In practice, one would use a much smaller value of e, such as 1.0001. In this case, the sum over all Life patterns is: 1.278763170138119 * 10^11222 Mathematica code: Module[{k = 800}, Fold[Times, 1, Flatten[Table[1 + 1.0001^(-i^2 - j^2), {i, -k, k}, {j, -k, k}]]]] Sincerely, Adam P. Goucher ----- Original Message ----- From: "Adam P. Goucher" <apgoucher@gmx.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Monday, June 25, 2012 5:30 PM Subject: [math-fun] Very nice measure for Life patterns
I have a more easily-computable measure, which avoids those axis issues.
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Let P be a pattern, which is the disjoint union of cells (x1, y1), (x2, y2), ..., (xn, yn). Then, we define a constant:
S(P) = x1^2 + y1^2 + x2^2 + y2^2 + ... + xn^2 + yn^2.
This is simply the sum of squared distances from the origin. Clearly, we have S(P disjointunion Q) = S(P) + S(Q), which is rather cute.
Next, we want to handle translations. If our pattern P has centroid (x,y), and we translate it to form a pattern P' with centroid (x',y'), then the following identity holds thanks to the Huygens-Steiner theorem:
S(P) - n(x^2 + y^2) = S(P') - n(x'^2 + y'^2)
This gives us S(P') = S(P) + n(x'^2 + y'^2 - x^2 - y^2).
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We now define R(P) = exp(-S(P)), so that summing over all translates of a pattern (and, indeed, all patterns) yields a finite measure. Disjoint union is now multiplicative:
R(P disjointunion Q) = R(P) R(Q).
And the Huygens-Steiner translation formula becomes:
R(P') = R(P) exp(x^2 + y^2 - x'^2 - y'^2)^n.
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Suppose P has a centroid (x,y). Then, we can compute the sum over all translations of P:
Sigma(R(Translate[P, (i,j)])) = R(P) (LM)
where L = Sum[integer i] [exp(n x^2 - n (x + i)^2)] and M = Sum[integer j] [exp(n y^2 - n (y + j)^2)]
[Aside: The maths is much nicer if we integrated over all *real* translations of P instead of summing over all integer translations. Then, L = Sqrt[Pi/n] exp(n x^2) and M = Sqrt[Pi/n] exp(n y^2).]
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Let's suppose we want to find the total value of R, summing over all Life patterns. We use the disjoint union formula:
R(P disjointunion Q) = R(P) R(Q)
where Q is just a single cell. Then, we have:
R(P) + R(P disjointunion Q) = R(P) (1 + R(Q))
So, the total value of R is equal to the infinite product:
Product[integer i, integer j] [1 + exp(-i^2-j^2)]
According to Mathematica, this evaluates to:
13.229455772329303730238455403464234174634112652323001 034270847525567752900584753556103622375911163436936902 250061828936681214420033729814493443269508795000058792 023403572348671341891245501751290476764
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Sincerely,
Adam P. Goucher
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