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miki
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 Posted: Thu Jul 03, 2008 5:40 am Post subject: A nice mathematical riddle |
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Hello All,
Here is a nice simple-looking riddle that leads to heavy mathematics.
Assume that there are 4 dogs in each corner of a square on a plane
(around the origin) at initial time.
Where the start gun is fired, each dog want to get his neighbor (lets
say, clock wise neighbor). The questions are, will the dogs get each
other? Will they stop runing some time if the dogs are infinitesimaly
small? What is the trajectory they will draw? How can you describe the
motion near the origin?
->-----------------
- \/
- -
- -
/\ -
-----------------<-
Let the fun begin :-)
Miki |
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miki
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| Posted: Thu Jul 03, 2008 10:28 am Post subject: Re: A nice mathematical riddle |
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On Jul 3, 11:10 am, "Philippe 92" <nos...@free.invalid> wrote:
| Quote: |
Fabian wrote :
On Wed, 2 Jul 2008 22:40:09 -0700 (PDT), miki <miki.li...@gmail.com> wrote:
Hello All,
Here is a nice simple-looking riddle that leads to heavy mathematics.
Assume that there are 4 dogs in each corner of a square on a plane
(around the origin) at initial time.
Where the start gun is fired, each dog want to get his neighbor (lets
say, clock wise neighbor). The questions are, will the dogs get each
other? Will they stop runing some time if the dogs are infinitesimaly
small? What is the trajectory they will draw? How can you describe the
motion near the origin?
->-----------------
- \/
- -
- -
/\ -
-----------------<-
Let us assume that the dogs move with constant velocity 1 (and I
prefer counter-clockwise chasing). We should use the symmetry of the
problem. Let us describe the plane by complex numbers. The initial
configuration (at time t=0) is given by
z_1(0)=1, z_2(0)=i, z_3(0)=-1, z_4(0)=-i
The symmetry group is generated by i (rotations by pi/2 around the
origin), there are no reflections allowed; in formulas (n is
modulo 4)
(a) i z_n(t) = z_{n+1}(t)
The equations of motion are given by
z'_n(t) = sign(z_{n+1}(t)-z_n(t))
= (z_{n+1}(t)-z_n(t))/|z_{n+1}(t)-z_n(t)|
= sign(i-1) z_n(t)/|z_n(t)| using (a)
(b) = e^{i 3pi/4} z_n(t)/|z_n(t)|
due to the symmetry, we have decoupled the differential equations.
We can solve the one for z_1(t) [the others will be generated by
multiplying by i]. Let us parameterize z_1(t) = r(t) e^{i p(t)} and
take real and imaginary part of (b) to obtain
(A) r'(t) = -1/sqrt(2)
(B) r(t) p'(t)= 1/sqrt(2)
Equation (A) can be solved by (with the initial condition r(0)=1)
r(t) = 1-t/sqrt(2)
so the dogs meet each other at time t=sqrt(2).
Equation (B) is fulfilled with (with p(0)=0)
p(t) = -log(r(t))
So, each of the dogs follows a logarithmic spiral.
Best regards,
Fabian
You did it the hard way.
Because of symetry, the dogs are at any time at vertices of a square,
the side of which decresaes at a rate which is the projection of
the speeds on the side. Hence (as a square has 90° angles) just the
speed of one dog.
Hence the 4 dogs will meet at center after side/speed = 1, not sqrt(2)
Hence you made some calculation mistake somewhere.
Once I did the full calculation for the differential equations, and
got MY value, not yours.
I don't want to do it again (a long time since I solved diff eqns...)
Regards.
--
Philippe Ch., mail : chephip+n...@free.fr
site :http://mathafou.free.fr/ (recreational mathematics)- Hide quoted text -
- Show quoted text -
|
Well, now that I got one elegant solution, I'd like to propose mine.
Following the equations of Fabian
| Quote: |
z'_n(t) = sign(i-1) z_n(t)/|z_n(t)|
|
it can be easily shown that the vector field is homogeneous. To say,
for every k, if (t, z_n) is a solution of the equation then also (k*t,
k*z_n).
Since the vector field directions are showing contraction (lets say,
for the first infinitesimal
movement) then there exist a disk contained in the sqaure in which the
solution will enter
and stay there forever. Now, since the vector field is homogeneous
then there is a smaller disk
in which the solutions will enter and stay there forever. Repeat this
argument inifinte times.
It can be proved that the solution will hit the origin. BUT, because
the origin is a singular point
then the solution there is a of sliding mode (in the Filippov sense).
Physically, the dogs will hit each other and their motion will be an
angular motion with infinite
angular velocity.
Regards,
Miki |
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Philippe 92
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| Posted: Thu Jul 03, 2008 11:01 am Post subject: Re: A nice mathematical riddle |
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Fabian wrote :
| Quote: |
On Wed, 2 Jul 2008 22:40:09 -0700 (PDT), miki <miki.livne@gmail.com> wrote:
Hello All,
Here is a nice simple-looking riddle that leads to heavy mathematics.
Assume that there are 4 dogs in each corner of a square on a plane
(around the origin) at initial time.
Where the start gun is fired, each dog want to get his neighbor (lets
say, clock wise neighbor). The questions are, will the dogs get each
other? Will they stop runing some time if the dogs are infinitesimaly
small? What is the trajectory they will draw? How can you describe the
motion near the origin?
->-----------------
- \/
- -
- -
/\ -
-----------------<-
Let us assume that the dogs move with constant velocity 1 (and I
prefer counter-clockwise chasing). We should use the symmetry of the
problem. Let us describe the plane by complex numbers. The initial
configuration (at time t=0) is given by
z_1(0)=1, z_2(0)=i, z_3(0)=-1, z_4(0)=-i
The symmetry group is generated by i (rotations by pi/2 around the
origin), there are no reflections allowed; in formulas (n is
modulo 4)
(a) i z_n(t) = z_{n+1}(t)
The equations of motion are given by
z'_n(t) = sign(z_{n+1}(t)-z_n(t))
= (z_{n+1}(t)-z_n(t))/|z_{n+1}(t)-z_n(t)|
= sign(i-1) z_n(t)/|z_n(t)| using (a)
(b) = e^{i 3pi/4} z_n(t)/|z_n(t)|
due to the symmetry, we have decoupled the differential equations.
We can solve the one for z_1(t) [the others will be generated by
multiplying by i]. Let us parameterize z_1(t) = r(t) e^{i p(t)} and
take real and imaginary part of (b) to obtain
(A) r'(t) = -1/sqrt(2)
(B) r(t) p'(t)= 1/sqrt(2)
Equation (A) can be solved by (with the initial condition r(0)=1)
r(t) = 1-t/sqrt(2)
so the dogs meet each other at time t=sqrt(2).
Equation (B) is fulfilled with (with p(0)=0)
p(t) = -log(r(t))
So, each of the dogs follows a logarithmic spiral.
Best regards,
Fabian
|
You did it the hard way.
Because of symetry, the dogs are at any time at vertices of a square,
the side of which decresaes at a rate which is the projection of
the speeds on the side. Hence (as a square has 90° angles) just the
speed of one dog.
Hence the 4 dogs will meet at center after side/speed = 1, not sqrt(2)
Hence you made some calculation mistake somewhere.
Once I did the full calculation for the differential equations, and
got MY value, not yours.
I don't want to do it again (a long time since I solved diff eqns...)
Regards.
--
Philippe Ch., mail : chephip+news@free.fr
site : http://mathafou.free.fr/ (recreational mathematics) |
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Fabian
Guest
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| Posted: Thu Jul 03, 2008 11:01 am Post subject: Re: A nice mathematical riddle |
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On Wed, 2 Jul 2008 22:40:09 -0700 (PDT), miki <miki.livne@gmail.com> wrote:
| Quote: |
Hello All,
Here is a nice simple-looking riddle that leads to heavy mathematics.
Assume that there are 4 dogs in each corner of a square on a plane
(around the origin) at initial time.
Where the start gun is fired, each dog want to get his neighbor (lets
say, clock wise neighbor). The questions are, will the dogs get each
other? Will they stop runing some time if the dogs are infinitesimaly
small? What is the trajectory they will draw? How can you describe the
motion near the origin?
->-----------------
- \/
- -
- -
/\ -
-----------------<-
|
Let us assume that the dogs move with constant velocity 1 (and I
prefer counter-clockwise chasing). We should use the symmetry of the
problem. Let us describe the plane by complex numbers. The initial
configuration (at time t=0) is given by
z_1(0)=1, z_2(0)=i, z_3(0)=-1, z_4(0)=-i
The symmetry group is generated by i (rotations by pi/2 around the
origin), there are no reflections allowed; in formulas (n is
modulo 4)
(a) i z_n(t) = z_{n+1}(t)
The equations of motion are given by
z'_n(t) = sign(z_{n+1}(t)-z_n(t))
= (z_{n+1}(t)-z_n(t))/|z_{n+1}(t)-z_n(t)|
= sign(i-1) z_n(t)/|z_n(t)| using (a)
(b) = e^{i 3pi/4} z_n(t)/|z_n(t)|
due to the symmetry, we have decoupled the differential equations.
We can solve the one for z_1(t) [the others will be generated by
multiplying by i]. Let us parameterize z_1(t) = r(t) e^{i p(t)} and
take real and imaginary part of (b) to obtain
(A) r'(t) = -1/sqrt(2)
(B) r(t) p'(t)= 1/sqrt(2)
Equation (A) can be solved by (with the initial condition r(0)=1)
r(t) = 1-t/sqrt(2)
so the dogs meet each other at time t=sqrt(2).
Equation (B) is fulfilled with (with p(0)=0)
p(t) = -log(r(t))
So, each of the dogs follows a logarithmic spiral.
Best regards,
Fabian |
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Narasimham
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| Posted: Fri Jul 04, 2008 12:06 am Post subject: Re: A nice mathematical riddle |
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On Jul 4, 2:58 am, Narasimham <mathm...@hotmail.com> wrote:
| Quote: |
On Jul 4, 1:26 am, miki <miki.li...@gmail.com> wrote:
Well, the discussion here is very interesting, especially Fabian's
solution which is quite elegant and general. But no one commented on
the the most problematic point which is the origin. Namely, the
solution to the equation of motion was solved at the origin as it has
a solution there.
But, unfortunately, the differential equation is not defined at the
origin. At least no in the usual sense. Although the vector field
direction is toward the origin, yet the solution there must be
formally defined.
Can anyone comment this?
Miki
|
Say similarly four or any number of ships making 45 degrees to sphere
meridian along loxodromes [z = tanh(th) , r = sech(th)] arrive at the
center to rotate indefinitely together as a block of points at north
and south poles with a uniform twist.
| Quote: |
The solution of y' = (y - x)/(y + x) is easy in terms of log and
arctan, center has a singularity. Somewhat like vortex flow at
center. ;) It would be like the dog going in circles trying to catch
its own tail, and anyway the point of origin cannot accommodate four
dogs.
|
The equation of above DE in polar cords is the log spiral:r/sqrt(2) =
Exp (-(th- pi/4) ).
Narasimham |
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David R Tribble
Guest
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| Posted: Mon Jul 07, 2008 5:07 am Post subject: Re: A nice mathematical riddle |
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David R Tribble wrote:
| Quote: |
This is an old riddle.
The path followed by each dog is a logarithmic spiral.
The interesting part, though, as you asked in the second half
of your poser, is that they traverse infinite spirals around the
origin (or whatever point they are equidistant from), approaching
it asymptotically at ever-smaller distances but never reaching it.
|
Peter Webb wrote:
| Quote: |
No. They meet in the middle at time=1, and hence it cannot be a logarithmic
spiral.
I posted a very simple proof above.
Basically, by symmetry you can show the direction motion of dog n+1 at any
time must be perpendicular to dog n; therefore the relative motion of dog
n+1 does not affect the time it takes dog n to reach it. This is the same
distance and hence time it takes dog n to traverse one side of the square.
So it can't be a logarithmic spiral, as they clearly all meet in the centre
in finite time (the same time as it would take to traverse one side).
|
Obviously, I remembered it wrong.
See:
http://www.cut-the-knot.org/Curriculum/Geometry/FourTurtles.shtml |
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Peter Webb
Guest
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| Posted: Mon Jul 07, 2008 5:48 am Post subject: Re: A nice mathematical riddle |
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"David R Tribble" <david@tribble.com> wrote in message
news:743e8c23-8869-4e83-b674-7591e3634d18@m45g2000hsb.googlegroups.com...
| Quote: |
miki wrote:
Here is a nice simple-looking riddle that leads to heavy mathematics.
Assume that there are 4 dogs in each corner of a square on a plane
(around the origin) at initial time.
Where the start gun is fired, each dog want to get his neighbor (lets
say, clock wise neighbor). The questions are, will the dogs get each
other? Will they stop runing some time if the dogs are infinitesimaly
small? What is the trajectory they will draw? How can you describe the
motion near the origin?
->-----------------
- \/
- -
- -
/\ -
-----------------<-
This is an old riddle.
The path followed by each dog is a logarithmic spiral.
The interesting part, though, as you asked in the second half
of your poser, is that they traverse infinite spirals around the
origin (or whatever point they are equidistant from), approaching
it asymptotically at ever-smaller distances but never reaching it.
|
No. They meet in the middle at time=1, and hence it cannot be a logarithmic
spiral.
I posted a very simple proof above.
Basically, by symmetry you can show the direction motion of dog n+1 at any
time must be perpendicular to dog n; therefore the relative motion of dog
n+1 does not affect the time it takes dog n to reach it. This is the same
distance and hence time it takes dog n to traverse one side of the square.
So it can't be a logarithmic spiral, as they clearly all meet in the centre
in finite time (the same time as it would take to traverse one side). |
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...
Guest
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| Posted: Mon Jul 07, 2008 10:45 am Post subject: Re: A nice mathematical riddle |
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On Mon, 7 Jul 2008 10:48:11 +1000, "Peter Webb"
<webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
| Quote: |
"David R Tribble" <david@tribble.com> wrote in message
news:743e8c23-8869-4e83-b674-7591e3634d18@m45g2000hsb.googlegroups.com....
miki wrote:
Here is a nice simple-looking riddle that leads to heavy mathematics.
Assume that there are 4 dogs in each corner of a square on a plane
(around the origin) at initial time.
Where the start gun is fired, each dog want to get his neighbor (lets
say, clock wise neighbor). The questions are, will the dogs get each
other? Will they stop runing some time if the dogs are infinitesimaly
small? What is the trajectory they will draw? How can you describe the
motion near the origin?
[snip]
This is an old riddle.
The path followed by each dog is a logarithmic spiral.
The interesting part, though, as you asked in the second half
of your poser, is that they traverse infinite spirals around the
origin (or whatever point they are equidistant from), approaching
it asymptotically at ever-smaller distances but never reaching it.
No. They meet in the middle at time=1, and hence it cannot be a logarithmic
spiral.
|
Yes it can. For more, see
<http://www.cut-the-knot.org/Curriculum/Geometry/FourTurtles.shtml>
<http://www.mathpages.com/home/kmath492/kmath492.htm>
<http://mathworld.wolfram.com/MiceProblem.html>
[snip] |
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Peter Webb
Guest
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| Posted: Mon Jul 07, 2008 11:02 am Post subject: Re: A nice mathematical riddle |
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"..." <...@domain.invalid> wrote in message
news:lva374l2pips0136stp1rg3qvheja0mesi@4ax.com...
On Mon, 7 Jul 2008 10:48:11 +1000, "Peter Webb"
<webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
| Quote: |
"David R Tribble" <david@tribble.com> wrote in message
news:743e8c23-8869-4e83-b674-7591e3634d18@m45g2000hsb.googlegroups.com...
miki wrote:
Here is a nice simple-looking riddle that leads to heavy mathematics.
Assume that there are 4 dogs in each corner of a square on a plane
(around the origin) at initial time.
Where the start gun is fired, each dog want to get his neighbor (lets
say, clock wise neighbor). The questions are, will the dogs get each
other? Will they stop runing some time if the dogs are infinitesimaly
small? What is the trajectory they will draw? How can you describe the
motion near the origin?
[snip]
This is an old riddle.
The path followed by each dog is a logarithmic spiral.
The interesting part, though, as you asked in the second half
of your poser, is that they traverse infinite spirals around the
origin (or whatever point they are equidistant from), approaching
it asymptotically at ever-smaller distances but never reaching it.
No. They meet in the middle at time=1, and hence it cannot be a logarithmic
spiral.
|
Yes it can. For more, see
<http://www.cut-the-knot.org/Curriculum/Geometry/FourTurtles.shtml>
<http://www.mathpages.com/home/kmath492/kmath492.htm>
<http://mathworld.wolfram.com/MiceProblem.html>
[snip]
************************
Hmmm ... I stand partially corrected. I hadn't realised that a finite
section of a log spiral can have an infinite number of turns.
However, the dogs do meet at the centre at t=1. As I noted above, the
standard logarithmic spiral over R has no point r = 0. You need to extend
the function by including -oo in the domain to also get the centre point,
where the dogs meet at a very finite t=1 but theta = oo. |
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