At my various jobs, I've always introduced what I call the "dot game" to my team. Start by placing three dots on a whiteboard (or paper, of course).

Draw a line between any two dots. Place a new dot anywhere on the line.

A dot can only have three lines touching it. (Explanation: start with just two dots. Draw a line between them. Each dot now has one line. Draw your new dot on the line. That new dot is touched by TWO lines. The new dot in effect cuts the one line in two. The new 3rd dot can only have one more line drawn to/from it.)

No line can intersect another.

You can draw a line out from and back to the same dot in a little loop, realize though that the dot would be touched TWICE if you do.

Lines can wrap AROUND a dot, encircling it. That's important, since no lines can intersect.

To help me with game strategy, I've looked at the math and realized that:

Max_Possible_Dots = (Starting_Dots * 4) - 1

My challenge to the group is to explain why that equation works.

(If you know how many dots are possible, you know if you want to play first or last. Knowing that, you can try to eliminate potential dots by limiting all possible moves, somehow.)

12 Years
Discussion Span
Last Post by server_crash

Each time I play(with 3 dots), I get a lot more than 11 dots. Am I doing something wrong? Can you show a simple picture of you playing the game, say in paint?

By the way, that seems like a pretty cool game even though I can't seem to get it right.


Sure. The original 3 dots I've colored blue. Each new dot is formed by connecting two other dots. I've numbered the dots in sequence, so you can see how I drew the lines. The 8th dot ends the game. Whoever drew the 8th dot is the winner, since it's the last dot. Why is it last? Because there are no other "open" dots to which it could connect. You can see that all dots have 3 lines except the 8th.

8 moves, 11 dots.



Thanks. I got it now. I was doing something wrong when I first tried it.

At first I was thinking that equation was coincidental and had really no relevance other than finding the correct number of dots, but since it works for all of them, there must be some kind of "strategy" this equation follows.

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