Complete Structures in the Octet Truss

Brian GoodhueDaniel Koch

Mary SpuchesAndrew ParetJacob Girard

Thomas Dickenson

April 13th, 2010

1 Introduction

1.1 DNA Nanostructures

Self-assembling DNA nanostructures is a new science where DNA is shaped in such

a way that wire-frame structures are created. The DNA is connected into tiles, which

are single vertex constructions with incomplete arms coming off of it. When a group

of arms are assembled together a complete structure is formed. Like said before, this

structure would be a wire-frame construct without solid edges. These DNA nanostruc-

tures are becoming more and more useful in biology and medicine.

Each individual tile, however, is expensive and there is a great risk that the tiles

will not assemble in the desired fashion. Our research involves looking into a generic

structure form, the octet truss, and finding structures that can be formed the most effi-

ciently. In order to find these, we are searching for structures that can be formed using

only one or two types of tiles, or tile types. By minimizing the tile types used, creating

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structures becomes more realistic because there would be less of a chance that arms

would bond in undesirable ways.

We used graph theory, geometry, and algebra to approach solve this problem.

Through solving the math portion of the problem, chemists and biologists will have

an easier approach to creating these DNA nanostructures. There are many possibilities

in what could be constructed, so our research focused on the octet truss and put in spe-

cific constraints we have created for our problem.

1.2 The Octet Truss

The octect truss is made up of three intersecting planes, which all come together at one

point. Each of the three planes has four arms, all at 90 degree angles from each other.

In order to describe the tile types, each arm in the octect truss must have a consistent

label. In this, we name each of the planes the alpha plane, the beta plane, and the

gamma plane. The four arms in the alpha plane then becomeα1, α2, α3, andα4. The

same happens to the beta plane and the gamma plane. Generally when describing tile

types, we will assume that one arm is theα1 arm. From there, we will describe the

best lexicographical form of the arms, starting with the alpha plane, and then moving

to beta and gamma planes if necessary.

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Arm X Vector Y Vector Z Vectorα1 0 1 0α2 -1 0 0α3 0 -1 0α4 -1 0 0β1 -1/2 1/2 -

√2/2

β2 -1/2 1/2√

2/2β3 1/2 -1/2

√2/2

β4 1/2 -1/2 -√

2/2γ1 1/2 1/2 -

√2/2

γ2 1/2 1/2√

2/2γ3 -1/2 -1/2

√2/2

γ4 -1/2 -1/2 -√

2/2

1.3 Summary of Design Constraints

In order to make sure that our results would be pertinent, we needed to create specific

design constraints. These constraints solidified our project and made sure our results

were consistent with each other.

1. Arms are straight and rigid.

2. The positions of the arms are fixed.

3. Arms are all of unit length

4. The arms do not experience twist strain.

5. Each arm is of unit length.

6. No molecule has more than twelve arms or less than two arms.

7. Final DNA structures must be complete.

8. No design may allow structures smaller that the target structure to form.

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2 Preliminary Work

2.1 Notation

Certain notation must be assigned in order to keep things straight and organized. Since

each cohesive end has a specific layout and can only connect to one possible arm type,

one arm will be denoted as ”a” and its complement as ” ˆa”. If there comes a need for

another, second type of arm ending, then it will be denoted as ”b” and its complement,

” b”.

2.2 Orientation

When two tiles are combined, it is important that we are able to describe the orientation

in which they connect. For this, we want to calculate the bond angle between the two

tiles.

For this, we say we are calculating the bond angle for some set (T1,ε1), (T2,ε2)

whereT1, T2 are two different tiles andεi is an element of [α1,...,α4,β1,...,β4,γ1,...,γ4]

for i=1,2. The set of allεi represent all of the possible arm positions on the octet truss.

We first take the two tiles,T1, T2, and pick an edge from each tile that bond together.

For instance arm c from tile 1 and arm ˆc from tile 2 will be a set of bonding arms.

Them choose the lexicographical minimal edge on each tile, referred to asε1 andε2.

This is the edge that comes first in the setεi . The only constraint here is thatε1 and

ε2 are not antipodal toc andc. This means that the two edges (c,ε1) and (c, ε2) do not

form a 180 degree angle between them. The final step is to look at the two planes that

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are formed by (c,ε1) and (c,ε2). We calculate the angle between those two planes and

call it the bond angle. Note that connecting two different original edges for teh same

set of tiles may give a different bond edge.

2.3 Convex Theorem

To help refine our work and make it more time efficient, we created a theorem that

would help lower the number of possible, usable tile types. This theorem would elim-

inate the tiles that would never be able to create complete structures on their own and

allows us to focus on only the tiles that we could potentially use.

Theorem 1. If G is a complete complex constructed from rigid tiles types, then at least

one tile type must have a geometric configuration such that the convex hull Hv formed

by the vertex v and the end points of the half edges has v as a corner point.

Proof. Consider the convex hullHc of all the vertices of G. Since G is a complete

complex, with straight edges, G⊆ Hc. This follows because if any edge of G extends

beyondHc, since the edges are straight, the endpoint vertex of e would lie outsideHc,

a contradiction. Now consider a corner ofHc. This must be a vertex of G. Since G⊆

Hc, the convex hullHc ⊆ Hc. Since c is a corner ofHc, it follows that c is a corner of

Hc.

Corollary 2. Corollary: If we wish to build a complete complex out of tile types all

having the same geometric configuration, we need only consider geometric configura-

tions where the vertex is a corner of the convex hull Hv.

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3 Tile Types

In order to attack this problem, we decided to create a list of the possible tile types.

By finding all the fundamental building blocks, we could look at each one individually

and see what could be created with it. For this, a computer program was created.

4 Generating n-armed tiles

4.1 Introduction

In order to ensure that in the building of DNA nanostructures each and every structures

is found, it is first necessary to find all lexicographically minimal tiles. Lexicographi-

cally minimal tiles refer to the tiles which are the canonical representatives of group of

tiles which are isomorphic based upon rotational symmetries with the lowest ordering

of arms. Prior to being able to achieve that first we must generate every singlen-armed

tile in order to sort them into their lexicographically minimal orders. The simplest

way to achieve this is to generate each and every tile using a computer program. The

way that this problem was approached was to solve the Hamming distance problem for

n-armed tiles.

4.2 Hamming Distance Problem

The Hamming distance between two strings is the number of corresponding positions

where the strings differ. Given that tiles can be represented as bitstrings of length 12,

the problem of generating alln-armed tiles is the same as the problem of finding all

length 12 bitstrings with a Hamming distance ofn from the length 12 string of all 0s

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(representing a 0-armed tile, with each 0 representing a location of an arm; the firstα1,

secondα2, thirdα3, fourthα4, fifth β1, etc.). This problem can be solved recursively by

finding all strings with a Hamming distance of 1 from some starting position (initialized

to the length 12 string of all 0s), with several additional constraints. Bits may never

be reset, and only bits to the right of already set bits may be newly set. Repeating this

process on each string in the set of output strings, will eventually output the set of all

strings with a Hamming distance of n from the length 12 string of all 0s, and thus the

set of alln-armed tile configurations.

4.3 Results

This process yields the set of alln-armed tiles. In this case the problem requires that

the tiles be grouped according to their rotational symmetries and find the canonical

(lexicographically minimal) representative of each group.

5 Using sets of pairwise arm angles to group tiles

5.1 Using the Lookup Table

In order to reduce the number of rotational comparisons that must be performed to

determine rotational symmetry, we first group the tiles according to the set of angles

between individual arm-pairings on the tile. To facilitate this process, a 12 x 12 lookup

table of all arm pairs and the angles between them was created.

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5.2 Lookup Table

The angle arm pairings are all shown in the table below.

5.3 Pairwise arm angle algorithm

In order to find determine isomorphisms using angles the algorithm requires two tiles

to be inputted at a time. It will then compare the arms of tiles and check to see if the

pairings have matching angles. If so it outputs true and groups them together, otherwise

it outputs false and leaves them in separate groups.

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6 Using Rotations to group tiles

6.1 Purpose of Rotations

Now we have a partially grouped list of tiles, however there are still some distinct tiles

within each group that have not been distinguished from each other. Notably, mirrored

tiles will have the same set of angles between arm-pairings without being rotationally

isomorphic. Therefore, rotational comparisons must be performed as a sort of tie-

breaker operation. The partial groupings served merely to reduce the number of such

comparisons which must be performed.

6.2 Labeling of Rotations

Since rotations are a necessary operation we need a standard for labeling them. Ro-

tations are labeled by a vertex or face on the cuboctahedron. The axis of rotation is

always through the center of the aforementioned face or vertex and the centroid of the

cuboctahedron. The labels vary based on the degree of the rotation. There are three

primary types of rotations, the first is a 3 arm face that forms a triangle on the cuboc-

tahedron and is rotated by multiples of 120◦ e.g. α1β1γ1 @ 120◦. The second type

of rotation is a 4 arm face that forms a square on the cuboctahedron and is rotated by

multiples of 90◦ e.g.β2β3γ2γ3 @ 90◦. The final type is a 180◦ rotation that is centered

on a vertex of the cuboctahedron, e.g.α1 @ 180◦. In addition it should be noted that as

a standard all rotations are to be made counterclockwise. In addition it should be noted

that the rotations compose a pure rotational subgroup of the octahedral group, with or-

der 24. Inversions do not function within the scope of our research as a molecule can

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not be easily inverted in solution, but is free to rotate.

6.3 Generators and Axes

Once again there is an issue of overhead on a program that has to generate and perform

thousands of operations so it is best to simplify the processes where possible. This is

the central concept behind the generators for the axes that were designed. Each of the

24 axes of rotation can be simulated with a permutation of rotations around 3 standard

axes (the generator rotations).

1.α1β1γ1 @ 120◦ or (α1β1γ1)

2. α1 @ 180◦ or (α1)

3. β2β3γ2γ3 @ 90◦ or (β2β3γ2γ3)

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Figure 1: Caption for axisgen

6.4 Axis/Generator Correspondence

Figure 1 demonstrates the permutations of the generators required to make any member

of the rotational subgroup of the octahedral group.

6.5 Additional Rotations

After having checked the results we found that there is a strange case where four addi-

tional rotations outside of the order 24 pure rotational subgroup of the octahedral group

were necessary in order to completely group each and every tile based upon its correct

lex-minimal form. These four rotations only apply in the case where a tile lies entirely

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in one of the four hexagonal planes which intersect the alpha plane, and consists of a

rotation of 60◦ or 120◦ aroundα1 or α2.

7 Exhaustive Case Finder

7.1 Introduction

The Exhaustive Case Finder is the name of our overall program and having explained

the processes taking place in the code; this next section will give a brief pseudo code

and will give our results for each group ofn-armed tiles, namely how many there are

and each lex-minimal representative tile.

7.2 Pseudo Code

In order to accomplish the goals that we set out to accomplish the code is comprised of

4 classes. These four classes each play a specific role in the creation of the result and

are built around each other.

Combo Input: Some integern in the open interval [0, 12]

Generates every possiblen-armed tile. Makes calls to CuboctahedronLUT and Cuboc-

tahedronSymmetry in order to group tiles. Sorts out lexicographically minimal tiles

from the groups.

Output: List of lexicographically minimal tiles.

CuboctahedronLUT

Input: List ofn-armed tiles.

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Lookup table used to group the tiles generated by Combo based on the angles between

arm-pairings. Makes a call to the AngleWrapper class in order to provide hashing and

equality functions for sets of angles.

Output: Hashtable of tiles grouped according to the angles between arm-pairings.

CuboctahedronSymmetry

Input: Two tiles to be compared rotationally.

This performs the generator rotations in order to further group distinct tiles not distin-

guished between by the CuboctahedronLUT class.

Output: Whether or not the two tiles are rotationally isomorphic, and optionally, infor-

mation on the orbit and stabilizers of the second tile.

AngleWrapper

Input: (none)

Simply a wrapper class used for the CuboctahedronLUT class. Wraps arrays repre-

senting angles between arm-pairs to override Java’s built-in array comparison behavior

(performs a shallow element comparison rather than a simple object-reference compar-

ison)

Output: (none)

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Figure 2: Caption for programflow

7.3 Correctness

We have shown experimentally, through use of the orbits and stabilizers theorem (need

citation??), that the sum of order of the orbits of then-armed lex-minimal tiles is equal

to 12 choosen. This shows that our rotations are correct.

8 Results

The results of the program were used in the rest of the paper. The complete results are

listed at the end of the paper.

9 Two Armed Tile Types Configurations

There are only four possible tile types that use two arms in the octet truss that do not

repeat themselves. These tiles are of 60, 90, 120, and 180 degrees. All other possible

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two arm configurations are just different rotations of these in the octet truss.

9.1 Structures that can be made in a plane

To find which types of structures can be made, we will use the Interior Angle Formula.

The Interior Angle Formula states:

(n−2)180= A (1)

in which n is the number of sides andA is the sum of the angles. IfA is the sum of the

angles, thenA/n is the average measurement, or the measure of each individual angle in

the polygon. For us,A/n=a. From here,A=anso therefore we will replace our formula

with

(n−2)∗180= a∗n. (2)

9.2 60 degrees

The lowest possible angle we have is 60 degrees, so we start with that.

Leta= 60;(n−2)180= 60n (3)

180n−360= 60n (4)

n = 3 (5)

With algebra we find thatn=3. So if the angle is 60 degrees and a polygon is being

formed in a plane with only that tile, a triangle will be formed with 3 sides.

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9.3 90 and 120 degrees

After the 60 degree angle, we can move on to 90 and 120 degree angle. Using the

algebra from before, and a little bit of common knowledge, we find that a 90 degree

angle will only be able to form a 4 sided polygon. This, of course, is a square. The 120

degree angle will form a 6 sided hexagon.

9.4 180 degrees

For a=180 we begin to have a problem.

Leta= 180 (6)

(n−2)∗180= 180∗n (7)

(180∗n)− (180∗n) = 360 (8)

0 = 360 (9)

By using the formula, we end up with 0=360. As a result of this conflict, it becomes

apparent that no polygon can be formed using 180 degree angles. Although it has

already been proven algebraically, it is clear that it is impossible to form a complete

structure using only 180 degree angles. The angles would never wrap around to the

beginning. This concept is important later on with the three armed tiles that are planar,

because disproving that they can form a complete structure is based on this fact as well.

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9.5 Results

Through the use of the Interior Angle Formula, we have found that angles 60, 90, and

120 form the structures of a triangle, square, and hexagon respectively. This is proven

because the angles and the number of sides are biconditional and only have one partner.

A certain angle must create a certain polygon and the Interior Angle Formula showed

us which structure each tile type will create

9.6 Structures outside of the plane

Complete structures with only a single tile type with only 2 arms cannot be made in

three dimensions. Once a structure with a single tile type has left the plane of origin,

then a spiral will be created. The spiral will continue to get further and further away

from the origin and never reconnect. Since it will never reconnect, there will never be

a complete structure formed. As a result, there are no three dimensional structures that

can be made with a two-armed tile type.

9.7 Conclusion

There are only three possible structures to be formed in the octet-truss using one, 2-

armed tile type. Those structures are a triangle, a square, and a hexagon. This was

proven through use of the Interior Angle Theorem. This was used because the angles

were given and biconditional to what sort of structures could be formed using them.

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Table 1:There are ten possible unique configurations of three arms incident to a singlevertex in the octet truss. Note that the armα1 is assumed to be one of the arms in eachof the configurations.

1. α2 β1

2. α2 β2

3. α4 β1

4. α4 β3

5. α2 α3

6. α3 β1

7. β1 γ1

8. β1 γ3

9. β3 γ4

10. β1 γ2

10 Three Armed Tile Type Configurations

10.1 Tile Types

There are 10 different types of three arm configurations in the octet-truss. These have

been proven through by use of a computer program. The tile types are listed in Table 1

with all tiles assumed to haveα1 as its first arm.

All other tile types are just rotations of these that are similar because they have the

same measured angles.

10.2 Parity Theorem

In order for a graph to be complete, there must be the same number of ”sticky” ends and

complement ends. This makes sense because if there are more ”a” arms then ”a” arms,

then they will not all be able to connect to something. This is all explained through the

parity theorem.

As a result of this, we will never be able to build a complete structure using only

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one, odd-armed tile type. With an odd number of arms, there would always be an

imbalance between the sticky arms and complement arms. We therefore must use two

tile types when creating three armed tiles. Both tiles types will be the same geometric

orientation, however, there will be two different layouts for sticky and complement

arms.

10.3 Constructions

Since there are only ten tile types, an it is possible to go through each tile individually

in order to see if a complete structure can be created. There are only a certain number of

possible ’sticky’ and ’cohesive’ ends of the arms of tiles that can match and be creating

a list of these, an exhaustive proof can be done by tile types to arm-end pairs. This

will give a more concrete number of potential structures, provide proof of which tile

types can make constructions, and provide the proof that certain tile types cannot form

complete structures.

10.4 Tile Type Partners

In order to do this cross-section technique, there needs to be a list of all the possible

tile type partnerships. We start with the assumption that all of the ends are ”a”’s, that

means an ”a” or its partner ”a”. The formula used to find the possible partners is :

r1A11+ r2A21 = r1A12+ r2A22 (10)

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In this equation,r1 represents the quantity of the base tile we start with andr2 repre-

sents the quantity of its partner tile.A11 stands for the number of ’a’ ends on the first

tile andA12 stands for the number of complimentary ends. Likewise,A21 stands for the

number of ’a’ ends on the second tile andA22 stands for the number of complimentary

ends on the second tile. When this equation is solved with a ratio forr1 to r2 is created,

then as long as the ratio is maintained, there will always be exactly the same amount of

’a’ ends and compliment ˆa ends so that a complete structure can be formed.

10.5 Tiles with only ’a’ ends

The first tile to start with is (a3,0), in which all three ends are ”a”. As a result,A11=3

andA12. We plug these values into the equation and solve for the possible values of the

other variables:

A11 = 3,A12 = 0 (11)

r1∗3+ r2∗A21 = r1∗0+ r2∗A22 (12)

3∗ r1 = r2∗A22− r2∗A21 (13)

3∗ r1 = r2∗ (A22−A21) (14)

I f r1 = 1 and r2 = 1 , then A21 = 0 and A22 = 3 (15)

I f r1 = 1 and r2 = 3 , then A21 = 1 and A22 = 2 (16)

Using this same formula and similar steps, the next job was to solve for (a2, ˆa). With

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this initial tile type, the only possible partners were (0, ˆa3) and (a, ˆa2). These results are

somewhat similar to the results from (a3,0). When the second tile is (0, ˆa3), it is found

that r2=3*r1. When the second tile is (a, ˆa2), r1=r2, which means that the two tiles must

have the exact same quantities.

(a3,0) and (a2, ˆa) are the only two possible tile types you can start with that have

only ’a’ ends. This is because any other configurations, like (a, ˆa2) or (0,a3) would just

repeat the already discovered ones. With this, we can move onto ’a’s and ’b’s.

10.6 Tiles with both ’a’ ends and ’b’ ends

Since there can be both ’a’ ends and ’b’ ends, we need to be able to find all the pairs

with both ’a’ and ’b’ in them. To do this we use the original formula from before:

r1A11+r2A21=r1A12+r2A21. This will solve for the ’a’ aspect of the problem; however,

we also need to solve for the ’b’ aspect with the formula r1B11+r2B21=r1B12+r2B22.

The first tile type (a, ˆa,b,b) we use is (a2,0,b,0). When plugged into the two formulas

there is only one result that comes out to work for both the ’a’ and ’b’ aspects of the

problem. The only tile to use is (0, ˆa2, 0,b). In this pairing r1=r1, so there must be the

same quantity of tile types r1 and r2 in order to form a complete structure and for all

ends to be connected.

The other possible first tile is (a, ˆa,b,0). From here, there are three possible comple-

ment tile types. These are (0,0,0,b3), (0,0,b,b2), and (a, ˆa,0,b). There are so many po-

tential partners mainly because the first equation ends with A11=A12. This means that

simply as long as there are the same number of ’a’ and ’a*’ ends in the complement,

the ’a’ portion of the problem is satisfied. From there, we can do any combination as

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long as B22¿B21.When the complement is (0,0,0,b3), r1=3r2. For all other cases r1=r2.

These are the only pairs using only ’a’ and ’b’ tile types because any other combi-

nation would be just repeating a previously found configuration.

10.7 Tile with ’a’, ’b’, and ’c’ ends

Lastly, we need to look at the tile types that have three completely different ends, ’a’,

’b’, and ’c’. For this, we need to add a third formula into the mix: r1C11+r2C21=r1C12+r2C22.

The tile type (a, ˆa,b,b,c,c) for r1 to be (a,0,b,0,c,0). With the formula, we find the only

possible partner to be (0, ˆa,0,b,0,c). This seems the rational result because for each

’a’,’b’, and ’c’, there must be a ’ ˆa’, ’ b’, and ’c’ respectively.

10.8 Crossing configurations with tile types

Now that there is a list of all the possible configurations and all the possible tile types

partners, we now know all possible tile types that can be made with three arms. From

here, table can be created. On the left will be the possible configurations and on the

top the possible tile type partners. By trying each configuration with partner combi-

nation, we can see all the possible structures that can be created and get rid of all the

configurations and partner combinations that cannot make anything.

10.9 Conclusions on Three-Armed Tile Types

The table is read by finding the tile type on the left and looking across the top to find

a arm pairing. By finding where they cross, you can read the result. An ’x’ shows that

no complete structure was made. An ’o’ shows that a complete structure was made,

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r1 3a 3a 2a,a 2a,a 2a, b a, a, b a, a, b a, a, b a, b, cr2 3a a, 2a 3a a, 2a 2a, b a, a, b 2b, b 3b a, b, c

α2, β1 x x x x x x x x xα2, β2 x x x x x x x x xα4, β1 x x x x x x x x xα4, β3 TrOct TrOct TrOct TrOct TrOct TrOct TrOct TrOct TrOctα2, α3 P P P P P P P P Pα3, β1 P P P P P P P P Pβ 1, γ 1 o Tetra Tetra Tetra o o o Tetra oβ 3, γ 3 o TrTetra TrTetra TrTetra TrTetra TrTetra TrTetra TrTetra oβ 3, γ 4 P P P P P P P P Pβ 1, γ 2 x x x x x x x x x

but the arm pairing did not fit the structure. A ’P’ shows that the tile was either planar

or non-convex, and therefore was eliminated. As for the structures that were actually

formed, ’TrOct’ stands for truncated octahedron, ’Tetra’ stands for tetrahedron, and

’TrTetra’ stands for truncated tetrahedron.

From these results we can determine that there are only three possible structures that

can be made using one tile type with two different arm pairings. The three possible

structures are the truncated octahedron, the tetrahedron, and the truncated tetrahedron.

These structures are made using tiles with the tile types [a4,b3],[b1,g1], and [b3,g3],

assuming that the first arm is a1. All other tile types with three arms can not complete

a structure in our problem restraints and are not needed.

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Convex Non-Convexα2,β1,γ4 α2,γ3,γ4α2,β2,γ3 α3,β1,β3α2,β1,γ1 α2,α3,β1α2,β2,γ1 α2,α3,β2α2,β1,γ2 α2,α3,β3α2,β2,γ2 α2,α3,β4α2,β1,β2 α3,β1,γ1α2,β1,β4 α3,β1,γ4α2,γ1,γ2 α2,β1,β3

α2,β2,β4α2,α3,α4α2,β1,γ3α2,β2,γ4α3,β1,γ2α3,β1,γ3α3,β2,γ4α2,γ1,γ3

11 Four-Armed Configurations

11.1 Tile Types

Through the a program, we were able to determine that there are 26 possible, non-

repeating four arm tile types in the octect truss; however, not all of them are useful

to us. To lower the number of tiles for the exhaustive proof, we can quickly get rid

of them using the Convex Theorem from earlier in the paper. This means that we can

get rid of all the tiles that contain a 180 degree angle or do not come to a point. The

following table displays which tiles are convex (useful) and which are non-convex, all

assuming a1 is used.

As a result of 17 of the tile types being non-convex, we are able to narrow our

search down to 9 tile types. From here we can go through individually and see which

ones can make complete structures.

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11.2 Tile Type Partners

The next step to the proof is figuring out the possible tile type partners for the arms.

This, however, is made easy because each vertex has an even number of arms. With

an even number of arms, the minimum number of possible tile types in a complete

structure becomes one, instead of requiring two like in the three armed tile types. The

only two possible arm pairings are now (a,a, ˆa,a) and (a, ˆa,b,b) because each arm must

have have the same number of sticky and cohesive ends so that all arms have pairs.

(a,a,b,b) is just a more specific way to write (a,a, ˆa,a), so we will just assume that all

tile types have the tile type partners (a,a, ˆa,a).

11.3 Structures and Conclusion

Only two possible tile types are usable out of the nine convex ones. The two usable ones

are (α1,α2,β1,β4) and (α1,α2,γ1,γ2). These tiles, when assembled into a complete

structure, from a cuboctahedron and an octahedron respectively. The rest of the four

arm tiles do not form complete structures and can therefore be discarded.

12 Greater Armed Tiles

12.1 Five Arms

Looking at the five armed tile types, we only find five convex tiles. With such a low

number, we were able to spend serious time looking at each tile individually; however,

none of these tiles were able to build a complete structure. The five armed tiles were in-

teresting, however, in building some incomplete structures that were essentially a series

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of smaller constructions put together. One example of this was a series of tetrahedrons

connected together to form an unending structure.

12.2 Six Arms

After looking at all the possible six armed tile, only one of them was convex. This

tile was (α1,α2,β1,β2,γ1,γ2). We were unable to find any complete structures using

only this tile. We spent sometime outside the problem set and combined this six arm

tile with other, lesser armed tiles. With these combinations, we were able to find some

interesting constructions.

12.3 Seven and Greater Armed Tiles

Adding a seventh arm to a tile will always mean there is at least one angle of pi radians.

This straight line will mean that none of the tiles with seven or greater arms will be

convex. As a result, we know for a fact that none of them can produce a complete

structure. This conclusion eliminates all seven or greater armed tiles from our problem

set.

13 Final Results

With the use of exhaustive proof methods and the convex theorem, we were able to

look at each case of possible arms in the octet truss. We found all of the possible

constructions using only a single tile type and have closed this specific problem set.

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13.1 Results of Code

The following lists provide the output of the code.

There is 1 unique tile type with 1 arm

α1,

There are 4 unique tile types with 2 arms

α1, β3,

α1, β1,

α1, α3,

α1, α2,

There are 10 unique tile types with 3 arms

α1, α2, α3,

α1, α3, β1,

α1, α2, β1,

α1, α2, β2,

α1, β1, γ1,

α1, α2, γ3,

α1, β1, γ2,

α1, β1, γ3,

α1, β3, γ4,

α1, α2, γ1,

27

There are 26 unique tile types with 4 arms

α1, α2, γ3, γ4,

α1, α3, β1, β3,

α1, α2, α3, β1,

α1, α2, α3, β2,

α1, α2, α3, β3,

α1, α2, α3, β4,

α1, α2, β1, γ4,

α1, α2, β2, γ3,

α1, α3, β1, γ1,

α1, α3, β1, γ4,

α1, α2, β1, γ1,

α1, α2, β2, γ2,

α1, α2, β1, γ2,

α1, α2, β2, γ1,

α1, α2, β1, β3,

α1, α2, β2, β4,

α1, α2, γ1, γ3,

α1, α2, β1, β2,

α1, α2, β1, β4,

α1, α2, γ1, γ2,

α1, α2, α3, α4,

α1, α2, β1, γ3,

28

α1, α2, β2, γ4,

α1, α3, β1, γ2,

α1, α3, β1, γ3,

α1, α3, β2, γ4,

There are 37 unique tile types with 5 arms

α1, α2, β1, β3, γ1,

α1, α2, β2, β4, γ1,

α1, α3, β1, β3, γ1,

α1, α3, β1, β3, γ2,

α1, α2, β1, β4, γ2,

α1, α2, α3, β1, γ3,

α1, α2, α3, β2, γ4,

α1, α2, β1, β2, γ3,

α1, α2, β1, β2, γ4,

α1, α2, α3, β1, β3,

α1, α2, α3, β2, β4,

α1, α2, β1, γ1, γ3,

α1, α2, β1, γ2, γ4,

α1, α2, β2, γ1, γ3,

α1, α2, β2, γ2, γ4,

α1, α2, α3, β3, β4,

α1, α2, α3, β1, β4,

29

α1, α2, α3, β2, β3,

α1, α2, β1, β4, γ3,

α1, α2, β1, γ3, γ4,

α1, α2, β2, γ3, γ4,

α1, α2, α3, β3, γ2,

α1, α2, β1, β4, γ1,

α1, α2, β1, γ1, γ2,

α1, α2, β1, β3, γ3,

α1, α2, β2, β4, γ3,

α1, α2, α3, α4, β1,

α1, α2, β1, β2, γ1,

α1, α2, β1, β2, γ2,

α1, α2, α3, β1, γ1,

α1, α2, α3, β2, γ2,

α1, α2, α3, β3, γ1,

α1, α2, α3, β4, γ2,

α1, α2, α3, β1, γ2,

α1, α2, α3, β1, γ4,

α1, α2, α3, β2, γ1,

α1, α2, α3, β1, β2,

There are 48 unique tile types with 6 arms

α1, α2, β1, β2, γ1, γ3,

30

α1, α2, β1, β2, γ2, γ4,

α1, α2, β1, β4, γ1, γ3,

α1, α2, β1, β4, γ2, γ4,

α1, α2, β1, β2, γ3, γ4,

α1, α2, α3, α4, β1, γ2,

α1, α2, α3, α4, β1, γ3,

α1, α2, α3, β1, β2, β3,

α1, α2, α3, β1, β2, β4,

α1, α2, α3, β1, β3, γ3,

α1, α2, α3, β1, β3, γ4,

α1, α2, α3, β1, γ2, γ4,

α1, α2, α3, β2, β4, γ4,

α1, α2, α3, α4, β1, β3,

α1, α2, β1, β3, γ1, γ3,

α1, α2, β2, β4, γ1, γ3,

α1, α2, α3, β1, β3, γ1,

α1, α2, α3, β1, β3, γ2,

α1, α2, α3, β2, β4, γ1,

α1, α2, α3, β2, β4, γ2,

α1, α2, α3, β1, β4, γ4,

α1, α2, α3, β1, γ1, γ2,

α1, α2, α3, β1, γ1, γ4,

α1, α2, α3, β2, γ1, γ2,

31

α1, α2, β1, β2, γ1, γ2,

α1, α3, β1, β3, γ1, γ3,

α1, α3, β1, β3, γ2, γ4,

α1, α2, α3, α4, β1, β2,

α1, α2, α3, α4, β1, β4,

α1, α2, β1, β3, γ3, γ4,

α1, α2, β2, β4, γ3, γ4,

α1, α2, α3, α4, β1, γ1,

α1, α2, α3, β1, β3, β4,

α1, α2, β1, β3, γ1, γ2,

α1, α2, α3, β1, β4, γ1,

α1, α2, α3, β2, β3, γ2,

α1, α2, α3, β3, β4, γ1,

α1, α2, α3, β3, β4, γ2,

α1, α2, β1, β2, γ1, γ4,

α1, α2, β1, β2, γ2, γ3,

α1, α2, α3, β1, β2, γ3,

α1, α2, α3, β1, β2, γ4,

α1, α2, α3, β1, β4, γ3,

α1, α2, α3, β2, β3, γ4,

α1, α2, α3, β1, β2, γ1,

α1, α2, α3, β1, β2, γ2,

α1, α2, α3, β1, β4, γ2,

32

α1, α2, α3, β2, β3, γ1,

There are 38 unique tile types with 7 arms

α1, α2, α3, α4, β1, β4, γ1,

α1, α2, α3, α4, β1, β2, γ1,

α1, α2, α3, α4, β1, β2, γ2,

α1, α2, α3, β1, β2, γ1, γ2,

α1, α2, α3, β1, β2, β3, γ4,

α1, α2, α3, β1, β2, β4, γ3,

α1, α2, α3, β1, β3, γ1, γ2,

α1, α2, α3, β2, β4, γ1, γ2,

α1, α2, α3, β1, β4, γ1, γ2,

α1, α2, α3, β2, β3, γ1, γ2,

α1, α2, α3, β1, β4, γ1, γ4,

α1, α2, α3, β3, β4, γ1, γ2,

α1, α2, α3, β1, β2, β3, γ1,

α1, α2, α3, β1, β2, β4, γ2,

α1, α2, α3, β1, β3, β4, γ3,

α1, α2, α3, β1, β3, γ1, γ3,

α1, α2, α3, β1, β3, γ2, γ4,

α1, α2, α3, β2, β4, γ2, γ4,

α1, α2, α3, α4, β1, β3, γ1,

α1, α2, α3, α4, β1, β3, γ3,

33

α1, α2, α3, β1, β2, β3, γ2,

α1, α2, α3, β1, β2, β4, γ1,

α1, α2, α3, β1, β3, β4, γ2,

α1, α2, α3, β1, β3, β4, γ4,

α1, α2, α3, α4, β1, β2, β3,

α1, α2, α3, α4, β1, β4, γ2,

α1, α2, α3, β1, β2, γ3, γ4,

α1, α2, α3, β1, β4, γ2, γ3,

α1, α2, α3, β2, β3, γ1, γ4,

α1, α2, α3, β1, β3, β4, γ1,

α1, α2, α3, β1, β3, γ1, γ4,

α1, α2, α3, β1, β3, γ2, γ3,

α1, α2, α3, β1, β4, γ2, γ4,

α1, α2, α3, β2, β3, γ2, γ4,

α1, α2, α3, β1, β2, γ1, γ3,

α1, α2, α3, β1, β2, γ2, γ4,

α1, α2, α3, β1, β2, γ1, γ4,

α1, α2, α3, β1, β2, γ2, γ3,

There are 27 unique tile types with 8 arms

α1, α2, α3, β1, β3, β4, γ1, γ3,

α1, α2, α3, β1, β3, β4, γ2, γ4,

α1, α2, α3, β1, β2, β3, γ1, γ4,

34

α1, α2, α3, β1, β2, β4, γ2, γ3,

α1, α2, α3, α4, β1, β2, β3, β4,

α1, α2, α3, β1, β2, β3, γ1, γ3,

α1, α2, α3, β1, β2, β3, γ2, γ4,

α1, α2, α3, β1, β2, β4, γ1, γ3,

α1, α2, α3, β1, β2, β4, γ2, γ4,

α1, α2, α3, α4, β1, β2, γ1, γ2,

α1, α2, α3, α4, β1, β2, γ1, γ4,

α1, α2, α3, α4, β1, β2, β3, γ2,

α1, α2, α3, α4, β1, β2, β3, γ3,

α1, α2, α3, α4, β1, β4, γ1, γ4,

α1, α2, α3, β1, β2, γ1, γ2, γ3,

α1, α2, α3, β1, β2, γ1, γ2, γ4,

α1, α2, α3, β1, β3, β4, γ1, γ2,

α1, α2, α3, β1, β3, β4, γ1, γ4,

α1, α2, α3, α4, β1, β3, γ1, γ3,

α1, α2, α3, α4, β1, β2, γ1, γ3,

α1, α2, α3, α4, β1, β2, γ2, γ4,

α1, α2, α3, α4, β1, β3, γ1, γ4,

α1, α2, α3, α4, β1, β2, β3, γ1,

α1, α2, α3, α4, β1, β2, β3, γ4,

α1, α2, α3, α4, β1, β4, γ2, γ3,

α1, α2, α3, β1, β2, β3, γ1, γ2,

35

α1, α2, α3, β1, β2, β4, γ1, γ2,

There are 13 unique tile types with 9 arms

α1, α2, α3, β1, β2, β3, γ1, γ2, γ4,

α1, α2, α3, β1, β2, β3, γ1, γ3, γ4,

α1, α2, α3, β1, β2, β4, γ2, γ3, γ4,

α1, α2, α3, β1, β3, β4, γ1, γ2, γ4,

α1, α2, α3, α4, β1, β2, β3, β4, γ1,

α1, α2, α3, α4, β1, β2, β3, γ1, γ3,

α1, α2, α3, α4, β1, β2, β3, γ2, γ4,

α1, α2, α3, α4, β1, β2, β3, γ2, γ3,

α1, α2, α3, α4, β1, β2, β3, γ1, γ4,

α1, α2, α3, α4, β1, β2, β3, γ1, γ2,

α1, α2, α3, α4, β1, β2, β3, γ3, γ4,

α1, α2, α3, β1, β2, β3, γ1, γ2, γ3,

α1, α2, α3, β1, β2, β4, γ1, γ2, γ4,

There are 5 unique tile types with 10 arms

α1, α2, α3, α4, β1, β2, β3, β4, γ1, γ2,

α1, α2, α3, α4, β1, β2, β3, β4, γ1, γ3,

α1, α2, α3, α4, β1, β2, β3, γ1, γ2, γ4,

α1, α2, α3, α4, β1, β2, β3, γ1, γ3, γ4,

α1, α2, α3, α4, β1, β2, β3, γ1, γ2, γ3,

36

There is 1 unique tile type with 11 arms

α1, α2, α3, α4, β1, β2, β3, β4, γ1, γ2, γ3,

37