纸艺社 折纸理论 折纸数学论文 Origmi Reserch

Greece Meets Japan The Mathematics behind Origami Cody Narber Computer Science and Mathematics Bloomsburg University 1 Introduction Origami is the ancient art of folding paper into artistic forms which can take on a variety of shapes from animals to tessellations from the natural to the geometric It is no wonder that mathematicians are fascinated with mathematics in origami as am I There is really only one possible way to create a fold that is when you fold paper you create a crease that is a straight line By varying the sizes and position of these creases a model of intricate details can be composed In order to differentiate some of these folds origami artists have come up with the terminology of Mountain folds M vs Valley folds V A Valley fold is a fold such that when both sides are collapsed towards each other the crease is farthest from the folder Mountain fold is when the fold is closer to the folder Figure 1 Figure 1 Mountain Fold Valley Fold 2 Flat Origami Mathematics has many ties to Origami several of which will be discussed in brief for the reader to get an understanding of all the different topics available for research The first being flat foldability Flat foldability is the study in origami that focuses on models that will fold flat i e lie in a single plane Flat foldability looks at issues such as given a crease pattern CP with M V assignments is it possible for it to flat fold Given a CP without M V assignments that is known to flat fold how many different M V assignments will allow it to fold flat First of all a CP is a graph where the edges are the creases and the vertices are where the creases meet M V assignment is when we assign the creases to be either M or V We can associate weights with M and V assignments as well with M having a weight of 1 and V a weight of 1 Then we would have a weighted graph Now that we have a clear understanding of what it means to flat fold and what a CP is we can begin to discuss how we can tell whether a M V CP can flat fold For a CP to flat fold the degree of a vertex must be even More than that the weight of the vertex must be 2 The reason for this can be seen in Maekawa s Theorem s Proof 1 This condition is necessary and sufficient for single vertices When looking at the entire model for multiple vertices the question becomes much more complicated in fact Bern and Hayes have shown that it is NP Hard Counting valid M V assignments is also a difficult problem for multiple vertices so I will only go over a recursive technique for single vertices developed by Tom Hull 2 Which has several cases Given a subset of S with the following property i 1 i and i i 1 we can use the following recursive statement C S niii aaaaaC 2111 2 Given a proper subset E of the set S composed of equal angles i i 1 i k 1 i k we can use the following recursive statements to calculate the number of possible mountain valley assignments if k is even C S nkiii k aaaaaC k 2111 2 2 2 if k is odd C S nkii k aaaaC k 2111 2 1 2 if S is entirely composed of equal angles C S 1 2 2 n n for these recursive cases always look at the smallest angle in the set S and use the recursive statement that it corresponds to 3 Folding and Cutting Problem This next topic answers the question Can we fold a piece of paper so that with a single cut we can cut out any polygon This question is referred to as the Folding and Cutting Problem For a polygon to be cut out of a piece of paper the paper must be folded in such a way that all the edges of a polygon lie along a single line the cut line and the paper must flat fold so that cutting can take place Since fold lines reflect one side of the paper onto the other if angle bisectors are folded at every vertex of the polygon the edges should align along the cut line This extension of angle bisectors is referred to as the straight skeleton and finding the straight skeleton is done by essentially shrinking the polygon and connecting the vertices Figure 2 Eric Demaine has described this process in several of his papers 3 4 5 Finding the straight skeleton is also an interesting computer science problem as well The first algorithm created ran at O n2 log n The straight skeleton should once folded put all the edges on the cut line however there are polygons that will not flat fold using only the straight skeleton a triangle being the simplest case since the straight skeleton generates a CP with a single vertex of degree 3 By adding perpendiculars to the edges we can preserve the cut line alignment and add another crease to the crease pattern and thus satisfy the flat foldability condition As polygons get more complicated more and more perpendiculars are needed and even some cases where a new vertex is created from adding perpendiculars Figure 2 The dashed lines show the shrunken polygon at stage 1 and 2 The darker lines represent the straight skeleton Note that this skeleton will not flat fold and needs perpendiculars to be added 4 Tree Theory and Model Design One of the major interests of origami designers is in tree theory one of the most notable origami designers who uses tree theory is Robert Lang A tree is used to describe the desired basic shape of the model wished to be created For instance a human has 2 arms and 2 legs and a head By drawing a stick figure with only lines representing where the legs arms and head are a tree is created Robert Lang has developed several algorithms given a tree to create a CP to generate a simple folded model called a base with flaps having the same length and placement as the tree did From there it is the origami artist s job to incorporate details and give the model some life This may not seem that important for simple objects but it is extremely powerful for multilegged organisms like insects Using this technique it is even possible to generate a CP for a hundred legged centipede There is another area of mathematics involved in model design and that is circle packing This type of circle packing is not the same as traditional circle packing that mathematicians are used to In traditional circle packing the objective is to optimize the radius of the circles so that the circles are entirely encompassed by the shape usually a square With origami the objective is still to optimize the circle s radius however the center of the circle needs enclosed in the shape not the entire circle This adds some interesting new twists since putting larger circles near the corners will not take up as much surface area The reason for using circles can best be described using the folding and cutting problem for a single vertex If a single vertex was folded and it had degree 4 with each of the creases representing an angle bisector if the shape was cut we should have a square Similarly looking at a vertex of degree 8 there would be an octagon resulting By taking the limit of this process a circle should result Thus circles are used to represent the flap in its unfolded state Robert Lang has created a program 6 for translating an input of a tree into an efficient CP That and his book 7 are excellent resources for anyone interested in designing their own models 5 Modular Origami Modular origami is a type of origami that uses several pieces of paper of the same folding sequence called modules The model is comprised of either entirely the same module or of a few different ones but the modules needed to construct the entire model can range into the hundreds even the thousands The study here in mathematics is of the polyhedra that can be created and how to construct a single module that can be used to construct the model desired I have not done a lot of research on this topic but there are several resources available for those interested 8 9 10 6 Geometric Constructions This has been my major point of interest in the creation of this paper Geometric constructions tell us what kinds of creases can be made given only a unit length First of all we will need to know what kinds of folds can be made with origami These operations are known as the Huzita Hatori axioms 11 Axiom 1 Given two points P1 and P2 fold a line passing through both P1 and P2 Axiom 2 Given two points P1 and P2 fold a line placing P1 onto P2 Axiom 3 Given two lines L1 and L2 fold a line placing L1 onto L2 Axiom 4 Given a point P1 and a line L1 fold a line passing through P1 and perpendicular to L1 Axiom 5 Given two points P1 and P2 and a line L1 fold a line placing P1 onto L1 and passing through P2 Axiom 6 Given two points P1 and P2 and two lines L1 and L2 fold a line placing P1 onto L1 and P2 onto L2 Axiom 7 Given a point P1 and two lines L1 and L2 fold a line placing P1 onto L1 and perpendicular to L2 Given these axioms it is possible to map a given length anywhere else on its extended line Locality map 1 Given a length of and a target point on a line fold Pt2 onto the target point Axiom 2 creating a perpendicular to the line This is so that is placed left of the target point Using Pt1 will place on the left hand side of the target point 2 Fold an angle bisector Axiom 3 3 Create similar triangles by folding perpendiculars to the angle bisector so that the creases pass through Pt1 and Pt2 Axiom 4 twice These similar triangles create the length on the vertical crease 4 Repeat step three on the other side of the vertical crease resulting in the length being mapped to the target point It is also possible to map a length onto any other line on the plane Orientation Map 1 As in steps 2 and 3 of the locality map First find the angle bisector and then create similar triangles so that will get mapped onto the other line Using both of these mappings it is possible to map a given length anywhere else on the plane given that we know where we need to put it This allows us to map one length onto the tip of another length in 2 different ways which are the operations of addition and subtraction Thus given a unit length it is possible to construct any length of integer values In order to multiply two lengths we will use the following folding sequence 1 Set up the unit length and so that there are aligned at the origin point and that the lengths are perpendicular Although not necessary to be perpendicular it is the easiest to do 2 Map the other length onto the same line as the unit length 3 Connect the two end points of and the unit length to create a triangle Axiom 1 4 Fold the line created in 3 back onto itself to create a perpendicular Axiom 4 there is no point for the line to go through however a perpendicular is needed and its position isn t critical 5 Create a perpendicular crease to the line from step 4 This perpendicular must pass through the endpoint of Axiom 4 Thus creating a triangle with exact slope as the original triangle This can be seen in the ratios 1 as Division is done in the exact same manner except that the ratios of the triangles are different to accommodate division as 1 Now we have shown that origami can construct the field of rational numbers Another interesting operation that can be done is that of constructing a right triangle with the base a given height and the hypotenuse to be another given length so long as is longer than This will be used to show that square roots of lengths can be constructed 1 Set up the base of the triangle to be and have a perpendicular set up for the right triangle is somewhere on the plane For the next step map one of the endpoints of onto the endpoint of 2 Place the other endpoint of onto the perpendicular so that the crease passes through the endpoint of Axiom 5 3 Fold a perpendicular to the crease created in step 2 so that the new line passes through the endpoint of Axiom 4 4 The resulting line from the point created in step 3 is of length by similar triangles Thus we have created a right triangle with base of length and hypotenuse of length Square roots can be found by setting up a right triangle with base of 4 1 and hypotenuse of 4 1 The height d of the resulting triangle can be seen to be by using the Pythagorean Theorem 22 2 4 1 4 1 d 16 1 2 1 16 1 2 1 222 d 2 1 2 1 2 d 2 d d Cube roots of lengths can even be constructed using Axiom 6 Given below is the folding sequence that I have come up with to use to construct3 By setting the lengths as such we can see that the slope of the newly created crease is 3 If we look at the line perpendicular to the crease which passes through the point 0 we can get the equation x m y 1 thus we can find the intersection of this line and the crease line I will call the x value of this intersection point x0 We can see that m m b xo 1 Since the reflected x value of 0 across the crease line is 2 x0 and we know that it lies on the line x 2 we can get the equation 2 1 2 m m b by solving that equation for b we can see that m mb 1 I will refer to this as equation 1 which I will be using again a little later Similarly with the point 1 0 we can find the reflected y value of the point generating the equation 2 1 1 2 m m m b Once again we will solve this equation for b This results in equation 2 mmb 2 By setting equation 1 equal to equation 2 mm m m 2 1 and then solve for m we do indeed see that 3 1 m We can then easily map that line in such a way to construct a length of 3 This shows that origami can construct a field of numbers that is closed under square and cube root which means that it is more powerful than traditional straight edge and compass constructions and that it can solve problems that could not be solvable with a straight edge and compass such as trisecting an angle and doubling the cube Roger Alperin has shown that the set of origami constructible numbers is the smallest field having those properties 12 Aside from constructing these folding sequences on my own I have written a program that wil