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Posers and Puzzles

Posers and Puzzles

  1. 17 Jun '05 14:40
    A new species of bacteria called blue and red behave in the following way: after one minute every blue bacterium turns into 6 red ones, and every red bacterium turns into 1 blue and 1 red bacterium. During an experiment you put 1 blue and 1 red bacterium into a test-tube, so that after one minute there were 1 blue bacterium and 7 red ones; and after two minutes there were 7 blue ones and 13 red ones. Prove that after n minutes there will be:

    3/5 * 3^n + 2/5 * (-2)^n
    Blue bacteria in the test-tube.

    I wont provide any hints at the moment but if later no one makes any headway, which i doubt, then i'll post some hints.
  2. Subscriber sonhouse
    Fast and Curious
    17 Jun '05 16:45
    this is a version of the game of "life". Anyone remember that one? It
    was in a Scientific American article many years ago. Used a board
    something like a Go board, but with a theoretically infinite number
    of intersections but a Go board works for most iterations.
    You have black and white stones, put down some initial pattern
    then some stones die, some reproduce and the pattern changes
    in response to a simple set of rules. Each intersection has 8 surrounds
    and depending on the count of black and white stones in those
    surrounds, the pieces live or die. Like if there is only one of its color
    in the 8 surrounds, it dies of loneliness, if there are two and only two
    it reproduces, etc. Each generation produces a new pattern that,
    depending on the original configuration, can make a self reproducing
    shape that eventually runs off the board. Some shapes sit there
    and go back and forth between two configurations, a stalemate
    and others dribble into a static state or dies off completely.
  3. 19 Jun '05 10:05 / 1 edit
    Originally posted by sonhouse
    this is a version of the game of "life". Anyone remember that one? It
    was in a Scientific American article many years ago. Used a board
    something like a Go board, but with a theoretically infinite number
    of intersections but a Go boa ...[text shortened]... e
    and others dribble into a static state or dies off completely.
    Ok here's a hint:

    This is a proof by Induction.

    Also note that,

    b1 = 1, r1 = 7
    b2 = 7, r2 = 13
    b3 = 13, r3 = 55

    See the pattern? Now you just need to prove it.
  4. 19 Jun '05 10:13
    Originally posted by elopawn
    Ok here's a hint:

    This is a proof by Induction.

    Also note that,

    b1 = 1, r1 = 7
    b2 = 7, r2 = 13
    b3 = 13, r3 = 55

    See the pattern? Now you just need to prove it.
    Well, induction means showing that it is true in one case, then assuming it is true for a particular case (n=k), then showing it is true for the next case (k + 1).

  5. Standard member orfeo
    Missing 285 + 1
    20 Jun '05 03:40
    Originally posted by elopawn
    Ok here's a hint:

    This is a proof by Induction.

    Also note that,

    b1 = 1, r1 = 7
    b2 = 7, r2 = 13
    b3 = 13, r3 = 55

    See the pattern? Now you just need to prove it.
    I can see more than one possible pattern, I will have to do the next step to decide which pattern it actually is.

    Or you can save me the trouble...
  6. Standard member PBE6
    Bananarama
    20 Jun '05 18:36
    This is a toughie. The induction isn't that hard, but how did you come up with the formula? The neatest thing I found was that the ratio of red to blue approaches 3. Not too helpful here.
  7. 21 Jun '05 11:14 / 1 edit
    Let b_t = number of blues after t minutes
    Let r_t = number of reds after t minutes

    What is b_(t+1) in terms of b_t and r_t?
    What is r_(t+1) in terms of b_t and r_t?

    Ok, lets see,

    b_t = r_(t-1)
    r_t = 6b_(t-1) + r_(t-1)

    b_(t-1) = r_(t-2)
    r_(t-1) = 6b_(t-2) + r_(t-2) = 6b_(t-2) + b_(t-1)

    Thus: b_t = b_(t-1) + 6b(t-2)

    Now, it should be easier.
  8. Standard member PBE6
    Bananarama
    21 Jun '05 14:17
    Originally posted by elopawn
    Let b_t = number of blues after t minutes
    Let r_t = number of reds after t minutes

    What is b_(t+1) in terms of b_t and r_t?
    What is r_(t+1) in terms of b_t and r_t?

    Ok, lets see,

    b_t = r_(t-1)
    r_t = 6b_(t-1) + r_(t-1)

    b_(t-1) = r_(t-2)
    r_(t-1) = 6b_(t-2) + r_(t-2) = 6b_(t-2) + b_(t-1)

    Thus: b_t = b_(t-1) + 6b(t-2)

    Now, it should be easier.
    I got that far already, but converting that into the exponential formula is giving me the runs. Help!
  9. 25 Jun '05 12:54
    Originally posted by PBE6
    I got that far already, but converting that into the exponential formula is giving me the runs. Help!
    hmm, i'm still working on it as well, i have a feeling it has got to do with the recurrance relation.