This month we had two new attendees, as well as some regulars. We talked about lots of different things, although one recurring theme was the Crisp Tube Enigma machine, which we were using to send coded messages to Newcastle MathsJam. There will shortly be a video chronicling our achievements, and I’ll post a link to the video and writeup here once it’s ready.

We also played some games of SET, and a few games of Rubik’s Race – this is a board game based on the Rubik’s Cube (although mathematically unrelated), in which two players compete to assemble a given 3×3 configuration of squares in the Rubik’s colours, using a similar mechanism to sliding rearrangement puzzles. Whoever completes this task first, and slams down the lid on their side is declared the winner. The configuration to aim for is determined by shaking a clear box containing nine six-sided dice, and the colours on the top faces give the pattern, similar to the way Boggle gives you a set of letters to anagram. Since there are only four squares of each colour in the grid, is there a way to colour the dice so that the configuration will always be possible, i.e. there won’t be more than four of any one colour?

We had a copy of an Oxford entrance maths paper (the equivalent of Cambridge’s STEP), and had a look at some of the questions – one which we thought was nice was the following: four crosses and twelve noughts are arranged in a 4×4 grid. Firstly, how many ways are there to do this; secondly, how many ways are there such that one of the crosses lies in each row; and thirdly, how many ways so that one cross lies in each row and column?

When trying to find some scrap paper to discuss Edinburgh’s question about a three-way duel (you have 33% accuracy, shooter B has 50% and shooter C has 100% – who should you shoot if you go first and then B and then C?; we weren’t clear if you got to go again, or what C’s motivation is, and didn’t get a reply until the next morning) we found a printout of the wikipedia page for the Hardest Logic Puzzle Ever, which we discussed briefly.

We also reminded ourselves of the Pizza Theorem, which we’d remembered as being slightly more broad a result, but in fact is just the case where a point is chosen on a pizza and it’s cut into 2n (n>4) segments with an equal angle; in this case, the sum of the areas of alternate segments is equal (a nice way to split a pizza with someone if it’s already got a cut in it that doesn’t cut it in half?)

That’s pretty much all I can find a record of, other than the Enigma fun, which will be written up separately.