Can we make 101 almost perfect banknotes from 100?

Disclaimer. The practical execution of the algorithm in question might be illegal in certain jurisdictions, and is thus strongly discouraged by the poser of the problem.

This recent post on the Muffin problem made me think of the following question.

Can we cut 100 banknotes into pieces of size at least $10\%$ each, and reassemble them into 101 banknotes of size $100\pm2\%$ each?

So each original banknote is cut into at most $10$ pieces of substantial size, and each new banknote also consists of at most $10$ pieces. The patterns on these newly formed banknotes should match, so we also demand, say, that no part of a banknote appears twice on a new banknote. Of course, these numbers are quite ad hoc, I'm happy to see any similar result. Note that if we don't require each piece to be at least $10\%$, then it is easy to make the trick by cutting each banknote into only two (sometimes very unequal) parts. I also wonder if non-vertical cuts might help, but I would like to keep the pieces simply connected regions bounded by Jordan curves.

Also, is there some implication between this question and the Muffin problem?

• I'm not sure what you mean exactly. I've tried solving the problem... – domotorp Jun 28 '18 at 20:36
• Have you tried going at it with a pair of scissors and some tape? – Asaf Karagila Jun 28 '18 at 21:46
• Note that non-vertical cuts are necessary to solve the problem. If only vertical cuts are allowed (so the banknotes are essentially 1D segments), then you cannot make cuts in the leftmost 10% of each banknote, and hence you have at most 100 "leftmost 10%" parts (and each banknote must have one). – Federico Poloni Jun 28 '18 at 21:52
• Somewhat similar: this MSE question – Robert Israel Jun 29 '18 at 7:27
• +1 for the "Illegal" tag :D – Qfwfq Jun 29 '18 at 9:43

This (top half) is a way to cut 49 banknotes into pieces of at least 10% each and (bottom half) reassemble them to 50 "98%" banknotes. Let $w$ denote the width and $h$ the height. The bills below have an aspect ratio of 1:2 but this isn't necessary.

(SVG source code)

I hope that it's clear how the pattern continues in the center; simply use each leftover piece together with the next one (which is matching in color) to form a 98% bill; the size of the square gap is $\frac{w}{10}$ by $\frac{h}{5}$.

The two outer columns on both sides are trickier, because it's impossible get rid of the corners with the same scheme as used in the center. Therefore, we cut strokes with width $\frac{w}{5}$, moving $\frac{h}{10}$ downwards each time. Reassembling them gives 'inlets' of $\frac{w}{5}$ by $\frac{h}{10}$.

Note that all bills get cut into two pieces, except for the two with the large light blue part (top left and top right) which are cut into three pieces.

• this doesnt really make sense because 1. thats not 50 notes (or 49 even) and 2. you dont show the reuse of the 2 block sections, which would be required for a valid solution – Steven Penny Jun 29 '18 at 2:21
• @StevenPenny I too was confused by the figure at first. The image shows the 49 notes before cutting and the colors show the way they are reassembled into 50 notes. The hatched squares, as explained in the text, do not indicate missing or discarded paper, but are instead meant to be divided half-and-half into the colored regions above and below. Presumably this will be fixed in the next version of this answer. – j.c. Jun 29 '18 at 3:37
• @StevenPenny The center column of rectangles with dashed outlines is supposed to indicate that some additional columns (4 of them if I'm counting correctly) are not shown; there is a not-too-complicated pattern of cutting and gluing shown in the 3rd and 5th columns which can be extended to those hidden columns. I do believe that the answer should include more textual explanation for the image as I did spend quite a bit of time figuring all of this out. – j.c. Jun 29 '18 at 5:34
• To be absolutely explicit: the new version of the image now shows both the pre-cutting bills (top 5 rows) and the post-cutting reassembled bills (bottom 5 rows). Four columns of bills are omitted in both and replaced by the central column of dashed rectangles. The answer and images, I think, are much clearer now. – j.c. Jun 29 '18 at 13:53
• Actually, I think that the middle part is a bit over complicated. After you manage to get the two side-strips, you can just finish off by vertical cuts. Isn't that simpler? – domotorp Jun 29 '18 at 19:34

This video http://thekidshouldseethis.com/post/62804856022 shows a possible solution (well, not for banknotes but for chocolate and one piece is only about 4% in size). This is a variant of Paul Curry's paradox described in Martin Gardner's book "Mathematics, Magic and Mystery": http://store.doverpublications.com/0486203352.html In fact some real money was done from this paradox: a plastic version of Gardner's square was made in China and sold in stores. See Gardner's interview https://www.jstor.org/stable/25653710

However, a better way to make money is provided by Banach-Tarski paradox: A gold sphere can be chopped into a finite number of chunks, and these chunks can then be put together again to yield two gold spheres, each of which has the same size as the one that just went into pieces. See https://arxiv.org/abs/math/0202309 (The Banach-Tarski paradox or what mathematics and religion have in common, by Volker Runde). The history of this theorem can be colourfully traced through the paper https://arxiv.org/abs/1710.05659 (From Poland to "Petersburg": The Banach-Tarski Paradox in Bely's Modernist Novel, by Noah Giansiracusa and Anastasia Vasilyeva).

P.S. Even if we forget about preserving the pattern on the banknote, the Banach-Tarski paradox doesn't work for planar banknotes. As mentioned in https://www.sciencedirect.com/science/article/pii/S016800720300126X (On the Warsaw interactions of logic and mathematics in the years 1919–1939, by Roman Duda):

This is the famous Banach–Tarski paradox. By the way, one can ask about money: can a banknote produce two of its kind? It is a problem in applied mathematics, but the answer is, unfortunately negative: no bounded set in the plane can have such a paradoxical decomposition [41, footnote 1 on p. 218].

And [41] is Adolphe Lindenbaums article "Contributions à l'étude de l'espace métrique": http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-fmv8i1p16bwm

• These well-known instances are not so related to my problem, because they don't preserve the pattern on the banknote in any way. I also wrote that the pieces should be simply connected regions bounded by Jordan curves to exclude BT type solutions. – domotorp Jun 29 '18 at 12:33
• The validity of the approach based on Banach-Tarski depends on the underlying model of set theory... – Dirk Jun 29 '18 at 16:18