4 meeting bugs

A forum for good logic/math puzzles.

Moderators: jestingrabbit, Moderators General, Prelates

4 meeting bugs

Postby Caesar » Thu Nov 11, 2010 1:44 pm UTC

Imagine a square with a side length of 1. In each corner sits a bug. The bug in the bottom left corner tries to meet up with the bug in the top left corner. That bug, however, tries to meet the bug in the top right corner, which tries to meet the bug in the bottom right corner, which tries to meet the bug in the bottom left corner, the one I started with. Every bug starts walking to its target at the exact same speed, adjusting its direction at all times.

Image

Question 1: How long (as in lenght) will they walk until they eventually meet?

Question 2, and this is something I'm personally wondering about: How big is the angle every bug will have turned until they all meet?
Caesar
 
Posts: 30
Joined: Fri Oct 24, 2008 12:34 am UTC
Location: Germany

Re: 4 meeting bugs

Postby jaap » Thu Nov 11, 2010 2:27 pm UTC

This is a classic puzzle problem,and I'm surprised I haven't been able to find a thread here that discusses it.

Caesar wrote:Question 1: How long (as in length) will they walk until they eventually meet?
Spoiler:
The situation is completely 4-fold rotationally symmetric. It therefore remains that way and the beetles will always be in a square formation (that shrinks and rotates as they move). The distance from one beetle to its target beetle is not affected by the movement of that target beetle because that movement is perpendicular to the line between them. Therefore each beetle travels the same distance it would have if its target were stationary, i.e. the side length of the square 1.
Caesar wrote:Question 2, and this is something I'm personally wondering about: How big is the angle every bug will have turned until they all meet?

I'd not thought about this one before.
Spoiler:
For simplicity, assume that the beetles travel a distance of 1 in 1 second.
When the distance between the beetles is d, then the angular velocity is 1/d radians per second.
At time t the distance is d=1-t, so the angular velocity at time t is 1/(1-t).
To get the actual rotation at a time T, we need to integrate that from 0 to T, which gives -ln(1-T). This goes to infinity as T reaches 1 so they rotate infinitely much.
User avatar
jaap
 
Posts: 1731
Joined: Fri Jul 06, 2007 7:06 am UTC

Re: 4 meeting bugs

Postby Veritas » Thu Nov 11, 2010 5:49 pm UTC

I've seen this one, and it has an interesting answer if you think of the 4 bugs as points, but I think it's even more interesting if you give the bugs a finite area. That way, the number of rotations does not approach inf since the center of the bugs can only reach a finite distance apart before they bump into eachother.

What if the square has length 1 and the bugs have radius 0.01. How many times do they circle before they meet?
Veritas
 
Posts: 24
Joined: Wed Aug 25, 2010 7:35 pm UTC

Re: 4 meeting bugs

Postby undecim » Thu Nov 11, 2010 11:32 pm UTC

Veritas wrote:I've seen this one, and it has an interesting answer if you think of the 4 bugs as points, but I think it's even more interesting if you give the bugs a finite area. That way, the number of rotations does not approach inf since the center of the bugs can only reach a finite distance apart before they bump into eachother.

What if the square has length 1 and the bugs have radius 0.01. How many times do they circle before they meet?

Spoiler:
Bugs will travel 1-2r. Using Jaap's formula -ln(1-t), we get -ln(2r) radians of rotation for bugs of radius r. At 0.01, they will rotate ~6.214 radians or .99 of a complete circle.

If the bugs have radius e^(-2*pi) or about 0.00186744, they will travel one complete rotation. e^(-2*n*pi) will travel n rotations.
Blue, blue, blue
User avatar
undecim
 
Posts: 286
Joined: Tue Jan 19, 2010 7:09 pm UTC

Re: 4 meeting bugs

Postby JamesD » Wed Oct 19, 2011 3:46 pm UTC

Brilliant puzzle - never seen that before. Am going to bore my friends with that for weeks!
Last edited by JamesD on Tue Oct 25, 2011 4:13 pm UTC, edited 1 time in total.
JamesD
 
Posts: 1
Joined: Wed Oct 19, 2011 11:57 am UTC

Re: 4 meeting bugs

Postby Qaanol » Wed Oct 19, 2011 6:04 pm UTC

Here is a thread about this on the Mathematics subforum. It includes the solution for n evenly-spaced bugs around an initial circle of radius r.
Small Government Liberal
User avatar
Qaanol
 
Posts: 2258
Joined: Sat May 09, 2009 11:55 pm UTC

Re: 4 meeting bugs

Postby mward » Thu Oct 20, 2011 11:29 am UTC

A simple way to calculate the amount of rotation of each bug:

Spoiler:
After the bugs have traveled half way, zoom in by a factor of 2: the bugs are in the same relative positions but travelling twice as fast. You can repeat this process infinitely often (taking half as long on each step). Since zooming does not affect the angles, the bugs must turn through an infinite angle.

They must be very dizzy by the time they meet! :D
mward
 
Posts: 53
Joined: Wed Jun 22, 2011 12:48 pm UTC

Re: 4 meeting bugs

Postby Adam H » Thu Oct 20, 2011 3:13 pm UTC

It's funny how all the mathematicians initially assume the bugs are infinitely small. What a bad assumption!
Goodbye
User avatar
Adam H
 
Posts: 685
Joined: Thu Jun 16, 2011 6:36 pm UTC

Re: 4 meeting bugs

Postby jestingrabbit » Fri Oct 21, 2011 10:38 am UTC

mward wrote:A simple way to calculate the amount of rotation of each bug:

Spoiler:
After the bugs have traveled half way, zoom in by a factor of 2: the bugs are in the same relative positions but travelling twice as fast. You can repeat this process infinitely often (taking half as long on each step). Since zooming does not affect the angles, the bugs must turn through an infinite angle.

They must be very dizzy by the time they meet! :D


Or you can just note that they're
Spoiler:
travelling along a loxodrome.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
User avatar
jestingrabbit
 
Posts: 5210
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

Re: 4 meeting bugs

Postby Kingreaper » Sat Oct 22, 2011 11:54 am UTC

jestingrabbit wrote:
mward wrote:A simple way to calculate the amount of rotation of each bug:

Spoiler:
After the bugs have traveled half way, zoom in by a factor of 2: the bugs are in the same relative positions but travelling twice as fast. You can repeat this process infinitely often (taking half as long on each step). Since zooming does not affect the angles, the bugs must turn through an infinite angle.

They must be very dizzy by the time they meet! :D


Or you can just note that they're
Spoiler:
travelling along a loxodrome.

Spoiler:
I haven't googled loxodrome yet, but I'd be willing to bet that explaining to a layman what a loxodrome is, showing they're travelling on a loxodrome, and proving a loxodrome turns through an infinite angle would require at least three lines of text.
User avatar
Kingreaper
 
Posts: 159
Joined: Sun Jan 27, 2008 4:23 pm UTC

Re: 4 meeting bugs

Postby kalakuja » Fri Oct 28, 2011 9:57 pm UTC

Huu nice. I had been thinking about a sort of similar case where there are planets orbiting other planets and it had similar solution of symmetric rotation.
What if the bugs are more spread so that the distances between them varies? For example the red bug is 1 unit more to the north.
Is it still a rotating polygon?
What if the bugs "attract" each other with different "speeds"? Mostly just interested in where to find the theory to read for solving those systems.
kalakuja
 
Posts: 40
Joined: Sun Feb 06, 2011 10:15 am UTC

Re: 4 meeting bugs

Postby tomtom2357 » Sat Jan 14, 2012 6:52 am UTC

What if they adjust their courses to an intercepting course, based on how the first and the second derivative of its position?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357
 
Posts: 433
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: 4 meeting bugs

Postby mfb » Sat Jan 14, 2012 3:33 pm UTC

That requires to know these derivatives to derive themself. I think there are consistent solutions, but it is not computable by iterations. The trivial solution is: All bugs go to the center in a straight line.
mfb
 
Posts: 803
Joined: Thu Jan 08, 2009 7:48 pm UTC

Re: 4 meeting bugs

Postby curtis95112 » Sun Jan 15, 2012 7:35 am UTC

tomtom2357 wrote:What if they adjust their courses to an intercepting course, based on how the first and the second derivative of its position?


I'd expect there to be infinitely many solutions depending on the starting conditions. The most trivial of which would be the one mfb mentioned. All bugs start with second derivative zero.
Also, this would probably open the door to assymetric solutions
addams wrote: There is no such thing as an Unbiased Jury.
curtis95112
 
Posts: 521
Joined: Thu Jan 27, 2011 5:23 pm UTC

Re: 4 meeting bugs

Postby tomtom2357 » Sun Jan 15, 2012 9:36 am UTC

Starting positions in the square formation that was originally stated, and 2nd derivative 0. What would be really interesting is if someone proved that there are no solutions. (I chose the second derivative is because if they have the same speed, then there is no intersection course for the first derivative model) :D
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357
 
Posts: 433
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: 4 meeting bugs

Postby mfb » Sun Jan 15, 2012 1:39 pm UTC

I already posted a solution.
Also note that a perfect intersection course does not need any second derivative, as the bug just heads towards the meeting point. Therefore, all solutions have a vanishing second derivative, as long as the initial velocity is not fixed - otherwise it may be impossible to derive (read: take the derivative of) their velocity.
But with 4 straight lines and constant velocity, the only possible meeting point is the center. Therefore, my trivial solution is the only solution to your problem unless you modify it in some way to get more interesting results.
mfb
 
Posts: 803
Joined: Thu Jan 08, 2009 7:48 pm UTC

Re: 4 meeting bugs

Postby tomtom2357 » Tue Jan 24, 2012 8:05 am UTC

What about the original problem, but the bugs can only turn a certain amount at a time, for example, 1rev/sec (assuming that the side length is 1 and they have speed 1) then they cannot possibly turn an infinite amount. Or, how about if the bug's maximum angular velocity is the reciprocal of its current speed, and the bugs must always face the bug that they are chasing. It will still be symmetrical, but this may change the solution.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357
 
Posts: 433
Joined: Tue Jul 27, 2010 8:48 am UTC


Return to Logic Puzzles

Who is online

Users browsing this forum: t21t3g5n and 4 guests