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.”

Answer.

See the Blockbusters Problem for solution.

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).”

See The Maths of Lviv