Day 10r, MAT115

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Announcements:
- The quiz today is 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 next week, so come with your questions!
- You have an assignment, which is a reading and some fun videos by Vi Hart.
Last time:
- I introduced the problem that gave rise to the "Fibonacci" numbers:
  Notice that we're counting pairs, not individual rabbits....
- The definition of the Fibonacci numbers: \[ \left\{ \begin{array}{cl} {F_1}&{=1}\cr {F_2}&{=1}\cr {F_{n}}&{=F_{n-1}+F_{n-2}} \end{array} \right. \] 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
- We discovered Fibonacci numbers in Pascal's triangle:
- Fibonacci numbers also generate a spiral that nature loves (details in Vi Hart videos): We simultaneously create a rectangle which gets successively closer to a "golden rectangle", one whose proportions are in the ratio of \[ \phi = \frac{1+\sqrt{5}}{2} \approx 1.618 \] The example we created, which is 34x21, has ratio of side lengths \[ \frac{34}{21} = 1.619 \approx \phi \] making it extraordinarily close to golden.
- The strategy for success in Fibonacci nim was relayed.
  It's based on this fact: 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
Today's Question of the Day:

What are some examples that show that nature loves Fibonacci numbers?
- Let's begin with a review of the secrets of Fibonacci Nim. If you play by this strategy, you can't lose. And you don't want to lose to me on the exam next week, so make sure that you understand....
  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.
- Who's going to beat me in "Fraudini nim"?
- Now on to some Fibonacci stories from nature:
  - Daisies:
  - Dandelions
  - Pine cones
  - Cactus skeletons
Come with your questions and concerns next week, and we'll review both Monday and Tuesday.
Time for a quiz!
Links:
- Hexagonal Paper

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.