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Constructible reals Instead of finding successive digits using systems like Turing machines, one can imagine constructing complete real numbers using idealizations of mechanical processes. … Linkages consisting of rods of integer lengths always trace out algebraic curves (or algebraic surfaces in 3D) and in general allow any algebraic number (as represented by Root ) to be constructed.
CTToR110[rules_ /; Select[rules, Mod[Length[#], 6] ≠ 0 &]  {}, init_] := Module[{g1, g2, g3, nr = 0, x1, y1, sp}, g1 = Flatten[ Map[If[#1 === {}, {{{2}}}, {{{1, 3, 5 - First[#1]}}, Table[ {4, 5 - # 〚 n 〛 }, {n, 2, Length[#]}]}] &, rules] /. a_Integer  Map[({d[# 〚 1 〛 , # 〚 2 〛 ], s[# 〚 3 〛 ]}) &, Partition[c[a], 3]], 4]; g2 = g1 = MapThread[If[#1 === #2 === {d[22, 11], s3}, {d[ 20, 8], s3}, #1] &, {g1, RotateRight[g1, 6]}]; While[Mod[ Apply[Plus, Map[# 〚 1, 2 〛 &, g2, 30] ≠ 0, nr++; g2 = Join[ g2, g1]]; y1 = g2 〚 1, 1, 2 〛 - 11; If[y1 < 0, y1 += 30]; Cases[ Last[g2] 〚 2 〛 , s[d[x_, y1], _, _, a_]  (x1 = x + Length[a])]; g3 = Fold[sadd, {d[x1, y1], {}}, g2]; sp = Ceiling[5 Length[ g3 〚 2 〛 ]/(28 nr) + 2]; {Join[Fold[sadd, {d[17, 1], {}}, Flatten[Table[{{d[sp 28 + 6, 1], s[5]}, {d[398, 1], s[5]}, { d[342, 1], s[5]}, {d[370, 1], s[5]}}, {3}], 1]] 〚 2 〛 , bg[ 4, 11]], Flatten[Join[Table[bgi, {sp 2 + 1 + 24 Length[init]}], init /. {0  init0, 1  init1}, bg[1, 9], bg[6, 60 - g2 〚 1, 1, 1 〛 + g3 〚 1, 1 〛 + If[g2 〚 1, 1, 2 〛 < g3 〚 1, 2 〛 , 8, 0]]]], g3 〚 2 〛 }] s[1] = struct[{3, 0, 1, 10, 4, 8}, 2]; s[2] = struct[{3, 0, 1, 1, 619, 15}, 2]; s[3] = struct[{3, 0, 1, 10, 4956, 18}, 2]; s[4] = struct[{0, 0, 9, 10, 4, 8}]; s[5] = struct[{5, 0, 9, 14, 1, 1}]; {c[1], c[2]} = Map[Join[{22, 11, 3, 39, 3, 1}, #] &, {{63, 12, 2, 48, 5, 4, 29, 26, 4, 43, 26, 4, 23, 3, 4, 47, 4, 4}, {87, 6, 2, 32, 2, 4, 13, 23, 4, 27, 16, 4}}]; {c[3], c[4], c[5]} = Map[Join[#, {4, 17, 22, 4, 39, 27, 4, 47, 4, 4}] &, {{17, 22, 4, 23, 24, 4, 31, 29}, {17, 22, 4, 47, 18, 4, 15, 19}, {41, 16, 4, 47, 18, 4, 15, 19}}] {init0, init1} = Map[IntegerDigits[216 (# + 432 10 49 ), 2] &, {246005560154658471735510051750569922628065067661, 1043746165489466852897089830441756550889834709645}] bgi = IntegerDigits[9976, 2] bg[s_, n_] := Array[bgi 〚 1 + Mod[# - 1, 14] 〛 &, n, s] ev[s[d[x_, y_], pl_, pr_, b_]] := Module[{r, pl1, pr1}, r = Sign[BitAnd[2^ListConvolve[{1, 2, 4}, Join[bg[pl - 2, 2], b, bg[pr, 2]]], 110]]; pl1 = (Position[r - bg[pl + 3, Length[r]], 1 | -1] /. {}  {{Length[r]}}) 〚 1, 1 〛 ; pr1 = Max[pl1, (Position[r - bg[pr + 5 - Length[r], Length[r]], 1 | -1] /. {}  {{1}}) 〚 -1, 1 〛 ]; s[d[x + pl1 - 2, y + 1], pl1 + Mod[pl + 2, 14], 1 + Mod[pr + 4, 14] + pr1 - Length[r], Take[r, {pl1, pr1}]]] struct[{x_, y_, pl_, pr_, b_, bl_}, p_Integer : 1] := Module[ {gr = s[d[x, y], pl, pr, IntegerDigits[b, 2, bl]], p2 = p + 1}, Drop[NestWhile[Append[#, ev[Last[#]]] &, {gr}, If[Rest[Last[#]] === Rest[gr], p2--]; p2 > 0 &], -1]] sadd[{d[x_, y_], b_}, {d[dx_, dy_], st_}] := Module[{x1 = dx - x, y1 = dy - y, b2, x2, y2}, While[y1 > 0, {x1, y1} += If[Length[st]  30, {8, -30}, {-2, -3}]]; b2 = First[Cases[st, s[d[x3_, -y1], pl_, _, sb_]  Join[bg[pl - x1 - x3, x1 + x3], x2 = x3 + Length[sb]; y2 = -y1; sb]]]; {d[x2, y2], Join[b, b2]}] CTToR110[{{}}, {1}] yields blocks of lengths {7204, 1873, 7088} .
This can be done for blocks up to length n in a 1D cellular automaton with k colors using ReversibleQ[rule_, k_, n_] := Catch[Do[ If[Length[Union[Table[CAStep[rule, IntegerDigits[i, k, m]], {i, 0, k m - 1}]]] ≠ k m , Throw[False]], {m, n}]; True] For k = 2 , r = 1 it turns out that it suffices to test only up to n = 4 (128 out of the 256 rules fail at n = 1 , 64 at n = 2 , 44 at n = 3 and 14 at n = 4 ); for k = 2 , r = 2 it suffices to test up to n = 15 , and for k = 3 , r = 1 , up to n = 9 .
The following generates explicit lists of n -input Boolean functions requiring successively larger numbers of Nand operations: Map[FromDigits[#, 2] &, NestWhile[Append[#, Complement[Flatten[Table[Outer[1 - Times[##] &, # 〚 i 〛 , # 〚 -i 〛 , 1], {i, Length[#]}], 2], Flatten[#, 1]]] &, {1 - Transpose[IntegerDigits[Range[2 n ] - 1, 2, n]]}, Length[Flatten[#, 1]] < 2 2 n &], {2}] The results for 2-step cellular automaton evolution in the main text were found by a recursive procedure.
Power cellular automata Multiplication by m in base k corresponds to a local cellular automaton operation on digit sequences when every prime that divides m also divides k . … The inverse rule, corresponding to multiplication by 1/m , can be obtained by applying the rule for multiplication by the integer k q /m , then shifting right by q positions.
(a) (Successive digits sequence) The sequence produced is repetitive, with the element at position n being black for n odd and white for n even. … (b) (Thue–Morse sequence) The color s[n] of the element at position n is given by 1 - Mod[DigitCount[n - 1, 2, 1], 2] . … (d) (Cantor set) The color of the element at position n is given by If[FreeQ[IntegerDigits[n - 1, 3], 1], 1, 0] , which turns out to be equivalent to If[OddQ[n], Sign[Mod[Binomial[n - 1, (n - 1)/2], 3]], 0, 1] There are 3 t elements after t steps, of which 2 t are black.
Other so-called Fourier series in which the coefficient of Sin[m x] is a smooth function of m for all integer m yield similarly simple results. … The curves obtained in this case show a definite nested structure, in which the value at a point x is essentially determined directly from the base 2 digit sequence of x .
The configuration of the tape on each side of the head in the Turing machine evolution can be obtained from the tag system evolution using Cases[history, x : {a[_], ___}  Apply[{#1, Reverse[#2]} &, Map[ Drop[IntegerDigits[Count[x, #], 2], -1] &, {_b, _d}]]]
So for example the equation a 2 + b 2  0 has solutions that are exactly those integers that satisfy the relation a  0 ∧ b  0 . … Given an integer a for which IntegerDigits[a, 2] gives the cell values for a cellular automaton, a single step of evolution according say to rule 30 is given by BitXor[a, 2 BitOr[a, 2a]] where (see page 871 ) BitXor[x, y]  BitOr[x, y] - BitAnd[x, y] and a is assumed to be padded with 0's at each end. … One can reduce such an exponential equation to a pure polynomial equation by encoding powers using integer equations.
.) • Is there a solution of a certain size to an integer linear programming problem? … (In cases where numbers are involved, it is usually crucial that these be represented by base 2 digit sequences, and not, say, in unary.) … It is not known whether problems such as integer factoring or equivalence of networks under relabelling of nodes (graph isomorphism) are NP-complete.
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