Wednesday, April 18, 2007

Solution to The Clocks' Problem

Problem:

A photograph
is all you've got. It shows two clocks. Two digital clocks. Here's how it shows:

Clock 1
February 29, 2:07 pm

Clock 2
February 29, 2:09 pm

If you're asked, "By how much is the second clock faster than the first?", what would you say?

Solution:

There can be two extreme cases. In one of the cases, when the photograph is taken the time in the first clock is 2pm+7minutes+0.000 seconds (or delta1 seconds) and that in the second clock is 2pm+9minutes+59.999 seconds (or 10 minutes minus delta2 seconds). In this case, the time difference between the clocks becomes 2 minutes and 59.999 seconds : Or 3 minutes minus [delta1+delta2] seconds. As time is continuous, delta1 and delta2 may be infinitesimally small. So 3 minutes is the maximum possible time difference between the clocks.

In the other case, the time in the first clock may be 2pm+7minutes+59.999 seconds (or 8 minutes minus delta1 seconds) and that in the seconds clock may be 2pm+9minutes+0.000 seconds (or delta2 seconds). In this case, similarly, the time difference between the clocks is 1 minute and 0.0001 seconds : Or 1 minute plus [delta1+delta2] seconds. Again, delta1 and delta2 are infinitesimally small in the limiting case. So 1 minute is the minimum possible time difference between the clocks.

The time difference between the clocks, hence, can be equal to any value in the open interval (1 minute, 3 minute). Boundary values are excluded. So if you're asked that by how much is the second clock ahead of the first, you should say it's between one to three minutes.