It was reported last week that that 37 NYPD members died of covid-19. I assume that they were way below 65. It is known that the coronovirus death rate for people below 65 is a quarter of the total death rate. That means, 37 people in NYPD correspond to at least 150 people in general. Assuming that the mortality rate of coronavirus is 1 percent, the number of infected NYPD members a month ago was 15000.
By now, it could be that more than half of NYPD was infected.
NYPD members have to communicate with people a lot due to the nature of their work. That means they are more prone to being infected. At the same time, they transmit more than people in many other professions.
I can conclude, that about half of the people that are high transmitters in NY have antibodies by now. Assuming they are immune, the covid transmission rate in NY has to be down.
Assuming the immunity stays with people for a while, the second wave in NY can’t be as bad as the first one.
Suppose we want to pack a unit square with non-overlapping rectangles that have sides parallel to the axes. The catch is that the lower left corners of all the rectangles are given. By the way, such rectangles are called anchored. Now, given some points in the unit square, aka the lower left corners, we want to find anchored rectangles with the maximum total area.
When the given points are close to the right upper corner of the square, the total area is small. When a single point is in the bottom left corner of the square, we can cover the whole square. The problem becomes more interesting if we add one extra assumption: one of the given points has to be the bottom left corner of the square. In the 1960’s, it was conjectured by Allen Freedman that any set of points has an anchored rectangle packing with the area of at least one half. The problem is quite resistant. In 2011, Dumitrescu and Tóth showed that every set of points has a packing of area at least 0.09, which was the first constant bound found, and is the best bound currently known.
I gave this problem to my PRIMES student Vincent Bian. He wrote a paper, Special Configurations in Anchored Rectangle Packings, that is now available at the arxiv. When you look at this problem you see that the number of ways to pack depends on the relative coordinates of the points. That means you can view the points as a permutation. Vincent showed that the conjecture is true for several different configurations of points: increasing, decreasing, mountain, split layer, cliff, and sparse decreasing permutations.
An increasing permutation is easy. There are two natural ways to pack the rectangles. One way, when rectangles are horizontal and each rectangle reaches to the right side of the square (see picture above). Another way, when rectangles are vertical. When you take the union of both cases, the square is completely covered, which means at least one of the cases covers at least half of the square. The worst case scenario, that is, the case when the maximum possible area is the smallest is when your points are placed equidistantly on the diagonal.
Other cases are more difficult. For example, Vincent showed that for a decreasing permutation with n points, the worst case scenario is when the points are arranged equidistantly on a hyperbola xy = (1-1/n)n. The picture shows the configuration for 15 points. The total area is more than one half.
Unlike many other mathematicians I know, John Conway cared a lot about the way he presented things. For example, in the puzzle he invented—known as Conway’s Wizards—the wizards had to be riding on a bus. Why was the bus so important? You see, the numbers in the puzzle were related to the age of one of the wizards, the number of the bus, and the number of the wizard’s children. It was important to John that the readers be able to use a convenient notation a, b and c for these numbers and remember which number is which.
When I give my lecture on integers and sequences, I show my students a list of different famous sequences. The first question from the audience is almost always: “What are the Evil Numbers?” As you can guess the name for this sequence was invented by John Conway. This name was invented together with the name of another sequence which is called Odious Numbers. These two sequences are complementary in the same sense as even and odd numbers are complementary: every natural number is either evil or odious. The names are good, not only because they attract, but also because they help remember what the sequences are. Evil numbers are numbers with an even number of ones in their binary representation. I assume that you can interpolate what the odious numbers are.
When he was lecturing, John used all sorts of tricks to emphasize important points: From time to time I saw him shouting or throwing his shoes. Once I remember him staring at his statement written on the blackboard for a really long time. My neighbor in the lecture hall got uncomfortable. He assumed that John, who was at that time way over 70, was blanking out and had forgotten what he wanted to say. I calmed my neighbor down. It was my fourth time listening to the same lecture, including the same pause. John Conway didn’t forget.
Every year I review MIT mystery hunt from a mathematician’s point of view. I am way behind. The year is 2020, but I still didn’t post my review of 2019 hunt. Here we go.
Every year I review MIT mystery hunt from a mathematician’s point of
view. I am way behind. The year is 2020, but I still didn’t post my
review of 2019 hunt. Here we go.
Many puzzles in 2019 used two data sets. Here is the recipe for
constructing such a puzzle. Pick two of your favorite topics: Star Trek
and ice cream flavors. Remember that Deanna Troi loves chocolate sundae.
Incorporate Deanna Troi into your puzzle to justify the use of two data
On one hand, two data sets guarantee that the puzzle is new and fresh.
On the other hand, often the connection between two topics was forced.
Not to mention that puzzle solving dynamic is suboptimal. For example,
you start working on a puzzle because you recognize Star Trek. But then
you have to deal with ice cream which you hate. Nonetheless, you are
already invested in the puzzle so you finish it, enjoying only one half
Overall, it was a great hunt. But the reason I love the MIT mystery hunt
is because there are a lot of advanced sciency puzzles that can only
appear there. For example, there was a puzzle on Feynman diagrams, or on
characters of representations. This year only one puzzle, Deeply Confused, felt like AHA, this is the MIT Mystery hunt.
Before discussing mathy puzzles I have to mention that my team laughed at Uncommon Bonds.
I will group the puzzles into categories, where the categories are obvious.
There was a strong hint that the extraction step was also mastermind.
My team spent some time trying to mastermind the ending, until we
backsolved. The extraction step was not mastermind. The final grid in
the puzzle had the word CODE written in red. It corresponded to letters
CDEO found at that location. Given that the letters were not in
alphabetical order, it gave the ordering, which didn’t exist in the
puzzle. Anyway, you can see that I have a grudge against this puzzle.
This could have been a great puzzle. But it wasn’t.
Schematics—Tons of Nikoli puzzles of different types.
Every day I check coronavirus numbers in the US.
Right now the number of deaths is 288 and the number of recovered is
171. More people died than recovered. If you are scared about the
mortality rate, I can calm you and myself down: our government is
incompetent—the testing wasn’t happening—that means the numbers do not
show people who had mild symptoms and recovered. The real number of
recovered people should be much higher.
Scientists estimated the mortality rate of coronavirus as being between 1
and 3.5 percent. Also, they say that it usually takes three weeks to
die. That means three weeks ago the number of infected people in the US
was between 8,000 and 29,000. The official number of cases three weeks
ago was 68. I am panicking again—our government is incompetent—three
weeks ago they detected between 0.25 and 1 percent of coronavirus cases.
If this trend continues, then the official 19,383 infected people as of
today means, in reality, somewhere between 2 million and 8 million
I can calm you and myself down: the testing picked up pace. This means,
the ratio of detected cases should be more than 1 percent today.
Probably the number of infected people today in the US is much less than
8 million. I am not calm.
My former student, Dai Yang, sent me the following cute puzzle:
Puzzle. You are playing a game with the Devil. There are n
coins in a line, each showing either H (heads) or T (tails). Whenever
the rightmost coin is H, you decide its new orientation and move it to
the leftmost position. Whenever the rightmost coin is T, the Devil
decides its new orientation and moves it to the leftmost position. This
process repeats until all coins face the same way, at which point you
win. What’s the winning strategy?
My friend Zeb, aka Zarathustra Brady,
invented a new game that uses chess pieces and a chessboard. Before the
game, the players put all chess pieces on white squares of the board:
white pieces are placed in odd-numbered rows and black pieces are in
even-numbered rows. At the beginning all white squares are occupied and
all black squares are empty. As usual white starts.
On your turn, you can move your piece from any square to any empty
square as long as the number of enemy neighbors doesn’t decrease. The
neighbors are defined as sharing a side of a square. Before the game
starts each piece has zero enemy neighbors and each empty square has at
least one white and one black neighbor. That means that on the first
turn the white piece you move will increase the number of neighbors from
zero to something.
As usual, the player who doesn’t have a move loses.
As you can immediately see, that number of pairs of enemy neighbors is
not decreasing through the game. I tried to play this game making a move
which minimizes the increase of the pairs of neighbors. I lost, twice. I
wonder if there is a simple strategy that is helpful.
It is important that this game is played with chess pieces in order to
confuse your friends who pass by. You can see how much time it takes
them to figure out that this game is not chess, but rather a Chessnot.
Or you can enjoy yourself when they start giving you chess advice before
realizing that this is not chess, but rather a Chessnot.
I heard this puzzle many years ago, and do not remember the origins of it. The version below is from Peter Winkler’s paper Seven Puzzles You Think You Must Not Have Heard Correctly.
Jan and Maria have fallen in love (via the internet) and Jan wishes to
mail her a ring. Unfortunately, they live in the country of Kleptopia
where anything sent through the mail will be stolen unless it is
enclosed in a padlocked box. Jan and Maria each have plenty of padlocks,
but none to which the other has a key. How can Jan get the ring safely
into Maria’s hands?
I don’t know whether this puzzle appeared before the Diffie-Hellman key
exchange was invented, but I am sure that one of them inspired the
other. The official solution is that Jan sends Maria a box with the ring
in it and one of his padlocks on it. Upon receipt Maria affixes her own
padlock to the box and mails it back with both padlocks on it. When Jan
gets it, he removes his padlock and sends the box back, locked only
with Maria’s padlock. As Maria has her own key, she can now open it.
My students suggested many other solutions. I wonder if some of them can be translated to cryptography.
Jan can send the ring in a padlock box that is made of cardboard. Maria can just cut the cardboard with a knife.
Jan can use the magic of the Internet to send Maria schematics of the
key so she can either 3d print it or get a professional to forge it. If they are
afraid of the schematics getting stolen Jan can send the schematics after the
package has been delivered.
Jan can use a digital padlock and send the code using the Internet.
Jan can send it in a secret puzzle box that can be opened without a key.
Maria can smash the padlock with a hammer.
Now that we’ve looked at the Padlock Puzzle, let’s talk about
cryptography. I have an imaginary student named Charlie who doesn’t know
the Diffie-Hellman key exchange. Charlie decided that he can adapt the
padlock puzzle to help Alice send a secret message to Bob. Here’s what
Suppose the message is M. Alice converts it to binary. Then she creates a
random binary key A and XORs it with M. She sends the result, M XOR A,
to Bob. Then Bob creates his own random key B and XORs it with what he
receives and sends the result, M XOR A XOR B, back to Alice. Alice XORs
the result with her key to get M XOR A XOR B XOR A = M XOR B and sends
it to Bob. Bob XORs it with his key to decipher the message.
Each sent message is equivalent to a random string. Intercepting it is
not useful to an evil eavesdropper. The scheme is perfect. Or is it?
Puzzle. You are visiting an island where all people know each
other. The islanders are of two types:
truth-tellers who always tell the truth and liars who always lie. You
meet three islanders—Alice, Bob, and Charlie—and ask each of
them, “Of the two other islanders here, how many are truth-tellers?”
Alice replies, “Zero.”
Bob replies, “One.” What will Charlie’s reply be?
The solution proceeds as follows. Suppose Alice is a truth-teller. Then
Bob and Charlie are liars. In this situation Bob’s statement is true,
which is a contradiction. Hence, Alice is a liar. It follows, that there
is at least one truth-teller between Bob and Charlie. Suppose Bob is a
liar. Then the statement that there is one truth-teller between Alice
and Charlie is wrong. It follows that Charlie is a liar. We have a
contradiction again. Thus, Alice is a liar and Bob is a truth-teller.
From Bob’s statement, we know that Charlie must be a truth-teller. That
means, Charlie says “One.”
But here is another solution suggested by my students that uses meta
considerations. A truth-teller has only one possibility for the answer,
while a liar can choose between any numbers that are not true. Even if
we assume that the answer is only one of three numbers—0, 1, or 2—then
the liar still has two options for the answer. If Charlie is a liar,
there can’t be a unique answer to this puzzle. Thus, the puzzle
question implies that Charlie is a truth-teller. It follows that Alice
must be lying and Bob must be telling the truth. And the answer is the
same: Charlie says, “One.”