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flashlight following path traced out by Peano-type curve, Kakeya needle problem
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David Bernier
Apr 21, 2013, 11:01:10 PM4/21/13
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The 2D Kakeya needle problem asks how "small" a compact set
K in R^2 can be if K contains a unit length line segment
[A, B] in every direction in the plane.
The smallness can be evaluated by the Lebesgue measure of
K, or some finer notion of smallness (Hausdorff dimension).
For example, with a circle of radius 100 and a point Q
on the circle, we can put the needle (line segment)
with its mid-point M at Q and with
the needle tangent to the circle,
and then slide the needle along the circumference
while keeping it tangent to the circle, with contact
point M = midpoint of needle, till the needle is
back at the starting point Q on the circle.
Then the figure (region) occupied by the needle as
it does around seems to be a thin annulus
bounded on the "inside" by the circle of
radius 100, and on the outside by a circle
of radius slightly larger than 100:
http://en.wikipedia.org/wiki/Annulus_%28mathematics%29
Anway, it seems the area is pi/4, same as for a circle
of unit diameter ...
So, "very small" plane Kakeya sets (in area) are not
really easy to think of.
There's been work on lower bounds for the Hausadorff dimension
of plane Kakeya sets and also 3-D and above.
I browsed the paper:
"An Improved Bound on the Minkowski Dimension of Besicovitch Sets in R^3" by
Nets Hawk Katz, Izabella Laba and Terence Tao,
Annals of Math., Sec. Ser., Vol. 152, No. 2 (Sep., 2000), pp. 383-446.
from here:
http://www.math.ubc.ca/~ilaba/preprints/besicovitch.ps
They show that in 3D, a Besicovitch set (satifies
analogous unit length needle problem in 3D, encompassing
a unit segment in any direction in space, R^3),
any Besicovitch set has a Minkowski dimension of
at least 5/2 + epsilon, for some epsilon>0 that they
put at about 10^(-10) or more. Minkowski dimension
is a notion of fractal dimension, just as the
Hausdorff dimension is also. (based on coverings by
balls in R^3, or small neighborhoods of point set ,etc.)
I can decribe two "animations" based on a flashlight (needle)
following the path traced out by a Hilbert curve (square filling
curve) in the unit time interval [0, 1]:
https://en.wikipedia.org/wiki/Hilbert_curve
for t in [0, 1], suppose (x(t), y(t)) traces out
the Hilbert curve in [0,1]x[0,1] as t goes from 0
to 1 (there's maybe a canonical parameterization ...).
In Animation 1,
the plane containing the tracing of the curve is
z=5, and we translate the square so that it is centered
at (0, 0, 5).
That way,
X(t) = x(t) - 1/2
Y(t) = y(t) - 1/2
Z(t) = 5
for 0<=t<= 1.
We imagine a flashligth of unit length (a needle) with
one end at (0,0,0) and the other end having always
z-corrdinate > 0 such that at time 't',
the flashligth points at ( X(t), Y(t), Z(t) ),
for t in [0, 1].
As on end is fixed, the "flashlight" traces out
a part of a square-shaped pyramid with base
with vertices at (1/2, 1/2, 5), (-1/2, 1/2, 5)
(-1/2, -1/2, 5) and (1/2, -1/2, 5)
and sumit at (0,0,0). It's the part
of the prism that lies within the unit ball
centered at (0,0,0) because the mobile end
of the flashlight lies on the unit sphere centered
at (0,0,0).
That figure has unit segments pointing in any direction
within (say) one degree of the (positive) z-direction,
along the vector (0,0,1).
But it has Lebesgue 3-D measure > 0.
In my second animation,
the lower end of the flashlight is at (0,0, t) at
time 't' and the upper end of the flashlight (needle)
points in a direction parallel
to the the flashlight (line segment) of Animation 1
at time 't'.
So for Animation 2, the lower point at time 't'
is at (0, 0, t)
and the upper point is at:
X_1 (t) = x(t) - 1/2
Y_1 (t) = y(t) - 1/2
Z_1 (t) = t.
Compared to animation 1, in animation 2 we add a velocity
of 1 unit/(time unit) in the positive z-direction
to the flashlight (needle) in animation 1.
With a "normal" curve (not the Hilbert or Peano curve),
the flashlight following a curve as in animation 1,
and then + 1 foot/second in z-direction in animation 2
would trace out a smooth "normal" well-behaved
surface (dimension 2) in unit time.
For a square-filling curve like a Hilbert curve,
the path or region occupied by the flashlight
as time goes from 0 to 1 is hard to imagine.
The Kakeya conjecture is that in 'd' dimensions,
any Besicovitch set has full dimension.
[ref.: N. H. Katz, Laba and Tao paper]
The authors say there that the Kakaya conjecture
has been proved for d=2, and remains open for d>2.
I'm not completely sure what "full dimension" means ...
If Animation 2 produces a "surface" of Hausdorff
dimesion close to 3 (as implied by the Kakyeya conjecture),
maybe there's a way of seeing that intuitively.
Anyway, that's all.
David Bernier
--
Jesus is an Anarchist. -- J.R.
K in R^2 can be if K contains a unit length line segment
[A, B] in every direction in the plane.
The smallness can be evaluated by the Lebesgue measure of
K, or some finer notion of smallness (Hausdorff dimension).
For example, with a circle of radius 100 and a point Q
on the circle, we can put the needle (line segment)
with its mid-point M at Q and with
the needle tangent to the circle,
and then slide the needle along the circumference
while keeping it tangent to the circle, with contact
point M = midpoint of needle, till the needle is
back at the starting point Q on the circle.
Then the figure (region) occupied by the needle as
it does around seems to be a thin annulus
bounded on the "inside" by the circle of
radius 100, and on the outside by a circle
of radius slightly larger than 100:
http://en.wikipedia.org/wiki/Annulus_%28mathematics%29
Anway, it seems the area is pi/4, same as for a circle
of unit diameter ...
So, "very small" plane Kakeya sets (in area) are not
really easy to think of.
There's been work on lower bounds for the Hausadorff dimension
of plane Kakeya sets and also 3-D and above.
I browsed the paper:
"An Improved Bound on the Minkowski Dimension of Besicovitch Sets in R^3" by
Nets Hawk Katz, Izabella Laba and Terence Tao,
Annals of Math., Sec. Ser., Vol. 152, No. 2 (Sep., 2000), pp. 383-446.
from here:
http://www.math.ubc.ca/~ilaba/preprints/besicovitch.ps
They show that in 3D, a Besicovitch set (satifies
analogous unit length needle problem in 3D, encompassing
a unit segment in any direction in space, R^3),
any Besicovitch set has a Minkowski dimension of
at least 5/2 + epsilon, for some epsilon>0 that they
put at about 10^(-10) or more. Minkowski dimension
is a notion of fractal dimension, just as the
Hausdorff dimension is also. (based on coverings by
balls in R^3, or small neighborhoods of point set ,etc.)
I can decribe two "animations" based on a flashlight (needle)
following the path traced out by a Hilbert curve (square filling
curve) in the unit time interval [0, 1]:
https://en.wikipedia.org/wiki/Hilbert_curve
for t in [0, 1], suppose (x(t), y(t)) traces out
the Hilbert curve in [0,1]x[0,1] as t goes from 0
to 1 (there's maybe a canonical parameterization ...).
In Animation 1,
the plane containing the tracing of the curve is
z=5, and we translate the square so that it is centered
at (0, 0, 5).
That way,
X(t) = x(t) - 1/2
Y(t) = y(t) - 1/2
Z(t) = 5
for 0<=t<= 1.
We imagine a flashligth of unit length (a needle) with
one end at (0,0,0) and the other end having always
z-corrdinate > 0 such that at time 't',
the flashligth points at ( X(t), Y(t), Z(t) ),
for t in [0, 1].
As on end is fixed, the "flashlight" traces out
a part of a square-shaped pyramid with base
with vertices at (1/2, 1/2, 5), (-1/2, 1/2, 5)
(-1/2, -1/2, 5) and (1/2, -1/2, 5)
and sumit at (0,0,0). It's the part
of the prism that lies within the unit ball
centered at (0,0,0) because the mobile end
of the flashlight lies on the unit sphere centered
at (0,0,0).
That figure has unit segments pointing in any direction
within (say) one degree of the (positive) z-direction,
along the vector (0,0,1).
But it has Lebesgue 3-D measure > 0.
In my second animation,
the lower end of the flashlight is at (0,0, t) at
time 't' and the upper end of the flashlight (needle)
points in a direction parallel
to the the flashlight (line segment) of Animation 1
at time 't'.
So for Animation 2, the lower point at time 't'
is at (0, 0, t)
and the upper point is at:
X_1 (t) = x(t) - 1/2
Y_1 (t) = y(t) - 1/2
Z_1 (t) = t.
Compared to animation 1, in animation 2 we add a velocity
of 1 unit/(time unit) in the positive z-direction
to the flashlight (needle) in animation 1.
With a "normal" curve (not the Hilbert or Peano curve),
the flashlight following a curve as in animation 1,
and then + 1 foot/second in z-direction in animation 2
would trace out a smooth "normal" well-behaved
surface (dimension 2) in unit time.
For a square-filling curve like a Hilbert curve,
the path or region occupied by the flashlight
as time goes from 0 to 1 is hard to imagine.
The Kakeya conjecture is that in 'd' dimensions,
any Besicovitch set has full dimension.
[ref.: N. H. Katz, Laba and Tao paper]
The authors say there that the Kakaya conjecture
has been proved for d=2, and remains open for d>2.
I'm not completely sure what "full dimension" means ...
If Animation 2 produces a "surface" of Hausdorff
dimesion close to 3 (as implied by the Kakyeya conjecture),
maybe there's a way of seeing that intuitively.
Anyway, that's all.
David Bernier
--
Jesus is an Anarchist. -- J.R.
William Elliot
Apr 21, 2013, 11:30:43 PM4/21/13
to
On Sun, 21 Apr 2013, David Bernier wrote:
> The 2D Kakeya needle problem asks how "small" a compact set
> K in R^2 can be if K contains a unit length line segment
> [A, B] in every direction in the plane.
A ball of diameter one.
> The 2D Kakeya needle problem asks how "small" a compact set
> K in R^2 can be if K contains a unit length line segment
> [A, B] in every direction in the plane.
> The smallness can be evaluated by the Lebesgue measure of
> K, or some finer notion of smallness (Hausdorff dimension).
> For example, with a circle of radius 100 and a point Q on the circle, we
> can put the needle (line segment) with its mid-point M at Q and with the
> needle tangent to the circle, and then slide the needle along the
> circumference while keeping it tangent to the circle, with contact point
> M = midpoint of needle, till the needle is back at the starting point Q
> on the circle.
pi.(sqr(r^2 + (1/2)^2))^2 - pi.r^2 = pi/4.
David Bernier
Apr 21, 2013, 11:49:13 PM4/21/13
to
Z_1(t) = t + sqrt(1 -X_b(t)^2 -Y_b(t)^2 )
by Pythagoras,
where X_b(t) and Y_b(t) are respectively the
x- and y- coordinates at time 't' of
the --upper end-- of the flashlight in
Animation 1, i.e. the one where the lower
end of the flashlight remains fixed at
(0, 0, 0).
> Compared to animation 1, in animation 2 we add a velocity
> of 1 unit/(time unit) in the positive z-direction
> to the flashlight (needle) in animation 1.
>
> With a "normal" curve (not the Hilbert or Peano curve),
> the flashlight following a curve as in animation 1,
> and then + 1 foot/second in z-direction in animation 2
> would trace out a smooth "normal" well-behaved
> surface (dimension 2) in unit time.
David Bernier
Apr 22, 2013, 12:05:30 AM4/22/13
to
On 04/21/2013 11:30 PM, William Elliot wrote:
> On Sun, 21 Apr 2013, David Bernier wrote:
>
>> The 2D Kakeya needle problem asks how "small" a compact set
>> K in R^2 can be if K contains a unit length line segment
>> [A, B] in every direction in the plane.
>
> A ball of diameter one.
Yes, I think that works in any dimension d ...
> On Sun, 21 Apr 2013, David Bernier wrote:
>
>> The 2D Kakeya needle problem asks how "small" a compact set
>> K in R^2 can be if K contains a unit length line segment
>> [A, B] in every direction in the plane.
>
> A ball of diameter one.
>
>> The smallness can be evaluated by the Lebesgue measure of
>> K, or some finer notion of smallness (Hausdorff dimension).
>
>> For example, with a circle of radius 100 and a point Q on the circle, we
>> can put the needle (line segment) with its mid-point M at Q and with the
>> needle tangent to the circle, and then slide the needle along the
>> circumference while keeping it tangent to the circle, with contact point
>> M = midpoint of needle, till the needle is back at the starting point Q
>> on the circle.
>
> The area covered for a circle of radius r >= 0 would be
> pi.(sqr(r^2 + (1/2)^2))^2 - pi.r^2 = pi/4.
"In 1928, Besicovitch showed that that in
fact one could rotate a unit needle using an
arbitrarily small amount of positive area."
Ref.:
http://terrytao.wordpress.com/tag/extremal-combinatorics/
For pictures of smaller area Besicovitch sets
(sometimes called Kakeya sets):
http://en.wikipedia.org/wiki/Kakeya_set
cf. the Deltoid is easy to visualise..
[...]
William Elliot
Apr 22, 2013, 3:24:20 AM4/22/13
to
On Mon, 22 Apr 2013, David Bernier wrote:
> On 04/21/2013 11:30 PM, William Elliot wrote:
n> >
> On 04/21/2013 11:30 PM, William Elliot wrote:
> > > The 2D Kakeya needle problem asks how "small" a compact set
> > > K in R^2 can be if K contains a unit length line segment
> > > [A, B] in every direction in the plane.
> >
> > A ball of diameter one.
>
> Yes, I think that works in any dimension d ...
Actually, half of that to have a line segement point in
> > > K in R^2 can be if K contains a unit length line segment
> > > [A, B] in every direction in the plane.
> >
> > A ball of diameter one.
>
> Yes, I think that works in any dimension d ...
every direction. On the other had, were you to have
meant a directed line segement, then pi/4 instead of pi/8.
Even so, pi/4 is too large.
Draw a circle through (0,0), (0,1) and (1/2,1/2)
and another through (0,1/2), (0,1/2) and (1/2,0).
What's the common area A, inclosed by the circles?
Cleary A < 1/2 < 3/4 < pi/4.
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