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Okay, welcome to Games without Chance. None of that.
Okay, I'm Tom Morley and let's see where we are this week, we're doing the
following. Some numbers are games and that's where we
are right now. Some numbers are games.
Okay, so let's look at Hackenbush. This was a game we looked at from the
beginning. Here's our ground.
Here's a blue edge. Blue is left.
Left can cut blue. Here's a red edge.
Red edges can be cut by right. Now, this was a game that we looked at and
we computed a while back, that two of these, copies of these the other with one
red is 0. Whoever moves first in this game loses.
If blue moves first, cutting blue edges, one at a time, and alternates with right,
who cuts red edges then if left moves first, left loses, if right moves first,
right loses. This is a zero game.
This is minus 1 here, so might as well call this 1 half.
1 half plus 1 half minus 1 is 0. So this game here, the number.
Now, suppose we have two of these. Two, we have blue, but two red on top.
Now there's no reason for right ever to cut the bottom red, so right might as well
cut the top red. And it turns out that, it turns out that
it's a four-letter acronym four-letter acronym.
It turns out that this behaves like 1 4th, you take two of these and they add up to
minus a half. There's a half minus a half is reverse
colors. Now in general, we have the following
games, so, so here's 0. There's 0.
There's the zero game, right there. Okay.
A picture of the zero game. I should hang this up on my wall.
There, there's the zero game. Let's see.
Here's so that's zero, this is 1, this is 2.
Run out of ground. Let's put some more ground over here.
Here's 3, one, two, three. We have the negatives of these.
This, for instance, is minus 2. We have some fractions here's 1 half.
I'm going to do a bunch of these wait. Here's a half.
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This is a quarter. This is an eighth.
This is a sixteenth. And, you can guess that when you have one
blue and a bunch of red on top, you end up with a 1 over a power of 2.
Now, once you add these all up, so you can take any of these and add them up, what
you can get by adding up these games is anything of the form p over two to the n.
P is any integer, n is in the integer and these are called the dyadic, dyadic
rationals. So all the dyadic rationals turn out to be
games that, they all turn out to be expressible in terms of Hackenbush.
So, let's look at a problem. You will contrive.
What number is this? These two games.
This is one game that's the sum of two games.
What number is this? Okay.
Here is the solution. This is one fourth.
This is two. And 1 4th plus 2 is 2 and 1 4th, also
known as, what C, two is eight, no, 9 4ths so either answer would be correct.
And that's the end of Module 1.
Dr. Tom MorleyProfessor
