An Origami-inspired adventure in Mathematics Jeanine Meyer Purchase College/SUNY http://faculty.purchase.edu/jeanine.meyer Jeanine.meyer@purchase.edu

The adventure … Mark Kennedy led people coming together to wait for free tickets to Shakespeare in the Park in NYC and fold. He taught the Dollar Bill Rosette: Martin Kruskal’s variation (22 panels) on Paul Jackson’s model (16 panels) I revisited the model and figured out things (on my own). I gave a talk to our Number Theory class. My Chair suggested we offer a GEN ED math class on origami: mostly basic algebra, geometry, trig. We did! The class was a success. We are repeating it and I am working on a book!

Overview We will make the Dollar Bill Rosette Procedure for marking 1/11, 2/11, etc. What does procedure do? Iterates to improve original estimate: E  0 Marks off ALL of intermediate sections For what numbers besides 11 does the procedure work? Python Program generating numbers (under 1000) that work Used this set to identify class of primes Proof

Dollar bill rosette first step Fan fold of 22 parts starts with need to divide bill into 11ths. Start with guess for 1/11. Make a pinch at your estimate of 1/11. Dollar bill is divided into 1 part and 10 parts, an even part and an odd part. Divide the even part in half by folding to your estimate to get 6 parts and 5 parts. Continue. 1 and 10 6 and 5 3 and 8 7 and 4 9 and 2 10 and 1 Keeping going. Get back to 1 and 10. So now have pinch mark at each place 1/11, 2/11, etc.

Argument on error Let the length of the edge be L. Let the correct measure of the estimate be A. Say the estimate is off by error term e. It should be A, but instead is A+e. (Can assume it is on one side.) This divides the bill into two parts: one measuring A+e from one end (call it the left) and the other measuring L – (A+e) from the other end (call it the right). Assuming that folding a section in half is accurate and in this case the next step is dividing the L-(A+e) section in half, that step produces a mark at ( L-(A+e) )/2 from the right. The new mark is off by e/2. The error has been halved! Let’s consider the other case: halving the A+e section produces something at A/2 + e/2. Again, the error e is halved. Continuing through the 10 steps, when back to the first mark (around the first mark), each step halves the error so, the new mark is off by e/210

Continuing the folding With improved estimate, go through one more time making complete folds, all valleys Divide each of the 11 sections by 2, making mountain fold, to get 22 panels in a fan fold Divide into two Fold out 3 panels on each side Fold at the midpoint Make tabs to fold in to attach the two parts. Pop open to be a rosette

Rosette procedure Besides improving the estimate, the procedure gets back to the start (1, N-1), and hits all the divisions in-between. For what numbers does the procedure work? Must be odd number Must be prime Proof? What kind of prime

Suggestion that it must be prime Suppose P is not prime, and equal to F * G, F and G each >1. Note: neither F or G can be even. In the rosette procedure, if procedure works, at some point, the pair of numbers is G and (F-1)*G. What is the next step? G+((F-1)/2)*G and ((F-1)/2)*G Claim: Always G factor in both numbers in pair: never get to 1 and (F*G) -1.

All primes? Does procedure hit all numbers for 17 1 and 16 9 and 8 13 and 4 15 and 2 16 and 1 8 and 9 4 and 13 2 and 15 Only 1 + 8 steps. Not all the intermediate places.

What primes? Adventure continues… After learning the model, a long time ago, a mathematician colleague at IBM Research told me that it probably was certain category of numbers, but I forgot… So I wrote a Python program to generate numbers that work Just learned Python by taking a Coursera course Python has arbitrary precision for integers. Knew it was used in the Number Theory course

Experiment Wrote program to invoke the testing program for all numbers May have checked for primes first … from 3 to 1000

def tryProcedure(N): count = 1 # start with 1 and N-1 currentpos = 1 remainder = N-currentpos while True: if (isEven(currentpos)): # decide which side is even currentpos = currentpos//2 # integer division remainder = N - currentpos else: currentpos = currentpos + remainder//2 # integer division count = count + 1 if (currentpos==1): # back to the start break. # leave the while True if (count==N): # did this go through N steps? print(" ",N,end="") # good, so print it out return

Numbers from 3 to 1000 3 5 11 13 19 29 37 53 59 61 67 83 101 107 131 139 149 163 173 179 181 197 211 227 269 293 317 347 349 373 379 389 419 421 443 461 467 491 509 523 541 547 557 563 587 613 619 653 659 661 677 701 709 757 773 787 797 821 827 829 853 859 877 883 907 941 947

Experimental computational mathematics Generated list of the numbers that work up to 1000. Took those numbers and put into Google…. Got to https://en.wikipedia.org/wiki/Full_reptend_prime

What is reptend prime base 2? Numbers P for which 2 is a primitive root modulo P This is a procedure, also. …. Number P for which 2 raised to the power N, N going from 0 to P-2, produces all the numbers (not necessarily in order) 1 to P-1, modulo P

What is modulo (mod) Two numbers are equal mod X if the difference is a multiple of X. M = N mod P if and only if (this is the definition of mod) M-N is a multiple of P. Putting it another way: M = N + a*P where a is an integer This sometimes is called clock arithmetic. For clock arithmetic, the P is 12.

Let's try 11 20 is 1 = 1 mod 11 21 is 2 = 2 mod 11 22 is 4 = 4 mod 11 23 is 8 = 8 mod 11 24 is 16 = 5 mod 11 24 = 5 + 1*11 25 is 32 = 10 mod 11 25 = 10 + 2*11 26 is 64 = 9 mod 11 Do the check: 64-9 is 55 and yes, that is a multiple of 11 26 = 9 + 5*11 27 is 128 = 7 mod 11 28 is 256 = 3 mod 11 29 is 512 = 6 mod 11

Interesting…. Yes, all the numbers 1 to 10 are produced. The Reptend process hits all the intermediate numbers AND the numbers appear to be the numbers in the dollar bill rosette procedure going backwards.

Plan to prove … that the numbers that can work using the folding procedures are the Reptend Primes Base 2, one needs to prove that the numbers for which the Folding procedure hits all the intermediate numbers are the same as the numbers for which the Reptend Process (raising 2 to powers from 0 to number -2 and then taking mod) hits all the intermediate numbers. Instead, I decided to try for a stronger result: the two procedures are the same procedure, though in opposite order.

Reverse folding process Need to determine, for number P: if F and P-F go to G and P-G, then starting at G, how to get back to F? Consider cases: Is(was) F odd or even The pair G and P-G came from F and P-F either by halving F or halving P-F, so Consider 2*G. If 2*G > P, then need to consider 2*(P-G) If 2*G > P, then 2*(P-G) <P. Doubling G and comparing to P will be what I use. Two numbers adding up to P. They are different (note P is odd so they can't be equal) so one must be less than ½ P and the other must be more.

Stronger result The two procedures are the same procedure, though need to treat folding in reverse order. Need to prove that for any k, the first number of the M, P-M pair by (reverse) folding is 2k mod P Note: both processes can continue past P-2. Proof by induction Initial case Induction case Assume true for k, prove for k+1

Initial case Reptend and reverse folding start out with 1 Reptend: 20 =1, so 20 =1 mod P Reverse folding starts with 1.

Induction step Can assume G = 2k mod P which means G = 2k + a*P where a is an integer What is the next case for reverse folding? Two cases: 2*G< P and 2*G > P. Case 1: 2*G< P. So F = 2*G. Expand F F=2*(2k + a*P) = 2k+1 + 2*a*P so F = 2k+1 mod P

Continuing with 2nd case Case 2: 2*G>P. So F and P-F produced next step by halving the P-F term. So the P-F term is equal to 2*(P-G). F = P-2*(P-G) F = P – 2*P + 2*G rearranging terms F = 2*G – P now substitute G F = 2*(2k + a*P) – P F = 2k+1 + 2*a*P – P F = 2k+1 + (2*a-1)*P F = 2k+1 mod P

Conclusion of proof The Reptend process of raising 2 to a power and then calculating mod P produces the same numbers as the reverse folding process. Because both processes yield the same results at each step, both either satisfy criteria of hitting all the intermediate points between 1 and P-1 for a number P or neither does. Note: the fact that folding resembled Reptend process for 11 and that the numbers under 1000 match does not constitute a proof that the two sets of numbers are the same! Something could happen above 1000. What I meant by a stronger result

Recap Origami inspired adventure folding improving approximations programming online investigation Proof establishing connection of folding procedure to class of primes

Materials Jeanine Meyer Academic Activities: http://faculty.purchase.edu/Jeanine.meyer There is a link to a page with origami materials, including this presentation. My programming books frequently contain origami-related material Programming 101 (Processing), Apress, June, 2018 Chapter 6: Origami flower; Chapter 8 (extra) Directions. HTML5 and JavaScript Projects, Apress, 2nd Edition published 2018 Chapter 7: Using HTML & JavaScript for Origami Directions New book project, working title: Origami with Explanations Comments welcome: jeanine.meyer@purchase.edu