Post without words #1

Posted on May 24, 2011 by Brent

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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13 Responses to Post without words #1

  1. JM says:
    May 24, 2011 at 6:17 pm

    This is somehow related to binary…and yet it’s also related to the Fibonacci Sequence. I’ll have to think about it.

  2. xander says:
    May 24, 2011 at 7:32 pm

    Well, there goes my afternoon. 😉

  3. Lucas Brown says:
    May 25, 2011 at 3:21 pm

    It’s almost a cellular automaton with a shift: add a column of green down the right hand side and you get a trace of Wolfram’s Rule # 106/120/169/225 (depending on which color you interpret to be 0 or 1, and on the order in which the cells are traced) with the output shifted one cell to the side at each step.

    • Brent says:
      May 25, 2011 at 4:35 pm

      Very interesting observation! You say it’s very close to a trace of those rules but not exactly. Which makes me wonder whether some sort of cellular automaton can be devised which would give exactly this as its trace — or more likely, would include this as a subtrace, possibly with some extra intermediate states in between.

  4. Lucas Brown says:
    May 25, 2011 at 3:23 pm

    Edit to the comment I just made: add that column down the right hand side and you get something /very close to/ a trace of one of those rules I mentioned.

  5. Pratik Poddar says:
    May 26, 2011 at 4:36 am

    I think I have figured out the pattern and I can make pictures like xander has made. But I do not understand the math behind the pattern. Help? Also, why is is tagged “partition”?

    • xander says:
      May 27, 2011 at 11:21 am

      If I recall correctly from undergraduate analysis, a partition of a set A is a collection of subsets such that the union of all of the subsets is A, and the intersection of any two subsets is empty. That is, the subsets completely cover the original set, but they do not overlap anywhere.

      In the original image, imagine that you are laying stones for a garden path. You have blue stones, which are 1 foot square; and green stones, which are 1 foot by 2 feet. The path is 7 feet long. One question that you might ask is “How many ways can I lay stones to build my path?”

      Each possible arrangement of stones represents one partition of the path—the stones completely cover the path, but do not overlap. What I find interesting is that the number of possible partitions of the type pictured is a Fibonacci number (specifically, the nth Fibonacci number, where n is the length of the garden path).

      This can be thought of inductively (or recursively). Let us suppose that there are F_n ways to lay a path that is n feet long. First, note that there is only one way to pave a 1 foot long path, and that there are two ways to pave a 2 foot long path. Thus

      F_1 = 1
      F_2 = 2.

      How many ways are there to lay stones for a path that is k feet long? That is, what is F_k? You can either lay a path that is k-1 feet long, and put a 1 foot stone at the end; or you can lay a path that is k-2 feet long, and put a 2 foot stone at the end. This gives us the relation

      F_k = F_{k-1} + F_{k+2}

      which is the Fibonacci sequence.

      As an added bonus, it seems that if you add a third type of stone that is 1 foot by 3 feet, you get a sequence defined by

      S_1 = 1
      S_2 = 1
      S_3 = 2
      S_n = S_{n-1} + S_{n-2} + S_{n-3}

      where S_n is the number of possible partitions for a path of length n.

      xander

  6. Jonathan says:
    May 30, 2011 at 3:52 pm

    Instant recognition. I use this problem with students (from middle school through teacher training)

    Ghost the Bunny hops up a flight of 7 steps. Ghost can hop up one or two steps at a time, and never hops down. In how many ways can Ghost reach the top step? (3 double hops then a single is not the same as two doubles, a single, and another double)

    • Eric Fleming says:
      June 7, 2011 at 6:22 am

      Yeah, I also instantly thought of the same bijection. I call it the “stair-step”problem.

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