2012-03-17

A Childish Problem

Bengt has no children of his own, for various reasons. Mainly because he's such a big baby himself, most likely. He does occasionally enjoy the company of children, though. Particularly the geeks. From them, he has now learned a classic old problem. The task is to continue the sequence:
1
11
21
1112
3112
211213

  1. What is the next element in the sequence? It's probably easier if you think like a child.
  2. We can imagine three possibilities for the long term behaviour of the sequence: either it settles on one element, repeating that, or it goes into a loop, repeating a subsequence, or it continues forever without looping. Which one is it?
  3. We could start the sequence with some other start value. Which of the three behaviours are possible?

2012-03-09

Misunderstanding Maths

Bengt and Alban are doing some calculations. Actually, Alban is doing some sort of homework, and Bengt is playing around with his calculator. Bengt tries to draw a graph, of sin2(x), but he has done something terribly wrong. The curve he gets looks a lot like k*sin(x), where k ≈ 0.841470985.
"This calculator is stupid." says Bengt.
"I think it's probably you who's made a mistake." replies Alban.
"I know that, you silly engineer. But it has a resolution of 96x64 pixels. I'm pretty sure my digital watch from the eighties has a higher resolution."
"Yeah, that is stupid. Anyway, I'm having some problems too."
"Good ones?"
"Maybe to you. It says here that the results are in the interval k ± s. But when I look at them, it seems that only a fraction f of them are."
"Interesting. What's f?"
"It's two thirds, I think… or maybe just a little more."

  1. What has Bengt done wrong?
  2. Is the curve that Bengt has found actually a constant times sin(x)?
  3. What is causing Alban's confusion? And what is f?

2012-03-02

Hole Puncher of Horrors

Bengt is building a machine. It is a most unusual kind of hole puncher. It consists of a holder for the papers, and a rather nasty-looking wheel with n sharp spikes pointing outwards at regular intervals. As the machine operates, the holder is pressed against the spike wheel, so that one spike makes a big hole in the paper. (One big hole is more effective than several small, says Bengt.) Occasionally, a spike will break. This is where it gets really clever, or so Bengt thinks. The broken spike falls off completely, making that side of the wheel lighter. If the centre of mass of the wheel is not in the axis, the wheel will spin to orient itself with the centre of mass downwards. That way, Bengt figures, the broken spike will automatically be replaced - at least for a while, until enough of them have been broken.Ideally, we would like to design the device so that this can happen as many times as possible before there is no longer a spike pointing at the paper.

    At which angle on the wheel, relative to the downward angle, should the holder be located? Are some choices of n better than others? Obviously more would be better, but aside from that?