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Slaying the dragon

We looked at some fun card tricks with maths behind them.

This highly addictive shape game is brilliant fun and has some interesting mathematical ideas to explore. We started with drawing all the one block, two block, three block and four block shapes we could think of. No repeats (reflections, rotations) were allowed – for example these two shapes of four blocks are effectively the same:

There are 9 different shapes, and the game contains each of these possibilities. There are 7 dice which you throw to generate where to put 7 wooden pieces. For example the game could start like this:

The values on the dice are:

Once you have the wooden pieces in place, it is a two player race to fit all the coloured pieces on the grid. Given the dice configuration above, how many possible games are there to play? The makers of the game have ensured all these games are possible in at least one way.

After we played a few times, we tried to make up an impossible configuration of wooden pieces (ignoring the dice). Here is a trivially impossible one (because there is only one blue one block piece):

But can you make up one that looks like it could work but then doesn’t? Ghazi even found a way of generating impossible solutions and proving they were impossible.

Puzzle 1 – The classic

You are on one side of a river, and with you, there is a wolf, a goat and a cabbage. You have one boat, and can only take one living thing at a time. The goat cannot be left alone with the cabbage and the wolf cannot be left alone with the goat. How many journeys must you do in minimum to get all the objects to the other side of the river? In how many different ways can you do it?

See this video for a brilliant extension leading to an explanation of Vertex Cover.

Puzzle 2 – The answer is smaller than you think … 

There are 4 people (A, B, C and D) who want to cross a bridge in night.

1. A takes 1 minute to cross the bridge.
2. B takes 2 minutes to cross the bridge.
3. C takes 7 minutes to cross the bridge.
4. D takes 10 minutes to cross the bridge.

There is only one torch with them and the bridge cannot be crossed without the torch. There cannot be more than two persons on the bridge at any time, and when two people cross the bridge together, they must move at the slower person’s pace.

What is the quickest way for all four to cross the bridge?

Puzzle 3 – Lions and Wildebeests

3 lions and 3 wildebeests need to cross a river using a raft.

The raft needs at least one animal to paddle it across the river, and it can hold at most two animals.

If the lions ever outnumber the wildebeest on either side of the river (including the animals in the boat it’s on that side) they will eat the wildebeest.

The animals can’t just swim across.

How many crossings does it take to get the animals across the river?

For the solution see this nice video.

A great game from NRich.

In a Latin Square each symbol or colour occurs exactly once in each row and exactly once in each column.

Can you complete this one? It is from Brilliant.org where you can do it interactively.

Now for the hard challenge …

Thanks to NRich for this great puzzle

Today we tried to get an insight into qubits – which can be 0,1 or a superposition of both simultaneously. Complex problems can theoretically be solved much faster using qubits instead of bits.

Terry Rudolph has written an excellent book suitable for high school students on this subject. We went through a few excerpts collated here, and then looked at pages 27 – 31 of Part 1 of his book.

Today we solved the first three puzzles from The Alan Turing Cryptography Competition 2022. You can see a sample archived competition here.

We were inspired by this Numberphile video by Neil Sloane today.

Starting with 2 ones, what is the maximum number you can place on an infinite grid according to the rule:

“You can place a number if all the numbers immediately around it add up to itself”

Here is a way to get up to 4:

But then we are stuck as there is nowhere to put the 5. Can you do better than 4?

You can choose where to place the 2 ones to start with, for example:

or

This puzzle can be tried with any number of 1s to start with.

What is the maximum number you can place starting with three 1s?

If the maximum number you can reach is a(n) for n starting 1s, an interesting question is what is the best lower bound we can find for a(n).

Imagine this diagram continuing …

If we split it into sections we have demonstrated that a(n)>=3n-3

Can you find a higher lower bound for a(n) by drawing a different pattern that would continue indefinitely?

Today we explored one of the amazing ways that pi crops up where you wouldn’t expect it.

We were inspired by this video to write a program in Trinket that simulated a random walk.

The rules are that you start at the origin and toss a coin. If you get a Head you go to the right by one, if you get a Tail you go to the left by one. If you did this for say 1000 steps, 1000 times, the average final position would be 0. But what would if we took the positive distance from the origin at each stage and averaged those values? It involves pi! Watch the video for more info.