Batch 2 - Class 290 - Game of Hex

• How can you divide a long paper strip into 5 equal parts?
• Answer: start with an approximation from one end. Then fold the far end into four parts. The new 1/5th estimation is better than the old one. Repeat few times to get a good approximation of 1/5

• Should we put out markers in connected positions, or is there a more efficient way?
• What are the kind of positions that your opponents can not prevent you from joining? For example:

• There are many such templates https://www.drking.org.uk/hexagons/hex/templates.html
• What happens if you are following your opponent to prevent them from getting through ("ladders"). What happens at end of ladder, and what do you need to do to stop a ladder? Blocking step-by-step sometimes doesn't work, it may help to move a bit away (and in a position closer to your edges to create winning options for yourselves)

• How do you place counters so that they can be useful more than once? How do you play with a long term view (for example ladder outcomes)?

• Who must win the hex game - first player or second player? Why?
• Explain the strategy stealing argument - an extra piece at a random position can't really hurt in Hex (note that this is not true for a chess game for example)
• What prevents the second player from stealing the strategy again?
• If after the first move, the second player had a winning strategy, shouldn't the second player then win?
• Can the Hex game end in a draw? Let us try to construct boards where the final position is a draw and see what happens.
• Explain the graph example, without too much terminology. Explain the notion that paths can not cross over (by taking any "vertex" where three counters meet), and hence each "enclosure" must be the same color. A path entering at one corner must exit at an adjacent corner (and not across otherwise paths would cross), and hence there is a path left in the middle. Islands don't matter because due to non-crossing, they always leave a path around them

• Note that the above proof doesn't actually give us a winning strategy, it only says that the game can't end in a draw. Combined with strategy stealing argument, the first player must win (though we don't know how)

• MartinShCol - 1.14 - There are bowling pins of red color and black color. Can you select 10 pins from these and place them in position of bowling pins, so that no equilateral triangle has all vertices of the same color?
• Answer: This is not possible. Progressively placing the pins starting from the center position helps arrive at the conclusion.