Tanya Khovanova's Math Blog

I know how to walk. Everyone knows how to walk. Or so I thought. Now I am not sure any more.

I’ve been taking ballroom dance lessons on and off for many years. But at some point I stopped progressing. I got stuck at the Silver level. I know many steps and am a good follower, but I often lose balance and my steps are short.

Then I met Armin Kappacher, an unusual dance coach for the MIT Ballroom dance team. I would like to share some of his wisdom with you, but Armin doesn’t have much presence on the web. He only wrote one article for Dance Archives: A Theoretical and Practical Approach to “Seeing The Ground of a Movement.”

Although I’ve been attending his classes for several years, I haven’t been able to understand a word. He might say, “Your right arm is disconnected from your chest center.” But what does that mean? Others seemed to understand him, because they greatly improved under his guidance. But I was so out of touch with my body that I couldn’t translate his words into something my body could understand. Being a mathematician, I lived my whole life in my brain. I never tried to listen to my body. I was never aware of whether my forehead was relaxed or tensed, or if my pelvic floor was collapsed. I grasped that it was my own fault that I failed to understand Armin. I stuck with his classes.

After three years of group classes, I asked Armin for a private lesson with an emphasis on the basics. He instructed me to take three steps and quickly discovered my issues, which included:

• I was too collapsed.
• My spine was too curved.
• My butt stuck out too much.
• My weight was not forward enough.
• My head was too forward.
• I didn’t sway.

So now I am taking walking classes from Armin. I am slowly starting to feel what Armin means when he says that my pelvic floor is collapsed. I feel better now. Whenever I pay attention to how I am walking, my posture improves. As a result, I feel more confident, my mood approves, and I feel like more oxygen is getting to my brain. My friends have noticed a change. For example, my son Sergei got married recently, and I was sitting under the Chuppah during the ceremony (see photo). Afterwards, several friends told me that I looked like a queen.

I have to give some credit to my earrings. They were too long and were getting caught on my dress. So I was constantly trying to wiggle my head up—using Armin’s techniques.

I proudly brought this photo to Armin to show him my queen-like posture. He told me that I look okay above the chest center. But below the chest center my spine is still collapsed. Next time I will take sitting lessons with Armin.

Skyscraper puzzles are one of my favorite puzzles types. Recently I discovered a new cute variation of this puzzle on the website the Art of Puzzles. But first let me remind you what the skyscraper rules are. There is an n by n square grid that needs to be filled as a Latin square: each number from 1 to n appears exactly once in each row and column. The numbers in the grid symbolize the heights of skyscrapers. The numbers outside the grid represent how many skyscrapers are seen in the corresponding columns/row from the direction of the number.

The new puzzle is called Skyscrapers (Sum), and the numbers outside the grid represent the sum of the heights of the skyscrapers you see from this direction. For example, if the row is 216354, then from the left you see 8(=2+6); and from the right you see 15(=4+5+6).

Here’s an easy Skyscrapers (Sum) puzzle I designed for practice.

The Art of Puzzles has four Skyscrapers (Sum) puzzles that are more difficult than the one above:

• Skyscrapers (Sum) by Grant Fikes
• Skyscrapers (Sum) by Thomas Snyder
• Dr. Sudoku Prescribes #146 — Skyscrapers (Sum)
• Dr. Sudoku Prescribes #69 — Skyscrapers (Sum)

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A button of unknown functionality should be pressed an even number of times.

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When I tell you that I am closer to 30 than to 20, I mean to tell you that I am 42.

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If a car with a student-driver sign gets its windshield wipers turned on, then the car is about to turn.

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I always learn from the mistakes of people who followed my advice.

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A traffic policeman stops a car:
—You’re going 70 in a 35 miles-per-hour zone.
—But there are two of us!

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The most popular tweet, “Live your life so that you do not have time for social networks.”

I recently wrote a post about the ApSimon’s Mints problem:

New coins are being minted at n independent mints. There is a suspicion that some mints might use a variant material for the coins. There can only be one variant material. Therefore, fake coins weigh the same no matter where they’ve been minted. The weight of genuine coins is known, but the weight of fake coins is not. There is a machine that can precisely weigh any number of coins, but the machine can only be used twice. You can request several coins from each mint and then perform the two weighings in order to deduce with certainty which mints produce fake coins and which mints produce real coins. What is the minimum total of coins you need to request from the mints?

The post was accompanied by my paper Attacking ApSimon’s Mints.

Unfortunately, both the post and the paper contain wrong information. They both state that the number of coins as a sequence of the number of mints is 1, 2, 4, 8, 15, 38, 74. This is wrong. I took this data from the sequence A007673 in the OEIS database. The sequence had incorrect data lying dormant for 20 years. I believe that the sequence was generated from the paper of R. K. Guy and R. J. Nowakowsky, ApSimon’s Mints Problem, published in Monthly in 1994. To the credit of Guy and Nowakowsky, they never claimed to find the best solution: they just found a solution, thus providing a bound for the sequence. Someone mistook their solution for the optimal one and generated the sequence in the database.

After my post, my readers got interested in the problem and soon discovered the mistake. First Konstantin Knop found a solution for 6 mints with 30 coins and for 7 mints with 72 coins. Konstantin is my long-time collaborator. I trust him so I was sure the sequence was flawed. Then someone located a reference to a paper in Chinese A New Algorithm for ApSimon’s Mints Problem. Although none of my readers could find the paper itself nor translate the abstract from Chinese. But judging from the title and the formulae it was clear that they found better bounds than the sequence in the database. My readers got excited and tried to fix the sequence. David Reynolds improved on Konstantin’s results with a solution for 6 mints with 29 coins and for 7 mints with 52 coins. David did even better on his next try with 28 and 51 coins correspondingly. He also found a solution with 90 coins for 8 mints. Moreover, his exhaustive search proved that these were the best solutions.

Now the sequence in the database is fixed. It starts 1, 2, 4, 8, 15, 28, 51, 90.

For future generations I would like to support each number of the sequence by an example. I use the set P(Q) to represent the sequence of how many coins are taken from each mint for the first(second) weighing. For one mint, only one coin and one weighing is needed. ApSimon himself calculated the first five values, so they were not in dispute.

• a(2) = 2: P=(1,0) and Q=(0,1). Found by ApSimon.
• a(3) = 4: P=(0,1,2) and Q=(1,1,0). Found by ApSimon.
• a(4) = 8: P=(0,1,2,3) and Q=(1,0,2,2) or P=(0,1,1,4), Q=(2,0,1,1). Found by ApSimon.
• a(5) = 15: P=(0,1,1,4,5) and Q=(2,1,2,5,0). Found by ApSimon.
• a(6) = 28: P=(1,2,2,5,5,0) and Q=(0,1,2,1,8,10). Found by Robert Israel, Richard J. Mathar, and David Reynolds,
• a(7) = 51. P=(2,3,7,2,8,12,0) and Q=(0,2,1,7,7,12,12). Found by David Applegate and David Reynolds.
• a(8) = 90. P=(4,6,6,7,3,13,15,3) and Q=(4,0,1,6,12,12,1,27). Found by David Applegate and David Reynolds.

There is a solution for nine mints using 193 coins that is not confirmed to be optimal. It was found by David Reynolds: P=(1,2,4,12,5,4,20,39,43) and Q=(0,1,3,3,25,33,34,18,27). In addition, David Reynolds provided a construction that reduces the upper bound for n mints to (3⁽ⁿ⁺¹⁾−2n−3)/4. The following set of coins work: P=(1,3,7,15,…,2ⁿ−1) and Q=(1,4,13,40,…,(3ⁿ−1)/2).