Each and every cut is characterized by the presence of a single leaf or subpat in one of the two pats. This single subpat will have a hinge difference across it equal to the order of the cut flex, be it a flex or a tubulation, since the order of a flex is defined as being equal to the value of the hinge difference across any pat involved in the flexing operation. Since in following the cut cycle each subpat is left alone in a pat in the order that the subpat turns up in the plan, the plan for one cycle of an improper flexagon may be constructed by following the cut cycle while noting what order flex is required to reach each face from the previous one. The results may be checked by orienting the leaf polygons and noting exactly what the hinge difference across each single subpat really is. Care must be taken, however, that the orientation is such that a forward flex uses a number 1 hinge, not a hinge.
Let us see how this works with the pentaflexagon pictured in figure 9.9. Let us start at face 14. The first cut flex is between side 1 and side 2. This flex removes all leaves except leaf 1, 2 from the left hand pat. We note that this is a flex, for it uses hinges which are in a 1 position, and further, that the hinge difference across the single leaf 1, 2 is +1. The next cut flex is from 2 to 3. This also is a flex, and we note that leaf 2, 3 has a hinge difference across it of +1. The cut between 3 and 4 is a tubulation with flex value of 2. Leaf 3, 4 has a hinge value of +2. To reach side 5, we must back flex, or flex ( flex). Leaf 4, 5 is 1 (but note that side 5 faces away). From here, we may flex to 1, noting that leaf 5, 1 is +2. From this information we may obtain the number and sign sequences, and then the plan:
+1  +1  +2  1  +2 
(1)  3  (3)  5  5 
2  (2)  4  (4)  1 
It should be noted that before proceeding with the determination of the plan, it was necessary first to decide which way around the cut cycle we were to travel, and second, just how to orient the leaf polygon. The orientation was determined by choosing arbitrarily a forward flex and assigning to it a positive sense. This may be represented on the map as in figure 9.9b. In such a flexagon, any flex represented by a solid vector is a flex. If the path is traversed in reverse, that is, against the vector, a hinge is required. Thus, although traveling from 5 to 4 requires a flex, traveling from 4 to 5 requires a flex or a flex. Here, then, is a method for determining the signs for the hinge differences across all leaves with a +1 or hinge: A direction about the map is designated as +, and one of the cut flexes is assigned a vector in this + direction. The flex cycle is then followed, and vectors are drawn on all lines so that one vector enters and one vector leaves every vertex of the map polygon. Those flex vectors that are of the cut cycle may be given +1 or values depending on whether they point with or against the preassigned + direction. If the values of any tubulations on the cut cycle are known, they may also be assigned vectors in accordance with their + or sense (see figure 9.9b). It should be noted that a flex is congruent to a flex mod. .
Here we have the true meaning of the polygon system. For any given proper flexagon, the orientation of the map polygon is opposite for adjacent cycles, and thus the signs in the plan are reversed. The convenience of turning in a + or direction when traveling around the map polygons works with proper flexagons but not with improper ones, in which signs other than are used.
With the knowledge we now have we may find the plan of any given proper and many of the improper tetraflexagons. By assigning a positive direction and by fixing one vector, all of the other cut flexes may be given a direction for any proper tetraflexagon and for any improper tetraflexagon, in which each cycle has only flexes in common with the neighboring cycles (those flexagons in which the cycles are attached by tubulations will be discussed later). All tubulations in tetraflexagons and pentaflexagons will signify hinge differences of +2 or 2, but in all tetraflexagons these are indistinguishable as far as the plan is concerned. Thus, given a tetraflexagon as shown in figure 9.10a, we assign a + direction (counterclockwise), draw vectors, and then construct the polygon system (see figure 9.10b.). As the polygon system touches each path in turn, we look at the type of flex; if it is a tubulation (flex) we can simply write (+2), whereas if it is a flex we compare the vector with the preassigned positive direction. If the vector goes with the + direction the vector is +1 and if it goes against the + direction the vector is . In improper flexagons with cycles higher than four we have a problem, for we have no way of reading the value of the tubulations in the cut cycle nor can we tell the sign associated with the tubulation at a glance, directly from the map. In order to calculate the value of the tubulations, we must employ a new principle. This principle states that the sum of the hinge differences encountered in flexing from a given point to another given point is constant regardless of the paths followed. To demonstrate this let us look again at the flexagon in figure 9.9. If we are at side 3 with 2 on the back and we want to flex to side 5, there are a number of paths we can use. Let us consider the two paths and . If we flex from 3 to 5 and don't rotate, we find that the hinge difference associated with the flex is +1 and that this hinge difference is arrived at by adding the hinge differences in the leaves which comprise the left pat. It is the hinge difference across these two leaves in which we are interested. This hinge difference, being +1 means that the hinge difference across the leaf 3,4 plus that across 4,5 must be equal to +1. In this case and and the sum is +1. Now if we go to side 5 by the paths , we would expect to first encounter a hinge difference across leaf 3,4 and then a hinge difference across leaf 4,5. Since these are unchanged for a single leaf, we would expect to encounter first a +2 flex and then a flex. The sum of these two flexes is +1, as predicted.
In general, if a shortcut flex between two sides removes all but leaves from one of the pats, then in traversing the cut cycle between these two sides, each one of the leaves must take a turn at being the only leaf in a pat. Since the hinge difference across the entire degree pat is equal to the sum of the hinge differences of each leaf in the pat, this sum must equal the sum of the hinge differences of the operations encountered in the cut cycle.
Now we have a valuable tool with which we can determine the nature of any tubulation we wish to. In a given map such as the one in figure 9.11, we may orient all the flexes and assign a positive direction. All flexes on the cut cycle should be assigned + or  values according to whether their vectors go with or against the positive direction. Each cut greaterthan1flex may be assigned values by finding another path consisting of 1flexes by which the gap across the tubulation may be bridged. In figure 9.11a, side 2 may be reached from side 3 either by the tubulation 32 or by the sequence of 1flexes 3, 1, 2. If this 1flex path goes with the vectors drawn on the map, the hinge difference for each 1flex is +1; if it goes against the vector, the hinge difference is 1. The hinge differences for all the 1flexes are then added up and the sum is equal to the hinge difference across the tubulation. Care must be taken with the direction of traversing the tubulation. In our example, the 32 flex is in a negative direction, and since it is equal to the two +1flexes, 31 and 12, the value of the tubulations is 2. The other tubulations are handled similarly. (see figure 9.11b). Now with all of the values known for the cut cycle, the polygon network can be drawn, and as the network touches each face in turn, the hinge value is recorded as being identified with that particular point.
Example:
To find the plan of the octaflexagon pictured in figure 9.12, we will arbitrarily orient the 1flexes by drawing vectors as in the figure, and we will arbitrarily assign counterclockwise a positive sense. Now we must look at each 0cut 1flex and discover whether it goes with or against this arbitrary positive sense. When the sign of these 1flexes is being determined, the 1flex in question must be referred to the map as a whole; not to the individual cycle of which that 1flex is a part. The 1flex values for this map are as follows:
32  1  1011  +1  
45  +1  1312  1  
67  +1  114  1 
Now we tackle the 0cut tubulations. First let us consider the tubulation from 1 to 2. There are two possible 1flex sequences to travel from 1 to 2: 1, 3, 2 and 1, 8, 5, 4, 7, 6, 2. In the first sequence, the flexing proceeds in the direction of the preassigned vectors, so all 1flexes are positive. There are two such 1flexes, 13 and 32, so since the hinge difference across the tubulation must equal the sum of the hinge differences of the 1flexes encountered, the flex 12 is a 2flex. Furthermore, since the vector 12 is in a counterclockwise direction, the tubulation is a +2flex when traversing the cut cycle in the assigned positive (counterclockwise) sense. In the second sequence , the direction of flexing is against the vectors, so the flexes are flexes. By this scheme, the flex 12 is a flex. This, however, is identical with the results obtained from the first sequence, for a flex is identical with a flex.
All the rest of the 0cut tubulations may be evaluated in the above manner. Following is a chart of the results obtained:
0cut  Shortcut  Hinge  Sense of  0cut 
flex  sequence  value  Tubulation  value 
34  3,2,6,7,4  +4  +  +4 
56  5,4,7,6  3  +  3 
78  7,4,5,8  +3  +  +3 
89  8,1,14,9  +3  +  +3 
109  10,12,13,9  3    3 
1211  12,10,11  +2    2 
1413  9,13  +2    2 
The first column specifies the tubulation under consideration and the direction the flex is to proceed (thus 34 signifies the flex from 3 to 4). The second column designates the sequence of shortcut flexes used to avoid the tubulation. The third column tells the sum of the hinge values of the shortcut flexes. The sum is positive if the flexing proceeds with the preassigned vectors and negative if against them. The fourth column tells whether the direction in which the 0cut flex was effected was with (+) or against () the assigned positive sense. The last column gives the final hinge value for the 0cut flex if flexing were proceeding in a consistently positive direction along the 0cut cycle. Now, with the value of each 0cut flex known, the polygon network may be drawn and followed. As each face is visited in turn, its constant order number and hinge value may be noted down. The plan may be constructed with this information:
 

+ 

+ 

    + 



(1)  3  (3)  5  (5)  7  (7)  1  (13)  13  (11)  11  (9)  7  
2  (2)  4  (4)  6  (6)  8  (14)  14  (12)  (10)  10  (8)  8 
Each operation shown by the map may now be given flex values by this same method. To find the value of a tubulation from any side to any side in a given cycle all we have to do is add up the value of the flexes along a 0cut cycle from to . Thus the value of the tubulation from 1 to 4 in figure 9.11b is . To do this systematically for every path on the map we write all the 0cut values for each side in order. Then we systematically iterate all possible paths on the map. In the case of the flexagon in figure 9.11 these possibilities are 12, 13, 14, 15, 16, 23, 24, 25, 26, 34, 35, 36, 45, 46 and 56. The value of the path 12 is found by looking at the hinge difference between 1 and 2; that of path 13 by adding the hinge difference of 12 to the hinge difference of 23, that of 14 by adding 12, 23, and 34, etc. (see figure 9.13). In the manner above described it is possible to build and predict the properties of any proper or improper flexagon we have so far considered.
Now, at last, it is possible to see why there are two possible ways of attaching two given improper cycles by any two given 0cut tubulations of the same order. Let us go back to the flexagons in figure 9.6b. The right hand cycle may be determined by arbitrarily assigning a + direction and 1flex vector (see figure 9.14a). The tubulation in common with the two cycles is then determined as when going from 3 to 5. This tubulation then determines the two 1flexes of the second cycle which also bridge the gap between 3 and 5. If in traveling across from 3 to 5 the tubulation is , then in traveling from 3 to 4 to 5 the sum of the hinge values for those flexes must be +2, or each single flex must be +1. These flexes may now be given vectors (see figure 9.14b). When the other two sides are added to complete the new cycle, the two 1flexes which bridge the tubulation have a choice as to which position they take (see figure 9.14 c & d). In both cases, the 1flex which remains a 0cut goes in a positive direction, thus satisfying all requirements. In figuring out the plans to improper flexagons with cycles attached to the tubulations, one cycle at a time must be figured out so that the value of the tubulation in common with two cycles is known first and can be used to determine the other cycle. It should be emphasized, however, that two cycles can only be attached by a flex whose absolute value is the same for both cycles. Thus a flex in one cycle of a hexaflexagon cannot be used to attach it to a flex in another (see figure 9.15a).
Example:
To find the plan of the hexaflexagon in figure 9.16, we assign a
positive direction, and arbitrarily orient the 1,2,3,6,5,4, cycle, as
in figure 9.16a. The values for the flexes in this cycle may then
be determined, and from them, the two 0cut tubulations. The value
of the flex from 6 to 1 may be determined as
from the 1flex cycle
6,5,4,1. Now, in the second cycle, the 1flex cycle 6,8,9,1 also
spans the tubulation. The sum of these three flexes must equal
the tubulation value +3. However, either three +1flexes or three
flexes will work since for hexaflexagons a +3flex is the same as
a flex. So, we have a choice for orienting the second cycle.
Let us choose the orientation in figure
9.16b. We can now specify values for
all the 0cut flexes and in particular,
we find that 9 to 14 is a +2flex.
Since 91214 is the only set of two
flexes of the third cycle bridging
this tubulation, they must be +1flexes,
and again we can determine all the 0cut flexes. The number and sign
sequences may now be written down:
+  + 

     


+    +  
(1)  3  (3)  5  (5)  1  (9)  11  (11)  13  (13)  9  (7)  7  (1)  
2  (2)  4  (4)  6  (14)  10  (10)  12  (12)  14  (8)  8  (6)  2 