还记得克罗内克的信条吗?"上帝创造了整数,其他都是人类的杰作"。如果你愿意,Ω 根本不是一个实数,它是关于某些二叉方程的事实;它只与整数、正整数有关!因此,你不能把停止概率 Ω 的比特乃不可还原的数学真理这一事实推卸掉,因为这可以被重新解释为关于二阶方程的陈述。
Chaitin (1987):
Exponential Diophantine Equation #1
In this equation n is a parameter,and k,x, y,z, . . . are the unknowns:
L(n,k,x,y,z,...)=R(n,k,x,y,z,...). It has infinitely many positive-integer solutions if the nth bit of Ω is a 1.
It has onlyfinitely many positive-integer solutions if the nth bit of Ω is a 0.
Ord,Kieu (2003):
Exponential Diophantine Equation #2
In this equation n is a parameter,
and k,x,y,z,... are the unknowns:
L(n,k,x,y,z,...) = R(n,k,x,y,z,...).
For any given value of the parameter n,it only has finitely many positive-integer solutions.
For each particular value of n:
the number of solutions of this equation will be odd if the nth bit of Ω is a 1,and
the number of solutions of this equation will be even if the nth bit of Ω is a 0.
如何构造这两个二元一次方程?嗯,细节有点乱。下面的方框给出了总体思路;它们总结了需要做的事情。正如你所看到的,我们之前讨论过的Ω近似值的可计算序列起着关键作用。同样重要的是,特别是对于 Ord 和 Kieu (2003),要记住这些近似值是一个非递减的有理数序列,它们越来越接近 Ω,但始终小于 Ω 的值。
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