David Cowan from Bessemer Venture Partners recently posted a tough logic puzzle that Nivi couldn’t solve.
Marginal Revolution has another cool puzzle with a simple answer:
One morning, exactly at sunrise, a Buddhist monk began to climb a tall mountain. The narrow path, no more than a foot or two wide, spiraled around the mountain to a glittering temple at the summit. The monk ascended the path at varying rates of speed, stopping many times along the way to rest and to eat the dried fruit he carried with him. He reached the temple shortly before sunset. After several days of fasting and meditation he began his journey back along the same path, starting at sunrise and again walking at variable speeds with many pauses along the way. His average speed descending was, of course, greater than his average climbing speed.
Prove that there is a spot along the path that the monk will occupy on both trips at precisely the same time of day.
Nivi will put the elegant and intuitive proof in the first comment to this post. In math mumbo-jumbo, this is known as a 1-dimensional fixed point theorem. Nivi couldn’t solve this problem either.
Moral of this post: Nivi is not so smart.
Actual moral of this post: Nivi shall blog in the third-person henceforth per Gabe Rivera’s suggestion.
Again, from Marginal Revolution: “Here is an intuitive proof of the monk problem. Imagine that there are two monks, one going down and one going up, each beginning on the same day at sunrise. At some point in the day the hiker’s must meet! QED.”
I’ve heard a two dimensional version of this problem: prove that there are always at least two antipodal points on the globe with exactly the same temperature and pressure. I don’t have a solution for this – do you know if it’s possible to do without invoking crazy topology?
this is one a puzzle by Martin Gardener
Hey Nivi, I saw this post, followed it to the puzzle and decided to try and solve it. Not as elegant as the solution you posted for the monk problem though.
Enjoy reading your posts, especially regarding “web 2.0″.
Let the first day’s velocity be v1 and the second’s be v2 and let the distance to the top be d. What we’re requiring is that there exist an elapsed time tau so that v1tau = d – v2tau. Clearly tau = d/(v1+v2) has solutions for all reasonable values of d, v1 and v2.
The lesson is, express problems as equations. The advantage of math is that you can solve problems essentially without thinking. When you’re done (usually with far more difficult manipulations than shown here), you can consider whether the answer makes sense, especially in limiting cases. Barbie was wrong: math is your friend!
Alison, the monk’s velocity is not a constant and is a function of time.
Sometimes the monk even stops for a break!
So the term v1*tau in your equation should be v1(tau) and that makes your equation unsolvable!
That is why the proof I discussed is so cool.
Cool! I didn’t read your solution, but I found the same answer (the intuitive one). I just imagined a map of their travels and superimposed the two paths in my head and imagined that they had to walk past each other at some point.
I used a graph: distance on the Y axis, time on the X axis. The monk’s trip up starts at (0,0) and ends up at (maxX, maxY). The trip down starts at (0, maxY) and ends at (maxX, 0). No matter how you draw the two lines between those points, they must intersect, which implies being in the same place at the same time.
Not as immediately obvious as your solution, but it worked pretty quickly for me.
Chad, that’s the first thing came to my mind too.
In my high school calculus text, this problem is in the section with the Intermediate Value Thm. Not that I’ve figured out how to solve it that way, but someone else might figure something out.
I do like Chad’s proof.