3 Puzzles and Plus 9781733374941, 9781733374989, 1733374949

Table of contents :
Twelve balls problem and obvious facts
The Sudoku puzzle and a master pattern
The Tower of Hanoi and a hidden pattern
Math and traits about the tower of Hanoi
Formula of sphere volume and algebra
Appendix: Algebra and the seesaw
Bibliography
Notes

Citation preview

3 PUZZLES AND PLUS NELSON DAVID K ENSYESS 3 T YLER ,T X

3 Puzzles and Plus Nelson David Published by Kensyess3 Copyright 2022 Kensyess3 All rights reserved. No text or illustration partially or in full of this book can be reproduced, or retransmitted in any form without the written permission from the publisher. Library of Congress Control Number: 2022909730 ISBN 978-1-7333749-4-1 (paperback) ISBN 978-1-7333749-8-9 (ebook)

To my wife Nelly, my daughter Yessika, to my mother and Luis memory, and to Mary, Xio, Eva, Eli, Roger, and Merly

Acknowledgments To Forrest Kaiser, an amazing person, for improving the fluidity of this book. To my daughter, Yessika Guerra, for her immeasurable patient. They made possible publishing this book *****

Introduction. The problem is never how to get new, innovative thoughts into your mind, but how to get old ones out. - Dee Hock

Following certain rules are often natural and easy because they make us comfortable and they generate feelings of certainty and consistency. However, some facts and patterns we take for granted were not obvious in the past, such as our need to have air to survive. In some cases, established and accepted “obvious” rules do not allow us to see, recognize or discover a law that remain completely or partially hidden to us. At one point in time there was the accepted rule that the earth was flat. This impeded the ability to accept that if a boat disappears in the distance, the earth has to be curved instead of flat. At times it may be difficult to break down some rules or obvious things, because unconsciously there is the tendency to follow other hidden laws and patterns that are intrinsic to humans, and sometimes just for selfish human reasons. Many times we do not want to hide things, facts, or traits but are forced

by simplifications, prejudices, or societal norms and/or beliefs and we end up hiding them. Unconscious, society, like a professional magician tends to conceal the transition from one movement to another so that the spectator cannot catch the trick behind the illusory acts, which seem close to reality. The more strategic the proverbial masquerade, the more likely the public will fall for the illusion. Three puzzles, namely the twelve balls problem, the Towers of Hanoi, and the very popular Sudoku puzzle, are not everything this book is about. They will be used to illustrate the influence of laws, patterns, and rules on man’s free will. They will first be presented as they are known, then their hidden patterns will be exposed. The twelve balls problem itself does not have a hidden pattern but is presented because it works well to display how prior knowledge may form bias in the brain. For clarity and a warm up for what is to come, consider these two riddles: 1) A man calls his dog who is on the other side of a river. The dog, completely dry, immediately arrives to his owner’s side. How is that possible? (Pavitt 77) 2) What is the only known fruit that has its big seed out of its pulp? Depending on background, it may be possible that a person could give the solution for these even before one is done reading it. For the same reason, however, one of the puzzles could prove a great challenge. Contradictory? The solutions will explain. 1) The dog arrived not wet because the river was frozen. 2) The fruit with an external seed is the cashew, its tree is typical of different tropical areas. As the three puzzles highlighted in this book stand alone, the reader can skip any of them. However, I encourage you to read the summary of each at the end of its description because it will help share how each is interconnected. Additionally, a process developed by the great scientist Archimedes over two millenniums ago will also be shared. While considered common

knowledge now, the procedure for getting the formula to calculate the volume of a sphere using only algebra and geometry reveals Archimedes’ greatness and shows what is possible when we look beyond what is already known. *****

3 PUZZLES AND PLUS *****

Twelve balls problem and obvious facts Nothing is more dangerous than an idea when it is the only one you have. Emile Chartier

In the twelve balls problem there are a dozen of balls of the same size, texture, and the same material. Eleven of the twelve have the same weight. Using only a twin pan scale, the puzzle consists of finding out which has a different weight and telling whether it is heavier or lighter than the rest. The task should be done in just three weighings. At first glance the solution appears so easy, but it is not. Or perhaps we may make it more difficult without knowing it. Depending on our aptitude for approaching obvious things, finding the solution of this puzzle may take only a few minutes or several days. The purpose of this section is to evaluate our perspective on how obvious facts and patterns unconsciously guide us throughout this world. If one already knows the solution, this topic may be skipped. However, understand that the solution may only be partial. The true solution shall always involve all twelve balls. For that reason, one may want to try to solve the puzzle again. One hint: the weighs shall be used in the most effective way by getting the maximum possible information out of the first two weighings. For the first, try at least two scenarios but do not be limited to them until you get a good feeling what advantage or disadvantages each of them offers. Once you get confident in which of them is more likely to solve the puzzle, go through all the possible scenarios of the second and third weigh. This was the approach used by the writer.

Although it is possible to find the ball on two weighs, it is likely to be just by luck. For instance, weighing one in each plate during the first weigh and finding that the scales are out of balance. It would mean that the unknown is on one of the two plates and can easily be identified on a second weigh. However, the actual solution shall be found by deduction or inference. All twelve balls should be considered. Even if you have never tried this puzzle, the hints, rules, or facts offered below may look very familiar and obvious to you. O1- On each weigh, each plate should have the same quantities of balls. O2- On a weigh, there are only two possibilities: the plates are either balanced (B) or unbalanced (U). O3- On a weigh, if the plates are balanced, meaning they remain at the same level, none of the balls on the plates is the unknown ball. O4- On a weigh, if the scale is balanced, the unknown ball is among the ones that are not on the plates. O5- If the plates are unbalanced, the unknown has to be on one of the plates. The following information can be gathered from these scenarios: (a) when the unknown is on the plate that goes up, it is lighter than the rest of the balls, and (b) when it is on the plate that goes down, it is heavier. O6 – If the scale is unbalanced after weighing one ball on each plate and the balls are swapped from the plates, the up and down movement of the plates will also be interchanged. O7 – It is possible to “discard” some balls after finding an unbalance after the first weigh, but they can be used to double check others. Some items of the above list appear redundant and for that reason we usually obviate them. However, some of them are powerful tools to solve this puzzle. None of them is minor. Are there other obvious facts regarding weighing using the two-plate scale? Yes, there is one more that is a hybrid of some of those listed above, and that one is the key as will be illustrated in the following paragraphs. To discard some possibilities, let us analyze the worst possible scenarios for the first weighing. First, on each plate there are two balls and the scale gets in balance. The unknown is out of the plates and among ten balls with

no additional information for the remaining two weighs. For the second, there are five balls on each plate and the scale gets unbalanced. In this scenario, although two are discarded the unknown remains hidden in the ten on the plates. They should be analyzed in the coming weighings by inferring the information shared in hint O5 listed above. Either of these two scenarios provide a hard task for the remaining two weighings because there are still many balls with little additional information known. Thus, for the first weighing, three or four balls on each plate seems to be more likely to produce a solution to this puzzle. This is because the quantities both on and off the plates are almost equal. The following paragraphs focus on the analysis of all scenarios that emerge from weighing four balls on each plate during the first weighing. For that reason, we are going to consider the only two possible scenarios of this first weigh: when the scale gets balanced (case #1), and when it is out of balance (case #2). For explanation purposes, let us: 1) call the group of balls that we know have the unknown ball as vying balls or group on probation. We determine this fact depending what happened on a previous weigh, and 2) use unknown to name the unknown ball. To provide a graphical example for the sketches shown on the figures below, the balls are painted with different colors and tagged with a number from 1 to 12 (See Figure T1 and Figure T2). These distinctions would be not necessary if we use marbles of the same size but of different colors like the ones in the front cover. The first case seems to be the easiest because in this scenario the eight balls on the plates are discarded and the unknown is among just the remaining four that were out of the plates. Figure T1 summarizes all the possible scenarios of this case. The comments of the sketch are summaries of the possible conclusions that could be inferred from a particular weigh. For example, notice the comment “Discarded balls 1 to 8”, means discard all the balls that are on the plates that we have labeled as one to eight. According to the example on the sketch for the second weigh, any three balls (9, 10, and 11) of the four balls remaining on probation can be weighed against any three (1, 2, and 3) of the eight discarded on the first weigh. This will simultaneously check if any of the three balls (9, 10, and 11) on the plates or the other one (ball number 12) off of the plates should remain as

part of the next probation group. Focus your attention on the next weighing that creates an unbalance and take note of the movement of each plate and what balls were on each plate. This obvious fact is very important on the analysis of the solution, and for that reason it stands out as a note at the bottom of both figures. In this case, the note is only needed if there is an unbalance in the second weigh. In the given example, three of the balls on probation were placed on the right plate.

Figure T1. Eight balls get balanced on 1st weigh

First let us consider all the possibilities on the second weigh if the plates are unbalanced (option U). In this scenario, ball number 12 is also discarded, and the unknown is among the three balls (9, 10 and 11) on the (right) plate where we just placed them. For the up or down movement of such plate, the ball weight (lighter or heavier) is known right away. Knowing what the weight is of the unknown, for the third weigh, we weigh any two (9 and 10) of three now on probation against each other (case U2 in weigh 3). If they do not balance, we observe or check which of the two plates moves on the same direction as the (right) plate where these two balls were on the second weigh (see note on the bottom of Figure T1 that is equivalent to obvious option O5 but adapted to this condition). The unknown (either 9 or 10) is the only one promoting and repeating this (up or down) movement on wherever plate it is. On the other hand, when the plates balance, the unknown is the one on probation (11) off the plates and its weight is inferred according to the note on the bottom of the sketch.

Now let us focus on the second scenario of case #1 or when on the second weigh the plates are balanced (option B). In this case, the unknown is the only one on probation (ball 12) that has not been on a plate. Its weight is found during the third weigh by weighing it against any discarded one (for instance against ball 1 as shown in case B2 of weigh 3). As an unbalance has to be reached in this case and because we know in which plate we have placed this ball, the weight is also inferred. If such plate goes up the ball is lighter than the rest; down, is heavier. Notice that in case #1 on the third weighing are studied in full the four balls on probation. Similarly, eight balls shall be considered on case #2. This makes this case more complex, thus its solution will be divided and detailed on two main stages: the analysis of the second and third weigh. Figure T2 summarizes all the possible scenarios to carry out the second weigh. On the sketch it is assumed that the left plate goes up and the right down, but it could be the other way around. During this first weigh, in addition to discarding four balls and limiting the contending ones to eight, we also will know which four balls on each plate move up and which four move down. This fact is summarized in the note at the bottom of the sketch. Are there other obvious facts regarding weighing using the two-plate scale? There are two more facts that happen to be well-known because they are part of the obvious hints with a slight variation. They are two of the most obvious already shared in Figure T1, and each subtly adjusted with the others. They may be so obvious that even when they are inside us, we usually do not see or recognize them as key facts to answer this puzzle. They seem to be hidden to us. It took days to simply visualize the combination of them. When they are combined, they build a new fact that is also obvious. Here are three steps to find it: 1st step- Re-weigh three balls (belonging to the probation group) on the same plate they were on the first weighed, but this time against three different balls which have been discarded. 2nd step- Re-weigh two balls (belonging to the probation group), one ball on each plate against themselves but after switching them to a plate

different to the one where they were in the previous weigh. 3rd step- Combine the two steps just listed above.

Figure T2. Eight balls get unbalanced on 1st weigh

As an example for the first step, see weigh “x” in Figure T2. According to this sketch, the three balls that were on probation and remain on the same left plate are 1, 2, and 3. We could use the right plate as reference, but the balls and conclusions need to be adjusted accordingly. These balls have to be weighed against any discarded ones, for instance 9, 10, or 11. For this scenario the conclusions are: (a) If the scale gets balanced (option B of potential weigh “x”), these three balls (1, 2, 3) on probation are discarded, and the probation group is reduced to five. This is the remainder of the eight balls that were on the plates during the first weigh. With this, the unknown ball should be either number 4 and be lighter, or be among 5, 6, 7, or 8 and be heavier because their plate went down during the first weigh. (b) If the scale gets unbalanced (Option U of potential weigh “x”), the unknown is one of the three balls on the left plate. In this case we are sure of two facts. The unknown is lighter and must be among ball 1, 2, and 3, and the left plate has to go UP again as it did in the first weigh according to this example scenario.

For the second step, see the potential weigh “y” shared in Figure T2. The two balls switched in this example are 4 and 5. The analysis is the same as when any other two balls are selected if they came from different plates with one exception: one of them has to be different from ones selected from the plate where three were taken to do potential weigh “x”. Doing that, the following four obvious and important facts may be inferred: 1) If the scale gets balanced (option B of weigh “y”), none of these two plate-switched balls is the unknown because such ball is unique and the only one that creates the imbalance. 2) If the plates get unbalanced again on this weigh (option U of weigh “y”), they also must switch their up-down movements regarding the first weigh because during the balls switching, we also switched their effects too. So according to the example sketch, the left plate (focusing on the right will not make difference on the analysis) now has to go DOWN because on the previous weigh it was up. Keep in mind the up-down movement direction of this plate on the first weigh. The next two conclusions are consequences of this one. 3) The unknown is one out of these two balls, and 4) If it were number 4 has to be lighter because the left plate was up on the previous first weigh, and heavier if it were 5. For the third step, notice the result of weigh 2 on Figure T2. It is a combination of the balls and their positions on the plates of the two steps described above. Keep in mind that if on this second weight the scale is unbalanced, the movement of the left plate has to be governed by the “x” and “y” potential weighs. Thus, the possible scenarios of this second weigh are: 1) If the plates balance (B option of weigh 2), all the balls on plates become discarded too and the unknown is among those in probation off the plates (6, 7, and 8), and it has to be heavier because when this group was weighed on the first weigh, they were on a plate that was down. 2) If the scale is unbalanced (U option of weigh 2), and regarding the movement of the left plate: only balls 1, 2, 3 make it moves UP, and only

either ball 4 or ball 5 makes it moves Down. So, when left plate goes: a) Down, the unknown is either ball 4 and lighter, or ball 5 and heavier. b) Up, the unknown is among 1, 2 and 3, and is lighter. In essence this last scenario (2U.b) is equal to the first scenario (2B). On both, the unknown is among three balls and its weight was just found. According to the analysis of this second weigh, five balls out of the eight on probation can be scrutinized on just one weigh, four on one plate and one on the other. Notice that all the three possible scenarios for the third weigh of this case #2 will be limited to the following only ones. a) The unknown is between two balls which relative weights have been identified (one scenario of two balls: case 2U-Down in Figure T2) b) The unknown is among three balls which relative weight have been identified (two scenarios of three balls each: 2B and 2U-Up) These scenarios to the third weigh of case #2 will not be exposed because they are practically identical to the third ones of case #1 (See Figure T1). For three balls on each plate during the first weigh, the worst scenario is when the scale is balanced, and the unknown ball is among the six balls that are off the plates. Is it possible, in just two weighs, to find the ball among all the possible scenarios? The analysis of this case is left to the reader; but as a warm-up it is suggested first to find the solution on just two weighs assuming that it is already known that the unknown ball weighs more or less than the rest. The combination of two scenarios to create another is known as superposition. This is the action of bringing two or more possible scenarios into same position to see how they coincide. This strategy has been used by mathematicians, physicists, and scientists for many years. It is also a very common and well-known problem-solving strategy frequently found in nature. For example: water is the superposition of two molecules, oxygen and hydrogen. This could also be dangerous in some cases as when oxygen and hydrogen are superposed it becomes a dangerously explosive mixture. "So explosive, in fact, that in liquid rocket engines the two components are

mixed only just before the engine is lit” (Gray 78). Another well-known example is you, the reader. Yes, you are the superposition of two very miniscule parts: your father’s sperm and you mother’s egg. According to these examples, in nature the result of a superposition is very complex. Were you and I at the instant of being conceived the result of complex superposition? Was it an explosive moment too? Was it a big bang singularity? According to Sebastian Seung, professor of neuroscience from MIT, who details the development of our brain just after our conception, the answer to these three questions, I think, is YES. The solution of this puzzle shows us that we unconsciously follow Newton’s law of inertia. It forces us to stick with the known obvious seven hints listed at the beginning of the section and prevents us from seeing other options or combining such hints. We are always looking for easy ways to solve problems. In the deepest of ourselves, we do not want even for a fraction of a second to be considered stupid or accept that an obvious thing has mislead our thinking. So, we shield it in our subconscious with amazing power. ******

The Sudoku puzzle and a master pattern If a cluttered desk is a sign of a cluttered mind, of what, then is an empty desk a sign? Albert Einstein

Sudoku is a very popular puzzle which popped up almost simultaneously with the advent of cellular phone. It consists of filling up digits numbered from 1 to 9 on a 9 row by 9 column matrix. In each row and column, no digit is repeated. Figure S1 shows two examples of this puzzle that are partially filled, as it is normally presented to be completed by a gamer. Each puzzle has only a single solution.

Figure S1. Examples of the Sudoku puzzle

If you have not tried this game before you may wish to play with one for a while for fun and to get accustomed to it. The person who created this puzzle fresh should start on a blank 9x9 cells square filling in the nine digits on one row and then completing the rest. It sounds easy, but when I tried it, I realized that it was not. It took the writer several days to get the one that inspired him to write about this puzzle. I wondered if that was not easy, how is it possible to find magazines full of Sudoku’s and none of them seem identical. Thus, it has to have at least one pattern or way to complete it. Finding it is the main purpose of this section. Are the puzzles of Figure S1 the same? At first glance they seem to be totally different, but once the solution for each of them has been found out, it may be easy to answer that question.

Figure S2. Solution for the Sudoku examples

Figure S2 shows the solutions for the Sudoku examples of Figure S1. They still look a bit different; however, the solution of Sudoku “a” becomes identical to the solution of Sudoku “b” by switching its first column with second one, and fifth with sixth. The rest of columns, rows, and sub-squares not affected by the switched columns, are identical for both solutions. The

solution shown on Figure S2a can be rearranged a little further to get the solution shown on Figure S3a. The first row shows a particular and wellknown sequence that on the second and sixth rows is shown slightly rearranged. The rest of the rows and all the columns of this “lead solution” seem chaotic and not following any pattern at all. If that is true, how it can be a lead solution? There is a well-hidden pattern on it: an algorithm behind the software developed to generate the examples shown on this section.

Figure S3. A random solution from a lead solution

Can the solution of puzzle of Figure S3b, the one presented to the gamer, be associated to its lead solution S3a? Even knowing the lead solution linked to the puzzle, the answer is NO or very close to it. Let us see why not. To generate a particular puzzle, we first create its solution and then from a copy of such solution we extract random numbers from different cells. The extraction is done randomly and on a non-fixed quantities that vary depending on the desired puzzle difficulty-level. Easy puzzles have few numbers extracted. More are extracted for medium puzzles and on hard levels almost no numbers remain on the big square. Before extracting the numbers and making a copy of the lead solution, the software that generated these puzzles makes random changes to the particular solution: (a) interchanged one pair, two pairs or none of columns, (b) interchanged one pair, two pairs or none of rows, and (c) interchanged groups of three columns or rows or none. All these changes were done to create the examples in Figure S1 and Figure S2. To get the puzzle in Figure S3b, an additional, critical, and more challenging random process was added, namely the sequence (1, 2, 3, 4 …) of the first row of the lead solution was systematically altered in a random way. At this point, it is almost impossible to associate the puzzle to its lead

solution because the first-row sequence is no longer known. The solution of Figure S3b that was originally generated from the lead solution of Figure S3a, can no longer be linked to its hidden solution, due to the scramble done on its master lead solution. Now it is unique. Human DNA and the Sudoku puzzles seem to share two things. They follow a master pattern, and later generated patterns are slightly varied before the result is shown. Most of the living species on earth, including human beings, follow a pattern similar to the generation of a Sudoku puzzle. Each of us is a unique person coming from two master patterns, our parents. When the DNA of a child is compared to the DNA of one of his parents, the comparison is between two structured frames and a slightly modified copy. It is not against the root-leader or master solution, the one that dictates how many hands, legs, eyes, heart, type of skin, etc. a being must have to be human. It does not matter how chaotic, unpredictable, different, and weird something looks or behaves, there is always a pattern, rule, or a law behind them. It may be difficult to detect or understand, because of variance. It may also be because the changes are so tiny or so huge, or because the changes have occurred too slowly or quickly to observe. However, because nature and the universe follow patterns, there is a chance to reveal one of these patterns that are hidden and embedded on the scrambling processes. Newton’s laws are ones of the few of these patterns that has been discovered, yet there are more that remain unknown. *****

The Tower of Hanoi and a hidden pattern If we were born knowing everything, what would we do with all this time on this earth? Nelly

This puzzle consists of a set of (perforated) disks or tokens of different 1 diameters arranged in a kind of tower and three poles . Initially, the disks are stacked on one pole in order of size, smallest to largest. The goal is to move the disks, one by one, to a target pole, using a third pole as a relay. A

disk can only be moved on top of a disk larger than itself. This puzzle has been known for more than a thousand years. Some historians believe that it was formulated in India and others in China. The quantity of movements required to move the tokens to the target pole increases fast as the quantity of tokens increases. For instance, if a group of people attempted to reach the end of a tower of only 64 tokens, and “were able to move disks at a rate of one per second, using the smallest number of moves it would take them 18.440.000.000.000.000.000 seconds or roughly 585 billion years to finish (Lewis and Loftus 600), which is about 42 times the current age of the Universe (Wikipedia)”. If instead of movements they were wheat grains to be collected on one pile for each movement, this number can be also stated as: “If doubling the amount sixty four times starting from one, all the fields of India were sown with wheat in two thousand centuries you would not harvest such number of grains of wheat” (Malba 113-114) The Towers of Hanoi puzzle is used in several books of computer science courses as a typical example of a recursive solution of a problem where the iterative solution is not known or quite difficult to implement (Lewis and Loftus 595). Recursion is used to solve a problem by solving a small instance of the same problem. In the process, the recursion reaches a point where the solution of the smallest problem is not recursive again, and that solution is known as special or base case. It is usually the simplest one. For this puzzle, a recursive solution is feasible, though not practical for an elevated number of tokens because the minimum required movements could be millions of millions as is the case for just 64 tokens. The game appears simple when played because it is simple. The token movements sequence may also seem easy for a few tokens, but why it is not well-known yet after all these years? It would seem the sequence should have at least one pattern. In this puzzle there are three special cases. The first is when the number of tokens to move is just one, so it is moved directly from the source to the target pole. The second is when the tokens to move are two, in this case three movements are needed. When B pole is the target and A the source, the movements are:

#1: Move the top token from A pole to relay C pole. #2: Move the second token from A pole to B pole. #3: Move the one now on C pole to B pole. The third special case will be revealed in the coming paragraphs.

Figure H1. Two positions of a tower of Hanoi

Figure H1 shows two stages of this puzzle for a tower of five tokens that are labelled by number for identification and explanation purposes. On (a), they are on the starting position; on (b), the lowest, biggest, or last one (token #5) has reached the ultimate target (B pole) and all the rest have been moved to the relay C. On this last configuration the process is set for re-starting, but now with C as the new source and A as the new relay pole. A pole is the temporal target pole where three tokens have to be moved to liberate the new last one (token #4). In other words, the poles have swung roles and will keep switching roles until all the tokens achieve the ultimate target B pole. The steps and pole switching needed for towers of one or two tokens suggest that it is possible to find a general procedure or pattern for moving any number of them. To chase such pattern, let us study some traits of this puzzle.

Figure H2. Movements, and tokens on poles

Figure H2 summarizes the seven (7) movements needed for a tower of three tokens and the fifteen (15) movements needed for the one of four. In this figure the tokens are labeled by consecutive numbers from top to bottom (see Figure H1a). The Tokens on pole column shows which token is on each pole just after performing the step listed on the Movement column. On the From subcolumn, the required token movement is listed from current pole a token is in to pole it is moving to. The Notes illustrate the numerical and graphical position of the tokens during the 13th movement. In this case, moving the token from “A to C” means moving token #1 from pole A to pole C. A procedure to know which token is on each pole for any number of movements without going through each step will be described later. The reader may wish to try these cases to get acquainted with the puzzle if it is unfamiliar. Comparing the required movements for a tower of three tokens with one of four may look different at first glance, but they are not. Let us find out why by comparing the initial seven steps of the 4-tokens tower against the same seven movements of the 3-tokens tower. First, by interchanging the positions of the B and C columns of the 4-tokens tower you will notice they are identical to those in the 3-token tower. Doing such comparison, it is easy to realize that the seven movements are the same even though one tower has odd tokens and the other even. Interchanging the A and C columns on the last seven steps (step 9 to step

15) of the 4-tokens tower and then comparing them again against the seven movements of the one of three’s, it is clear that those movements are also identical to the seven initial movements. Even on larger towers with more tokens, similar interchanges in groups of seven with an eight step between them the writer realized that this pattern continues. The repetition of these seven movements is what makes them the third special case. The last token to complete the re-grouping of tokens in the ultimate goal pole always is moved in the last movement of the last seven fixed movements. From the above paragraphs, it is inferred that the repetitive and fixed seven movements of the third special case are to be performed as many times as needed to move all the tokens to the ultimate target pole. Interspersed among these repetitions, are the eighth movements, and the constant switching of the roles (source, target, and relay) of the poles. On the last of the fixed seven movements, the only time that two poles have no tokens happens when all have been moved to the ultimate target pole. On any other seventh movement, only two options are possible, namely no tokens on one of the poles or tokens on all three poles. Figure H3 shows numerically and graphically two of these last seven movements, number 15th and 39th, extracted from the steps of a seven-token tower (see Figure M3 in Math and traits about the Tower of Hanoi). When the distribution of tokens on the poles is known on any of the intermediate last of the fixed seven movements, the eighth movement is based on only two possible options. Using the cases shown in Figure H3 as example, this 8th movement shall be as follows: When one pole is empty, as shown in Figure H3a, the bigger token (five on the sketch) out of the two tops (one and five) should be moved from its actual pole (from A) to the empty one (to B). When all the poles are occupied, as shown in Figure H3b, the mid-size token (four) out of the three tops (seven, four and one) should be moved from its pole (from B) to the pole having the biggest top (to pole A).

Figure H3. The eighth movement options

This procedure works well for a manual solution as we can see which tokens are on the poles and know when the actual or last 7th movement has been performed. What makes it difficult to program an iterative solution is to know when and in what direction to switch the pole roles. Switching movements are masked by the seven fixed movements that are similarly hidden between the eighth movement. Without knowing the sequence, this pattern of movements seems scrambled and may be hard to observe. You may notice a similarity to the scrambled nature of the Sudoku puzzle. Looking at Figure H2 and using the special cases described by the one and two-token towers, it is possible to figure out what movement should first when starting a tower of any number of tokens. It will depend on whether the number of tokens to move to the (ultimate or temporal) target pole, is odd or even. When this number is odd, the first movement of the top one shall be to the (ultimate –B- or temporal) target pole; and if the tower is even, to the current relay pole, whichever it is after the pole role switching. The iterative solution. Instead of being arranged on a straight line, let us fictitiously arrange the target B pole so that the poles form an imaginary equilateral triangle as shows Figure H4a.

Figure H4. Tokens moving over a triangular pattern

Now let us suppose that from an odd tower of more than three tokens (see Figure H2b), we have extracted the fourth to eighth token movements and combined them to get Figure H4b. When these movements are performed over the poles on a triangular arrangement, the odd tokens (one and three) seem to be moving on the counterclockwise direction and the even tokens (two and four) on the clockwise direction. It happens that whatever size an even-tower is, all and each of its odd tokens always move in a counterclockwise direction, and all its even tokens in a clockwise direction. For any odd-tower, the even and odd tokens swap such movement directions. The triangular arrangement reveals a very important issue: All of the odd tokens always move in one direction and all even tokens move in the opposite direction for any particular tower.

Figure H5. Faces of the 3-positions roulette dial

To represent this fact, imagine a circle on each side of a disk with three dots over it representing the poles. The dots will form the vertices of an equilateral triangle. Mark the vertices as S (Source), T (target), and R (Relay) in the clockwise direction. This side of the disk will be called “Odd”. On the other side of the disk, draw the same vertices in a counterclockwise direction and call this face “Even”. Each disk face should then be marked with directional arrows to show the named direction of how the STR vertices were arranged. Figure H5 shows the two faces of such a disk as a two-face roulette dial. The odd face shall be used for odd token towers and the even face for even. For towers of any size of three or more tokens, the roulette faces can be used to perform the seven-fixed movements, and to re-assign the role of each pole. Let us see how to use the roulette for these purposes.

Figure H6. The poles and the roulette

Imagine that it is placed between the poles and set so that the S vertex aligns with the A pole as Figure H6 shows. This is the starting position for either type of tower. If the “odd” face of the roulette is selected, the T vertex aligns with the B pole and R matches C. When the selected face is the “even” one, T vertex will match C pole and R will face B as Figure H7a shows. Figure H7c is the same as Figure H2b but now the “From” column is associated to the STR vertices of the roulette instead of the ABC poles. As this tower is even (four tokens) the even-face was selected. The roulette in the initial position (Figure H7a) is used to perform the first seven-fixed movements (one to seven). In this case, the first movement “From S to T” actually means from A to C, see Figure H7a for the equivalence.

Figure H7. Movements referred to the STR sequence

However, once the roulette has been advanced one step forward (S being moved to matching with C pole), it is set to guide the next seven-fixed

movements or from nine to fifteen. In this case the 9th movement, “From S to T”, now means from C to B, see Figure H7b. Notice that the sequence of the 1st to 7th movements are identical to the ones from the 9th to 15th compared to the STR vertices of the roulette because the roles of ABC poles were switched. But those two sets of movements are different related to the ABC poles. On each switching, a pole will take the role of the one in front of it. In other words, it means the actual source will be the relay, the actual target the new source, and the actual relay the new target. The changing roles of the ABC poles should be done just after an eighth movement has been carried out. This movement shall be according to the description given in Figure H3. Except for the eighth movement, the marks (R, S, & T) on each roulette face act as variables that command from what pole (A, B, or C) to what pole a token shall be moved. The fixed and repeating seven movements of the third special case described in the above paragraphs are the key base of the proposed iterative solution (known pattern) that is valid for towers of any size (see Figure H8). The step identified by 0* on the sketch is only needed for a programmed routine, it is not needed for a manual process. The steps for the solution are: Step # 0*- Calculate and get the total number of minimum movements (M) required to transport all the (n) tokens to the target pole. Step # 0- Depending if the number of tokens is odd or even, select the roulette face (see Figure H5) to be utilized, and align the roulette S vertex with the A pole. In the case of a computerized routine, assign the initial (S, T, R) roles to the poles and re-initiate the counters.

Figure H8. Tower Hanoi: iterative solution sketch

The above steps are listed as zeros as they are performed just one time. For the manual process, a rosary or a string of 8-beads having one bead bigger than the rest, can be used as a reminder to perform the eighth movements. Advance a bead for each token movement performed. Step #1 - Perform the seven repetitive movements grouped into this step. On the computerized process, increase a counter of overall movements (for example m) for each movement carried out. This counter is used to tell the computer when the whole process has to be stopped. Step #2 - Perform the eighth movement. Further detail about this step will be shared later for the computerized process. For the manual process, perform this step when dictated by the bigger bead of the rosary, and on the computer while the counter is lower than M. Step #3 - Switch the role of the poles. For the manual process, this is done by moving the roulette dial one step forward. For the computer, reassign the S-R-T roles to the poles and re-adjust a counter of a loop/subroutine that controls this movement. If all the tokens have not reached the target pole yet, return to step #1 to repeat the whole process. For a high number of movements in the manual process, it may be hard to remember the actual role of the three fixed poles A, B and C. This is the revolving roulette disk does for us. Once a pattern has been selected, the turning of the roulette shall always be in the same selected direction, counterclockwise or clockwise until all the tokens reach the target pole. In summary, for four or more tokens, any solution comprises two

repetitive basis movements (step #1) and one eighth movement (step # 2). This depends on two factors, namely knowing exactly when the seventh movement has been performed and knowing the size of its top tokens. Let us suppose that we want to know how many jumps token number five has done embedded into an unknown group of tokens that all together have been moved 17,093 times. We also want to know in which pole it should be located after its last movement. Detailed formulas and procedures to know the position of each token for any number of movements without performing all of them are described in the next section Math and traits about the Tower of Hanoi. However, this information also can be found by a manual procedure by creating a table similar to the one shown in Figure H9. According to this table, token number five has been moved or jumped 534 times and it has reached the A pole on its last movement. The procedure to build such table is as follows:

Figure H9. Position of 15 tokens after 17093 movements

In the Jumps column, each number is found by dividing the number of jumps in the cell above by two and rounding to an integer if needed. The first value, the one for row or token number one, is achieved by dividing the total number of movements (17,093 in this case) by two. When this quantity is odd, the result of this first division (17,093/2 = 8,546.5) always needs to be rounded up to the nearest integer (8,547). This column is completed when the last number is one or zero, and the sum of the number of jumps adds up to the number of total movements given. You will notice the sum of jumps in

the example table adds to 17,093. Once the Jumps column has been completed, the minimum number of required tokens to generate the total number of movements will be equal to the number of rows (15 on the example) on the table. Regarding the rounding process, any rounding up action is posted on the Round column with the symbol ↑ and the rounding down, should be reported on the table with the symbol ↓. Each of the up and down arrows is a remainder for a next rounding which should be done opposite to the previous one. The first rounding is always up. When a division has no remainder and requires no rounding, the cell is left blank or filled with “---“ to avoid confusion. That is how the jumps of token #2 (4,273) and the rest of the rounded values of the Jumps column were found. We now know how many jumps each token requires but still do not know what pole the token will be in. Let us finish the table to know how to get the position of each of the tokens in the poles. The cells of the Mod3 column are completed by dividing each Jumps value by three (the number of poles) and extracting and transforming the remainder into an integer. When we perform such division by hand, the only possible values of this remainder are: 0, 1, or 2. Using an electronic device (phone calculator, computer, etc.), the remainder of such divisions could be expressed as a decimal. However, it only has three possible values or options: nothing left over (or 0), .333…, and .666… Notice that the fraction 1/3 translates to the repeating decimal 0.333…, and the fraction 2/3 to 0.666… Thus, when the remainder is zero, input 0 in the cell, input 1 for .333…, and when it is .666… fill the cell with 2. For example, regarding token #2 with 4273 jumps that divided by three its remainder is 1 because 4273÷3 is equal to 1423.333. Assigning an integer to the remainder of a division of two integer numbers, is known in mathematics as the Modulo, often abbreviate as Mod. That is the reason why this column was called Mod3. Each remainder value represents one of the poles with 0 being equivalent to A pole. The input to the Move column is related to how a token is moving around the poles when they are arranged on an imaginary equilateral triangle (see Figure H6). Each token is always moving from one pole to

another in the same direction while it is guided toward the ultimate pole target. The direction of movement for token #1, the leader, depends on the total number of tokens. Its direction is clockwise or CW (shown on the table as →) when the tower is odd and counterclockwise (CCW or ←) if even. Odd tokens will be moved in the same direction as the leader, and even tokens in the opposite direction. Clockwise means that the jumping is from A, to B, then to C. For illustration, we will call that direction ABC, while for counterclockwise movement it will be ACB. This nomenclature coincides with the values (0, 1, and 2) reported in the Mod3 column and means that for the ABC direction, 0 is equivalent to A, 1 to B, and 2 to C; while for the ACB direction 0 is also equivalent to A, but 1 is to C and 2 to B. The Pole column is completed by combining the information in Mod3 column and Move column. For example, regarding token #2 for which the Mod3 cell value is 1 with a direction of movement of (←) counterclockwise or ACB will be at pole C when all 17093 movements have been done by the 15 tokens. Once the table is complete, it is possible to find all the tokens on each pole by simply looking down the Pole column for each. The Tkn# column will display the tokens (see Figure H9). By looking up all the tokens listed by letter in the Pole column, every token may be accounted for. For example, looking down the Pole column the first listing for A has a Tkn# of 1, the second listing for A is 4, and the third is 5. All the listed tokens for the A pole are 1,4,5,6, and 9. For the B pole the tokens are 3 and 15, and for the C pole the tokens are 2, 7, 8, 10, 11, 12, 13, and 14. This result is shown on the bottom of the table of Figure H9. This result is valid for odd-numbered towers of 15 or more tokens. It is important to note that if these 17093 movements are part of a 16 token or greater number even tower, the pole letters will be different because the direction of token number one (and all other odd tokens) will be opposite to the one shown in Figure H9. Nevertheless, the quantity and number of tokens will remain the same per pole. This means that on one of the poles will be tokens 1, 4, 5, 6, and 9, another will have 3 and 15, and the rest will be on the remaining pole.

By visualizing the poles as if they were in a circle and finding constant cycle of token movement allowed for the discovery of a simple and obvious pattern. As is clear now, this pattern is composed of two steps: a fixed 7movements pattern and one easy to carry on eighth movement. This pattern was discovered by the writer, and provides a simpler iterative solution instead of the recursive one used currently. The three puzzles described on the previous paragraphs demonstrate a simple and very important fact. As knowledge becomes more common over time there is a human inclination to accept the established status quo and a similar reluctance to challenge or question it. These realities make it difficult to modify and substitute old knowledge, especially when such knowledge is solidified thought our formative years as children, adolescence, and into adulthood. Substituting prior knowledge with new learning is hard for the human brain. It requires first erasing what we already believe to be true before making room for something different. In a way this feels like denying the truth as it creates an emptiness, a vacuum, or a blot in our brain. This can make us uncomfortable. What we forget is what it felt to be a newborn, looking at the world anew. ******

Math and traits about the tower of Hanoi Slow down and everything you are chasing will come around and catch you. John De Paola

A tower of Hanoi puzzle is any number of perforated disks or tokens initially stacked on one pole in order of size such that a small disk is on top of another bigger. In the following paragraphs it is assumed that in the initial arrangement, the token on the top of the tower is numbered one, the next under is numbered two, and so on to the last one as it is shown on Figure M1a. It is also assumed that each of them are moved over three (ABC) poles arranged in a triangular layout while following the path of minimum movements to reach the target pole.

Figure M1. A tower of Hanoi and the three poles

Once a moving direction is set up, odd tokens shall be moved following the direction preset at the start of movements, and the even tokens in the opposite direction. These moving directions are identified as clockwise (CW) or counterclockwise (CCW) (see Figure M1b.) For an odd-tower or those with an odd number of tokens, the top (number one) and other odd tokens shall move/turn on the clockwise direction, or following the ABC path over the poles. Compare this to the tower whose first and shortest path is from pole A (source pole) toward the goal pole B. On the contrary, the even tokens shall move on the opposite (counterclockwise) direction following the ACB path. However, on an eventower the turning rotation of its odd and even tokens are opposite to the odd-tower described above. In mathematical terms, the movement direction of a token T being part of a tower of n tokens, can be summarized into two parts as If n is even, and T is

If n is odd, and T is (M.1)

Where the rightwards arrow above T means that token is moving on clockwise direction; and leftwards arrow above, on the counterclockwise direction (see Figure M1b).

Figure M2. Total number of movements by formulas

The column labeled Total of Figure M2 summarizes the minimum quantity of movements needed to move each of the tokens listed on column Tkn Qty. The last two columns list how these totals can result using two different calculation methods, which generic formulas are displayed on the last row. These formulas actually are just one expressed in different ways but the one on the bottom right corner insinuates the seven-fixed movements and the eight movements. We already know that on odd and even towers of three or more tokens, there are seven fixed movements that are repeated after an eighth movement. They are hidden but they become “visible” once the location of the static ABC poles is swung after each eighth movement. After each swing, a pole will take the role of the one in front of it. In other words, it means the actual source will be the relay, the actual target the new source, and the actual relay the new target. The minimum number of movements Mn needed to move all the n tokens from the source A pole to the goal B pole (see Figure M2) is given by: (M.2)

The above equation can also be used to find an unknown minimum number no of tokens needed to get any number M of movements lower or equal than Mn, that is: (M.3)

Due to the logarithm, the result of the above equation is usually not an integer and needs to be rounded to a whole number because the tokens are complete units. For example, for a total movement M of 17093, what shall be the minimum number of tokens needed according to equation M.3? It shall be: n₀ ≈ 14.06

Let us check out some integers close to 14.06 using equation M.2:

As case (b) gave a value higher than M (17093), the unknown no is 15. For very high numbers of M equation M.3 is especially useful. For example if M is 333,344,445,555 (≃33.3 x 10¹⁰), the calculated no is 38.25, and the actual needed no is 38. Figure M3 shows that with the help of several of the first movements of a tower of seven tokens it is possible to find some equations associated to towers of any number of tokens. In this figure, notice that token three is moved for the first time during the fourth movement which is equivalent to 2(³-¹) = 4. Similarly, the first movement of token number four happens at the eighth movement that is equivalent to 2(⁴-¹) = 8. Whatever the tower size, the minimum number of movements needed for a certain token to be moved the first time is given by the power of two elevated at the token number minus one. In math terms, for any token T, the minimum movements MT,1 needed to move it the first time is given by: Also notice that token number four is moved or jumped for the second time at the 24th movement, and it is moved for the third time at the 40th movement. From these three movements, we get that the movements’ gap (24 - 8 = 16) between its second and first jump is the same as the gap (40 24 = 16) between its third and second jumps. This gap is equal to 2⁴ = 16.

Figure M3. First 48 movements of a 7-tokens tower

For token number three, the gap between consequent movements is eight. This is equivalent to 2³ = 8. In general, for a T token, the movements’ gap between its consequent jumps is the power of two elevated at its number (2ᵀ). So, for the second movement of each token, according to the path of minimum movements, is the gap plus the quantity of movements needed to get its first jump. Thus, the second jump of any token T happens at movement: or From Figure M3, the third jump of token number four (4) takes place at the 40th movement, which based on the above equation is equivalent to: In general, when a T token embedded into a group of other tokens has been moved J number of times, the partial movements MT,J of T and the other tokens of the tower that have been also moved so far add to:

MT,J also shows when T gives its last jump J. Simplifying and arranging a bit the above relationship, it becomes: (M.4)

Because both J and T are whole numbers, and because 2ᵀ is always an even number, the result of above equation is always a whole number also. From this equation, the total jumps J given for T within a partial movements MP of the tokens is

The result of the above equation is only a whole number in some cases when MP is an odd number. However, the jumps of a token are complete movements that are represented by a whole number. This means the result must be round off when needed. To represent this, the above equation needs to be transformed to: (M.5) 2

The two parallel bars (| |) mean that we must discard the decimals and keep the whole part of such number. To see how equation M.5 works, using Figure M3, let us find the net jumps (J) for token number five that are part of the 47 movements (MP = 47):

This result shows that token five has contributed one movement to the total of 47 movements given by all the tokens. Notice that when it gives its first jump (J = 1), the whole group has moved just 16 times. Per equation M.4: According to Figure M3, after this movement, token five keeps this position until the 47th movement. Now, let us find out how many jumps token five needs when it is part of a group of tokens that have moved a total

of 48 times. In this case, equation M.5 gives:

The results can be double checked using Figure M3. If we would use equation M.4 and find at what movement token five has its second jump, the result will be 48. This means that MP equals MT,ⱼ only when T is the last token moved in the group. According to Figure M3, all the tokens of the tower have been moved except for token seven. Using equation M.5, we can get the jumps of these six tokens (one to six) for 47 and 48 overall movements of such tokens and generate the two tables in Figure M4. In the two cases displayed by this figure notice that the sum of the jumps of all the tokens matches with the combined quantity of movements (MP) of all of them.

Figure M4. Jumps of first six tokens of a 7-tokens tower

According to the figure, the number of jumps of any of the tokens is near half of the ones of its previous token except for those of the top (number one) token. In Figure M4b the jumps of token five is the only one that is not half the jumps of a previous token because three jumps (of token four) divided by two equals one and a half (1.5). The result in this case seems to be rounded up to two (the nearest greater whole number) by “the two parallel bars” of equation M.5. However, in Figure M4a, this division was rounded down to one. The next division (1/2 = 0.5) for token number six was rounded up to one. If the overall number of movements (47) is divided by two, it equals 23.5 and is rounded up to 24. This is the value found by using equation M.5 for the jumps of the top token. In this table, this is the first rounding that happens to be the previous to the rounding associated to the

jumps of token five. Figure M4 summarizes these rounding steps with up and down red arrows on the right side of the tables. Based on these observations, it is possible to infer that the jumps of any token can be found without using equation M.5 and instead by dividing the jumps of its precedent token by two. If the jumps of the precedent token is an odd number, such division has to be rounded off to one of the two nearest whole numbers. The first rounding has to be up and the subsequent rounding, if any, shall be down. In general, when the rounding is performed more than one time, the rounding process shall alternate between up and down. . Now that the last movement (J) of a token can be calculated, let us find what pole (A, B, or C) it will reach when it has jumped certain number of times. As a token can be in only one of three (3) possible places or poles, its exact position can be found using the Mod of a number. That is given by the remainder of dividing the number of jumps (J) by the number of poles. As 3 the results of Mod3 are: 0, 1, and 2, the possible positions of token T moving in the counterclockwise direction and after its last jump J,is given by:

(M.6a)

Or when T is moving in the clockwise direction,

(M.6b)

Where Pc is the position of token T when it is moving on counterclockwise (←) direction and has been moved J times, and Pw is its position when moving on (→) clockwise direction. According to equation M.6, to get the position of T, two parameters shall be known, specifically the Mod3 of its jumps (J) and its moving or turning direction. Let us find how it works using the following example based on Figure M5: What shall be the position of token number six after 1005 combined movements when it is part of a 10-tokens tower? First, using equation M.5, let us determine how many jumps it has done

As the tower is even (n = 10) and the token is also even (T = 6), according to equation M.1a token number six is moving in the clockwise direction (→). Thus, as 16Mod3 equals to one and according to equation M.6b, token six shall be at the B pole after the 1005th movement (See Figure M5).

Figure M5. Few movements of a 10-tokens tower

To check equation M.4 for higher relative numbers, let us find out on what movement token six was moved by the 16th time. Using this equation, it happened at the 992th movement: Notice that this movement is the 124th (124 = 992/8) of the 8th movements. Using the equations and the procedure developed in the previous paragraphs it is possible to find out the position of each of the tokens on each of the poles after a large quantity of movements, such as required by a 64-tokens tower. This task shall be reasonably easy even for this tower because of the relative low quantity of tokens.

Figure M6. Tokens position at the 1005th movement

However, in some occasions when a computer is not available, such a task could be even easier to do by creating a table similar to the one shown in Figure M6. This table is presented for clarification purposes but will not be detailed because the procedure to develop it was exposed in the previous section summarized in Figure H9. Compare the result of this table for ten tokens with the tokens position after 1005th movement of Figure M5. The right side of the equation shown below (see Figure M2) consists of two main parts. They are the seven digits of the equation representing the last fixed seven movements, plus a term in parenthesis multiplied by eight. This multiplication represents the rest of movements composed by the other seven fixed movements plus the total of the eighth movement. The quantity of eighth movements (En) in a tower of n tokens is the term in parentheses in the above equation or: (M.7)

This equation is valid and gives positive values if n is greater than three. According to this equation, this quantity is 63 for a 9-tokens tower (see Figure M7), and 127 (2¹⁰-³ -1 = 127) for a 10-tokens tower. Figure M8 shows such movements where the last one occurs on the 1016th movement. In these figures, the pole at which a token arrives is shown mated with A, B or C. For instance, 6C means that token six is moved to the C pole. Notice that Tokens one, two, and three are moved on none of the eighth movements, so

only tokens bigger than three are moved in these movements.

Figure M7. 8th movements of a 9-tokens tower

Also notice that token number four not only is the one moved in the first and last of such eighth movements, it is also the only one moved on the odd-eighth movements. This means that its movements are half of all (En) the eighth movements. Thus, if the total of the eighth movements are divided by two, we get the eighth movements of tokens number four. That is:

As the tokens move from one pole to another, their movements have to be a whole number. So if we drop the half fraction (1/2), for tokens number four its quantity of eighth movements is:

Also, in Figure M7 and Figure M8 notice that the fives are between every other four. It means the quantity of fives are half the quantity of fours. Thus if we divide by two the above relationship, the quantity of eighth movements for token fives is:

From these two expressions for token four and five, it is possible infer that for token number T, its quantity of eighth movements is: (M.8)

Where En,T is the maximum quantity of jumps or eighth movements of

token T in a tower of n tokens. According to this equation token number five is moved 16 times in a 9-tokens tower (See Figure M7). The result of this formula is consistent including for the bottom token that only is moved one time. In this case T is equal n and the equation’s result equals one. Notice that equation M.8 and equation M.5 both give the total of jumps for a token. However, while the relationship M.5 is for any token, equation M.8 is just for those ones moved in an eighth movement in all the (Mn) movements of a tower. The eighth movements are unique for odd and even towers. For instance, on the 80th movement (or the 10th of the 8th movements) not only will token five be moved but it also must arrive to the A pole. Similarly, on the 320th movement, token seven will always arrive to the A pole. This fact means that only the tokens arriving to the B and C (or S and R on the STR sequences) poles will be the ones swapping the pole positions on even and odd towers.

Figure M8. 8th movements of a 10-tokens tower

Conclusions inferred from Figure M7 and Figure M8 that apply for other even and odd towers are: The odd and even tokens moved during the 8th movements also follow the ABC or ACB path imposed for the leader token number

one. Token number four is the first one moved within the 8thmovements. Token number five is the first odd token to move and it does it for the first time on the second eighth movement. In their first movement, all even tokens arrive at the same pole as the fourth. For instance, in Figure M8 tokens six, eight, and 10 arrive at the B pole in their first movement. In their first movement, all odd tokens arrive at the same pole as token five. Using the eighth movements of the 10-tokens tower shown in Figure M8 as example, it is possible to infer that the eighth movement sequence is symmetrical around the last token of a tower of n tokens. This n token also is moved in the middle movement of the Mn movements of such tower. And each of the two symmetrical sides about n is also symmetrical about token n-1 (nine). Also, each of the two symmetrical sides of n-1 is symmetrical about token n-2 (eight), and so on.

4

Figure M9. Last movements of a 7-tokens tower

Within a big tower of any (n) size are embedded n-2 sub-towers of m

tokens that depending on their size, can re-join or set in one or more times before all the n tokens reach the target pole. For example, the sub-tower of n-1 tokens always rejoins itself two times, namely the first time on the relay C pole at the movement just before the middle one, and the second time at the last movement when it rejoins with the rest of tokens at the ultimate target pole. Notice that in the 7-tokens tower, see Figure M3 and Figure M9, are embedded five sub-towers each of two, three, four, five and six tokens. For observations in such figures, it is possible to infer that the whole quantity of reintegrations, rejoins, or resettings of small sub-towers of m tokens that are part of a bigger main tower of n tokens is given by: (M.9)

Where is Rm,n is the times of resettings of a small tower of m tokens that is a sub-tower of a main tower of n tokens. For instance, in the main 7tokens tower, the three tokens sub tower (1,2,3) is re-integrated sixteen times (2⁷⁻³ = 16); the one of 4-tokens (1,2,3,4), eight times (2⁷⁻⁴ = 8). The number partial of movements Mm.r needed to move the sub-tower of m tokens within an n-tokens tower to certain reintegration r is given by: (M.10)

Where r varies from 1 to Rm,n. For instance, the sub-tower of three tokens rejoins for the 9th time (r = 9) at the 71st movement (71 = 9x2³ – 1) (See Figure M3 and Figure M9). Notice that in this case the sub-tower is alone. However, on its 12th rejoin that occurs at the 95th movement (95 = 12x2³ – 1), it rejoins forming a bigger sub-tower of five tokens. On the other hand, on its 13th resetting which happens at the 103th movement, it lay over other two tokens neither alone nor creating or complementing another bigger sub-tower.

Figure M10. Re-settings of different sub-towers

When the small sub-tower matches with the main tower, the re-joining happens at the target pole, at this condition (r = 2(n-m)) equation M.10 becomes equation M.2. In general, the number of re-settings Rm,n of a subtower is the addition of three items: (a) the times (Rs) it set into or becomes part of another bigger complete sub-tower that can be the main tower when it has been completely moved to the target pole, (b) the times (Ro) it rejoins over other tokens but is not forming a bigger sub-tower and (c) the times (Ra) the tower rejoins alone on a pole that could be also the source A pole Figure M10 shows details of these type of settings (sub-big tower, over, and alone). Figure M10a summarizes the quantities of rejoins of sub-towers of three and four tokens within main towers whose number of tokens vary from six to ten. In addition to the total for each sub-tower, this figure also shows how they are distributed into the different settings (alone - a, over o, and sub-big-towers - s). Figure M10b shows the first 16th rejoins out of 32 of a sub-tower of three tokens that is part of an 8-tokens tower. Figure M10c summarizes the rejoins of the 3-tokens sub-tower within a 7-tokens tower displayed in Figure M3 and Figure M9. The last two figures also show in which pole each of the re-joins takes place. *****

Formula of sphere volume and algebra

Mathematics reveals its secrets only to those who approach it with pure love, for its own beauty. Archimedes

Finding the formula of the volume of a sphere using just geometry and algebra is not an easy task (see Algebra and the seesaw appendix for a review of these topics). It required the ingenuity of the great Archimedes (287-212 B.C.) who was the first to find it using his postulate about levers. His method was as follows:

Figure V1. Sphere, cone, and cylinder dimensions

Imagine a sphere of radius R along with a cylinder and a cone that are fabricated of the same solid and homogeneous material. All of these shapes have the same height (2R), and the cylinder and the cone have a base whose diameter is equal to twice the diameter of the sphere (or 4R). These shapes can be seen in Figure V1. If we took a cross section of each of the bodies and drew (superimposed) them to coincide along their central vertical axes, they should adopt a shape similar to the one shown in Figure V2a. The illustration is turned 90 degrees to one side of the cylinder for explanation purposes

Figure V2. Sections of cylinder, sphere, and cone.

In Figure V2a an imaginary and horizontal line that “coincides” with the bodies’ centerline axes and passes through the O point, divides each of the

cross sections of the three bodies into two equal parts. This line divides the 5 triangle formed by the cone into two isosceles-rectangular triangles whose perpendicular sides (a common height and their bases) have a length 2R. In this figure, the two blue and two purple triangles are identical because their hypotenuses are a diagonal of a square of 2R sides. On the purple ones, the horizontal line and a vertical line (thin slice) at any distance x from the point O creates two sub-triangles that are also isosceles-rectangular triangles whose adjacent and identical sides (base and height) have the same length x. Thus, this cone-slice has a radius equal to x, or a diameter equal to 2x. On the other hand, this vertical line intersects the circle created from the sphere at a point that is at a distance b from such imaginary horizontal line. So, the green sphere-slice has a radius b or a diameter 2b. This vertical distance b is also a side of the internal and small rectangular triangle whose hypotenuse is the radius R of the sphere, and the other adjacent side is a. Per the Pythagorean Theorem (in appendix, see equation A.5 and Figure A4), the b value of such triangle is: However, according to the excerpt shown in Figure V2b, the distance a is equal to the distance x subtracted by the radius R, or in mathematics term “a = x − R”. After substituting this value into the above relationship, it becomes That in appendix Algebra and the seesaw, it is equation A.6 which is equivalent to equation A.6b, or Between two persons of the same sex and having the same height, the sturdy one (having more volume) weighs more than the slender one (having less volume). This is true if they have the same body (denseness) type. With 6 this in mind, it is possible to infer that the weight of an object depends on its volume and density. For instance, a liter (about a quarter of a gallon) of water weighs a kilogram, but a liter of mercury (a solid liquefied at room

temperature) weighs 13.6 kg because its density is 13.6 kg per liter while the water density is 1 kg per liter. In other words, when two bodies have the same density and volume, they must weigh the same weight whatever their shapes. For instance, if you have a ball made of Playdough, and you gently rammed it into a disk, the subsequent disk and original ball have the same weight. Let us imagine that by using a special slicing tool, the three solids in their actual upward position (Figure V1) are sliced horizontally, as thin as a paper sheet, into the same number of slices. For the circular sliced disks that are made of the same homogeneous material (same density), their weights are proportional to their volume. This weight equals its height times the surface area of its circular cross-section which is times its radius squared (A = πr²). Therefore, for each slice having a very small height of one unit, their weight or volume becomes equivalent/proportional to their areas. They are: For the cylinder-slice of radius 2R (see equation A.10),

For the cone-slice of radius x, For the sphere-slice of radius b, it is equal to

Which is also equal to equation A.9a, or Now let us suppose that at a distance x from the top of the sliced bodies (or from their common summit O line) one slice of each body is taken out. Imagine two slices, one from the cone and the other from the sphere, are set on the left side of a lever (seesaw), and the slice from the cylinder is set on the right side (see Figure V3). Notice that if the slices were taken from a place near the top where the

distance x is close to zero, the sphere and cone slices would be almost a tiny dot. This also means that the cylinder slice which always has the same diameter should be very close to the fulcrum of the seesaw. It is possible that the cylinder slice could be balanced by the other two smaller disks provided they are far enough from the fulcrum.

Figure V3. Circular slices hanging from a lever

In Figure V3 the cylinder-slice is hung on one side of a seesaw at a distance x from the fulcrum and the cone and sphere slices taken from the same distance x are hung on the other side at a distance D from the fulcrum. For this distribution, at what distance D do the bodies on the lever balance? This is similar to the kids and the father on a seesaw detailed in Appendix Algebra and the seesaw. They will balance when the two opposite turning effects at the two sides of the seesaw have the same magnitude. The lever will balance when the (weight) area-distance turning effects of the slices on both sides are equal (see Figure V3). That is After replacing the Asp, Aco, and Acy areas for their corresponding values calculated on previous paragraphs, the above relationship becomes Dropping the bracket after canceling out the πx² inside terms The distance x is always positive since when it is zero there is nothing to balance. Dividing by 2πxR both terms of the above relationship results in

So, the distance D at where the lever balances is always the same (2R) constant and does not depend on the variable x!!! This means that for whatever trio of slices extracted, the slices of the cone and the sphere, whatever their sizes, will always balance at the same distance 2R against the cylinder-slice, whose distance to the fulcrum shall be the extraction distance x measured from the top of the bodies. Notice the resemblance in the example (see Appendix) where the kids always sit at a fixed place and their father changes position on the other side of the seesaw to balance them. When all the slices have been extracted one by one from the top (when x is the first slice) to the bottom (last slice or when x is equal to 2R) and balanced, we have added both the entire weight of each body and the variable distance x. So, if we want to balance the entire bodies, the total weight of the cone and sphere has to be concentrated at the fixed distance 2R just as all their slices were. On the other seesaw side, as all the x distances add to the cylinder length or 2R, the cylinder mass can be concentrated (suspended) at the middle of this distance or at a distance R from the fulcrum. To visualize this, imagine this distance as a solid bar (like a pencil) of uniform cross section. Its “weight” or center of mass is concentrated at the midpoint of its length. For that reason, the three bodies shall balance as it is shown in Figure V4. This balance has been proven with actual bodies at the University of Texas at Austin (Starbird -Lecture 9).

Figure V4. Sphere and cone balancing against the cylinder

In Figure V4, while the sphere volume (Vsp) is unknown, the volume of the cone and cylinder solids can be computed by their known formulas. As

the radius and height of these two objects is 2R (see Figure V1), their volumes are: For the cylinder, it is the area of its circular base times its height, or (see development of equation A.12 in Appendix Algebra and the seesaw for details) For the cone, it is a third of the area of its circular base times its height, or (see development of equation A.13):

It is known that the volume of a body times its material density is equal to its mass, which multiplied by gravitational force is its weight. But as the density and gravitational force act simultaneously over the three hanging bodies at both sides of the lever, they cancel out leaving just the volume of each body. Thus, the balance is obtained when the distance-volume turning effects on both sides of the lever are equal, or when (see Figure V4): Dividing each side of this equation by 2R, and dropping the parentheses, we get:

Then subtracting Vco on both side of the expression, it becomes:

Inserting the cylinder and cone volumes, or values of Vcy and Vco, we can write it as

After adding a parentheses to extracting πR³, grouping and re-arranging the remaining factors or fractions in such parentheses (for details, in appendix Algebra and the seesaw see equation A.1, and Figure A3 Combining additions and multiplications), the equation becomes

7

Performing the multiplication and subtraction in the parentheses

This is the well-known and current formula for the volume of a sphere of radius R. After Archimedes, others followed who were able to find this formula using different approaches and without calculus but using the now wellknown work of Archimedes in generating isosceles-rectangular triangles from a cone and a cylinder. The generation of this triangle is possible when the cone and the cylinder have the same height and same base, but their diameters are twice the height as the ones shown in Figure V1. However, these “obvious” artifices (the lever, bodies’ proportions to create the isosceles-rectangular triangles, and small slices or differentials) required the true ingenuity of Archimedes. He linked the cylinder, the cone, and the sphere that appeared only to share being made of multiple circular disks. Archimedes’ procedure now may look easy and straightforward because the calculation for sphere volume is now already known but imagine figuring it out on your own without knowing it and without calculus. *****

Appendix: Algebra and the seesaw The reader that frequently uses or remembers algebra may skip this section. As everybody forgets things, let us start this section by reviewing the algebra associated with the upward and downward motions of a seesaw (teeter-totter), a long narrow board with a center pivot. These motions and the physics behind them are the most important facts to develop the formula of the volume of a sphere using the procedure devised by Archimedes and detailed in the section Formula of sphere volume and algebra. Imagine that three friends meet together at the house of one of them. By chance, each weight 100 kg (220 lb) and each is the father of two kids, a

boy and a girl. The weight of each of the six kids is unique. To teach a bit of the basic law of physics to their children, they decide to play with the kids on the backyard seesaw using the following rules: Each father shall balance only with his own children. The family with the lightest overall weight shall be the first to play and the heaviest family will be the last to play. On the seesaw, the children shall sit on each other’s laps at a distance of 5 meters (5.5 yards) from the seesaw’s fulcrum on the left side. The father shall sit on the other side at a distance from the fulcrum that makes the family achieve balance or equilibrium. To be balanced, nobody shall push off the ground (feet in the air), and the board shall be completely level and not moving. The seesaw is similar to an old traditional scale with two plates and bowls that are suspended at an equal distance from a fulcrum. At the time when this scale was widely used, the single-tray supermarket scale and electronic scale did not exist. For the traditional scale, one plate holds an object of known weight (or mass), for instance a kilogram (pound), while on the other plate items such as rice, cheese or other food were added to be purchased by the customer. When the bowls level off without moving, the two objects (bodies) on the scale have the same weight (mass). Another more modern version of this scale is found in some clinics and hospitals. On the platform (acting as a bowl) of this scale stands the patient while the nurse slides small cylinders of known weight on a calibrated bar until there is a balance. Notice that on the hospital scale, the unknown weight (patient) stands fixed on one plate while the known weight is moved until there is equilibrium. On the other hand, in our seesaw the known (father’s) weight is the one to be moved to achieve a static balance. Even with these differences, the two-plate scale, the hospital weighing scale, and the seesaw perform under the same basic principles. In the seesaw, is obvious that one end moves upward while the other does it downward, but it is also true that the two sides rotate around the fixed central pivot point. It could be hard to visualize the seesaw turning effect because the upward and downward motions alternate and happen on very short paths. To visualize this turning effect in the seesaw that occurs in

a vertical plane instead of the horizontal, consider a working wall clock with second markings. All the three hands of the clock move in a circle around the axle holding them. Watch the two ends of the seconds hand when is at or very close to the horizontal position. Notice in this position that while one end of the seconds hand moves upward the other goes downward. The hands in the wall clock always move in just one direction whereas the seesaw move upward and downward in alternate directions.

Figure A1. First family balanced

In physics, for two forces cancel out or balance one another, they have to meet three conditions: (1) be opposite, (2) have the same magnitude, and (3) be collinear or acting along the same straight line. In a seesaw when a father-kids family is on balance, none of these three conditions are met. However, father and kids balance each other on the seesaw because the children’s weight is turning around the fulcrum (point “o” in Figure A1) in a counterclockwise direction (red path on Figure A1), while their father’s weight does this effect in a clockwise direction (blue path on the sketch). They do not move because the product of their weight times the distance, or kilograms times meters (Kg x m or lb x yard) to the fulcrum on both side of 8 the seesaw are equal. This “Kg x m” product generates a turning effect known in physics as moment of a force that has the same magnitude and opposite directions at the two sides of the seesaw when it is in equilibrium. To consider this product using some numbers let us suppose that the weight of the girl and the boy of each family are: 20 Kg and 40 kg for the first family, 50 and 30 kg for the second, and 10 kg and 90 kg the third. Notice that the overall weight of the kids per family are 60 kg, 80 kg and 100 kg. For the first family, the turning effect of the kids is 300 (kg.m) which is the product of their combined weight (60 kg) times their distance (5m) to the pivot point “o” ( 60 times 5 equals 300). The turning effect on the other side is also 300 (kg.m) or the product of the father’s weight (100 Kg) times

his distance (3 m) to the fulcrum. As these two multiplications are equal, the kids and father of the first family are balanced in such positions (see Figure A1). Once all families have played, they create the results summarized on Figure A2.

Figure A2. Balanced families summary

We are using whole numbers for explanation purposes to make it easy to carry on the products, however whatever the weight of the persons on each side, it is always possible to get an equilibrium on the seesaw. For instance, if the weight of the two kids were just 10 kg, their turning effect would be 50 Kg.m (10 times 5), and their father only has to sit even closer to the fulcrum, or in this case at a half meter (0.5 x 100 = 50). Notice that the two tables in Figure A3 are simplified excerpts of Figure A2. Even when the multiplications in each row of the tables are different, the resulting products are the same. For instance, the products on the second row are 80 times five, and 100 times four; the results of these multiplications are 400. Notice also that the addition of the three product results (300 + 400 + 500) on each table is equal to 1200.

Figure A3. Combining additions and multiplications

These values are shown below the tables and expresses the result of the products of two factors (one unknown and one known). The known factor is the fixed value of each table. On the left column, it is the fixed distance to the fulcrum where the kids sit. For the right column, it is the weight of the fathers. The second factor, the unknown or variable, can be found by two different ways. One way is by a division: the product result (1200) divided by

the fixed and known quantity equals the unknown factor. For example, on the left table, 1200 divided by 5 equals 240, and on the right table, 1200 divided by 100 equals 12. The second way is by the addition of the variable factors. Notice that if we add 60, 80, and 100 the sum is 240. Similarly, if we add 3, 4, and 5 the sum is 12. Knowing this fact, is it possible determine the overall weight of just the three girls if we know the overall weight of the boys and the fathers? Yes, it is. Here is an example using the left table with the kids. According to the previous paragraph, we already can determine that the overall weight of all the boys and girls is 240. From this we know the overall weight of all the girls equals 240 minus the overall weigh of the boys. If in the Figure A3, we assign fonts to the variables (for instance a stands for 60, b for 80, and c for 100), and we call K the fixed constant (in algebra a symbol — a, b, K, α, takes the place of a number, or a constant that could be another value such as pi — π). Using these assignments for the left table, we can rewrite or combine the multiplications inside of such table (60x5 + 80x5 + 100x5 =1200) and the multiplication below this table {5x (60 + 80 +100) =1200}. As the right side of both multiplications are equal, their left sides must be also equal. Equalizing them, we find: (A.1) (A.1a) (A.1b)

The three above equations are really the same where the multiplication x symbol was substituted by either a dot (as in equation A.1a) or by leaving a blank space among letters and between letter and parenthesis (as in equation A.1b). These substitutions frequently are done to avoid confusion when one of the variables is x. In the next paragraphs when the number of two or more equations differ only in a low type letter like the three above, it means these equations are equivalent or it is a unique one expressed in different ways. The tables of Figure A3 suggest that the above relationship is valid in both directions (the left to right or the right to left). In other words, if we know the terms of the left side, we can generate the right term by placing

the constant and add a parenthesis with the variable inside. For example, if it is known that: (A.2)

According to equation A.1b, the parenthesis is eliminated if the constant π (pi) is multiplied for each term (variable or not) inside the parenthesis and we keep the same sign of each internal term. Then the above expression is equivalent to: (A.2a)

The next relationship is well-known in algebra: (A.3)

As an illustrative example let us find the square of seven. Let us make 7 the difference between 10 and three (7 = 10 − 3), meaning that a = 10 and b = 3. So, using the above equation, the square of seven would be:

This procedure associated to equation A.3 seems and it is more complicated than just using the definition of square of a number: multiply the number by itself (7² = 7 x 7 = 49, or R x R = R²). However, the next example shows a case where the equation A.3 is really useful. If a is equal to x, and b is equal to R, such equation becomes (A.4)

The Pythagorean theorem is applicable to rectangle triangles or those having two sides (a and b) that may have different length and are forming an 9 angle of 90 degrees (also known as right angle) as the one formed by the “internal” corner between a horizontal ceiling and a vertical wall. This theorem is known by c² = a² + b², but usually it is very useful to get the length of one side (a or b) of such triangles when the other side (b or a) and the hypotenuse (c) are known. When we want to know the b side, the theorem would become: (A.5)

Figure A4. A rectangle triangle

Figure A4 shows one of these cases where the hypotenuse is equal to R and the magnitude of the other (horizontal) side is equal to “x-R”, and we want to know how long the other (vertical or b) side is. In this case where c = R and a = x – R, and according to equation A.5, the square of such side is equal to (A.6)

After replacing the term in parenthesis (x – R)² by equation A.4, the above relationship A.6 becomes (A.6a)

In algebra when a negative sign precedes a parenthesis to be cancelled out, such a sign changes the sign of all the terms inside the parenthesis. Applying this rule to the above term in parenthesis, After that, if these three new terms are added to the other term (R²) on the right side of the above expression A.6a, the R² terms cancel each other out by having different signs. What it is left is just –x² + 2xR. When a mathematical expression has one negative term (-x²) and one positive (2xR), it is customary to first list the positive term and then the negative. Applying this conversion, the above equation A.6 becomes equivalent to: (A.6b)

During Archimedes’ time, the formula to compute the volume of a sphere was not known, but the ones for the volume of a cone and cylinder were known. For that reason, let us review such formulas along with the 10 surface area of a circle (A or Acir) that is known as:

(A.7)

Where r is the radius of the circle. It means that if such radius for any reason varies, it could be identified by any variable for instance x. In these cases, the area of a circle becomes: (A.8)

Let us see an easier case where such radius is equal to a variable already squared. For example, when the radius is equal to b whose square value is given by equation A.6b. In this case, the circle’s area will be equal to: (A.9)

As the right term of the above relationship is equal to equation A.2 that is also equivalent to equation A.2a, the above area could be rewritten as: (A.9a)

Or when the radius is for example 2R, such area is:

(A.10)

Another common practice in algebra is to place the number followed by a constant and/or variable as in the above equation. 11

The formula for the volume of a cylinder is equal to its height times the area of its base that is equal to the area of a circle (normally the height and the area are swapped to place the numbers and constant in this order at the beginning of the formula). It is: (A.11)

When its height is h, and r is the radius of its circular base, whose area is given by equation A.7 (Acir = πr²), the cylinder volume formula is the wellknown example:

When the height and radius of the cylinder are equal to 2R, the area of its base is given by equation A.10 (or A= 4πR²), and for being its height equal

to 2R, its volume or equation A.11 becomes: (A.12)

Notice that for only having one term inside, the parentheses can be cancelled out. Carrying on the multiplication among the like terms (4 x 2 = 8, and R² x R = R³), the above relationship becomes: (A.12a)

The volume of a cone is equal to a third (1/3) of the volume of a cylinder when both bodies have the same height and their bases have the same radius. If the height and radius of the base of the cone are the same as the previous cone or equal to 2R, its volume become a third of the above value or:

(A.13)

Although these volumes normally are known just as V, subscripts were added to differentiate one from another and to avoid confusion when these formulas are used simultaneously. ****733374989****

Bibliography Gray, Theodore. "Theo Gray's Mad Science." Gray, Theodore. Theo Gray's Mad Science Experiments You Can Do at Home- But Probably Shouldn't. New York: Black Dog & Leventhal Publishers, Inc., 2011. 239. Paper. Lewis, John and William Loftus. Java Software Solutions. 6Th Edition. Boston: Pearson - Addison Wesley, 2009. Malba, Tahan. The Man Who Counted, A collection of Mathematical Adventures. New York: W.W. Norton & Company, 1993. Pavitt, Neil. Brainhack - Tips and Tricks to Unleash Your Brains's Full Potential. The Atrium, Southern Gate, Chichester, West Sussex, United Kingdom: John Wiley and Sons Ltd, 2016.

Starbird, Michael. "Change and Motion: Calculus Made Clear, 2nd Edition." The Great Courses - Science & Mathematics. Vol. Disc 2. Chantilly, VA: The Teaching Company, 2006. Wikipedia. Tower of Hanoi. 6 February 2017. 7 March 2017. .

Notes [←1] The perforation and the poles are only really required to avoid the tokens fall when they number is high.

[←2] In math these bars are used to represent a different thing: the absolute value

[←3] The possible remainders of dividing a whole number by three are zero, one, or two

[←4] For the first 48 movements see Figure M3

[←5] An isosceles-rectangular triangle has two identical sides (same length) that intercept and form a 90 degrees (right or perpendicular) angle.

[←6] Actually, it is its mass. The weight of an object is equal to its mass times gravity.

[←7] The four coming from the 8/2 division multiplied by one expressed as three thirds (3/3), is equivalent (4 x 3/3) to twelve thirds or 12/3.

[←8] The turning effect generated by a force acting a certain distance of a point and not aiming to it is known as torque.

[←9] Angle is a measurement of the opening of either two lines at a meeting point or two (almost) flat surfaces like the jaws at the articulation.

[←10] Surface area is the extend of a region that could be cover with paint, tiles or any other very thin thing or layer.

[←11] Volume is the space of a body that if it were hollow could be filled with water, soup, or any other filling substance that completely fits the shape of the body.