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 >> >> > > >

**References**:**Re: A problem in Pi digits as Lattice space filling***From:*dh <dh@metrohm.com>