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GXWeb Paper Folding and Points in a Fold

Bunder, M, Tognetti, K and Bates, B (2022): Points in a Fold (preprint)

Bates, B, Bunder, M and Tognetti, K (2010): Mirroring and Interleaving in the Paperfolding Sequence

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Bates, B., Bunder, M. and Tognetti, K. (2010): Locating Terms in the Stern-Brocot Tree

Bates, B., Bunder, M. and Tognetti, K. (2010): Linking the Calkin-Wilf and Stern-Brocot trees

Bruce Bates (2014): The Stern-Brocot Continued Fraction

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GXWeb PaperFolds and Points in a Fold: YouTube Video Overview (19:36)

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Finding Points in a Fold And Here There Be Dragons The Paper Fold Challenges Model Paper Folding with GXWeb Mathematical ToolBox Share Your Results Explore Further Build your own Model Behind the Scenes

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Finding Points in a Fold

When a line joining 0 to 1 (or a rectangular piece of paper with [0;1] at its edge) is folded in half, right over left, the point \(\frac{1}{2}\) is at the fold. When the line is then folded in half again, right over left, the points \(\frac{1}{4}\) and \(\frac{3}{4}\) are at this second fold, in that order, from the outside of the fold to its interior.

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Proceeding in this way, the sequence of points from the outside of the third fold to the inside becomes

\[\frac{1}{8}, \frac{7}{8}, \frac{5}{8}, \frac{3}{8}\]

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And Here There Be Dragons

If, after folding, you open the page out again, you will observe the paperfold sequence - a series of downward (▼ 1) and upward (▲ 0) folds that may catch you by surprise! Perhaps also surprising: if each fold is opened out to create a right-angled corner, the resulting shape approaches the dragon curve fractal. For example, a page folded 4 times will show the folds as ▼ ▼ ▲ ▼ ▼ ▲ ▲ ▼ ▼ ▼ ▲ ▲ ▼ ▲ ▲ \[[1,1,0,1,1,0,0,1,1,1,0,0,1,0,0]\] Can you see how the sequence unfolds? (Hint: think about our value (k) from Challenge 1 following!)

The PaperFold Sequence is related to the Stern-Brocot Sequence and has applications to fields such as error-checking in communication systems - fundamental in cyber-security.

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The Paper Fold Challenges

1. Can you explain why, for \(k=0,1,2,3,...,2^{n-1}-1\), the points for each fold are of the form

   \[\frac{2\cdot k+1}{2^n}\]

   (Theorem 1)

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   Each numerator must be an odd number, since all even terms will have reduced to a lower denominator, and occurred already in a previous fold.

   Such fractions resulting from \(n\) folds will all have denominator \(2^{n}\)

    

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2. Can you predict the order that each fold value takes in the folded page?

   (Theorems 5-9)

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   A page is folded in half from right to left \(n\) times.

   How do we calculate each term of the folded sequence?

    

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   For \(n\) folds, the \(n\)-th fold will have \(2^{n-1}\) terms, and each will have denominator \(2^n\).

   The first and second terms (\(F(n,1)\) and \(F(n,2)\)) are easy: they will have numerators

   1 and \((2^{n-1}-1)\).

   So for \(4\) folds, the first two of the 8 \((2^{3})\) terms will be

   \[\frac{1}{16}, \frac{15}{16}\]

   In fact, each ODD-numbered term and the subsequent (EVEN-numbered) term will add to 1:

   \[\frac{1}{16} + \frac{15}{16} = \frac{16}{16} = 1\]

   The next term \(F(4,3)\) is more interesting. Begin by expressing the term number (3) as a sum of powers of 2:

   \([1,0] => (2^1 + 2^0) = 3\).

   Choose the smallest non-zero term (\(p = 1\)) and find the difference from the number of folds:

   \[4 - 1 = 3\]

   Take this number as a power of 2 and multiply by 3: \(3\cdot (2^3) = 24\).

   Finally, subtract the previous numerator (24 - 15): \(F(4,3)\) => \(\frac{9}{16}\)!

   AND as a bonus we also have \(F(4,4)\) : \(\frac{7}{16}\)! (Why?)

   Each EVEN-numbered term and the subsequent (ODD-numbered) term will add to

   \[\frac{2^{n-(p-1)}}{2^n} + \frac{2^{n-p}}{2^n}\]\[= \frac{1}{(2^{p-1})} + \frac{1}{(2^p)}\]\[= \frac{2}{(2^p)} + \frac{1}{(2^p)}\]\[ = \frac{3}{(2^p)} = 3\cdot (\frac{2^{n-p}}{2^n})\]\[\frac{15}{16} + \frac{9}{16} = 3\cdot \frac{2^{3}}{16}\]

   So for 4 folds, the first four terms (from the bottom) are:

   \[\frac{1}{16}, \frac{15}{16}, \frac{9}{16}, \frac{7}{16}\]

   Repeat for subsequent terms...

    

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3. Can you predict the fraction that occupies a particular position after n folds?

   (Theorems 5-9)

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   A page is folded in half from right to left \(n\) times.

   How do we calculate any term \(F(n,s)\) of the folded sequence? Consider, for example, \(n = 4\) and \(s = 6\).

    

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   For \(n\) folds, the \(n\)-th fold will have \(2^{n-1}\) terms, and each will have denominator \(2^n\).

   Begin by expressing the term number (6) as a list of powers of 2:

   \[6 = (2^2 + 2^1) = [2, 1]\]

   (Note: If odd, the last term will be 0.)

   Ignoring the last term (AND note the last non-zero term - call it \(m\). We will use this later for even-numbered terms!)

   1. Take the difference between \(n (4)\) and each remaining term:

      \[(4-2=2)\]

   2. Express each as a power of 2 and, if more than one term, take the SUM of these, and ADD 1.

      \[s=6 => 2^2 +1 = 5\]

   If your term number is ODD, then you already have the numerator of your fraction!

   For an EVEN term, subtract from \(2^{n-(m-1)}\) to get the numerator of your fraction.\[F(4, 6) = \frac{11}{16}\]

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4. And can you predict the particular position a fraction occupies after n folds?

   (Theorems 10-11)

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   How would you predict the particular position a fraction occupies after \(n\) folds?

   Begin by recalling that, from Theorem 1, a fraction in fold \(n\) at position \(j\) is of the form \[F(n,j) = \frac{2\cdot k+1}{2^n}\]

   From Theorem 10, however, ANY fraction between 0 and 1 can by situated within \(n\) folds by finding the term \(F(n,j)\) just below (or equal to) the given fraction.

    

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   Show First and Last

    

    

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   From Theorems 5-9 we can easily list the first two and last two fractions for any fold:

   • For \(j=1 (k=0)\) \[F(n,1) = \frac{1}{2^n} => F(4,1) = \frac{1}{16}, F(5,1) = \frac{1}{32}\]

   • For \(j=2\) \[F(n,2) = \frac{2^{n-1}-1}{2^n} => F(4,2) = \frac{15}{16}, F(5,2) = \frac{31}{32}\]

   • For \(j=2^{n-1}-1\) \[F(n,2^{n-1}-1) = \frac{2^{n-1}-3}{2^n} => F(4,7) = \frac{13}{16}, F(5,7) = \frac{29}{32}\]

   • For \(j=2^{n-1}\) \[F(n,2^{n-1}) = \frac{3}{2^n} => F(4,8) = \frac{3}{16}, F(5,16) = \frac{3}{32}\]

    

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   For other fraction values, from Theorem 11, we distinguish between EVEN and ODD values of \(k\).

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   Show Even k

    

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   For even terms, express \(k\) as a sum of powers of 2: For example, \[13 = [8+4+1] = [2^3+2^2+2^0] => [3,2,0] = [u_1,u_2,u_3]\]

   For each of these \(u_p\) terms, express the sum \(S(p) = 2^{(n-1-u_p)}\) for \(p = 1,2,3\), and position \(j\) follows directly: \[j = 1+S(p)\]

   For example: \[\frac{13}{16}\]\[16 = 2^4 => n = 4\]\[13 = 2\cdot k + 1 => k = 6\]\[6 = [2^2+2^1] => [2,1]\]\[S(p) = [(4-1-2),(4-1-1)] = [1,2] => [2^1+2^2] = 6\]\[j =1+6 = 7\]\[F(4,7) = \frac{13}{16}\]

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   Show Odd k

    

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   While we again use a sum of powers of 2 when \(k\) is ODD, we approach it from the other direction: begin with a power of 2 GREATER than \(k\) and subtract down to it!\[k = 2^{n-1-m}-1-S(q)\] where \(S(q) = [2^{u_1}+2^{u_2}...+2^{u_r}]\) is again a sum of powers of 2, and \(m\) allows for the fact that there are multiple powers of 2 greater than \(k\), so it is simplest (but not essential) to use the smallest of these.

   Then \[j = 2^{m+1} + S(n-1-{u_r}) r=1,2,...\]

   For \[\frac{11}{16}\]\[16 = 2^4 => n = 4\]\[11 = 2\cdot k + 1 => k = 5\]\[5 = 2^{4-1-0} - 1 - [2^1]\]\[= 8 - 1 - 2 : m = 0\]\[S(q) = (4-1-1) = 2 => [2^2] = 4\]\[j =2^{1+0}+S(q) = 2+4 = 6\]\[F(4,6) = \frac{11}{16}\]

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   And that curious \(m\) value? Consider \(\frac{11}{32}\) or even \(\frac{11}{64}\).

   Both equate to \(k = 5\) and both will deliver the correct result using \(m = 0\).

   However, using \(m = 1\) when \(n =5\): \[5 = 2^{5-1-1} - 1 - [2^1]\]\[5 = 2^{3} - 1 - 2\] rather than \[5 = 2^{5-1-0} - 1 - [2^3+2^1]\]\[5 = 2^{4} - 1 - 10\] for the first and \(m = 2\) when \(n = 6\): \[5 = 2^{6-1-2} - 1 - [2^1]\]\[5 = 2^{3} - 1 - 2\] rather than \[5 = 2^{6-1-0} - 1 - [2^4+2^3+2^2]\]\[5 = 2^{5} - 1 - 26\] for the second will deliver the result somewhat more easily!

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