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Re: A problem in Pi digits as Lattice space filling

  • To: mathgroup at smc.vnet.net
  • Subject: [mg93887] Re: A problem in Pi digits as Lattice space filling
  • From: dh <dh at metrohm.com>
  • Date: Fri, 28 Nov 2008 05:04:54 -0500 (EST)
  • References: <ggj7co$j2i$1@smc.vnet.net> <200811261222.HAA22459@smc.vnet.net> <gglss8$8fg$1@smc.vnet.net>


Hi Bob,

try:

d = RealDigits[Pi, 10, 10^6][[1]];

and make a histogram of it. Shure this is no proof, but it does 

certainly not look normal.

Daniel





DrMajorBob wrote:

>> first some picky things. The digits in Pi are not normal, but uniformly

>> distributed

> 

> I doubt that (uniformity) is actually known. Is it?

> 

> Bobby

> 

> On Wed, 26 Nov 2008 06:22:10 -0600, dh <dh at metrohm.com> wrote:

>>

>> Hi Roger,

>>

>> first some picky things. The digits in Pi are not normal, but uniformly

>>

>> distributed. Further you should have mentioned that you define the

>>

>> coordinate tuples by moving only one digit. Then, I can verify  the

>>

>> filling number for 1 dim. is 33, but for 2 dim. with 606 the duple {6,8}

>>

>> does not appear, you need 607 digits. For 3 dim. I need 8556, not 8554.

>>

>> Finally, here is a way to do it rather fast. I give the example for 3  

>> dim:

>>

>> n=8556;

>>

>> all=Flatten[Table[{i1,i2,i3},{i1,0,9},{i2,0,9},{i3,0,9}],2];

>>

>> d=Union@Partition[RealDigits[\[Pi],10,n][[1]],3,1];

>>

>> Complement[all,d]

>>

>> You increase n until the answer is empty: {}. Of course this can be

>>

>> automated by wrapping a binary search around the code above, but I let

>>

>> this for you.

>>

>> hope this helps, Daniel

>>

>>

>>

>>

>>

>> Roger Bagula wrote:

>>

>>> I need help with programs for  4th, 5th and 6th, etc.

>>> levels of lattice filling:

>>> The idea that the Pi digits are normal

>>> implies that they will fill space on different levels

>>> in a lattice type way ( Hilbert/ Peano  space fill).

>>> Question of space filling:

>>> Digit(n)-> how fast till all ten

>>> {Digit[n],Digit[n+1]} -> how fast to fill the square lattice

>>> {0,0},{0,9},{9,0},{9,9}

>>> {Digit[n],Digit[n+1],Digit[n+2]} -> how fast to fill the cubic lattice

>>> {0,0,0} to {9,9,9}

>>> I've answers for the first three with some really clunky programs.

>>> 33,606,8554,...

>>> my estimates for the 4th is: 60372 to 71947

>>> ((8554)2/(606*2)and half the log[]line result of 140000.)

>>> but it appears to  outside what my old Mac can do.

>>> I'd also like to graph the first occurrence to see how random the path

>>> between

>>> the lattice / space fill points is.

>>> The square:

>>> a = Table[Floor[Mod[N[Pi*10^n, 1000], 10]], {n, 0, 1000}];

>>> Flatten[Table[

>>>         If[Length[Delete[Union[Flatten[Table[Table[If[a[[n]] == k &&

>>> a[[n + \

>>> 1]] - l ==

>>>     0, {l, k}, {}], {k, 0, 9}, {

>>>           l, 0, 9}], {n, 1, m}], 2]], 1]] == 100, m, {}], {m, 600,  

>>> 610}]]

>>> {606, 607, 608, 609, 610}

>>> Table[Length[Delete[Union[Flatten[Table[Table[If[a[[n]] == k && a[[

>>>                   n + 1]] - l == 0, {l, k}, {}], {k, 0, 9}, {l,

>>>                 0, 9}], {n, 1, m}], 2]], 1]], {m, 600, 650}]

>>> {99, 99, 99,

>>>     99, 99, 99, 100,

>>>    100, 100, 100,

>>>     100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100,  

>>> 100,

>>> 100, 100,

>>> 100, 100, 100,

>>>    100, 100, 100, 100, 100,

>>>       100, 100, 100,

>>>         100, 100, 100, 100, 100,

>>>             100, 100, 100, 100, 100, 100, 100, 100, 100}

>>> Mathematica

>>> Clear[a, b, n]

>>> a = Table[Floor[Mod[N[Pi*10^n, 10000], 10]], {n, 0, 10000}];

>>> b = Table[{a[[n]], a[[n + 1]], a[[n + 2]]}, {n, 1, Length[a] - 2}];

>>> Flatten[Table[

>>>   If[Length[Union[Table[b[[n]], {

>>>       n, 1, m}]]] == 1000, m, {}], {m, 8550, 8600}]]

>>> {8554, 8555, 8556,

>>>        8557, 8558, 8559, 8560, 8561, 8562, 8563, 8564, 8565, 8566,

>>>   8567, 8568, 8569, 8570, 8571, 8572, 8573, 8574, 8575, 8576,

>>>             8577, 8578, 8579, 8580, 8581, 8582, 8583, 8584, 8585, 8586,

>>>   8587, 8588, 8589,

>>>     8590, 8591, 8592, 8593, 8594, 8595, 8596, 8597, 8598, 8599, 8600}

>>> The proof of the 33 is:

>>> Clear[a,b,n]

>>> a=Table[Floor[Mod[N[Pi*10^n,10000],10]],{n,0,10000}];

>>> Flatten[Table[If[Length[Union[Table[a[[n]],{n,1,

>>>         m}]]]==10,m,{}],{m,1,50}]]

>>> {33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}

>>> My own 4th level program that won't run on my older machine /older

>>> version of Mathematica:

>>> Clear[a, b, n]

>>> a = Table[Floor[Mod[N[Pi*10^n, 100000], 10]], {n, 0, 100000}];

>>> b = Table[{a[[n]], a[[n + 1]], a[[n + 2]], a[[n + 3]]}, {n, 1, Length[a]

>>> - 3}];

>>> Flatten[Table[If[ Length[Union[Table[b[[n]], {n, 1, m}]]] == 10000, m,

>>> {}], {m, 1, 50}]]

>>> Respectfully, Roger L. Bagula

>>> 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814

>>> :http://www.geocities.com/rlbagulatftn/Index.html

>>> alternative email: rlbagula at sbcglobal.net

>>

>>

> 

> 

> 




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