       Re: Re: A problem in Pi digits as Lattice space filling

• To: mathgroup at smc.vnet.net
• Subject: [mg93856] Re: [mg93829] Re: A problem in Pi digits as Lattice space filling
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Thu, 27 Nov 2008 05:30:40 -0500 (EST)
• References: <ggj7co\$j2i\$1@smc.vnet.net> <200811261222.HAA22459@smc.vnet.net>

```> 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][],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
>
>>
>
>
>

--
DrMajorBob at longhorns.com

```

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