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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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