2008年度冬の LA シンポジウム [11]

Complexity of pleat folding

Tsuyoshi Ito* Masashi Kiyomi\dagger Shinji Imahori\ddagger Ryuhei Uehara\S

Abstract

We introduce a new origami problemabout pleat foldings. For a given assign-ment of $n$ creases of mountains and val-leys, we make a strip of paper well-creasedaccording to the assignment at regular in-tervals. We use simple folding as a ba-sic operation. More precisely, we assumethat (1) paper has $0$ thickness and somelayers beneath a crease can be folded si-multaneously, (2) each folded state is flat,and (3) the paper is rigid except at the$n$ given creases. We also assume thateach crease remembers its last folded statemade at the crease. Wc aim to find ef-ficient ways of folding a given mountain-valley assignment in this model. We callthis problem unit folding problem for gen-eral patterns, and pleat folding problemwhen the mountain-valley assignment isMVMVMV $\cdots.$

” The complexity ismeasured by the number of foldings andthe cost of unfoldings is ignored. Tlrivially,we have an upper bound $n$ and a lowerbound $\log(n+1)$ . We first give some non-

$*$ School of Computer Science, McGill Univer-sity, tsuyoshiOcs.mcgill. ca

\dagger School of Information Science, Japan Ad-vanced Institute of Science and Technology(JAIST), mkiyomioj aist. ac. jp

\ddagger Graduate School of Information Scienceand Technology, The University of Tokyo,imahor$i0mi$st. $i$ . u-tokyo. ac. jp

\S School of Information Science, Japan Ad-vanced Institute of Science and Technology(JAIST), ueharaOjaist. ac. jp

trivial upper bounds: (a) any mountain-valley assignment can be made by $\lfloor\frac{n}{2}\rfloor+$

$\lceil\log(n+1)\rceil$ foldings, and (b) a pleat fold-ing can be made by $O(n^{\epsilon})$ foldings for any$\epsilon>0$ . Next, we also give a nontrivial lowerbound: (c) almost all mountain-valley as-signments require $\Omega(\frac{n}{\log n})$ foldings. Theresults (b) and (c) imply that a pleat fold-ing is easy in the unit folding problem.

1 lntroduction

Origami has recently attracted much at-tention as mathematics and as theoreti-cal computer science [3]. In the computa-tional origami, the best studied problem isthe characterization of flat foldable creasepatterns. When the paper and crease pat-tem are orthogonal, the problem is calledthe map folding problem, and the problemhas been well studied by Arkin et al. [2].Roughly speaking, (1) the problem in lDcan be solved in linear time; it can be de-termined in $O(n)$ time whether or not agiven strip of paper with $n$ creases is flatfoldable, and (2) the problem in $2D$ is NP-complete; to determine whether or not agiven zig-zag form paper with $n$ creases isflat foldable is NP-complete (see [2] for thedetails).

In th\’is paper, we deal with a simplerproblem. The input of our problem is astrip of paper of length $n+1$ with $n$ de-sired creases, and we assume that thesecreases are placed at regular intervals on

数理解析研究所講究録第 1649巻 2009年 66-72 66

Figure 1: An $t$ ‘angel” designed andfolded by Takashi Hojyo on the coverof [1]. (http: $//ww$ . origami. gr. $jp/$

Magazine/Index/103-108. html $)$

the paper. That is, we assume that thepaper corresponds to an interval $[0..n+1]$ ,$n$ creases are placed at each integer point$i$ with $1\leq i\leq n$ , and each integer pointis assigned to $M$ (mountain) or $V$ (valley)which is the target folded state. We aim tofold the paper and make it ”well-creased”according to the assignment. More pre-cisely, each integer point memorizes its lastfolded state $(M$ or $V)$ made at the point,and we have to make the creases’ foldedstates fit the given assignment.

A typical example is pleat folding that isone of basic tools for origami design (Fig-ure 1; the angel is made from just onesquare paper without any cutting). Themountain-valley assignment for the pleatfolding is in the form MVMV $\cdots.$

” Fromthe industrial point of view, pleat foldinghas many applications including clothes,curtains, etc.

We employ simple folding defined in [2]

(see also [3, Section 14.1]). More precisely,our origami model is as follows; the pa-per has $0$ thickness, some layers beneath acrease can be folded simultaneously, eachfolded state is flat, and the paper is rigidexcept at given $n$ creases. Clearly, thisfolding problem has a trivial solution (orupper bound of the number of foldings);we can fold any mountain-valley assign-ment by just folding $n$ creases indepen-dently. On the other hand, we have a triv-ial lower bound of the number of foldings;we have to fold at least $\lceil\log_{2}(n+1)\rceil$ timesto make $n$ creases (by folding repeatedly atthe center point of the paper).

For a given string $s$ of length $n$ in$\{M, V\}^{n}$ , the unit folding problem is to ob-tain the well-creased paper following themountain-valley assignment $s$ . We callit pleat folding problem if the mountain-valley assignment is MVMV $\cdots.$

” Thecomplexity is measured by the number offoldings, and the number of unfoldings isignored. We first show the following non-trivial upper bound of the number of fold-ings to solve any unit folding problem:

Theorem 1 The unit folding problemof $n$ creases can be solved by $\lfloor\frac{1}{2}n\rfloor+$

$\lceil\log(n+1)\rceil$ foldings for any mountain-valley assignmentl.

One may th\’ink that the pleat foldingproblem is the most dfficult in the unitfolding problems. However, this is not thecase. We show an efficient way to obtainthe pleat foldings.

Theorem 2 For any positive real num-$ber\epsilon>0$ , the pleat folding problem of$n$ creases can be solved by $O(n^{\epsilon})$ foldingsfor $s$ufficiently large $n$ .

On the other hand, we can show a lowerbound of the general unit folding problem.

lIn this paper, we assume the base of logarithmis two unless specified otherwise.

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Theorem 3 For the unit folding prob-lem, almost all mountain-valley assign-$m$ents With $n$ creases require $\Omega(\frac{n}{\log n})$ fold-ings.

Theorems 2 and 3 imply that the pleatfolding problem is not the most difficultunit folding problem.

In origami, “unfolding” is much easierthan “folding.” Hence we ignore the costof unfoldings in our model. However, evenif we count the number of unfoldings inaddition to foldings, all the results in thispaper hold with small changes within con-stant factors.

2 Models of foldings and unfoldings

The input to a unit folding problem con-sists of a string of length $n$ over an alpha-bet $\{M, V\}$ . The ith letter $(M$ or $V)$ cor-responds to the $ith$ assignment at the ithcrease which is placed at the integer point$i$ in the interval $[0..n+1]$ . In general, wehave several ways when we fold the paperat the point $i$ . The following simple foldmodels are proposed by Arkin et al. [2] (seealso [3, p. 225] $)$ :

One-layer simple fold model: We al-ways fold the top layer of paper.

All-layers simple fold model: Wesimultaneously fold all layers of paperunder the crease point.

Some-layers simple fold model: Wefold some layers beneath the creasepoint. (Note that we have to valleyfold inside consecutive layers.)

We measure the complexity of a folding al-gorithm by the number of simple foldingsmade by it, and ignore the cost of unfold-ings. In the one-layer simple fold model,we always fold exactly one crease. Thus,we camot improve the trivial upper bound$n$ for the unit folding problem, and hence

we have nothing to do. Therefore, here-after, we only consider the all-layers simplefolding and some-layers simple folding. Weadditionally introduce the following threeunfolding models, which define the allow-able unfolding operations.

All-unfold model: Once we decide tounfold, the paper is unfolded com-pletely. That is, we have the papercorresponding to the interval $[0..n+1]$

again.

Reverse-unfold model: We can rewindany number of the last folding stepsas far as we can. This model is use-ful to consider some upper and lowerbounds.

General-unfold model: For a foldedstate $s$ , we can obtain another foldedstate $t$ by the one general unfoldingoperation if $s$ can be obtained from$t$ by consecutive some-layers simplefoldings.

The models will be used to discuss bothof the upper and lower bounds of thealgorithms. Of course, the general-unfold model is the most natural and thestrongest model. This paper mainly stud-ies the first two unfold models, which arenatural restrictions of the general-unfoldmodel.

Without loss of generality, we assumethat the leftmost point of any folded paperis always placed at point $0$ . Each point onthe paper has its own label $i$ that is givenby the initial position in [1..n].

A folding algorithm will be called end-ffee if it can be applied on the paper with $n$

creases even if we extend both endpoints ofthe paper to infimty. Precisely, the papercan be corresponds to an interval $[-N..N]$for any large $N\gg n$ . This property willbe required when we use the algorithm re-cursively.

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3 Upper bounds

We first prove Theorem 1, which is the ba-sic tool of this paper.

Proof. (of Theorem 1) To simplify theproof, we first suppose that $n\simeq 2^{k}-1$ .The strategy is simple; (1) fold the paperin half $k$ times, (2) unfold all, and (3) fixeach crease if it is not folded correctly.

On step (1), the folding direction is de-termined by the majority at the point.That is, we see the point $i$ , which is thecenter of the current paper, check the alldirections of the desired creases on thepoint $i$ , and take majority. (We note thathalf creases are turned over from the origi-nal direction, but fixing this is easy; thecreases are alternately turned over andthat can be checked from 1 to $n.$ ) Hencestep (1) makes at least half of creases tobe folded correctly. Thus in step (3), thenumber of foldings is bounded above by

$L\frac{n}{2}\rfloor$ (the rounding comes from the firstfolding that always succeeds). Thereforewe have the theorem.

When $n\neq 2^{k}-1$ , we just add an imagi-nary paper to make the length of the paper$n’=2^{k}-1$ with $\min_{k}n<2^{k}-1$ , and ap-ply the algorithm on this paper of length$n^{t}$ . $\square$

We note that (a) the algorithm in theproof is end-free, (b) it works in the all-layers simple fold model and hence some-layers simple fold model, and (c) it worksin all the unfold models we described sinceall unfoldings in steps (2) and (3) can bedone by all-unfoldings.

We call a unit folding problem mountainfolding problem if the mountain-valley as-signment is $MMM\cdots.$ ” We next show asimple proposition which gives us a strongintuition to consider the pleat folding.

Proposition 4 Let $A’$ be any algorithmfor the mountain folding problem in anymodel, and $f_{0}(n)$ be the number offoldings

by $A’$ to make $n$ mountain foldings. Thenwe can solve the pleat folding problem inthe model with $f_{0}(Ln/2\rfloor)+f_{0}(\lceil n/2\rceil)$ fold-ings.

Proof. For the mountain-valley assign-ment MVMV $\cdots,$

” we first fold all$f_{0}(\lceil n/2\rceil)$ mountains by $A’$ . Next we re-verse the paper and fold all $f_{0}(\lfloor n/2\rfloor)$ val-leys by $A’$ again.

Now we turn to the upper bounds of thenumber of foldings for solving the pleatfolding problem. First, we focus on the all-layers simple fold and reverse-unfold (orall-unfold) model. In this weak model, weshow a weaker upper bound. By Propo-sition 4, we can focus on the mountainfolding problem instead of the pleat fold-ing problem. This will help us to have anintuition.

Lemma 5 The mountain folding problemof $n$ creases can be solved by $O(n^{\log\phi})\sim$

$O(n^{0.69})$ foldings in the all-layers simplefold and all-unfold model, where $\phi=\frac{1+\sqrt{5}}{2}$ .

Proof. To simplify the proof, we supposethat $n=2^{k}$ . The basic strategy is simi-lar to the algorithm in the proof of The-orem 1. In step (0), we fold the paper atpoint 1 and make the length of the pa-per $2^{k}$ . The next two steps (1) and (2)are the same; we fold the paper in half $k$

times and unfold completely. Now we fo-cus on the mountain-valley pattem madeby the two steps (1) and (2). First, we foldthe paper in half, and obtain the correctcrease at the center point. Second, we foldthe paper in half, and obtain one correctcrease, and one incorrect crease. Third,we fold the paper in half, and obtain twocorrect creases, and two incorrect creases.In ith folding with $i>1$ , we obtain $2^{i-2}$

correct creases and $2^{i-2}$ incorrect creasesthat appear on the paper alternately. The$2^{i-2}$ incorrect creases are placed at regular

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(1)$-$(2) $\ovalbox{\tt\small REJECT}$

(3) $rightarrowrightarrowrightarrowarrow\overline{hl}\overline{\overline{V}}\overline{ht}\overline{\overline{V}}\overline{M}\overline{\overline{V}}\overline{M}MVMVhlV$

(4) $\overline{=-}\underline{=}$

Figure 2: Mountain folding by zig-zagmethod

intervals. Thus, we can use our end-freealgorithm recursively on these $2^{i-2}$ creasesin step (3). Let $f_{1}(n)$ be the number ofthe foldings of this recursive algorithm. Itis easy to see that $f_{1}(1)=1,$ $f_{1}(2)=2$ ,$f_{1}(3)=3$ , and $f_{1}(4)=4$ . By the aboveobservation, we have

$f_{1}(2^{k})$ $=$ $1+k+f_{1}(2^{k-2})+f_{1}(2^{k-3})+$

. . . $+f_{1}(2)+f_{1}(1)$ .

By the fact that $f_{1}(2^{k})-f_{1}(2^{k-1})$ equals$f_{1}(2^{k-2})+1$ , we have the following rela-tionship.

$(f_{1}(2^{k})+1)=(f_{1}(2^{k-1})+1)+(f_{1}(2^{k-2})+1)$ .

Thus $(f_{1}(2^{k})+1)$ gives the Fibonacci num-bers for $k$ . (We define the Fibonacci num-bers $F_{i}$ by $F_{0}=0,$ $F_{1}=1$ and $F_{i}=$

$F_{i-1}+F_{i-2}$ for $i>1$ . It is well knownthat $F_{i}= \frac{\phi^{n}-(-\phi)^{-n}}{\sqrt{5}}=O(\phi^{n})$ , where $\phi=$

$\frac{1+\sqrt{5}}{2}.)$ Using initial conditions $f_{1}(2^{0})=1$ ,$f_{1}(2^{1})=2$ , and $f_{1}(2^{2})$ $=4$ , we have$f_{1}(2^{k})+1=F_{k+3}$ . Hence we have $f_{1}(n)=$

$f_{1}(2^{k})=O(\phi^{k})=O(\phi^{\log n})=O(n_{\square }^{\log\phi)}$

which completes the proof.

Next, we show that the number of fold-ings can be reduced by using the strongermodel.

Lemma 6 For any fixed $\epsilon>0$ , the moun-tain folding problem of $n$ creases can besolved by $O(n^{\epsilon})$ foldings for suflicientlylarge $n$ in the some-layers simple fold andreverse-unfold models.

Proof. In the algorithm of the proof ofLemma 5, “folding in half” is the basicoperation. We change this to ”folding inzig-zag form of $p$ segments” (Figure 2).Roughly speaking, the algorithm first foldsthe paper into zig-zag form of $p$ segmentsof the same length by making $p-1$ al-ternating foldings at regular intervals (inFigure 2(1) and (2) for $p=7$). It re-cursively calls itself with the folded pa-per of length $n/p$ . After the recursivecall, it unfolds at the folded $p-1$ creasepoints. Then around $p/2$ segments havebeen mountain folded, and $p/2$ segmentshave been valley folded (in Figure 2(3),thick lines and dashed lines indicate thesegments of the mountain and valley fold-ings, respectively). The algorithm thenpiles up the valley segments exactly byfolding at the center points of the moun-tain segments (see Figure $2(4)$ ). The al-gorithm again calls itself recursively withthe local area of length $n/p$ that consistsof all and only the valley segments. By therecursive call, half of the valley segmentsare corrected, and still half of them arenot corrected. (The creases of the centerpoints of the mountain segments are fixedhere.) Hence the algorithm again piles upthe valley segments and recursively calls it-self again. The algorithm repeats this pro-cess and finally obtain the desired moun-tain folded state.

The algorithm runs in the some-layerssimple fold and reverse-unfold model, andthe correctness of the algorithm is easy.Hence we analyze its complexity. To sim-plify the proof, we suppose that $n=p^{k}-1$

and $p=2^{q}-1$ . (We let $q$ be an arbitrarypositive integer, and hence $p$ is odd and $n$

is even.) Let $f_{2}(n)$ be the number of fold-ings made by the algorithm. By a careful

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analysis, we have $f_{2}(0)=0,$ $f_{2}(x)\leq x$ and

$f_{2}(n)$ $=$ $q \cdot f(\frac{n+1}{p}-1)+p-2(q-2)$

$+ \frac{3(p+1)}{2}\sum_{=p2}^{q-1}\frac{1}{2^{\ell-1}}$ .

By a simple calculation, we obtain

$f_{2}(n)$ $<$ $q \cdot f(\frac{n+1}{p}-1)+\frac{5p+11}{2}$

$<$ $(5p+11)q^{k-2}+p-1$$=$ $O(p\cdot q^{k-2})$ .

Since $q=\log_{2}(p+1)$ and $k=\log_{p}(n+1)$ ,we have $p\cdot q^{k-2}\sim p\cdot nl\circ gp$ . Hence, letting

$X1010arrow R$

$p$ be a sufficiently large constant, we have$R^{l\circ}lo_{\log p}\ovalbox{\tt\small REJECT}<\epsilon$ and $p\cdot q^{k-2}=O(n^{\epsilon})$ , whichcompletes the proof. $\square$

By Proposition 4 and Lemma 6, we haveTheorem 2.

4 Lower bounds

We have to clarify the unfold model tostate a lower bound. In fact, Theorem 3holds for both of the all-unfold model andthe reverse-unfold model, but we have noresults about general-unfold model.

Lemma 7 All but $o(2^{n})$ mountain-valleyassignments of $n$ creases require $\Omega(\frac{n}{\log n})$

foldings in all of the three folding modelsand both of the all-unfold model and thereverse-unfold model.

Proof. The lemma is obtained by a sim-ple counting argument. Since the some-layers simple fold model is at least as pow-erful as the one-layer and the all-layerssimple fold models, we only consider thesome-layers simple fold model. Similarly,we only discuss the reverse-unfold model.Suppose that we make at most $k$ foldings.At each folding, there are two choices for

the direction of folding (mountain or val-ley). There are at most $n$ choices for theset of positions of the folding by consid-ering the bottom-most layer of a valleyfolding or the top-most layer of a moun-tain folding. There are at most $n+1$choices for how many unfolding steps werewind just after the last folding. In to-tal, the number of the possible mountain-valley assignments made by at most $k$ fold-ings is bounded by $\sum_{i=0}^{k}(2n(n+1))^{i}\leq$

$(2n(n+1)+1)^{k}$ $<$ $(2(n+1)^{2})^{k}$ . Ifwe let $k= L\frac{n}{4(1+\log_{2}(n+1))}\rfloor$ , then $(2(n+$

$1)^{2})^{k}<2^{n/2}=o(2^{n})$ . Therefore all but$o(2^{n})$ mountain-valley assignments requireat least $L\frac{n}{4(1+\log_{2}(n+1))}\rfloor=\Omega(\frac{n}{\log n})$ foldingsin the reverse-unfold model $($hence also inthe all-unfold model). $\square$

Lemma 7 implies Theorem 3. Lemma 7also implies that if we choose a mountain-valley assignment of $n$ creases uniformlyat random, the expected number of fold-ings required to make it is $\Omega(\frac{n}{\log n})$ in theall-unfold model and the reverse-unfoldmodel.

5 Conclusion

There are several open problems.By Theorem 3, a random mountain-

valley assignment requires $\Omega(\frac{n}{\log n})$ fold-ings with high probability. Finding a spe-cific assignment which requires $\Omega(\frac{n}{\log n})$

foldings will give some new insight onwhich assignments are easy to fold andwhich are not. Is there any mountain-valley assignment that requires $\Omega(n)$ fold-ings?

Lemma 7 deals with general mountain-valley assignments. When we focus on thepleat folding problem, is there a nontriv-ial lower bound for this specific problem?Especially, can we make pleat foldings inpoly-log foldings?

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IFlrom the practical point of view, devel-oping better algorithm for the pleat fold-ing problem is interesting, especially, ingeneral-unfold model. Up to now, we haveno idea for dealing with general-unfoldmodel.

Acknowledgments

The authors thank to Takashi Hojyo whogives a permission to use his “angel” for ourpresentation of this research. The authors aregrateful to Erik Demaine, Joseph O’Rourke,and Ron Graham for their helpful commentand suggestion at CCCG 2008.

References

[1] Origami Tanteidan, Number 107. JapanOrigami Academic Society, 2008.

[2] E. Arkin, M. Bender, E. Demaine,M. Demaine, J. Mitchell, S. Sethia, andS. Skiena. When Can You Fold a Map?Computational Geometry: Theory andApplications, 29(1):23-46, 2004.

[3] E. D. Demaine and J. O’Rourke. Ge-ometric Folding Algorithms: Linkages,Origami, Polyhedra. Cambridge Univer-sity Press, 2007.

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