For one step in rule 30, for example, this yields {{1, 0, 0}, {0, 1, 1}, {0, 1, 0}, {0, 0, 1}} , as shown on page 616 . … To reduce the representation, one must introduce "don't care" elements _ ; in this example the final minimal form consists of the list of 3 so-called implicants {{1, 0, 0}, {0, 1, _}, {0, _, 1}} . … Given an original DNF list s , this can be done using PI[s, n] : PI[s_, n_] := Union[Flatten[ FixedPointList[f[Last[#], n] &, {{}, s}] 〚 All, 1 〛 , 1]] g[a_, b_] := With[{i = Position[Transpose[{a, b}], {0,1}]}, If[Length[i]  1 && Delete[a, i] === Delete[b, i], {ReplacePart[a, _, i]}, {}]] f[s_, n_] := With[ {w = Flatten[Apply[Outer[g, #1, #2, 1] &, Partition[Table[ Select[s, Count[#, 1]  i &], {i, 0, n}], 2, 1], {1}], 3]}, {Complement[s, w, SameTest  MatchQ], w}] The minimal DNF then consists of a collection of these prime implicants.

Implementation [of repetitive array] The color of a cell at position {x, y} in the pattern shown is given by Extract[{{1, 0, 1}, {0, 1, 0}}, Mod[{y, x}, {2, 3}] + 1] .

But in updating networks a particularly straightforward implementation of one scheme can be obtained if one uses instead a more explicit symbolic representation such as u[1  v[2, 3, 4], 2  v[1, 3, 4], 3  v[1, 2, 4], 4  v[1, 2, 3]] This allows one to capture the basic character of networks by Attributes[u] = {Flat, Orderless}; Attributes[v] = Orderless Updating rules can then be written in terms of ordinary Mathematica patterns. … The rule at the top of page 509 must therefore be written out as and this corresponds to the Mathematica rule u[i1_  v[i2_, i3_, i4_], i3_  v[i1_, i5_, i6_], i4_  v[i1_, i7_, i8_]]  u[i1  v[i2, new[1], new[2]], new[1]  v[i1, new[2], i3], new[2]  v[i1, new[1], i4], i3  v[new[1], i5, i6], i4  v[new[2], i7, i8]] (Strictly there also need to be additional rules to cover where for example nodes 3 and 4 are actually the same.)

In the continued fraction for a randomly chosen number, the probability to find a term of size s is Log[2, (1 + 1/s)/(1 + 1/(s + 1))] , so that the probability of getting a 1 is about 41.50%, and the probability of getting a large term falls off like 1/s 2 . … The sequence of odd numbers gives the continued fraction for Coth[1] ; the sequence of even numbers for BesselI[0, 1]/BesselI[1, 1] . … Fairly large terms are sometimes seen quite early: in 5 1/3 term 19 is 3052, while in Root[10 + 8 # - # 3 &, 1] term 34 is 1,501,790.

[PDEs in] higher dimensions The pictures below show as examples the solution to the wave equation in 1D, 2D and 3D starting from a stationary square pulse. In each case a 1D slice through the solution is shown, and the solution is multiplied by r d-1 . … In 1D and 3D, the value at the origin quickly becomes exactly 0; in 2D it is given by 1-t/Sqrt[t 2 -1] , which tends to zero only like -1/(2t 2 ) (which means that a sound pulse cannot propagate in a normal way in 2D).

, a]  1 ∧ a > 1) a  BitAnd[c, d] ∧ b  BitOr[c, d] ↔ ( σ [c, a] ∧ σ [d, a] ∧ σ [b, c] ∧ σ [b, d] ∧ a + b  c + d)/. σ [x_, y_]  Mod[Binomial[x, y], 2]  1 where the last encoding uses the result on page 608 . … The corresponding form for rule 110 is BitXor[BitAnd[a, 2a, 4a], BitOr[2a, 4a]] The final equation is then obtained from {1 + x 4 + x 12  2 (1 + x 3 ) (x 1 + 2 x 3 ) , x 2 + x 13  2 x 1 , 1 + x 5 + x 14  2 x 1 , 2 x 3 x 5 + 2 x 1 + 2 x 3 x 6 + 2 x 1 + x 3 x 15 + x 16  x 4 , 1 + x 15 + x 17  2 x 3 , 1 + x 16 + x 18  2 x 3 , 2 1 + x 3 (1 + x 1 + 2 x 3 ) (-1 + x 2 ) - x 10 + x 11  2 x 4 , x 7  BitAnd[x 6 , 2 x 6 ] ∧ x 8  BitOr[x 6 , 2 x 6 ], x 9  BitAnd[x 6 , 2 x 7 ] ∧ x 19  BitOr[x 6 , 2 x 7 ], x 10  BitAnd[x 9 , 2 x 8 ] ∧ x 11  BitOr[x 9 , 2 x 8 ]} where x 1 through x 4 have the meanings indicated in the main text, and satisfy x i ≥ 0 . … The simplest known way of doing this (see note below ) involves a degree 8 equation with 60 variables: a  b c ↔ α [d, 4 + b e, 1 + z] ∧ α [f, e, 1 + z] ∧ a  Quotient[d, f] ∧ α [g, 4 + b, 1 + z] ∧ e  16 g(1 + z) λ [a_, b_, c_] := Module[{x}, 2 a + x 1  c ∧ (Mod[b - a, c]  0 ∨ Mod[b + a, c]  0)] α [a_, b_, c_] := Module[{x}, x 1 2 - b x 1 x 2 + x 2 2  1 ∧ x 3 2 - b x 3 x 4 + x 4 2  1 ∧ 1 + x 4 + x 5  x 3 ∧ Mod[x 3 , x 1 2 ]  0 ∧ 2x 4 + x 7  b x 3 ∧ Mod[-b + x 8 , x 7 ]  0 ∧ Mod[-2 + x 8 , x 1 ]  0 ∧ x 8 - x 11  3 ∧ x 12 2 - x 8 x 12 x 13 + x 13 2  1 ∧ 1 + 2 a + x 14  x 1 ∧ λ [a, x 12 , x 7 ] ∧ λ [c, x 12 , x 1 ]] (This roughly uses the idea that solutions to Pell equations grow exponentially, so that for example x 2  2y 2 + 1 has solutions With[{u = 3 + 2 √ 2 }, (u n + u -n )/2] .)

Implementation [of proof example] Given the axioms in the form s[1] = (a_ ⊼ a_) ⊼ (a_ ⊼ b_)  a; s[2, x_] := b_  (b ⊼ b) ⊼ (b ⊼ x); s[3] = a_ ⊼ (a_ ⊼ b_)  a ⊼ (b ⊼ b); s[4] = a_ ⊼ (b_ ⊼ b_)  a ⊼ (a ⊼ b); s[5] = a_ ⊼ (a_ ⊼ (b_ ⊼ c_))  b ⊼ (b ⊼ (a ⊼ c)); the proof shown here can be represented by {{s[2, b], {2}}, {s[4], {}}, {s[2, (b ⊼ b) ⊼ ((a ⊼ a) ⊼ (b ⊼ b))], {2, 2}}, {s[1], {2, 2, 1}}, {s[2, b ⊼ b], {2, 2, 2, 2, 2, 2}], {s[5], {2, 2, 2}}, {s[2, b ⊼ b], {2, 2, 2, 2, 2, 1}}, {s[1], {2, 2, 2, 2, 2}}, {s[3], {2, 2, 2}}, {s[1], {2, 2, 2, 2}}, {s[4], {2, 2, 2}}, {s[5], {}}, {s[2, a], {2, 2, 1}}, {s[1], {2, 2}}, {s[3], {}}, {s[1], {2}}} and applied using FoldList[Function[{u, v}, MapAt[Replace[#, v 〚 1 〛 ] &, u, {v 〚 2 〛 }]], a ⊼ b, proof]

Arithmetic systems [emulating register machines] Given the program for a register machine with nr registers in the form on page 896 , an arithmetic system which emulates it can be obtained from RMToAS[prog_, nr_] := With[{p = Length[prog], g = Product[Prime[j], {j, nr}]}, {p g, Sort[Flatten[MapIndexed[ With[{n = First[#2] - 1}, #1 /. {i[r_]  Table[n + j p  (1 + n Prime[r] (-n + #) &), {j, 0, g - 1}], d[r_, k_]  Table[n + j p  If[Mod[j, Prime[r]]  0, -1 + k + (-n + #)/Prime[r] &, # + 1 &], {j, 0, g - 1}]}] &, prog]]]}] The rules for the arithmetic system are represented so that the system from page 122 becomes for example {2, {0  (3 #/2 &), 1  (3 (# + 1)/2 &)}} . If the register machine starts at instruction n with values regs in its registers, then the corresponding arithmetic system starts with the number n + Table[Prime[i]^reg 〚 i 〛 , {i, nr}] p - 1 where p = Length[prog] .

Equation for the background [in my PDEs] If u[t, x] is independent of x , as it is sufficiently far away from the main pattern, then the partial differential equation on page 165 reduces to the ordinary differential equation u''[t]  (1 - u[t] 2 )(1 + a u[t]) u[0]  u'[0]  0 For a = 0 , the solution to this equation can be written in terms of Jacobi elliptic functions as ( √ 3 JacobiSN[t/3 1/4 , 1/2] 2 ) / (1 + JacobiCN[t/3 1/4 , 1/2] 2 ) In general the solution is (b d JacobiSN[r t, s] 2 )/(b - d JacobiCN[r t, s] 2 ) where r = -Sqrt[1/8 a c (b - d)] s = (d (c - b))/(c (d - b)) and b , c , d are determined by the equation (x - b)(x - c)(x - d)  -(12 + 6 a x - 4 x 2 - 3 a x 3 )/(3a) In all cases (except when -8/3 < a < -1/ √ 6 ), the solution is periodic and non-singular. For a = 0 , the period is 2 3 1/4 EllipticK[1/2] ≃ 4.88 . For a = 1 , the period is about 4.01; for a = 2 , it is about 3.62; while for a = 4 , it is about 3.18.

Implementation [of texture perception model] The exact matches for a template σ in data containing elements 0 and 1 can be obtained from Sign[ListCorrelate[2 σ - 1, data] - Count[ σ , 1, 2]] + 1