Eastern religions have however normally tended to take a different view, and to consider humans as just one of many elements that make up the universe as a whole.

If there is a tree of possible replacements (as in "A"  "AA" ), then the sequential substitution system in a sense does depth-first recursion in the infinite tree, never returning from the single path it takes.

Mechanical randomness It takes only small imperfections in dice or roulette wheels to get substantially non-random results (see page 971 ). … In processes like stirring and shaking it can take a long time for correlations to disappear—as in the phenomenon of long-time tails mentioned on page 999 .

Given two numbers x and y their product can be computed in base k by ( FromDigits does the carries) FromDigits[ListConvolve[IntegerDigits[x, k], IntegerDigits[y, k], {1, -1}, 0], k] For numbers with n digits direct evaluation of the convolution would take about n 2 steps. … And this implies that to find a particular digit of m t in base k will take altogether about t Log[t] 2 steps. … However, the straightforward method for converting a t -digit number x to base k takes about t divisions, though this can be reduced to around Log[t] by using a recursive method such as FixedPoint[Flatten[Map[If[# < k, #, With[ {e = Ceiling[Log[k, #]/2]}, {Quotient[#, k e ], With[ {s = Mod[#, k e ]}, If[s  0, Table[0, {e}], {Table[0, {e - Floor[Log[k, s]] - 1}], s}]]}]] &, #]] &, {x}] The pictures below show stages in the computation of 3 20 (a) by a power tree in base 2 and (b) by conversion from base 3.

Undecidability [of cellular automaton classes] Almost any definite procedure for determining the class of a particular rule will have the feature that in borderline cases it can take arbitrarily long, often formally showing undecidability, as discussed on page 1138 .

One can take the original stem to extend from the point -1 to 0; the rule is then specified by the list b of complex numbers corresponding to the positions of the new tip obtained after one step.

For any input x one can test whether the machine will ever halt using u[{Reverse[IntegerDigits[x, 2]], 0}] u[list_] := v[Split[Flatten[list]]] v[{a_, b_: {}, c_: {}, d_: {}, e_: {}, f_: {}, g___}] := Which[a == {1} || First[a]  0, True, c  {}, False, EvenQ[Length[b]], u[{a, 1 - b, c, d, e, f, g}], EvenQ[Length[c]], u[{a, 1 - b, c, 1, Rest[d], e, f, g, 0}], e  {} || Length[d] ≥ Length[b] + Length[a] - 2, True, EvenQ[Length[e]], u[{a, b , c, d, f, g}], True, u[{a, 1 - b, c, 1 - d, e, 1, Rest[f], g, 0}]] This test takes at most n/3 recursive steps, even though the original machine can take of order n 2 steps to halt.

With vectors of length n it generically takes about n 2 steps to compute u given v , and a little less than n 3 steps to compute v given u (the best known algorithms—which are based on matrix multiplication—currently involve about n 2.4 steps). … It still takes only about n 2 steps to compute u given v , but it becomes vastly more difficult to compute v given u , taking perhaps 2 2 n steps.

The rules in each case take a cell to become black if the specified number of its neighbors (including diagonals) on a square grid are black on the step before.

In general, negation is just some operation that takes a string and yields another, giving back the original if it is applied a second time.