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Yih or Luk Tsut K'i or Three Men's Morris

Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and knot arithmetic.


Stage: 1, 2 and 3 Challenge Level: Challenge Level:1
We had a few ideas sent in for this game, firstly from Arran, Alec, Martin, Tom and Saul at St. Nicholas Junior School in Newbury:

Subtract counters to make 3 on your opponent’s last go because then if they take 2, you take 1 and win and if they take 1 you take 2 and win!
The first person needs to take one counter to win. Also, if you are playing a good opponent, you need to go first and take 1. Then if they take 1, you take 2 leaving 3 and you win, but if they take 2 on their first go, you take 1 leaving 3 again and you can win.
If your opponent is not as experienced as you and they go first and take 1, then you are done for unless they make a mistake! If they take 2 on their first go, then you can take 2 leaving 3 and you will win!

We tried playing Nim-10 using 10 counters and found that the number you have to get to is 6 and before that 9 and then you’ll win. 

We looked at taking more counters e.g. 1, 2 or 3 and the number of counters that you need to leave on your opponent’s last go is always the next number from the most counters that you are allowed to pick up. For example, if you can take 1, 2 or 3 counters, then the number of counters to leave is 4.

Thank you for that excellent subission. You have included a really good explanation.
Then Tom and Jack who are from Canberra Grammar School in Australia sent in this table of their results from playing:

They go on to say;

When the starter takes takes two then you win.

Rebekah (helped by her dad) sent in this:

The first player can always win:
  - on the first turn she should take one counter
  - on the next turn she should take enough counters to reduce the pile to three
  - the other player will be unable to stop her taking the last counter.

In general, with N counters we think that:
  a. if N is a multiple of three the second player can always win
  b. otherwise the first player can always win

The winning strategy is always to reduce the remaining pile to a multiple of three (3, 6, 9, etc.).

In case (b) the first player can always get to a multiple of three by taking one or two. Then the second player will not be able to get to a multiple of three. 
Since zero is a multiple of three (!) the first player can get there and win.

What a great explanation.  I like the way you have thought about the game no matter how many counters you have in total.  Generalising like that is what maths is all about!