Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

There is a clock at the bottom of the hill and a clock at the top of the hill. The clock at the bottom of the hill works fine but the clock at the top doesn't. How will you synchronize the two clocks. Obviously, you can't carry either of the clocks up or down the hill! And you have a horse to help you transport yourself. And, the time required for going up the hill is not equal to the time required to go down the hill.

share|improve this question

closed as off-topic by Grigory M, anorton, daw, user1729, amWhy Jul 15 at 12:03

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "This question is not about mathematics, within the scope defined in the help center." – Grigory M, user1729
  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – anorton, daw
If this question can be reworded to fit the rules in the help center, please edit the question.

What exactly do you mean when you say that the clock at the bottom doesn't work fine. Does it mean - it's loses time incrementally or it's set wrong originally but increments correctly or whatever else? –  user93353 Oct 24 '12 at 12:05
I decided to Google this question, since it looked like it was copied/pasted from somewhere. Turns out, that the first 3+ pages of hits were word-for-word matches for this question. Some examples of answers are here, here, here. –  Douglas S. Stones Oct 24 '12 at 12:32
I downvoted this because I think questions where someone doesn't even take the minimal effort to state the puzzle in their own words shouldn't be encouraged. (Let alone "What have you tried?") –  ShreevatsaR Oct 24 '12 at 14:19
I seriously doubt if any of the downvoters had a slightest clue of whats going on here. Firstly, the link posted by Douglas are simply questions and not answers so it's very likely that the OP comes here to find an answer.Secondly, how do you improve upon a 4 liner question which is a puzzle after all and why would you ignore the fact that he might have tried. Wouldn't it be easy to ask the question as it is and not confuse by manipulating it in own words. This is a simple misuse of the power of downvoting else they should come up with an answer to this question. –  Arham Apr 16 '13 at 3:51
Easy. Synchronize your watch with the clock at the bottom of the hill. Then synchronize the clock at the top of the hill with your watch. If you don't have a watch, then use your phone. Problem solved. –  T-Wayne May 28 '13 at 5:10

4 Answers 4

When you say that the clock at the top of the hill doesn't work, I assume you're saying that the clock at the top runs at a rate that is constant but incorrect. By "synchronize," I assume you mean that you want to find the constants $a$ and $b$ in the calibration function $t=aT+b$, where $t$ is the reading on the bottom clock, $T$ the top.

This is similar to, but not equivalent to, the problem of Einstein synchronization in a static spacetime. The reason it's not equivalent is that in relativity, we essentially define the horse as having the same velocity in both directions.

You can determine the constant $a$ so as to calibrate away the error in frequency. Send the horse up, down, up, and down again, giving a sequence of observations {$t_1$, $T_1$, $t_2$, $T_2$}. Then $a=(t_2-t_1)/(T_2-T_1)$.

However, your problem has no solution because it's not possible to determine the phase difference $b$ between the two clocks. Say the horse takes time $\Delta t_{up}$ to go up, and $\Delta t_{down}$ to come down. Suppose that $a=1$, so that $t=T+b$. Then under the transformation $b\rightarrow b+c$, $\Delta t_{up}\rightarrow\Delta t_{up}+c$, $\Delta t_{down}\rightarrow\Delta t_{down}-c$, we get exactly the same sequence of observations {$t_1$, $T_1$, $t_2$, $T_2$} (assuming that in both cases we make the same arbitrary choice of $t_1$ as the time to start the process). This shows that synchronization is impossible for $a=1$, which means that it's not possible in general.

share|improve this answer

You cannot synchronize these clocks without breaking them. They go at different rates due to gravitational time dilation.

share|improve this answer
Gravitational time dilation is irrelevant both in theory and in practice. It's irrelevant in practice because we're talking about a mule going up and down a hill, so relativistic effects are on the order of nanoseconds. It's irrelevant in theory because there doesn't seem to be any other way of interpreting the word "broken" in the question than as a statement that it runs at the wrong rate. Since it already runs at the wrong rate for nonrelativistic reasons, the difference in rate due to relativity can just be absorbed into that. Rate matching is only impossible in a non-stationary spacetime. –  Ben Crowell Oct 24 '12 at 15:23
@Ben: This answer was half earnest and half in jest, inspired by the OP's lack of effort. But about the serious half: The word "broken" doesn't occur in the question; the word "synchronize" does. To my mind, to synchronize two clocks would mean to make them show the same time. Since the clocks wouldn't go at the same rate when working properly, we can only make them show the same time by breaking at least one of them -- or, as far as the question implies that one of them is already broken, by leaving it broken. –  joriki Oct 24 '12 at 15:55

Here we can assume that down clock has time T and the difference between upper clock and down clock is t. Now

TUH = Uphill time with horse

TDH = Downhill time with horse

TUF = Uphill time without horse

TDF = Downhill time without horse

Firstly we will go to with horse then we know about t+TUH Then we will come down we know about : TUH+TDH

Without horse up : t+TUF

withour horse down : TUF+TDF

There are 4 equation 5 unknowns but we can make 2 more equations We go up with horse and come wihout horse then we know about TUH+TDF and similary TUF+TDH

6 equations 5 unkown. Ok?

share|improve this answer
These six equations are all linear combinations of four quantities: TUH+t, TDH-t, TUF+t, TDF-t. So we can measure all these quantities, from which we can figure out things like TUH+TDF, but we can't get t. –  Peter Shor Jun 5 '13 at 19:30

This is a broken problem. As Ben's answer shows, there is no solution.

In the original problem, as given in this 1983 book (courtesy of Google books), you didn't have a horse but a wife to assist you. The solution given in this book is easy, though ... it assumes there is a way to communicate instantaneously from the top of the hill to the bottom. You can then have one person time the other climbing the hill, which makes it trivial to synchronize the clocks.

If you are allowed to train the horse so that it comes when you call (so again you have instant communication) you can also solve this problem, even if the horse travels at a different speed depending on whether or not you are riding it.

  1. Train the horse to come to you when you blow a whistle.

  2. Leave the horse on the top of the hill, walk down the hill, blow the whistle and see how long the horse takes to go down the hill.

  3. Leave the horse at the top of the hill, walk down, then blow the whistle and start walking up the hill. Mark the spot where you meet the horse.

You can measure the ratio of the distance to the marked spot and the distance to the top of the hill: time your round trip to the marked spot, time your round trip to the top of the hill, and divide.

This gives you the ratio of the speed of the horse coming down the hill to your speed going up the hill. From this, you can calculate how long it takes to climb the hill, which gives you enough information to synchronize the clocks.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.