox{mod } 360^{\mbox{o}} )$. The reason for this was said to be that when $D\not\equiv 0^{\mbox{o}}$, binding results during the flexing operation. This is still true. Most compound flexagons (those whose faces are compound) can be operated only with difficulty, if at all. In fact, they must be made of rather pliable material, as with polycyclic flexagons of cycle greater than 4, since rigid leaves would make flexing totally impossible. The reason for this annoying complication is rather easier to observe than to explain. If the hinges entering and leaving the folded-together unit do not coincide, a certain amount of twist (varying with $D$) will be imparted to the flexagon when each unit is folded completely together. When we attempt to fold all the units together simultaneously, this twist builds up and unwinds part of the flexagon before we are finished. If, then, we try to flex one unit at a time, we encounter trouble with the twist given the flexagon by flexing, itself. However, this twist problem is of less magnitude than the preceding one, so that it is practical, in most cases, to strain the flexagon to this extent and flex each unit separately. Either this, or we can eliminate the twist problem entirely by not folding the unit completely together during flexing. In this way, the flexagon will flex all at once, but the pats will be found to interfere with one another, thus binding the flexing operation. In most cases the best procedure is to keep in mind what one is attempting to do, and then simply feel around for the easiest method of attaining the desired structure.

Why do we trouble ourselves with compound flexagons at all, if they do not flex in a simple, respectable manner? It is because there are so many compound flexagons. There are far more compound flexagons than noncompound, of course, but there are also far more compound flexagons than there are heterocyclic flexagons. To find the compound flexagons made up of one type of cycle only, we return to our proof that $D \equiv 0^{\mbox{o}} (\mbox{mod } 360^{\mbox{o}} )$ in flexagons other than heterocyclic flexagons. This proof depends upon the sign sequence substitution of groups of signs having equivalent sums, mod $360^{\mbox{o}}$. The fact is that the substitution process, basic as it is, depends only upon the changes that occur in the number sequence. These alter the flexagon's structure; sign sequence changes alter only the positions of the hinges and the shapes of the leaves. These latter have been shown to have little relevance to the former; the two can actually proceed independently.

Therefore, why can we not use arbitrary signs in the sign sequence corresponding to a given number sequence, so long as hinges are not allowed to intersect? The reasons that we have created rules to prevent doing this are twofold: first, zero-faces might result. This would happen whenever the leaf polygons would be made to coincide and complete cycles would be used. Second, $D$ might not be congruent to zero. Both of these complications have finally been accepted as members of the flexagon family, so that the objections are overruled. With only one necessary sign sequence restriction remaining, the shape of the flexagon can take on unlimited variety.

The main difference between compound and non-compound flexagons, as far as general appearance is concerned, is that compound flexagons do not lie about the center of the flexagon in an even ring of pats. Alternate pats lie closer to the center than the others. The reason for this will be seen by the reader as we progress. Compound flexagons usually require more units than other flexagons. To determine how many units will be used in a specific case, we must find the angle between the hinges entering and leaving each unit. If the face degrees of the two pats are $a$ and $b$, then this angle, $c$, is less than or equal to $360^{\mbox{o}} -a-b$. It may be less than this value if the pats are folded partly together (see fig. 13.2). We may note that if the polygon $ABDCE$ in the fig. 13.2 is concave at $D$ or $E$, a correction of $-180^{\mbox{o}}$ for each concave angle must be added to the expression above.

The number of units used must then be at least $360^{\mbox{o}} -a-b$ and at least 2. Since this number does not contain the constant $D \equiv (b-a)(\mbox{mod } 360^{\mbox{o}} )$, the number of units depends upon the face chosen for the determination. We therefore choose the smallest face degree possible in the flexagon as $a$. Notice that $b$ is congruent to the sum of the signs of the leaves in that pat, while $a$ is congruent to minus the sum of the signs of the leaves in its pat.

In order to give an indication of the variety of compound flexagons, we will arbitrarily limit ourselves in this discussion to flexagons made up of coinciding regular leaves. These flexagons are perhaps the most picturesque of the compound flexagons. To begin, we examine the possible faces for $K=3$. Other than $D\equiv 0^{\mbox{o}}$, we have here only one possibility: $D\equiv 120^{\mbox{o}}$. Then the face degrees will be either $120^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$ or $240^{\mbox{o}} \mbox{--- } 120^{\mbox{o}}$. (Faces one of which is the negative of the other, such as $240^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$ and $120^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$ , are not counted separately here. For this reason, we limit ourselves to just one of the two possible values thus presented for $D$; $D = -120^{\mbox{o}}$ is not considered.) For the $120^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$ face, we need $\frac {360^{\mbox{o}} }{180^{\mbox{o}} -a-b}=6$ units (see fig. 13.3).

These lie in a hexagon, with flaps between the units (see fig. 13.4). Clearly, whenever one pat has face degree $0^{\mbox{o}}$, the flexagon will require twice as many units as the corresponding non-compound flexagon with face degree $\neq 0^{\mbox{o}}$, yet look the same. The $240^{\mbox{o}} \mbox{--- } 120^{\mbox{o}}$ face (fig. 13.5) apparently needs an infinite number of units, since its hinges are parallel.

When this is the case, we merely give the flexagon two units and do not permit it to lie flat. This face then would look like an octahedron with two opposite faces removed.

When $K=4$, we have the possibilities $D=90^{\mbox{o}}$, $D=180^{\mbox{o}}$ and thereby the faces $90^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$, $180^{\mbox{o}} \mbox{--- } 90^{\mbox{o}}$, $180^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$ and $270^{\mbox{o}} \mbox{--- } 90^{\mbox{o}}$. From parallel cases already considered we can eliminate $90^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$, $180^{\mbox{o}} \mbox{--- } 0^{\mbox{o}}$ and $270^{\mbox{o}} \mbox{--- } 90^{\mbox{o}}$. As for $180^{\mbox{o}} \mbox{--- } 90^{\mbox{o}}$, we need $\frac{360^{\mbox{o}} }{360^{\mbox{o}} -270^{\mbox{o}} }= 4$ units, which will be arranged in the hollow square shown in fig. 13.6.

The flexagon $\begin{array}{ccc\vert c} 1 & 3 & 3 & 2 \\ \hline 2 & 2 & 1 & 1 \\ +90^{\mbox{o}} & +90^{\mbox{o}} & +90^{\mbox{o}} & \end{array},$ $G=3$, has three faces of this interesting type, and is the simplest interesting compound flexagon, in its construction. For $K=5$, all cases but $144^{\mbox{o}} \mbox{--- } 72^{\mbox{o}}$ and $216^{\mbox{o}} \mbox{--- } 72^{\mbox{o}}$ cannot lie flat: $\frac {360^{\mbox{o}} }{360^{\mbox{o}} - 216^{\mbox{o}} }=\frac 5 2$ units. But the face $216^{\mbox{o}} \mbox{--- } 72^{\mbox{o}}$ will lie flat, with five units, in the handsome star shape of fig. 13.7. As before, we can make a flexagon of this type with all ``+'' signs, and therefore with all like faces, having $G=3$. In fact, for $K$ equal to any integer $n>3$ the flexagon with the sign sequence $+\frac {360^{\mbox{o}} }{n}$, $+\frac {360^{\mbox{o}} }{n}$, $+\frac {360^{\mbox{o}} }{n}$ will have three like faces the degree of which will be $(n-2)\frac {360^{\mbox{o}} }{n}-\frac {360^{\mbox{o}} }{n}$, and all of which will lie flat. If $n=6$, the flexagon is shown in fig. 13.8. The other noteworthy faces for $K=6$ are $120^{\mbox{o}} \mbox{--- } 60^{\mbox{o}}$, $180^{\mbox{o}} \mbox{--- } 60^{\mbox{o}}$ and $180^{\mbox{o}} \mbox{--- } 120^{\mbox{o}}$. These have the forms shown in fig. 13.9.

The face degree $180^{\mbox{o}} \mbox{--- } 60^{\mbox{o}}$ is that of all the faces of the flexagon with the sign sequence $+60^{\mbox{o}} ,\;+60^{\mbox{o}} ,\;+60^{\mbox{o}} ,\;+60^{\mbox{o}}$, except the cuts across the diagonals of the square map. These faces are of degree $240^{\mbox{o}} \mbox{--- } 120^{\mbox{o}}$, so that they look like octahedrons with 2 opposite faces removed and the corners clipped (fig. 13.10). The compound flexagons already given indicate what will be found when we increase $K$ further.

One very important question that remains unanswered is how we can tell the face degree of a compound flexagon without actually experimenting. This problem is easily dispelled when we remember that $D=b-a$ is a constant, so that a computation of each value of $a$ or of $b$ would give each value of the other. If we are given the sign sequence and its map, we can compute $D$ immediately. To distinguish between the two different face degree values, we can place them on opposite sides of path in the map. To determine the face degrees, we begin by placing the sign sequence terms on the outside of the map, next to their corresponding faces. The remainder of the face degree numbers are computed by the rule that the face degree term on the outside of a path is equal to the sum $(\mbox{mod } 360^{\mbox{o}} )$ of the sign sequence terms lying on that side of the path, taken in the direction indicated by the path vectors. Hence, the term across a path from a given sign sequence term will be $(s-D) \bmod 360^{\mbox{o}}$ where $s$ is the term. This rule is merely an adaptation of the old method of computing face degrees, pointing out that the two possible ways to calculate each face correspond to the two possible values for each unit. It follows inmediately from the fact that those leaves making up faces lying on one side of a given face line in the map make up one pat of this face; those lying on the other side make up the other pat. In computing face degrees, careful track must be kept of the directions of the face vectors. Around the outside edge of the map, these are understood to travel in one direction about the map. Why have we said that the term across a path from a sign is $(s-D) \bmod 360^{\mbox{o}}$, when it is to be equal to the sum of the signs other than $s$? It is because of the direction of the face vectors. Another way of viewing this is that the difference of the two face degrees of each face must be $D$: $s - (s - D) = D$.

Three examples are shown in fig. 13.11: one for a flexagon with re-gular coinciding leaves worked in degrees, one for a flexagon with regular coinciding leaves worked on in terms of a definite $K=3$, and one for a heterocyclic flexagon made from regular leaves.

This method at last gives us a strong tool for working effectively with heterocyclics in which $D\not\equiv 0^{\mbox{o}}$. We can now accurately predict such otherwise confusing data as which faces will be nearly or exactly zero faces and which faces will not lie flat. We can also compute how many units will be required to make any given face lie flat.