# Box Code Puzzle

This is an intriguing puzzle from Futility Closet.

“In Robert Chambers’ 1906 novel The Tracer of Lost Persons, Mr. Keen copies the figure below from a mysterious photograph. He is trying to help Captain Harren find a young woman with whom he has become obsessed.

‘It’s the strangest cipher I ever encountered,’ he says at length. ‘The strangest I ever heard of. I have seen hundreds of ciphers—hundreds—secret codes of the State Department, secret military codes, elaborate Oriental ciphers, symbols used in commercial transactions, symbols used by criminals and every species of malefactor. And every one of them can be solved with time and patience and a little knowledge of the subject. But this … this is too simple.’

The message reveals the name of the young woman whom Captain Harren has been seeking. What is it?”

As is usual with these types of puzzles, I felt foolish that I couldn’t see the immediate, simple interpretation of the boxes—after a great deal of effort.  So I solved it using the usual cryptographic methods that rely heavily on logic and letter frequencies, though the message is a bit short for that.

See Box Code Puzzle for solutions.

# Logical Dead End

One is reduced to hysterical laughter to try to maintain a modicum of sanity.

Our Senate at work: Republican Mitch McConnell said (Dec 6) “Legislation that doesn’t include policy changes to secure our borders will not pass the Senate.”  Republican Trump said (Feb 3) the Senate should not pass legislation that includes border security.  Let P be the statement “Senate legislation should include border security.” and let Q be the statement “Senate should pass legislation.”  Then we have the Republicans saying

(~P ⇒ ~Q) ˄ (P ⇒ ~Q)

Show that this is equivalent to ~Q, that is, “The Senate should not pass legislation.”—basically stop working.

It looks like the Republicans in the House are doing the same thing:

Old Codger Rant, with Update (4/24/2024):  Continue reading

# Peirce’s Law

The June 2023 Carnival of Mathematics # 216 at Eddie’s Math and Calculator Blog has the rather arresting item concerning Peirce’s Law from the American logician Charles Sanders Peirce (1839 – 1914).

“Peirce’s Law:  Jon Awbrey of the Inquiry Into Inquiry blog

This article explains Pierce’s Law and provides the proof of the law.  The proof is provided in two ways:  by reason and graphically.  Simply put, for propositions P and Q, the law states:

P must be true if there exists Q such that the statement “if P then Q” is true.  In symbols:

(( P ⇒ Q) ⇒ P) ⇒ P

The law is an interesting tongue twister to say the least.”

Perhaps another way of saying it is “if the implication P ⇒ Q implies that P is true, then P must be true.”  Still, it sounds weird.

See Peirce’s Law

(Update 6/20/2023)  Appendix: Valid Argument Continue reading

# Blockbusters Problem

For his Monday Puzzle in the Guardian Alex Bellos provided a seemingly impossible puzzle from the 1983 British teenager quiz show Blockbusters.

“In the much-missed student quiz show Blockbusters, teenagers would ask host Bob Holness for a letter from a hexagonal grid. How we laughed when a contestant asked for a P!  Holness would reply with a question in the following style: What P is an area of cutting edge mathematical research and also a process in the making of an espresso? The answer is the subject of today’s puzzle: percolation.

Today’s perplexing percolation poser concerns the following Blockbusters-style hexagonal grid:

The grid above shows a 10×10 hexagonal tiling of a rhombus (i.e. a diamond shape), plus an outer row that demarcates the boundary of the rhombus. The boundary row on the top right and the bottom left are coloured blue, while the boundary row on the top left and the bottom right are white.

If we colour each hexagon in the rhombus either blue or white, one of two things can happen. Either there is a path of blue hexagons that connects the blue boundaries, such as here:

Or there is no path of blue hexagons that connects the blue boundaries, such as here:

There are 100 hexagons in the rhombus. Since each of these hexagons can be either white or blue, the total number of possible configurations of white and blue hexagons in the rhombus is 2 x 2 x … x 2 one hundred times, or 2100, which is about 1,000,000,000,000,000,000,000,000,000,000.

In how many of these configurations is there a path of blue hexagons that connects the blue boundaries?

The answer requires a simple insight. Indeed, it is the insight on which the quiz show Blockbusters relied.

For clarification: a path of hexagons means a sequence of adjacent hexagons that are the same colour.”

See the Blockbusters Problem for solution.

# “Fermat’s Last Theorem” Puzzle

Here is a mind-numbing logic puzzle from Futility Closet.

“A puzzle by H.A. Thurston, from the April 1947 issue of Eureka, the journal of recreational mathematics published at Cambridge University:

Five people make the following statements:—

Which of these statements are true and which false?  It will be found on trial that there is only one possibility.  Thus, prove or disprove Fermat’s last theorem.”

Normally I would forgo something this complicated, but I thought I would give it a try.  I was surprised that I was able to solve it, though it took some tedious work.  (Hint: truth tables.  See the “Pointing Fingers” post regarding truth tables.)

One important note.  The author is a bit cavalier about the use of “Either …, or …”.  In common parlance this means “either P is true or Q is true, but not both” (exclusive “or”: XOR), whereas in logic “or” means “either P is true or Q is true, or possibly both” (inclusive “or”: OR).  I assumed all “Either …, or …” and “or” expressions were the logical inclusive “or”, which turned out to be the case.

See the Fermat’s Last Theorem Puzzle

# Who Did It?

This is a fun logic problem from the Mathigon math calendar for December 2022.

1. Alice said Bob did it.

2. Bob said Alice did it.

3. Carol said Alice didn’t do it.

4. Dan said it was either Alice or Carol.

Only one person is telling the truth.  Who did it?

See Who Did It? for a solution.

# The Maths of Lviv

Unfortunately Ukraine has receded from our attention under the threat from our own anti-democratic forces, but this Monday Puzzle from Alex Bellos in March is a timely reminder of the mathematical significance of that country.

“Like many of you I’ve hardly been able to think about anything else these past ten days apart from the war in Ukraine. So today’s puzzles are a celebration of Lviv, Ukraine’s western city, which played an important role in the history of 20th century mathematics. During the 1930s, a remarkable group of scholars came up with new ideas, methods and theorems that helped shape the subject for decades.

The Lwów school of mathematics – at that time, the city was in Poland – was a closely-knit circle of Polish mathematicians, including Stefan Banach, Stanisław Ulam and Hugo Steinhaus, who made important contributions to areas including set-theory, topology and analysis. …

Of the many ideas introduced by the Lwów school, one of the best known is the “ham sandwich theorem,” posed by Steinhaus and solved by Banach using a result of Ulam’s. It states that it is possible to slice a ham sandwich in two with a single slice that cuts each slice of bread and the ham into two equal sizes, whatever the size and positions of the bread and the ham.

Today’s puzzles are also about dividing food. The first is from Hugo Steinhaus’ One Hundred Problems in Elementary Mathematics, published in 1938. The second uses a method involved in the proof of the ham sandwich theorem.

1. Three friends each contribute £4 to buy a £12 ham. The first friend divides it into three parts, asserting the weights are equal. The second friend, distrustful of the first, reweighs the pieces and judges them to be worth £3, £4 and £5. The third, distrustful of them both, weighs the ham on their own scales, getting another result. If each friend insists that their weighings are correct, how can they share the pieces (without cutting them anew) in such a way that each of them would have to admit they got at least £4 of ham?_
2. Ten plain and 14 seeded rolls are randomly arranged in a circle, equidistantly spaced, as below. Show that using a straight line it is possible to divide the circle into two halves such that there are an equal number of plain and seeded rolls on either side of the line.

Show there is always a diameter that cuts the circle into two batches of 12 rolls with an equal number of plain and seeded.

Question 2 is adapted from Mathematical Puzzles by Peter Winkler, who gives as a reference Alon and West, The Borsuk-Ulam Theorem and bisection of necklaces, Proceedings of the American Mathematical Society 98 (1986).

# Pinocchio’s Hats

This problem in logic from Presh Talwalkar recalled an article I wrote a while ago but did not publish.  So I thought I would post it as part of the solution.

“Assume that both of the following sentences are true:

1. Pinocchio always lies;
2. Pinocchio says, “All my hats are green.”

We can conclude from these two sentences that:

• (A) Pinocchio has at least one hat.
• (B) Pinocchio has only one green hat.
• (C) Pinocchio has no hats.
• (D) Pinocchio has at least one green hat.
• (E) Pinocchio has no green hats.”

Actually, the question is which, none or more, of statements (A) – (E) follow from the two sentences?

See Pinocchio’s Hats for solutions.

# Shared Spaces Puzzle

This is a nice puzzle from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2008.  It is more a logic puzzle than a geometric one.

“In the diagram, each question mark represents one of six consecutive whole numbers. The sum of the numbers in the triangle is 39, the sum of those in the square is 46 and the sum of those in the circle is 85.  What are the six numbers?”

See the Shared Spaces Puzzle

# Date Night

This is a fairly straight-forward logic puzzle from Alex Bellos’s Monday Puzzle in The Guardian.

“When it comes to the world of mathematical puzzles, Hungary is a superpower. Not just because of the Rubik’s cube, the iconic toy invented by Ernő Rubik in 1974, but also because of its long history of maths outreach.

In 1894, Hungary staged the world’s first maths competition for teenagers, four decades before one was held anywhere else. 1894 also saw the launch of KöMaL, a Hungarian maths journal for secondary school pupils full of problems and tips on how to solve them. Both the competition and the journal have been running continuously since then, with only brief hiatuses during the two world wars.

This emphasis on developing young talent means that Hungarians are always coming up with puzzles designed to stimulate a love of mathematics. (It also explains why Hungary arguably produces, per capita, more top mathematicians than any other country.)

I asked Béla Bajnok, a Hungarian who is now director of American Mathematics Competitions, a series of competitions involving 300,000 students in the US, whether he knew of any puzzles that originated in Hungary. The first thing he said that came to mind was the ‘3-D logic puzzle’, a type of logic puzzle in which you work out the solution in a three dimensional box, rather than (as is the case with the standard version) in a two-dimensional grid. He said he had never seen this type of puzzle outside Hungary.

Below are two examples he created. You could solve these using an extended two dimensional grid. It’s more in the spirit of the question, however, to draw a three-dimensional one, like you are looking at three sides of a Rubik’s Cube.

Date night

Andy, Bill, Chris, and Daniel are out tonight with their dates, Emily, Fran, Gina, and Huong. We have the following information.

1. Andy will go to the opera
2. Bill will spend the evening with Emily,
3. Chris would not want to go out with Gina,
4. Fran will see a movie
5. Gina will attend a workshop.

We also know that one couple will see an art exhibit. Who will go out with whom, and what will they do?

See Date Night