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Can you get the value of 'pi' by throwing darts?You can get very close to the value of 'pi' if you throw over a thousand darts.. or you can let the computer throw (simulate) the darts for you in a fraction of a second...how? just read on.. |
![]() Figure 1 |
The figure 1 on the left, is simply a unit circle circumscribed by a square. We could examine this problem in terms of the full circle and square, but it's easier to examine just one quadrant of the circle, as in the figure 2, on the right. If you are a very poor dart player, it is easy to imagine
throwing darts randomly at Figure 2, and it should be apparent
that of the total number of darts that hit within the square,
the number of darts that hit the shaded part (circle quadrant)
is proportional to the area of that part. In other words,
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![]() Figure 2 |
If you remember your geometry, it's easy to show that
If each dart thrown lands somewhere inside the square, the ratio of "hits" (in the shaded area) to "throws" will be one-fourth the value of pi. If you actually do this experiment, you'll soon realize that it takes a very large number of throws to get a decent value of pi...well over 1,000. To make things easy on ourselves, we can have computers generate random numbers.
If we say our circle's radius is 1.0, for each throw we can generate two random numbers, an x and a y coordinate, which we can then use to calculate the distance from the origin (0,0) using the Pythagorean theorem. If the distance from the origin is less than or equal to 1.0, it is within the shaded area and counts as a hit. Do this thousands (or millions) of times, and you will wind up with an estimate of the value of pi. How good it is depends on how many iterations (throws) are done, and to a lesser extent on the quality of the random number generator. Below is Flash version of this with Action Script (Flash scripting language) as follows:
Sample computer code. The one below is Flash 'Action Script' code, but it can be adapted to just about any programming language: hits = 0; |
Web author: Tug Sezen