Standard operations in logic can be generalized as Not[a_] = 1 - a , And[a_, b_] = Min[a, b] , Or[a_, b_] = Max[a, b] , Xor[a_, b_] = Abs[a - b] , Equal[a_, b_] = 1 - Abs[a - b] , Implies[a_, b_] = 1 - UnitStep[a - b](a - b) .

Thus in case (f), for example, the number of positive and negative fluctuations appears on average to be equal even after a million steps.

And in this case the best model is again straightforward to find: it simply takes the probabilities for different blocks to be equal to the frequencies with which these blocks occur in the data.

The color of a particular cell is then found by evolving for a number of steps equal to the length of these input digit sequences.

The operators shown are And , Equal , Implies and Nand .

Arithmetic coding Consider dividing the interval from 0 to 1 into a succession of bins, with each bin having a width equal to the probability for some sequence of blocks to occur.

Unsolved problems [in number theory] Problems in number theory that are simple to state (say in the notation of Peano arithmetic) but that so far remain unsolved include: • Is there any odd number equal to the sum of its divisors?

In 1909 Emile Borel had formulated the notion of normal numbers (see page 912 ) whose infinite digit sequences contain all blocks with equal frequency. And in the 1920s Richard von Mises —attempting to capture the observed lack of systematically successful gambling schemes—suggested that randomness for individual infinite sequences could be defined in general by requiring that "collectives" consisting of elements appearing at positions specified by any procedure should show equal frequencies.

To get some idea about the origin of this behavior, one can assume that successive values of n are randomly even and odd with equal probability. … In case (b), the number of steps is equal to the number of base 2 digits in n , while in case (c) it is determined by the number of 1's in the base 2 digit sequence of n .

If the last 2 axioms are dropped any statement can readily be proved true or false essentially just by running rule 110 for a finite number of steps equal to the number of nested ↓ plus 〈 … 〉 in the statement.