Day 10, MAT115

Last Time

Next Time

Announcements:
- The quiz today will be over the Babylonian and Mayan number systems; also the class agendas, and the readings.
- Just a reminder that we have our first exam next week, Thursday, over everything up to Fibonacci numbers. We will review Tuesday.
- You have an assignment, which is a reading and some fun videos by Vi Hart.
Last time:
- We reviewed the quiz, and emphasized some of the cool number patterns in Pascal's triangle;
- we practiced writing numbers in the Babylonians' and Mayans' number systems; and
- then I taught you how to win a million dollars....
Today's Question of the Day:

Why does nature love Fibonacci numbers?
Let's begin with the story that gave rise to the Fibonacci numbers. Once upon a time there was a pair of rabbits....
"A certain man had one pair of rabbits together in a certain enclosed place, and one wishes to know how many are created from the pair in one year when it is the nature of them in a single month to bear another pair, and in the second month those born to bear also." (source).
(Oh no! A story problem! Here's Fibonacci's solution.)
But I prefer a tree:

Notice that we're counting pairs, not individual rabbits....
How are Fibonacci numbers defined? \[ \left\{ \begin{array}{cl} {F_1}&{=1}\cr {F_2}&{=1}\cr {F_{n}}&{=F_{n-1}+F_{n-2}} \end{array} \right. \] for $n=3, 4, 5, \ldots$. So we're given $F_1$ and $F_2$; and then we generate all the rest using the rule, e.g. \[ F_3 = F_2+F_1 = 1+1 = 2 \]
The first few Fibonacci numbers (those below 100):
1,1,2,3,5,8,13,21,34,55,89
It turns out that the Fibonacci numbers appear in Pascal's triangle in a systematic way, which we'll illustrate with the hexagonal graph paper, which is best for showing this relationship.
Fibonacci numbers also have a connection to a beautiful concept called the golden ratio, and the golden rectangle. We'll use ordinary graph paper to illustrate that.
We're going to create what's known as a "Fibonacci Spiral".
1. Start with a small square (which is also a rectangle); then
2. append the largest square you can to the current rectangle.
3. Do it again, and again, proceeding in a consistent sense (e.g. clockwise)
In the end, we draw quarter circles in each box, to make a beautiful spiral -- a Fibonacci spiral.
Now let's get back to "Fraudini Nim" for the exciting conclusion and for the strategy.
- Rules for "Fraudini Nim":
  - There are two players, Player 1 and Player 2.
  - An arbitrary number of counters is thrown down (e.g. toothpicks, candies, pennies).
  - Objective: The player who picks up the final counter wins.
  - Player 1 will go first. On the first play, Player 1 must pick up at least one counter, but may not pick up all the counters (otherwise this would be a silly game!).
  - Player 2 must then pick up at least one counter, and not more than twice the number of counters picked up on Player 1's turn.
  - The players then alternate turns, subject to these same conditions: pick up at least one, no more than twice what the preceding player took.
  - The player who legally picks up the final counter on their turn wins.
- Okay, so what's the strategy? How does one win at Nim? How could I be so confident that I was going to beat anyone?
  We tried to figure out the strategy, by creeping up on it. When you confront a big problem, start with some really simple examples.
  - We looked at some of the simplest cases, and this is what we concluded:
    
    Number of counters 1 2 3 4 5 6 7 8
    
    Winner (1 or 2? -- if they play right!) X 2 2 1 2 1 1 2
  - The first few Fibonacci numbers (those below 1000): 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,....
  - What's the winning strategy in Fraudini Nim? It all hinges on this, the third of our big factorizations:
    The Fibonacci factorization
    
    Every natural number is either
    1. Fibonacci, or
    2. it can be written as a sum of non-consecutive Fibonacci numbers in a unique way.
    Examples?
    1. 89 is Fibonacci
    2. 24 = 21 + 3
    3. 67 = 55 + 8 + 3 + 1
  - Now, for the secrets of Fibonacci Nim. You need to learn this strategy, and you can't lose.
    Suppose there are $n$ counters on the table to start.
    1. First or Second? To be "in the driver's seat", you should
      - arrange to go second if $n$ is Fibonacci;
      - go first otherwise.
    2. Your move? If you're in the driver's seat, then you'll be able to do this:
      - First of all, if you can legally take all the counters, do so! Then you win. Otherwise,
      - Write $n$ as the unique sum of non-consecutive Fibonacci numbers; call the smallest Fibonacci in the sum $m$.
      - Take $m$ counters.
      You'll be guaranteed a victory.
      You'll know that you're not in the driver's seat if you can't follow this strategy. The reason why? Remember that you can only take up to twice what your opponent took. Although you might write the number of counters remaining as a sum of Fibonacci numbers, you will find that you are not allowed to take even the smallest of the Fibonaccis -- it will be more than twice what your opponent picked up, so that would be illegal.
    Let's look at some examples:
    - 29 counters, and you're first. You're in the driver's seat, since 29 isn't Fibonacci.
      29 = 21 + 8
      so you take 8.
      I'm up, and looking at 21. It's already Fibonacci. I can't take the smallest in the "sum" -- 21. So I'm out of luck. I can't follow "the strategy". Whatever I take, you'll be able to beat me.
      To slow things down, I take 1.
      It's now 20 to you. 20 = 13 + 5 + 2. So you take 2 (that's legal). And so on, and you're going to beat me. Argh!
    - 34 counters, and you're second. You're in the driver's seat, since 34 is Fibonacci.
      Suppose I go first, and take 4.
      Then it's 30 to you. 30 = 21 + 8 + 1, so you take 1. Now it's 29 to me.
      You might think "29 to Long, and he can pick first -- he's going to win!" But that's wrong. I'd like to take 8, as before, but I can't -- because 8 is more than twice what you took.
    Now here's another key point: when you're not in the driver's seat, take 1 -- the smallest legal move. That's because you want to slow down the game, and let your opponent make a mistake and put you back in the driver's seat.
  - So basically, the whole things comes down to this. If you can choose whether to go first or second, do it and you're guaranteed a victory.
    1. Go first for non-Fibonacci, second for Fibonacci.
    2. Of course, take all of the counters if that's a legal move.
    3. Whatever the number of counters on the table, write it as a sum of non-consecutive Fibonacci numbers;
      1. take the smallest Fibonacci, if you can; you're now in the driver's seat, and should win.
      2. if you can't take the smallest (because it's more than twice your opponent's move), take 1 -- it will slow the game down, and hopefully your opponent will make a mistake.
Time now for a quiz....
Links:
- Hexagonal Paper
- Fibonacci numbers in Pascal's triangle:

Website maintained by Andy Long. Comments appreciated.

Fibonacci's solution in Liber Abaci,, chapter 12:
Because the above written pair in the first month bore, you will double it; there will be two pairs in one month. One of these, namely the first, bears in the second montth, and thus there are in the second month 3 pairs; of these in one month two are pregnant and in the third month 2 pairs of rabbits are born, and thus there are 5 pairs in the month; ...
there will be 144 pairs in this [the tenth] month; to these are added again the 89 pairs that are born in the eleventh month; there will be 233 pairs in this month.
To these are still added the 144 pairs that are born in the last month; there will be 377 pairs, and this many pairs are produced from the abovewritten pair in the mentioned place at the end of the one year.
You can indeed see in the margin how we operated, namely that we added the first number to the second, namely the 1 to the 2, and the second to the third, and the third to the fourth and the fourth to the fifth, and thus one after another until we added the tenth to the eleventh, namely the 144 to the 233, and we had the abovewritten sum of rabbits, namely 377, and thus you can in order find it for an unending number of months.

Number of counters	1	2	3	4	5	6	7	8
Winner (1 or 2? -- if they play right!)	X	2	2	1	2	1	1	2