Lycee Francais SAMI Maths Club

This week we solved a tricky puzzle involving swords and trying to sit in the right position of a circle to survive!

A powerpoint, created at the Ghana Maths Camp in 2015, explaining the puzzle and with solution is here.

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.

Today we played this game from our SAMI Virtual Maths Camp Cards. An interesting variant is to turn it into a pure strategy game where everyone has just the numbers 1 to 9.

Today we looked at three conjectures – one was easily proven false, one took some computer power to decide, and one … no one knows …!

Conjectures

Try out the game here.

Can you adapt the Mastermind code below to make the game above?

Starting with the following trinket can you draw a simple square and hexagon?

Can you write a program that asks for the number of sides, and dependent on your answer, produces a drawing like this:

By following a simple iterative formula and using complex numbers, the fantastic shape above can be created. Some starting instructions are here.

Today we played a geometric game and tried to solve a couple of maths questions around it. See here for more details.

We played a “perfect strategy” game today where everyone theoretically has an equal chance of winning.

The game is called 91 and the rules are below.

We also played a couple of “collaborative games” where all players work together to try and beat the cards. The two we played were German games called The Game and Hanabi.

Rules for 91:

Number of players: 2-7

Materials: 1 card deck for 2-3 players; 2 card decks for 4-7 players (no jokers).

Players and Cards: The game is for two to seven players, using a complete suit from a standard 52-card pack for each player plus one extra suit. Two decks are needed for four or more players. Cards rank Ace (low), 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K (high). As a prize, the Ace is worth 1 point, cards 2-10 face value, Jack 11, Queen 12 and King 13.

Setup: The cards are sorted into suits. One suit (traditionally diamonds) is shuffled and stacked face down as a prize pile. Each of the other players takes one complete suit.

Play: The top card of the prize pile is turned face up (so you see its value). Then each player selects a card from their hand with which to bid for it and places it face down (so you can’t see its value). When all players are ready, the bid cards are revealed simultaneously, and the highest bid wins the prize card. The bid cards are then discarded and the prize card is placed beside the player who won it. The next card of the prize pile is turned face up and players bid for it in the same way. If two players put the highest bit, the bid cards are discarded but the prize card remains on offer. A new prize card is turned face up and all players then make their next bid for the two prize cards together, then for three prize cards if there is another tie, and so on. If any of the players’ last bid cards are equal, the last prize card (and any others remaining from immediately preceding tied bids) are not won by any player.

Scoring: When all players run out of bid cards, the play ends. Each player totals the value of the diamonds they have won in bids (Ace=1, 2- 10 face value, J=11, Q=12, K=13) and the greater total wins the game.

We ran through a whole series of puzzles today starting with the obvious and getting into some challenging maths.

The general concept was trying to guess what hat you are wearing based on info you can see of those around you and on a careful strategies decided on beforehand. You can see the list of challenges here.

Today we played a game where everyone simultaneously guesses a number between 0 and 100. The person who is closest to the average (mean) wins. What number would you choose?

The game is played in multiple rounds. It is interesting to see how strategies change and how the mean changes over time.

One fun variation is that the winner is the person who is closed to 2/3 of the mean – an analysis is here.

We played a game that John H. Conway he co-invented with Michael S. Paterson while they were both at Cambridge.

It is called Sprouts, and the rules are summarised by Nrich here.

You can try and discover some of the maths behind it by working through this article. We managed to prove the maximum and minimum number of moves for a game starting with three points.

We started with this warm up puzzle from Nrich:

The interactive version is here.

We then used this worksheet to explore how a disease could spread through a network. This was adapted from this set of resources from Nrich.

Take 7 playing cards.

Put them into any number of piles and arrange in descending order. For example:

Take one card from each pile to form a new pile.

Put the piles in descending order.

Keep repeating this process.

What do you notice?

Try all different combinations of 7 cards and try and represent your findings on paper. What is the longest sequence you can find?

Then try with 6 cards. What is different this time?

For more questions and solutions see our SAMI Maths Club app.

I have a 10-sided dice with sides numbered: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

I roll the dice many times until a same number has come up twice.

Example: 3, 1, 5, 6, 8, 4, 5

5 has come up twice so I stop. It took 7 rolls for this to happen.

1. How many times do I need roll the dice until I am certain to roll a same number twice?
2. What is the smallest number of times I can roll the dice before I could get a same number twice?
3. How many times do I need to roll the dice until there is a 50% chance of getting a same number twice….? Try to do an experiment to find out….

Experiment: Use your calculator (RanInt#(1,10)) to simulate a dice roll. Keep rolling until you get a same number twice. Repeat this experiment 50 times in total and complete the table below.

Or write a program in Python to do this for you.

(Using a simulation to learn about probabilities is called a Monte Carlo Simulation)

See here for a Python program and discussion of the solution and extensions …

We had a great session today working together with the Kongali Community Library Maths Camp on modulo arithmetic.

Here are the slides (as a pdf), the 1st and 2nd explorations and a fantastic geogebra app.

 

Challenge

Can you draw this curve smoothly using only a straight edge and a pencil?

You can do it by continuing this pattern:

You could try it on Python from this starting point. Note that the co-ordinates are not correct for the first segment yet!

You could even try to do it in all four quadrants:

Or at an angle:

There is an ancient art called cross stitching that uses these ideas. Instead of stitching, we could make Christmas cards for the Bazar de Noël using Python and a 3D printer … let me know your designs at [email protected]

Welcome back to Maths Club! We are so happy to be able to get together again in person.

Today we tried this puzzle from the SAMI VMC playing card deck

For a hint, the solution and extensions see here.

In these problems, which we learnt about from Maths Pickle, you will see some shapes made out of hexagons.

It might look something like one of these:

And we’re basically going to try to fill in each of the hexagons in one of three colours (red, orange or green).

Green means that from the starting point you can ALWAYS pass through all the hexagons (no matter what path you take). You can’t go back on yourself though.

Orange means that SOMETIMES you can pass through all the hexagons and other times you can’t (it depends on the path you take)

Red means that you can NEVER pass through all of the hexagons no matter what path you take.

Let’s look at an example:

Take this shape here:

If we start from the this hexagon:

We have two options of paths:

As you can see both paths pass through all the hexagons.

This means the top square is green:

Taking the same example. But starting from this hexagon:

There are 4 options of paths to take:

None of these paths pass through all the hexagons.

So the hexagon is red.

Finally if we start from this hexagon:

We have 3 options of paths:

Here some of the paths pass through all the hexagons and others don’t.

This means the hexagon is orange.

For the moment our shape looks like ths:

We want to fill each hexagon in in either red, orange or green.

Select the 16 Jacks, Queens, Kings and Aces from a pack of cards. Try to put them in a 4 by 4 square so that each rank (J, Q, K, A) and each suit (Clubs, Diamonds, Hearts, Spades) appears only once in each row and column.

This puzzle is an example of a Latin square. Latin squares are used in medical trials to ensure every participant is allocated to each treatment for the same time period to prove which is the best treatment. See these links for some more info – mathsisfun, wikipedia and nrich.

We are really missing having maths club in person, and we missed the maths camps in Africa in the summer … but we are so happy that circumstances have meant the launch of the Virtual Maths Camp. The puzzle above is going to appear on the 6 of spades in our card deck! Please check out the idea behind our maths club app  which can be accessed online here. We are very keen to translate our activities into French so that they can be used in countries like Togo. If you would like to help with translation, please get in touch with Emily Fleming at [email protected].

Go to this https://bitstorm.org/gameoflife/ and colour in a 2 by 2 square on the screen like this:

Press start. Nothing should happen to the square. Why?

Well, because each yellow square has exactly 3 neighbours so all survive. There are no other cells that have exactly 3 neighbours so no new cells are born.

Now try with a three by three cell:

Press start. What happens?

Try to work out on paper what you think would happen for a 4 x 4 shape.

Then test your answer.

How about 5 x 5, 6 x 6 and 7 x 7?

Have a play around and see what fun patterns you can make.

You might like to read the article from Plus Maths or even try this question from the 1996 British Informatics Olympiad.