The experiment starts with a two tailed CA. At t = 35 the lefttmost bit of the left tail was injured ( = 0), and the CA lost its tail. From now on the CA attempts to regenerate its tail by recalling its previous experience. A similar experiment was described before when the tail was cut off, and later grew again. In the present experiment the CA has to regenerate its tail despite continuing injury at t = 35. During regeneration it recalls its previous states, and their respective age distributions.
The following graph
summarizes the regeneration efforts. It depicts the deviation of the center
of mass of the regenerating CA from the center of mass of an uninjured one. A perfect regeneration ought to be symmetric, and its center
of mass, ought to equal that of an uninjured CA (deviation = 0). Only one CA (t = -16)
meets these requirements. The
CA succeeds to regenerate its tail even when injury is three bits deep.
Following more extended injury the CA fails to regenerate its tail.
The CA never experienced a similar injury before. Now when attempting to regenerate its tail, it examines its past experience and
comes up with an adequate solution. Although state 19 (-16) was never injured, it knows how to regenerate an injured tail. The CA has an innate
knowledge how to repair injured tails. Its solution is creative. Kant would call
such a knowledge 'synthetic
a priori'.
This kind of regeneration was described in newts by M. Singer. When a leg of a newt is cut off, the wound is infiltrated with stem cells called blastema, which gradually regenerate a new leg. State 19 is a CA blastema.
Further reading:
Singer M. Neurotrophic control of limb regeneration
in the newt
Ann. N.Y. Acad. Sci. 228: 308-312,1974.
The experiment illustrates that CA memory is distributed Unlike
memories in conventional computers
it does not store images or data, but actions,
like how to regenerate a tail. Similar
information in a conventional computer
would require several lines of instructions. Here it is stored in state 19. Plant a zygote specified
by {rule = #600, age distribution
at t > 19} Let it live more than 35 time units. At t = 35 injure its
leftmost bit. Its state 19
will regenerate two perfect tails.
Setup
injurytime = 35; injuryrange =1-3; statetime=
34; prevstate = *; agetime = 34; prevage = *; preva[[1,*]] = a[[1,*]]; prevage[[1,*]] = age[[1,*]]; effect[1,
1, 25];