## Borders of Strips

Why would I complicate my life by crocheting colored borders onto different strips?

Answer: I wanted to emphasize their borders.

Do you recognize the objects in the picture? The leftmost one is a Möbius strip. I made it by crocheting a long rectangle. Then, instead of connecting the short sides to form a cylinder, I twisted one side 180 degrees before stitching them together. For the other two objects, I made 360 and 540 degree twists, respectively.

I used green yarn for the internal part of the strips. When the twist in the strip is a multiple of 360 degrees, the resulting surface is orientable and has two borders. I used two different colors to emphasize this fact. In other cases, the resulting surface is not orientable and has only one border, so I only used one color for the border.

The point of using extra colors for the borders is to make them more prominent. For example, it is easy to see that the Möbius strip’s border is a circle. The border of the piece in the middle consists of two loops, and the different colors make it obvious that the two borders are linked. The last object has one border, and the color helps you notice that its border is a trefoil knot!

What would happen with the borders if we increase the number of degrees in a twist? Can you figure it out? Are you willing to take up crocheting to solve this puzzle?

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## Some More Math Jokes

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A math problem is the only place where a person buys 7744 watermelons for dinner, but no one knows why!

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Today I saw a tweet from someone I knew in middle school. He tweeted, “I turned my life around 360 degrees!” Now do you see why it is important to study math?

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Looking for energy? Multiply time by power!

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The mom of a third grader calls her friend, “Lucy, did you do your son’s math homework?”
“I did.”

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If money is measured in piles, then I have a pit.

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My girlfriend is the square root of −100. She’s a perfect 10, but purely imaginary.

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A mathematical collapse: while cutting a worm, you divide it by 2 and multiply it by 2, simultaneously!

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## Nostalgia

I used to love math problems about weighing things. But then I got distracted by my own personal scale with its slowly rising numbers. However, having recently lost a few pounds, I want to get back to other scales!

Puzzle. You have a balance scale that is broken in a consistent way: if you put two objects on its two pans, the scale will show you that the left pan is heavier, lighter, or the same weight as the right pan, but it may be wrong. However, it will give the same answer each time you repeat this test with the same two weights. You have a bag of flour and a 1-kilo weight. How can you use this scale to measure out 1 kilo of flour?

Puzzle. This time, your scale is not broken, and, moreover, it is not a balance scale but a digital one that tells you the weight of the objects you put on it. The scale does have a quirk. It can only measure two objects at a time. You have 13 coins of potentially different weights. How can you figure out the total weight of the 13 coins in 8 measurements?

The next puzzle was sent to me by Konstantin Knop, a coin puzzle master. This time there is no physical scale involved; rather, some sort of god answers your questions.

Puzzle. 26 identical-looking coins are arranged in a circle. Two of the coins which are next to each other are fake. You are allowed to pick any set of coins and ask how many fake coins are in the set. What is the smallest number of questions you need to find both fake coins if you only get the answers after you have posed all your questions?

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A link is defined as two closed curves in three-dimensional space. The first picture shows an example of a link with one yellow curve and one blue. The linking number is a simple numerical invariant of a link. Intuitively, it represents the number of times that each curve winds around the other. For example, if it is possible to pull the two curves apart, the linking number is zero.

When I studied the linking number, I would look at a picture of a link trying to calculate this number. It was confusing. It only became easy after I started crocheting. For example, the second picture shows the same link as the first but slightly rearranged. I simply slid the yellow loop along the blue one until I could clearly see a piece of the blue loop as a straight segment and the yellow loop circling around it. Now, it is easy to see that the yellow loop winds around the blue one 3 times, making the linking number 3.

The only thing to remember is that while counting the number of windings, I need to consider the direction. It is possible for a loop to wind clockwise and then counterclockwise. In this case, the linking number is the difference between the two.

I crocheted a lot of links, and now my students and I have no problem calculating the linking numbers.

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## The Power of a Computational Proof: Uncrossed Knight’s Tours Continued

In December 2022, I wrote a blog post Uncrossed Knight’s Tours about Derek Kisman’s amazing achievement of calculating the largest uncrossed knight’s tours on rectangular chessboards on sizes M-by-N, where M is small, and N can be very very large.

The data showed some asymptotic periodicity, and I wondered how to prove it mathematically. I didn’t realize that Derek already proved it. In my ignorance of programming, I assumed that programs just spewed out the data and didn’t think they could prove anything. I was wrong. It appears that no other proof is needed. Derek tried to explain the details to me using the terminology of dynamic programming, but I am not sure I can reproduce it here.

Let’s recall the problem. Consider an M-by-N chessboard and a knight that moves according to standard chess rules: jumping one square in one direction and two squares in an orthogonal direction. The knight must visit as many squares as possible, without repeats, and then return to its starting square. In addition, the knight may never cross its own path. If you imagine the knight’s path consisting of straight line segments connecting the centers of the squares it visits, these segments must form a simple polygon. To summarize, given M and N, we want to calculate the longest uncrossed knight’s tour length.

To be clear: the programs, their output data, proven answers, and images are by Derek Kisman. I am just a humble messenger showing my new appreciation of the power of a computational proof.

The image shows Derek’s solution for a 3-by-13 chessboard. There is a repeating 3-by-4 pattern marked by dashed lines. The same tour works for boards of lengths 10, 11, and 12. Thus, for chessboards of width 3 and length from 10 to 13 inclusive, the longest uncrossed knight’s tour is length 10. We can write the answers for 3-by-N chessboards as a sequence with index N, where -1 means the tour is impossible. The sequence starts with N = 1: -1, -1, -1, 4, 6, 6, 6, 6, 10, 10, 10, 10, 14, 14, ….

We can prove that this sequence is correct without programming. Suppose the tour starts in the leftmost column. If we start in the middle of the column, the whole tour ends as a rhombus and a tour of length 4, which, by the way, is the longest tour for N = 5. Thus, for a larger N, we have to start in a corner. From there, there are only two possible moves. We can see that the continuation is unique and that, asymptotically, we gain one step per extra column. That is, asymptotically, the length of the longest tour divided by N is 1.

Derek uses an additional notation in the following sequence: each cycle is in brackets. Any two consecutive cycles differ by the same constant. So to continue the sequence indefinitely, it is enough to know the first two cycles.

Closed tours: 3xN (asymptote 1):  -1, -1, -1, -1, 4, [6, 6, 6, 6], [10, 10, 10, 10], …

I continue with other examples Derek calculated:

Closed tours: 4xN (asymptote 2): -1, -1, -1, [4], [6], …

Closed tours: 5xN (asymptote 2 3/5):  -1, -1, 4, 6, [8, 12, 14, 18, 20, 22, 24, 28, 30, 34], [34, 38, 40, 44, 46, 48, 50, 54, 56, 60], …

Closed tours: 6xN (asymptote 4): -1, -1, 6, 8, 12, 12, 18, 22, 24, 28, 32, 36, [38], [42], …

Closed tours: 7xN (asymptote 4 10/33):  -1, -1, 6, 10, 14, 18, 24, 26, 32, 36, 42, 44, 48, 54, 58, 62, 66, 72, 74, 80, 84, 88, 94, 98, 100, 106, 112, 114, 118, 124, 128, 130, [136, 140, 144, 148, 154, 158, 162, 166, 170, 176, 180, 184, 188, 192, 196, 200, 204, 210, 214, 218, 222, 226, 232, 236, 240, 244, 248, 254, 256, 260, 266, 270, 274], [278, 282, 286, 290, 296, 300, 304, 308, 312, 318, 322, 326, 330, 334, 338, 342, 346, 352, 356, 360, 364, 368, 374, 378, 382, 386, 390, 396, 398, 402, 408, 412, 416], …

Closed tours: 8xN (asymptote 6): -1, -1, 6, 12, 18, 22, 26, 32, 36, 42, 46, 52, 58, [64, 70, 76, 80, 88, 92], [100, 106, 112, 116, 124, 128], …

Closed tours: 9xN (asymptote 6 6/29): -1, -1, 6, 14, 20, 24, 32, 36, 42, 50, 56, 60, 68, 74, 80, 86, 94, 98, 106, 114, 118, 126, 132, 136, [144, 150, 156, 162, 168, 174, 180, 186, 192, 200, 206, 212, 218, 224, 230, 236, 242, 250, 254, 262, 268, 274, 280, 286, 292, 300, 304, 312, 318, 324, 330, 336, 342, 348, 354, 360, 366, 372, 378, 386, 392, 398, 404, 410, 416, 422, 428, 436, 440, 448, 454, 460, 466, 474, 478, 486, 492, 498], [504, 510, 516, 522, 528, 534, 540, 546, 552, 560, 566, 572, 578, 584, 590, 596, 602, 610, 614, 622, 628, 634, 640, 646, 652, 660, 664, 672, 678, 684, 690, 696, 702, 708, 714, 720, 726, 732, 738, 746, 752, 758, 764, 770, 776, 782, 788, 796, 800, 808, 814, 820, 826, 834, 838, 846, 852, 858], …

Closed tours: 10xN (asymptote 8): -1, -1, 10, 16, 22, 28, 36, 42, 50, 54, 64, 70, 78, 84, 92, 100, [106], [114], … I do not have an image for this case.

As you might have noticed, for an even M, the asymptote equals M-2. The asymptote for an odd M is slightly greater than the asymptote for M-1.

Derek also calculated the longest open knight’s tours: the tours where the knight doesn’t have to return to its starting position.

Open tours: 2xN (asymptote 1/2): -1, -1, [2, 2], [3, 3], …

Open tours: 3xN (asymptote 1): -1, 2, 3, 5, 6, 7, [9], [10], …

Open tours: 4xN (asymptote 2): -1, 2, 5, [6], [8], …

Open tours: 5xN (asymptote 3): -1, 3, 6, 8, 11, 15, 17, 20, 23, 26, 29, [32, 35, 38, 41, 44, 46], [50, 53, 56, 59, 62, 64], …

Open tours: 6xN (asymptote 4): -1, 3, 7, 10, 15, 18, 22, 26, 30, 33, [36], [40], …

Open tours: 7xN (asymptote 5): -1, 4, 9, 12, 17, 22, 25, 31, 36, [40], [45], …

Open tours: 8xN (asymptote 6): -1, 4, 10, 14, 20, 26, 31, 36, 43, 48, 54, 60, [64], [70], …

Open tours: 9xN open (asymptote 7): -1, 5, 11, 16, 23, 30, 36, 43, 48, 56, 62, 68, 75, [82, 88, 94], [103, 109, 115], … I do not have an image for this case.

There are a lot of interesting new sequences in this essay that were very nontrivial to calculate. I hope someone adds them to the OEIS database.

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## Balls in Boxes

I stumbled upon a cute puzzle on Facebook which originally came from a new book, Creative Puzzles to Ignite Your Mind by Shyam Sunder Gupta.

Puzzle. We have four identical boxes. One of the boxes contains three black balls (BBB), another box has two black and one white balls (BBW), the third box has one black and two white balls (BWW), and the last box has three white balls (WWW). Four labels, BBB, BBW, BWW, and WWW, are put on the boxes, one per box. As is often the case in such puzzles, none of the labels match the contents, and this fact is common knowledge. Four sages get one box each. Each sage sees his label but doesn’t know the other’s labels. Without looking in the box, each sage is asked to take out two balls and guess the color of the third ball. All the sages are in the same room and can hear each other and see the colors of the balls that are taken out.

• The first sage takes out two black balls and says, “I know the color of the third ball.”
• The second sage takes out one black and one white ball and says, “I know the color of the third ball.”
• The third sage takes out two white balls and says, “I don’t know the color of the third ball.”
• The fourth sage says, without taking out any balls, “I know the color of all the balls in my box and also the content of all the other boxes.”

Can you figure out what’s in the boxes?

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## Klein Bottles without Holes, or How to Crochet Through

After I crocheted Whitney Umbrellas, I got excited that I figured out how to crochet through an existing layer of fabric. I decided to use my skills to crochet “correct-er” Klein bottles.

A Klein bottle is a cool mathematical surface with only one side, similar to the Möbius strip. Unlike the strip, the Klein bottle has no border making it a non-orientable manifold. The problem in making Klein bottles is that the Klein bottle can’t be embedded into 3D space. Thus, all 3D models of Klein bottles have to self-intersect. But all the models that I saw, including glass models and crocheted hats that you can buy at ACME Klein Bottle, have holes.

I realized that my method of crocheting through might allow me to make more accurate models of Klein bottles, ones without holes.

Now it is time to spill my secret. The idea is easy, the implementation is not. The two pictures show the same yellow cylinder crocheted through a green square, viewed from different angles. I crocheted the green square first, then half of the yellow cylinder. Afterwards, I had to pull the whole ball of yellow yarn through a tiny hole in the middle of the green square. Then, with my hook, I pulled each yellow stitch from one side of the green square to the other side through its own tiny hole and finished the stitch on the new side. In the third picture, you can see my Klein bottle made in two colors. You might be able to see the second color inside instead of a hole.

I invented this method while crocheting Whitney’s umbrellas. I had to pull the whole ball of yarn through a tiny hole once per row. I still remember the tediousness of it with dread.

After the bottles, I decided to try projective planes. In the fourth picture, you can see two projective planes and two projective planes with holes. For the former, I started with the bottom hemispheres, and for the latter with cylinders. I didn’t need to crochet through or pull yarns through tiny holes. I just crocheted one row across the other. I left the easiest crochet task for last!

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## The Odd-one-out Dilemma

I do not like odd-one-out puzzles. This old and famous example is one of the reasons why. When given the list: skyscraper, cathedral, temple, and prayer, people usually pick “prayer” as the odd-one-out because it’s not a building. On the other hand, when the order is prayer, temple, cathedral, and skyscraper, people would pick “skyscraper” as it is not related to religion.

I created several pairs of questions that might have a similar issue: the answer depends on how the list is ordered.

Odd-one-out dilemma. Pick the odd-one-out from:

• Pen, book, notebook, tissue.
• Tissue, notebook, book, pen.

Odd-one-out dilemma. Pick the odd-one-out from:

• Banana, apple, orange, grape, lime, nectarine, artichoke.
• Artichoke, nectarine, lime, grape, orange, apple, banana.

Odd-one-out dilemma. Pick the odd-one-out from:

• 6, 3, 15, 9, 5.
• 5, 9, 15, 3, 6.

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## A Bribe for a 5-Star Review

I buy almost everything on Amazon. Recently, I ordered a back stretcher. It took half an hour to assemble, and after the first use, it changed shape. However, this story is not about quality but about a card that was included with the item.

In the box, I found a gift card that wasn’t a gift card but rather a promise of a \$20 Amazon gift card for a 5-star review. Hmm! A bribe for a good review.

I looked at the card more closely, and it had the following text.

This is not only a bribe. It contains a threat.

Initially, I assumed it was Amazon, but it makes more sense that the company making this thingy is behind it. I gave Amazon the benefit of the doubt and went to their website to leave a 1-star review. I related the card’s story to warn others that the 5-star reviews can’t be trusted. Amazon rejected my review as it didn’t comply with their guidelines. Is Amazon in on it?

I wrote a different 1-star review, which did comply. It seems I can complain about the product, but I can’t complain about the bribe and threat.

I called Amazon’s customer service, and they promised to investigate. This was three months ago. This crappy product, with a stellar average 4.6 rating, is still out there.

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“Johnny, how much will your mom pay at the market for three pounds of apples if one pound is three dollars?”
“I do not know,” answers Johnny, “My mom loves bargaining, and she is good at it.”

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